McGILL UNIVERSITYFACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH 141 2010 01 CALCULUS 2 Information for Students (Winter Term, 2009/2010) Pages 1 - 20 of these notes may be considered the Course Outline for this course. The page numbers shown in the table of contents and in the upper right hand corners of pages are not the same as the numbers of pages in the PDF document. If you wish to print out specific pages, you should first view the relevant pages at your screen, and determine what are the numbers of the corresponding PDF pages. W. G. Brown April 17, 2010 Information for Students in MATH 141 2010 01 Contents 1 General Information 1.1 Force Majeure . . . . . . . . . . 1.2 Instructors and Times . . . . . . 1.3 Calendar Description . . . . . . 1.3.1 Calendar Description . . 1.3.2 Late transfer from MATH 151/MATH 152 . . . . . 1.4 Tutorials . . . . . . . . . . . . . 1.4.1 Tutorial Times, Locations, and Personnel (subject to change) . . . . . . . . . 1.4.2 Teaching Assistants (TA’s) 1.4.3 Friday, April 02nd, 2010 and Monday, April 05th, 2010 . . . . . . . . . . 1.5 Evaluation of Your Progress . . 1.5.1 Your final grade (See Table 3, p. 11) In the event of extraordinary circumstances beyond the University’s control, the content and/or evaluation scheme in this course is subject to change. . . . . . . . . 1.5.2 WeBWorK . . . . . . . 1.5.3 Written Submissions. . . 1.5.4 Quizzes at the Tutorials. 1.5.5 Final Examination . . . 1.5.6 Supplemental Assessments 1.5.7 Machine Scoring: “Will the final examination be machine scored?” . . . . 1.5.8 Plagiarism. . . . . . . . 1.5.9 Corrections to grades . . 1.6 Published Materials . . . . . . . 1.6.1 Required Text-Book . . 1.6.2 Optional Reference Books 1.6.3 Recommended Video Materials . . . . . . . . . . 1.6.4 Other Calculus Textbooks 1 1 2 2 2 3 3 3 3 2 4 5 3 5 6 7 7 8 4 8 9 9 10 10 10 10 12 13 1.6.5 Website . . . . . . . . . Syllabus . . . . . . . . . . . . . Preparation and Workload . . . 1.8.1 Prerequisites. . . . . . . 1.8.2 Calculators . . . . . . . 1.8.3 Self-Supervision . . . . 1.8.4 Escape Routes . . . . . 1.8.5 Terminology . . . . . . 1.9 Communication with Instructors and TA’s . . . . . . . . . . . . . 1.10 Commercial tutorial and exam preparation services . . . . . . . 1.11 Special Office Hours and Tutorials 13 14 15 15 15 16 17 18 Draft Solutions to Quiz Q1 2.1 Instructions to Students 2.2 Monday Versions . . . 2.3 Tuesday Versions . . . 2.4 Wednesday Versions . 2.5 Thursday Versions . . 2.6 Friday Versions . . . . . . . . . . 21 21 21 23 24 25 26 . . . . . . 28 28 29 31 33 35 38 1.7 1.8 . . . . . . . . . . . . Draft Solutions to Quiz Q2 3.1 Instructions to Students . . 3.2 Monday Versions . . . . . 3.3 Most Tuesday Versions . . 3.4 Most Wednesday Versions 3.5 Thursday Versions . . . . 3.6 Friday Versions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References 4.1 Stewart Calculus Series . . . . . 4.2 Other Calculus Textbooks . . . . 4.2.1 R. A. Adams . . . . . . 4.2.2 Larson, Hostetler, et al. . 4.2.3 Edwards and Penney . . 4.2.4 Others, not “Early Transcendentals” . . . . . . 4.3 Other References . . . . . . . . 18 19 20 201 201 202 202 203 203 204 204 A Timetable for Lecture Section 001 of MATH 141 2010 01 1001 Information for Students in MATH 141 2010 01 B Timetable for Lecture Section 002 of MATH 141 2009 01 2001 C Supplementary Notes for Students in Section 001 of MATH 141 2010 01 3001 C.1 Lecture style in Lecture Section 001 . . . . . . . . . . . . . . . 3001 C.2 Supplementary Notes for the Lecture of January 04th, 2010 . . . 3002 C.2.1 §5.1 Areas and Distances. 3002 C.3 Supplementary Notes for the Lecture of January 06th, 2010 . . . 3005 C.3.1 §5.1 Areas and Distances (conclusion). . . . . . . 3005 C.3.2 §5.2 The Definite Integral 3007 C.4 Supplementary Notes for the Lecture of January 08th, 2010 . . . 3013 C.4.1 Summary of the last lectures . . . . . . . . . . . 3013 C.4.2 §5.2 The Definite Integral (conclusion) . . . . 3014 C.4.3 §5.3 The Fundamental Theorem of Calculus . . . . 3019 C.5 Supplementary Notes for the Lecture of January 11th, 2010 . . . 3021 C.5.1 §5.3 The Fundamental Theorem of Calculus (conclusion) . . . . . . . . . 3021 C.5.2 §5.4 Indefinite Integrals and the “Net Change” Theorem . . . . . . . . . . 3023 C.6 Supplementary Notes for the Lecture of January 13th, 2010 . . . 3028 C.6.1 §5.4 Indefinite Integrals and the “Net Change” Theorem (conclusion) . . . 3028 C.6.2 §5.5 The Substitution Rule 3033 C.7 Supplementary Notes for the Lecture of January 15th, 2010 . . . 3036 C.7.1 §5.5 The Substitution Rule (conclusion) . . . . . . . 3036 C.7.2 5 Review . . . . . . . . 3049 C.8 Supplementary Notes for the Lecture of January 18th, 2010 . . . 3052 C.8.1 §6.1 Areas between Curves 3052 C.9 Supplementary Notes for the Lecture of January 20th, 2010 . . . 3061 C.9.1 §6.2 Volumes . . . . . . 3061 C.10 Supplementary Notes for the Lecture of January 22nd, 2010 . . . 3069 C.10.1 §6.3 Volumes by Cylindrical Shells . . . . . . . 3069 C.10.2 §6.4 Work . . . . . . . . 3075 C.11 Supplementary Notes for the Lecture of January 25th, 2010 . . . 3076 C.11.1 §6.5 Average value of a function . . . . . . . . . 3077 C.12 Supplementary Notes for the Lecture of January 27th, 2010 . . . 3082 C.12.1 §7.1 Integration by Parts 3082 C.13 Supplementary Notes for the Lecture of January 29th, 2010 . . . 3089 C.13.1 §7.1 Integration by Parts (conclusion) . . . . . . . 3089 C.13.2 §7.2 Trigonometric Integrals . . . . . . . . . . . 3092 C.14 Supplementary Notes for the Lecture of February 01st, 2010 . . . 3095 C.14.1 §7.2 Trigonometric Integrals (conclusion) . . . . 3095 C.15 Supplementary Notes for the Lecture of February 03rd, 2010 . . . 3101 C.15.1 §7.3 Trigonometric Substitution . . . . . . . . . 3101 C.16 Supplementary Notes for the Lecture of February 05th, 2010 . . . 3106 C.16.1 §7.3 Trigonometric Substitution (conclusion) . . 3106 C.17 Supplementary Notes for the Lecture of February 08th, 2010 . . . 3111 C.17.1 §7.4 Integration of Rational Functions by Partial Fractions . . . . . . 3111 Information for Students in MATH 141 2010 01 C.18 Supplementary Notes for the Lecture of February 10th, 2010 . . . 3117 C.18.1 §7.4 Integration of Rational Functions by Partial Fractions (conclusion) 3117 C.19 Supplementary Notes for the Lecture of February 12th, 2010 . . . 3124 C.19.1 §7.5 Strategy for Integration . . . . . . . . . . . 3124 C.19.2 §7.6 Integration Using Tables and Computer Algebra Systems (OMIT) . 3129 C.19.3 §7.7 Approximate Integration (OMIT) . . . . . 3129 C.20 Supplementary Notes for the Lecture of February 15th, 2010 . . . 3130 C.20.1 §7.8 Improper Integrals . 3130 C.21 Supplementary Notes for the Lecture of February 17th, 2010 . . . 3140 C.21.1 §8.1 Arc Length . . . . 3140 C.21.2 §8.2 Area of a Surface of Revolution . . . . . . 3146 C.22 Supplementary Notes for the Lecture of February 19th, 2010 . . . 3147 C.22.1 §8.2 Area of a Surface of Revolution (conclusion) 3147 C.22.2 §8.3 Applications to Physics and Engineering (OMIT) 3150 C.22.3 §8.4 Applications to Economics and Biology (OMIT) 3151 C.22.4 §8.5 Probability (OMIT) 3151 C.23 Supplementary Notes for the Lecture of March 01st, 2010 . . . . 3152 C.23.1 §10.1 Curves Defined by Parametric Equations . . 3152 C.23.2 §10.2 Calculus with Parametric Curves . . . . . . 3156 C.24 Supplementary Notes for the Lecture of March 03rd, 2010 . . . . 3158 C.24.1 §10.2 Calculus with Parametric Curves (continued) 3158 C.25 Supplementary Notes for the Lecture of March 05th, 2010 . . . . 3164 C.25.1 §10.2 Calculus with Parametric Curves (conclusion) 3164 C.25.2 §10.3 Polar Coordinates 3165 C.26 Supplementary Notes for the Lecture of March 08th, 2010 . . . . 3172 C.26.1 §10.3 Polar Coordinates (continued) . . . . . . . 3172 C.26.2 §10.4 Areas and Lengths in Polar Coordinates . . 3179 C.27 Supplementary Notes for the Lecture of March 10th, 2010 . . . . 3181 C.27.1 §10.4 Areas and Lengths in Polar Coordinates (continued) . . . . . . . . . 3181 C.28 Supplementary Notes for the Lecture of March 12th, 2010 . . . . 3195 C.28.1 §10.4 Areas and Lengths in Polar Coordinates (conclusion) . . . . . . . . . 3195 C.28.2 §10.5 Conic Sections . . 3196 C.28.3 §11.1 Sequences . . . . 3197 C.28.4 Sketch of Solutions to Problems on the Final Examination in MATH 141 2005 01 . . . . . . . . . . . . 3198 C.29 Supplementary Notes for the Lecture of March 15th, 2010 . . . . 3207 C.29.1 §11.1 Sequences (conclusion) . . . . . . . . . . 3207 C.29.2 §11.2 Series . . . . . . . 3210 C.30 Supplementary Notes for the Lecture of March 17th, 2010 . . . . 3213 C.30.1 §11.2 Series (conclusion) 3213 C.30.2 §11.3 The Integral Test and Estimates of Sums . 3217 C.31 Supplementary Notes for the Lecture of March 19th, 2010 . . . . 3219 C.31.1 §11.3 The Integral Test and Estimates of Sums (conclusion) . . . . . . . 3219 Information for Students in MATH 141 2010 01 C.32 Supplementary Notes for the LecC.39 Supplementary Notes for the Lecture of March 22nd, 2010 . . . . 3225 ture of Monday, April 12th, 2010 3301 C.32.1 §11.4 The Comparison C.39.1 Final Examination in MATH Tests . . . . . . . . . . 3225 141 2009 01 (Version 4, C.32.2 Sketch of Solutions to Probcontinued) . . . . . . . 3301 lems on the Final ExamC.40 Supplementary Notes for the Lecination in MATH 141 2006 ture of Wednesday, April 14th, 01 . . . . . . . . . . . . 3230 2010 . . . . . . . . . . . . . . . 3308 C.33 Supplementary Notes for the LecC.40.1 Final Examination in MATH ture of March 24th, 2010 . . . . 3245 141 2009 01 (Version 4, C.33.1 §11.5 Alternating Series 3245 concclusion) . . . . . . 3308 C.33.2 Solutions to Problems on D Problem Assignments from Previous Years 5001 the Final Examination in D.1 1998/1999 . . . . . . . . . . . . 5001 MATH 141 2007 01 . . 3249 D.1.1 Assignment 1 . . . . . . 5001 C.34 Supplementary Notes for the LecD.1.2 Assignment 2 . . . . . . 5001 ture of March 26th, 2010 . . . . 3265 D.1.3 Assignment 3 . . . . . . 5002 C.34.1 §11.6 Absolute ConverD.1.4 Assignment 4 . . . . . . 5002 gence and the Ratio and D.1.5 Assignment 5 . . . . . . 5002 Root Tests . . . . . . . . 3265 D.2 1999/2000 . . . . . . . . . . . . 5003 C.35 Supplementary Notes for the LecD.2.1 Assignment 1 . . . . . . 5003 ture of March 29th, 2010 . . . . 3268 D.2.2 Assignment 2 . . . . . . 5004 C.35.1 §11.6 Absolute ConverD.2.3 Assignment 3 . . . . . . 5006 gence and the Ratio and D.2.4 Assignment 4 . . . . . . 5007 Root Tests (conclusion) . 3268 D.2.5 Assignment 5 . . . . . . 5009 C.36 Supplementary Notes for the LecD.2.6 Assignment 6 . . . . . . 5010 ture of Wednesday, March 31st, D.3 2000/2001 . . . . . . . . . . . . 5012 2010 . . . . . . . . . . . . . . . 3272 D.4 2001/2002 . . . . . . . . . . . . 5012 C.36.1 §11.7 Strategy for TestD.5 MATH 141 2003 01 . . . . . . . 5012 ing Series . . . . . . . . 3272 D.6 MATH 141 2004 01 . . . . . . . 5012 C.37 Supplementary Notes for the LecD.7 MATH 141 2005 01 . . . . . . . 5013 ture of Wednesday, April 7th, 2010 3278 D.7.1 Written Assignment W1 5013 C.37.1 Final Examination in MATH D.7.2 Written Assignment W2 5014 141 2008 01 (one version) 3278 D.7.3 Written Assignment W 5016 3 C.37.2 Draft Solutions to the FiD.7.4 Written Assignment W 5017 4 nal Examination in MATH D.7.5 Written Assignment W5 5019 141 2009 01 (Version 4) 3293 D.8 MATH 141 2006 01 . . . . . . . 5021 C.38 Supplementary Notes for the LecD.8.1 Solution to Written Asture of Friday, April 09th, 2010 . 3297 signment W1 . . . . . . 5021 C.38.1 Final Examination in MATH D.8.2 Solution to Written As141 2009 01 (Version 4, signment W2 . . . . . . 5024 continued) . . . . . . . 3297 Information for Students in MATH 141 2010 01 D.8.3 Solutions to Written Assignment W3 . . . . . . D.8.4 Solutions to Written Assignment W4 . . . . . . D.8.5 Solutions to Written Assignment W5 . . . . . . D.9 MATH 141 2007 01 . . . . . . . E Quizzes from Previous Years E.1 MATH 141 2007 01 . . . . . . . E.1.1 Draft Solutions to Quiz Q1 E.1.2 Draft Solutions to Quiz Q2 E.1.3 Draft Solutions to Quiz Q3 E.1.4 Draft Solutions to Quiz Q4 E.2 MATH 141 2008 01 . . . . . . . E.2.1 Draft Solutions to Quiz Q1 E.2.2 Draft Solutions to Quiz Q2 E.2.3 Draft Solutions to Quiz Q3 E.2.4 Draft Solutions to Quiz Q4 E.3 MATH 141 2009 01 . . . . . . . E.3.1 Draft Solutions to Quiz Q1 E.3.2 Draft Solutions to Quiz Q2 E.3.3 Draft Solutions to Quiz Q3 F.8 5025 F.9 5028 5030 5032 5033 5033 5033 5043 5055 5070 5086 5086 5099 5110 5121 5134 5134 5140 5146 F Final Examinations from Previous Years 5152 F.1 Final Examination in Mathematics 189-121B (1996/1997) . . . 5152 F.2 Final Examination in Mathematics 189-141B (1997/1998) . . . 5153 F.3 Supplemental/Deferred Examination in Mathematics 189-141B (1997/1998) . . . . . . . . . . . 5155 F.4 Final Examination in Mathematics 189-141B (1998/1999) . . . 5156 F.5 Supplemental/Deferred Examination in Mathematics 189-141B (1998/1999) . . . . . . . . . . . 5158 F.6 Final Examination in Mathematics 189-141B (1999/2000) . . . 5160 F.7 Supplemental/Deferred ExaminaG tion in Mathematics 189-141B (1999/2000) . . . . . . . . . . . 5161 F.10 F.11 F.12 F.13 F.14 F.15 F.16 F.17 F.18 F.19 F.20 F.21 F.22 F.23 F.24 Final Examination in Mathematics 189-141B (2000/2001) . . . Supplemental/Deferred Examination in Mathematics 189-141B (2000/2001) . . . . . . . . . . . Final Examination in Mathematics 189-141B (2001/2002) . . . Supplemental/Deferred Examination in Mathematics 189-141B (2001/2002) . . . . . . . . . . . Final Examination in MATH 141 2003 01 . . . . . . . . . . . . . Supplemental/Deferred Examination in MATH 141 2003 01 . . . Final Examination in MATH 141 2004 01 . . . . . . . . . . . . . Supplemental/Deferred Examination in MATH 141 2004 01 . . . Final Examination in MATH 141 2005 01 . . . . . . . . . . . . . Supplemental/Deferred Examination in MATH 141 2005 01 . . . Final Examination in MATH 141 2006 01 (One version) . . . . . Supplemental/Deferred Examination in MATH 141 2006 01 . . . Final Examination in MATH 141 2007 01 (One version) . . . . . Supplemental/Deferred Examination in MATH 141 2007 01 (One version) . . . . . . . . . . . . . Final Examination in MATH 141 2008 01 (one version) . . . . . . Supplemental/Deferred Examination in MATH 141 2008 01 (one version) . . . . . . . . . . . . . Final Examination in MATH 141 2009 01 (one version) . . . . . . 5162 5164 5165 5167 5169 5171 5172 5181 5185 5189 5192 5195 5199 5203 5207 5217 5221 WeBWorK 6001 G.1 Frequently Asked Questions (FAQ) 6001 G.1.1 Where is WeBWorK? . 6001 Information for Students in MATH 141 2010 01 G.1.2 Do I need a password to G.1.14 How many attempts may use WeBWorK? . . . . 6001 I make to solve a particG.1.3 Do I have to pay an adular problem on WeBditional fee to use WeBWorK? . . . . . . . . . WorK? . . . . . . . . . 6001 G.1.15 Will all WeBWorK asG.1.4 When will assignments signments have the same be available on WeBWorK? 6002 length? the same value? G.1.5 Do WeBWorK assignG.1.16 Is WeBWorK a good inments cover the full range dicator of examination perof problems that I should formance? . . . . . . . . be able to solve in this course? . . . . . . . . . 6002 H Contents of the DVD disks for Larson/Hostetler/Edwards G.1.6 May I assume that the distribution of topics on quizzes and final examList of Tables inations will parallel the distribution of topics in 1 Schedule and Locations of Tuthe WeBWorK assigntorials, as of April 17, 2010. . . ments? . . . . . . . . . 6002 2 Tutors’ Coordinates, as of April G.1.7 WeBWorK provides for 17, 2010 . . . . . . . . . . . . . different kinds of “Dis3 Summary of Course Requirements, play Mode”. Which should as of April 17, 2010; (all dates I use? . . . . . . . . . . 6002 are subject to change) . . . . . . G.1.8 WeBWorK provides for 4 Some Antiderivatives . . . . . . printing assignments in 5 Very Short Table of Indefinite In“Portable Document Format” tegrals . . . . . . . . . . . . . . (.pdf), “PostScript” (.ps) and “TEXSource” forms. Which should I use? . . 6003 List of Figures G.1.9 What is the relation between WeBWorK and We1 The region(s) bounded by y = bCT? . . . . . . . . . . 6003 x2 and y = x4 . . . . . . . . . . G.1.10 What do I have to do on 2 The region(s) bounded by y = WeBWorK? . . . . . . 6003 sin x, y = sin 2x between x = 0 G.1.11 How can I learn how to and x = π2 . . . . . . . . . . . . use WeBWorK? . . . . 6004 3 The region(s) bounded by y = G.1.12 Where should I go if I 8 − x2 , y = x2 between x = ±3 . have difficulties with WeB4 The region(s) bounded by y = √ WorK ? . . . . . . . . . 6004 x + 2, y = x between x = 0 G.1.13 Can the WeBWorK sysand x = 4 . . . . . . . . . . . . tem ever break down or 5 Regions for Example C.31 . . . degrade? . . . . . . . . 6004 6005 6005 6005 6101 4 5 11 3025 3026 3053 3055 3056 3058 3061 Information for Students in MATH 141 2010 01 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 The region(s) bounded by x + y = 3 and x = 4 − (y − 1)2 . . . The curve x = cos θ+sin 2θ, y = sin θ + cos 2θ . . . . . . . . . . The cardioid with equation r = 2(1 − sin θ) . . . . . . . . . . . The lima¸con r = 1 − 3 cos θ, . . The spiral with equation r = θ, (θ ≥ 0) . . . . . . . . . . . . . . The spiral with equation r = θ, (θ ≤ 0) . . . . . . . . . . . . . . The full spiral with equation r = θ, −∞ < θ < +∞ . . . . . . . . The “4-leafed rose” with equation r = sin 2θ, . . . . . . . . . . The “5-leafed rose” with equation r = sin 5θ, . . . . . . . . . . The lemniscates r2 = sin 2θ, r2 = cos 2θ . . . . . . . . . . . . . . Intersecting polar curves r = 1+ sin θ, r2 = 4 sin θ . . . . . . . . Curves r = sin θ, r = cos θ . . . The strophoid r = 2 cos θ − sec θ Curves r = 2 + sin θ, r = 3 sin θ . Intersections of the limac¸on r = 1 − 2 cos θ with the circle r = 1 . Intersections of the curve r = sec θ with the circle r = 1 . . . . The curves with equations r = 1 − cos θ, (θ ≤ 0), and r = 1 + sin θ, and the point 12 , π2 . . . . The cardioids with equations r = 2 + 2 sin θ, r = 6 − 6 sin θ . . . . The region bounded by cardioids r = 2+2 sin θ, r = 6−6 sin θ and containing the point (r, θ) = (1, 0) The curves with equations r = 4 + 2 cos θ, r = 4 cos θ + 5 . . . The curves with equations r = 3 + 3 cos θ, (0 ≤ θ ≤ 2π), and r = 9 cos θ, (0 ≤ θ ≤ π) . . . . . The limac¸on r = 1 + 2 sin θ . . . 3070 3158 3170 3172 3174 3175 3176 3177 3178 3179 3182 3184 3186 3188 3191 3193 3243 3262 3263 3291 3304 5071 Information for Students in MATH 141 2010 01 1 1 General Information Distribution Date: January 04th, 2010 (all information is subject to change) Pages 1 - 20 of these notes may be considered the Course Outline for this course. These notes may undergo minor corrections or updates during the term: the definitive version will be the version accessible at http://www.math.mcgill.ca/brown/math141b.html or on myCourses, at http://www.mcgill.ca/mycourses/ 1 Students are advised not to make assumptions based on past years’ operations, as some of the details concerning this course could be different from past years. Publications other than this document may contain unreliable information about this course. All details of the course could be subject to discretionary change in case of force majeure. 1.1 Force Majeure In the event of extraordinary circumstances beyond the University’s control, all details of this course, including the content and/or evaluation scheme are subject to change. 1 Please note that the statements about MATH 141 in an SUS publication called Absolute Zero were not given to instructors of this course to check, and some of them may not be currently correct. Information for Students in MATH 141 2010 01 2 1.2 Instructors and Times INSTRUCTOR: LECTURE SECTION: CRN: OFFICE: OFFICE HOURS: (subject to change) TELEPHONE: E-MAIL:2 CLASSROOM: CLASS HOURS: Prof. W. G. Brown (Course Coordinator) 1 576 BURN 1224 W 15:45→16:45 F 10:00→11:00 or by appointment (514)-398-3836 BROWN@ MATH.MCGILL.CA ADAMS AUD MWF 11:35–12:25 h. Dr. S. Shahabi Dr. A. Hundemer 2 577 BURN 1243 F 09:30→11:30 (tentative) 3 578 BURN 1128 MW 15:30→16:25 (514)-398-3803 SHAHABI@ MATH.MCGILL.CA LEA 219 MWF 11:35–12:25 h. (514)-398-5318 HUNDEMER@ MATH.MCGILL.CA ADAMS AUD MW 16:35–17:55 h. 1.3 Calendar Description 1.3.1 Calendar Description MATH 1414 CALCULUS 2. (4 credits; 3 hours lecture; 2 hours tutorial. Prerequisites: MATH 139 or MATH 140 or MATH 150. Restriction: Not open to students who have taken MATH 121 or CEGEP objective 00UP or equivalent; not open to students who have taken or are taking MATH 122 or MATH 130 or MATH 131, except by permission of the Department of Mathematics and Statistics. Each Tutorial section is enrolment limited.) The definite integral. Techniques of integration. Applications. Introduction to sequences and series. Students Lacking the Prerequisite will, when discovered, be removed from the course. Students without the prerequisite (or standing in a course recognized by the Admissions Office as being equivalent to MATH 140) should not assume that, in possibly permitting MINERVA to accept their registration for MATH 141, the University was tacitly approving their registration without the prerequisite. In particular, students who obtained a grade of F in MATH 139/140/150 are expressly excluded from registration in MATH 141, even if they registered in the course before the failed or missed examination. 2 Please do not send e-mail messages to your instructors through the WebCT or WeBWorK3 systems; rather, use the addresses given in §1.2 on page 2. 3 E-mail messages generated by the Feedback command in WeBWorK should be used sparingly, and confined to specific inquiries about WeBWorK assignments. 4 The previous designation for this course was 189-141, and the version given in the winter was labelled 189-141B; an earlier number for a similar course was 189-121. Information for Students in MATH 141 2010 01 3 1.3.2 Late transfer from MATH 151/MATH 152 Some students from MATH 151 or MATH 152 may be permitted to transfer into MATH 141 after the end of the Change of Course Period. If your instructor in MATH 151 or 152 advises you that you are in this category, please send an e-mail message to Professor Brown as soon as your transfer has been approved.5 1.4 Tutorials 1.4.1 Tutorial Times, Locations, and Personnel (subject to change) Every student must be registered in one lecture section and one tutorial section for this course. Tutorials begin in the week of January 11th, 2010. The last tutorials in all Tuesday, Wednesday, and Thursday tutorial sections will be in the week beginning Monday, April 05th, 2010; the last tutorial in Monday tutorial sections will be on Monday, April 12th, 2010; the date of the last tutorial in Friday tutorial sections will be announced later in the term. Table 1 gives times, locations, and the tutor’s name for each of the tutorials; Table 2 gives the tutors’ coordinates. The information in these tables is subject to change. We try to publicize changes but sometimes we are not informed in advance.6 You are expected to write quizzes only in the tutorial section in which you are registered.7 You do not have a licence to move from one tutorial section to another at will, even if you find the time, location, or personnel of your tutorials either temporarily or permanently inconvenient; in the latter case the onus is on you to transfer formally to another tutorial section, to change your other classes, or to drop MATH 141 2010 01. Please remember that transfers must be completed by the Course Change (drop/add) deadline (January 19th, 2010), and are subject to the maximum capacities established for each tutorial section8 1.4.2 Teaching Assistants (TA’s) The tutors in MATH 141 2010 01 are graduate students in Mathematics and Statistics. Like you, they are students, albeit at the graduate level; they have deadlines and commitments and personal lives, and the time they have available for MATH 141 is limited and controlled by a collective agreement (union contract). Please respect the important functions that our tutors provide, and do not ask them for services they are not expected to perform: 5 This is to ensure that your WeBWorK account is opened, and that your date of entry to the course is recorded. The current room for your tutorial should always be available by clicking on “Class Schedule” on MINERVA FOR STUDENTS, http://www.mcgill.ca/minerva-students/. 7 In some time slots there may be several tutorial sections, meeting in different rooms. 8 Your instructors do not have the ability to change the maximum capacities of tutorials. 6 Information for Students in MATH 141 2010 01 4 # CRN Day Begins Ends Room T004 T005 T006 T007 T008 T009 T010 T011 T012 T013 T014 T015 T016 T017 T018 T019 T020 T021 579 580 581 582 583 584 585 586 587 588 589 590 591 2071 2072 8194 8840 8841 Fri Tue Tue Tue Thurs Thurs Thurs Mon Mon Mon Wed Wed Wed Mon Wed Tue Wed Fri 13:35 14:05 16:05 16:05 14:05 16:05 16:05 13:35 14:35 14:35 13:35 14:35 14:35 13:35 13:35 08:05 15:35 15:35 15:25 15:55 17:55 17:55 15:55 17:55 17:55 15:25 16:25 16:25 15:25 16:25 16:25 15:25 15:25 09:55 17:25 17:25 BURN 1B39 ARTS 260 BURN 1B23 BURN 1B24 BURN 1B39 BURN 1B39 BURN 1B23 ARTS W-20 BURN 1B36 BURN 1B24 ARTS W-20 LEA 14 BURN 1B36 ENGMD 276 BURN 1B24 BURN 1B23 ENGMD 279 ENGMD 256 Tutor J. Feys H. Bigdely L. Candelori F. Castella Y. Canzani X. Zhang A.-P. Grecianu J. Macdonald ´ M. Prevost J. Tousignant-Barnes J. Restrepo B. Taji A. Tcheng A. Tomberg P. Rempel Y. Rabhi A. Farooqui Y. Zhao Some of these room assignments could change before or early in the beginning of the term, as we have a pending request to upgrade some of the rooms. In any case, all assignments are subject to change. Table 1: Schedule and Locations of Tutorials, as of April 17, 2010. • Outside of the normal quiz times in their tutorials, tutors are neither expected nor authorized to administer a special quiz or a quiz that has already been administered to others. • Tutors in MATH 141 2010 01 are not permitted to offer paid, private tuition to students in any tutorial section of this course. 1.4.3 Friday, April 02nd, 2010 and Monday, April 05th, 2010 These two lecture/tutorial days are lost because of the Easter holidays. While the lectures will resume on Wednesday, April 07th, 2010, and the total number of lecture hours is similar to past years, there will be some disruption to Monday and Friday tutorials: the Monday tutorials will meet on Monday, April 12th, 2010, but there is no scheduled Friday available to complete the Friday tutorials; alternative arrangements for students in Friday tutorials will be announced later in the term. Information for Students in MATH 141 2010 01 Tutor E-mail address Bigdely, H. Candelori, L. Canzani, Y. Castella, F. Farooqui, A. Feys, J. Grecianu, A.-P. Macdonald, J. ´ Prevost, M. Rabhi, Y. Rempel, P. Restrepo, J. Taji, B. Tcheng, A. Tomberg, A. Tousignant-Barnes, J. Zhang, X. Zhao, Y.
[email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected] 5 Office BURN Day(s) Begins Office Hours Ends Day 1030 1032 1133 1008 1023 1020 1007 1030 1117 1021 1140 1117 1031 1029 1031 1032 1020 1034 M T T M TTh T Th M TTh M Th T F TTh W TTh M F 12:00 11:30 12:30 08:30 16:00 17:00 14:30 09:30 13:00 13:00 10:00 10:30 15:00 14:30 12:30 09:00 08:30 12:30 15:00 13:00 14:30 10:30 17:30 18:30 16:00 11:30 14:30 14:30 13:00 12:00 18:00 16:00 15:30 10:30 11:30 15:30 Begins Ends T Th M 18:30 13:00 14:30 20:00 14:00 15:30 W F M 09:00 13:00 15:30 10:30 14:30 16:30 T 10:00 11:30 T 14:30 16:00 During her/his office hours, a tutor is available to all students in the course, not only to the students of her/his tutorial section. For last minute changes, see myCourses (WebCT). Table 2: Tutors’ Coordinates, as of April 17, 2010 1.5 Evaluation of Your Progress 1.5.1 Your final grade (See Table 3, p. 11) In the event of extraordinary circumstances beyond the University’s control, the content and/or evaluation scheme in this course is subject to change. Your grade in this course will be a letter grade, based on a percentage grade computed from the following components: 1. Assignments submitted over the Web: Six9 (6) WeBWorK homework assignments — counting together for 10%. 2. Quizzes given at the tutorials: Four (4), counting together for 20%.10 9 Numbers of assignments, quizzes, etc., are as planned as of the date of this version of these notes. Students must be prepared for the possibility that it could be necessary to adjust these numbers during the term. If there are any changes, these will be announced on myCourses, by broadcast e-mail messages, or by announcements at the lectures. 10 But be warned: students who fail to write quizzes are often at risk in this course. The quizzes are mainly a learning, rather than a testing experience. You need the information that comes from writing quizzes in a Information for Students in MATH 141 2010 01 6 3. The final examination — counting for 70%. Where a student’s performance on the final examination is superior to her performance on the tutorial quizzes, the final examination grade will replace the quiz grades in the calculations; in that case the grade on the final examination will count for 90% of the final grade. It is not planned to permit the examination grade to replace the grades on WeBWorK assignments. 1.5.2 WeBWorK 1. The WeBWorK system, developed at the University of Rochester — is designed to expose you to a large number of drill problems, and where plagiarism is discouraged. WeBWorK is accessible only over the Internet. Details on how to sign on to WeBWorK are contained in Appendix G to these notes, page 6001. Only answers submitted by the due date and time will count. The WeBWorK assignments which count in your term mark will be labelled A1 , . . ., A6 . 2. Due dates and times for WeBWorK assignments. Most due dates for WeBWorK assignments will be on specified Sundays, about 23:30h; last minute changes in the due dates may be announced either on WeBWorK, on myCourses, or by an e-mail message11 As mentioned in the WeBWorK FAQ (cf. Appendix G), if you leave your WeBWorK assignment until the hours close to the due time on the due date, you should not be surprised if the system is slow to respond. This is not a malfunction, but is simply a reflection of the fact that other students have also been procrastinating! To benefit from the speed that the system can deliver under normal conditions, do not delay your WeBWorK until the last possible day! If a systems failure interferes with the due date of an assignment, arrangements may be made to change that date, and an e-mail message may be broadcast to all users (to the e-mail addresses on record), or a note posted in the course announcements on myCourses; but slowness in the system just before the due time will not normally be considered a systems failure.12 3. Numbers of permitted attempts at WeBWorK questions. While the number of times you may attempt each problem on WeBWorK An will be limited, there will be a companion “Practice” Assignment Pn (n = 1, 2, . . . , 6) with an unlimited number of attempts at similar problems, but in which the specific data may be different. Thus you have group, and observing whether your performance was at an appropriate level. Students who deny themselves this experience often undergo a rude awakening at the final examination. 11 Be sure that your e-mail addresses are correctly recorded. See 4, p. 19 of these notes. 12 Should you find that the system is responding slowly, do not submit your solutions more than once; you may deplete the number of attempts that have been allowed to you for a problem: this will not be considered a systems failure. Information for Students in MATH 141 2010 01 7 the opportunity to prepare yourself on the Practice assignment before attempting the actual assignment. The practice assignments DO NOT COUNT in your term mark, even though a grade is recorded.. Practice assignment Pn is normally due one week before assignment An . Another assignment which will not count will be Practice Assignment P0 , which is directed to students who are not familiar with the WeBWorK system. 1.5.3 Written Submissions. In accord with McGill University’s Charter of Students’ Rights, students in this course have the right to submit in English or in French any written work that is to be graded; course materials are normally provided only in English. Written Assignments. There will be no Written Assignments in MATH 141 2010 01.13 1.5.4 Quizzes at the Tutorials. 1. There will be 4 quizzes, numbered Q1 , Q2 , Q3 , Q4 , administered at the tutorials. These quizzes will be graded, and returned. The primary purpose of a quiz is to diagnose possible gaps in your understanding. In the grading formula the quiz component of the final grade will be replaced by the final examination grade, if that is to a student’s advantage. 2. Students may write a quiz only in the tutorial in which they are registered. 3. Medical absences. If you have missed or expect to miss a quiz for a valid reason (medical or otherwise), please communicate directly with Professor Brown, providing a copy of the medical or other supporting documents; do not contact your TA. Authorized medical absences can be accommodated only through averaging, as students in MATH 141 are never permitted to write a quiz in any tutorial section other than the one in which they are registered. We cannot offer “makeup” sessions for quizzes. 4. To prepare for a quiz you should be working exercises in the textbook based on the material currently under discussion at the lectures, and you should have attempted any open WeBWorK assignments. But, unlike the WeBWorK assignments — where the emphasis is on correct answers alone — students may be expected to provide full solutions to 13 While it is not required for grading purposes, students are urged to keep a systematic record of written solutions to problems in the textbook. This could be in the form of a workbook, or a file folder, but should be orderly enough that you can look back at a later time to see your solutions. You are invited to bring such a file to TA’s or instructors at their office hours, to receive advice about the quality and correctness of your solutions. Information for Students in MATH 141 2010 01 8 some or all problems on quizzes.14 The quizzes may examine on only a sampling of topics. Students should not assume that topics not examined are in any subsidiary parts of the syllabus. 5. Your tutors will normally bring graded quizzes to the tutorial to be returned to you. University regulations do not permit us to leave unclaimed materials bearing names and student numbers in unsupervised locations; you may be able to recover an unclaimed quiz from the tutor who graded it, during her/his regular office hours. Be sure to attend the tutorial following a quiz15 , as claims of incorrect recording of a quiz or assignment grade will need to be substantiated by a graded paper. 6. Your quiz grades on assignments and quizzes will be posted on myCourses within about 2 weeks after they become available. Your WeBWorK grades may not be transferred to myCourses until the end of the term, but will be visible on the WeBWorK site. 1.5.5 Final Examination A 3-hour-long final examination will be scheduled during the regular examination period for the winter term (April 15th, 2010 through April 30th, 2010). You are advised not to make any travel arrangements that would prevent you from being present on campus at any time during this period.16 1.5.6 Supplemental Assessments 1. Supplemental Examination. There will be a supplemental examination in this course. (For information about Supplemental Examinations, see http://www.mcgill.ca/artscisao/departmental/examination/supplemental/.) 2. There is No Additional Work Option. “Will students with marks of D, F, or J have the option of doing additional work to upgrade their mark?” No. (“Additional Work” refers to an option available in certain Arts and Science courses, but not available in MATH 141 2010 01.) 14 In Math 141 the general rule for quizzes is that full solutions are expected to all problems, unless you receive explicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in the Student Solutions Manual [9] to the textbook can serve as a guide to what should be included in a “full” solution. 15 The return of Quiz Q1 may be delayed to the 2nd week after the quiz was written. 16 Your instructors learn the date of your examination at the same time as you do — when the Provisional examination timetable is published. Information for Students in MATH 141 2010 01 9 1.5.7 Machine Scoring: “Will the final examination be machine scored?” Multiple-Choice Problems It is possible that the final examination, or part of it, could be machine scored. Multiple choice problems, possibly machine scored, could also appear on some quizzes. (Machine grading, if implemented in whole or in part, would be a change from the practice of past years in this course. Such a change was introduced into the most recent examination in MATH 140, which included a substantial number of machine-graded multiple choice questions.) Answer-Only Problems. Some of the problems on quizzes and/or the final examination — possibly a substantial number of them — may request that the answer only be given, and may not carry part marks which could be based on the work leading up to the answer. 1.5.8 Plagiarism. While students are not discouraged from discussing methods for solving WeBWorK assignment problems with their colleagues, all work that you submit must be your own. The Senate of the University requires the following message in all course outlines: “McGill University values academic integrity. Therefore all students must understand the meaning and consequences of cheating, plagiarism and other academic offences under the Code of Student Conduct and Disciplinary Procedures. (See http://www.mcgill.ca/integrity for more information). “L’universit´e McGill attache une haute importance a` l’honnˆetet´e acad´emique. Il incombe par cons´equent a` tous les e´ tudiants de comprendre ce que l’on entend par tricherie, plagiat et autres infractions acad´emiques, ainsi que les cons´equences que peuvent avoir de telles actions, selon le Code de conduite de l’´etudiant et des proc´edures disciplinaires. (Pour de plus amples renseignements, veuillez consulter le site http://www.mcgill.ca/integrity).” It is a violation of University regulations to permit others to solve your WeBWorK problems, or to extend such assistance to others; you could be asked to sign a statement attesting to the originality of your work. The Handbook on Student Rights and Responsibilities17 states in ¶A.I.15(a) that “No student shall, with intent to deceive, represent the work of another person as his or her own in any academic writing, essay, thesis, research report, project or assignment submitted in a course or program of study or represent as his or her own an entire essay or work of another, whether the material so represented constitutes a part or the entirety of the work submitted.” 17 http://upload.mcgill.ca/secretariat/greenbookenglish.pdf Information for Students in MATH 141 2010 01 10 You are also referred to the following URL: http://www.mcgill.ca/integrity/studentguide/ Other Fraud. It is a serious offence to alter a graded quiz paper and return it to the tutor under the pretense that the work was not graded properly. 1.5.9 Corrections to grades Grades will eventually be posted on myCourses. If you believe a grade has been recorded incorrectly, you must advise your tutor not later than 4 weeks after the grade has been posted, and not later than the day before of the final examination whichever of these dates is earlier. It is hoped that grades will be posted within 2 weeks of the due date. You will have to present the graded quiz to support your claim, which must be submitted to the tutor that graded the quiz. If he/she believes there has been an error, the tutor will advise Professor Brown. New corrections to the myCourses posting will appear the next time grades are uploaded to myCourses. 1.6 Published Materials 1.6.1 Required Text-Book The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01169-X, [1]. This book is the first half of J. Stewart, CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01166-5, [2]; this edition covers the material for Calculus 3 (MATH 222) as well, but is not the text-book for that course at the present time. The textbook will be sold in the McGill Bookstore bundled with its Student Solutions Manual (see below). The ISBN number for the entire bundle is 0-495-42966-X. 1.6.2 Optional Reference Books Students are urged to make use of the Student Solution Manual: • D. Anderson, J. A. Cole, D. Drucker, STUDENT SOLUTIONS MANUAL FOR STEWART’S SINGLE VARIABLE CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01240-8, [3]. This book is also sold “bundled” with the text book; we expect the Bookstore to stock the bundle numbered ISBN 0-495-42966-X [4]. The publishers of the textbook and Student Solutions Manual also produce UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 Item WeBWorK Assignments (cf. §1.5.2) 10% Quizzes (cf. §1.5.4) 20% or 0% Final Exam 70% or 90% Supplemental Exam (cf. §1.5.6.1) # P0 P1 A1 P2 A2 P3 A3 P4 A4 P5 A5 P6 A6 Due Date Q1 Q2 Q3 Q4 18–22 Jan 10 08–12 Feb 10 08–12 Mar 10 22–26 Mar 10 15–30 Apr 10 17 Jan 10 24 Jan 10 31 Jan 10 07 Feb 10 14 Feb 10 21 Feb 10 28 Feb 10 07 Mar 10 14 Mar 10 28 Mar 10 28 Mar 10 12 Apr 10 18–19 Aug 10 11 Details DOES NOT COUNT: introduces WeBWorK DOES NOT COUNT; practice for A1 DOES NOT COUNT; practice for A2 DOES NOT COUNT; practice for A3 DOES NOT COUNT; practice for A4 DOES NOT COUNT; practice for A5 DOES NOT COUNT; practice for A6 A1 –A6 count equally, but may have different numbers of problems. Quizzes Q1 — Q4 count equally, but the quizzes may be of different lengths. Date of exam to be announced by Faculty Only for students who do not obtain standing at the final. Supplemental exams count in your average like taking the course again; exam counts for 100%. Table 3: Summary of Course Requirements, as of April 17, 2010; (all dates are subject to change) • a “Study Guide”, designed to provide additional help for students who believe they require it: R. St. Andre, STUDY GUIDE FOR STEWART’S SINGLE VARIABLE CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0495-01239-4, [5]. (The “Study Guide” resembles the Student Solution Manual in appearance: be sure you know what you are buying.) • a “Companion” which integrates a review of pre-calculus concepts with the contents of Math 140, including exercises with solutions: D. Ebersole, D. Schattschneider, A. Sevilla, K. Somers, A COMPANION TO CALCULUS. Brooks/Cole (1995), ISBN 0534-26592-8 [39]. Information for Students in MATH 141 2010 01 12 1.6.3 Recommended Video Materials Use of the following materials is recommended, but is not mandatory18 . Text-specific DVDs for Stewarts Calculus, early transcendentals, 6th edition [videorecording]. The publisher of Stewart’s Calculus has produced a series of videodisks, [?]. These will initially be available for reserve loan at the Schulich Library. There may not be DVD viewing equipment freely available in the library; the intention is that interested students borrow disks for viewing on their own equipment at home. Disk 1 covers Chapters 1-6 of the textbook. Videotapes for Stewart’s Calculus The publisher of Stewart’s Calculus had earlier produced a series of videotapes, [14] Video Outline for Stewart’s Calculus (Early Transcendentals), Fifth Edition. These will be available for reserve loan at the Schulich Library. There may not be VCR viewing equipment in the library; the intention is that interested students borrow a tape for viewing on their own equipment at home. Larson/Hostetler/Edwards DVD Disks A set of video DVD disks produced for another calculus book, [28] Calculus Instructional DVD Program, for use with (inter alia) Larson / Hostetler / Edwards, Calculus of a Single Variable: Early Transcendental Functions, Third Edition [29] is produced by the Houghton Mifflin Company. A copy has been requested to be placed on reserve in the Schulich Library. In Appendix H of these notes there are charts that indicate the contents of these disks that pertain to MATH 141. Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 6th Edition, (similar to [15]) 19 This CD-ROM is included with certain new copies of the textbook. It contains, after an enlightening “pep-talk” by the author, a discussion of some of the worked examples in the text-book, followed by a quiz for each section in the book. Some students may find the animations of the examples helpful, although the examples are all worked in the book. You might wish to try some of the quiz questions using paper and pencil, and then check your answers with those given on the CD. It is not recommended that you attempt to enter your answers digitally, as this is a time-consuming process, and uses a different input method from your WeBWorK assignments, which serve the same purpose. 18 No one will check whether you have used any of these aids; a student can obtain a perfect grade in the course without ever consulting any of them. No audio-visual or calculator aid can replace the systematic use of paper and pencil as you work your way through problems. But the intelligent use of some of these aids can deepen your understanding of the subject. However, the most important aid is the Student Solutions Manual to the textbook! 19 The version of this CD-ROM for the 6th edition is being catalogued by the Library; it may not be available at the beginning of the term. Information for Students in MATH 141 2010 01 13 1.6.4 Other Calculus Textbooks While students may wish to consult other textbooks, instructors and teaching assistants in Math 141 will normally refer only to the prescribed edition of the prescribed textbook for the course. Other books can be very useful, but the onus is on you to ensure that your book covers the syllabus to at least the required depth; where there are differences of terminology, you are expected to be familiar with the terminology of the textbook.20 In your previous calculus course(s) you may have learned methods of solving problems that appear to differ from those you find in the current textbook. Your instructors will be pleased to discuss any such methods with you personally, to ascertain whether they are appropriate to the present course. In particular, any methods that depend upon the use of a calculator, or the plotting of multiple points, or the tabulation of function values, or the inference of a trend from a graph should be regarded with scepticism. 1.6.5 Website These notes, and other materials distributed to students in this course, will be accessible through a link on the myCourses page for the course, as well as at the following URL: http://www.math.mcgill.ca/brown/math141b.html The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader, which many users have on their computers. This free software may be downloaded from the following URL: http://www.adobe.com/prodindex/acrobat/readstep.html 21 The questions on some old examinations will also be available as an appendix to these notes on the Web.22 Where revisions are made to distributed printed materials — for example these information sheets — we expect that the last version will be posted on the Web. The notes and WeBWorK will also be available via a link from the myCourses (WebCT) URL: http://mycourses.mcgill.ca 20 There should be multiple copies of the textbook on reserve in the Schulich library. At the time of this writing the current version appears to be 8.n. 22 There is no reason to expect the distribution of problems on quizzes or in assignments and examinations from previous years be related to the frequencies of any types of problems on the examination that you will be writing at the end of the term. 21 Information for Students in MATH 141 2010 01 14 1.7 Syllabus Section numbers in the following list refer to the text-book [1]. The syllabus will include all of Chapters 5, 6, 7, 8, 10, 11 with omissions, as listed below.23 Chapter 5: Integrals. §§5.1 – 5.5. The Midpoint Rule, defined in §5.2, and appearing from time to time subsequently, is not examination material. Chapter 6: Applications of Integration. §§6.1 – 6.3; §6.5. (§6.4 is not examination material, but Science students are urged to read it.) Chapter 7: Techniques of Integration. §§7.1 – 7.3; §7.4, excluding the Weierstrass substitution [1, Exercises 57-61]; §7.5 §7.8. (§7.6, intended for use in conjunction with integral tables and/or computer algebra systems, is not examination material, but students are advised to try to solve the problems manually; §7.7 requires the use of a calculator or a computer, and consequently is not examination material.) Chapter 8: Further Applications of Integration. §8.1, §8.2 only. (§§8.3, 8.4 are not examination material, but students are urged to read the applications relevant to their course of study; §8.5 is not examination material.) Chapter 9: Differential Equations. (No part of this chapter is examination material; however, students are urged to read §9.4 Exponential Growth and Decay). Chapter 10: Parametric Equations and Polar Coordinates. §§10.1 – 10.4. (§§10.5, 10.6 are not examination material. Chapter 11. Infinite Sequences and Series. §§11.1 – 11.7. (§§11.8–11.12 are not examination material; however, students are urged to peruse these sections.) Appendices Appendix G contains material shifted from [22, §5.6]. Students are expected to know the properties themselves, as they were discussed in MATH 139 and MATH 140. After the class has studied Chapter 5, the definition of the natural logarithm ln x will Zx dt . thenceforth be taken to be that given in the Appendix, as t 0 23 If a textbook section is listed below, you should assume that all material in that section is examination material even if the instructor has not discussed every topic in his lectures; however, the instructors may give you information during the term concerning topics that may be considered subsidiary. Do not assume that a topic is omitted from the syllabus if it has not been tested in a WeBWorK assignment or a quiz, or if it has not appeared on any of the old examinations in the course! Some topics to not lend themselves to this type of testing; others may have been omitted simply because of lack of space, or oversight. By the same token, you need not expect every topic in the course to be examined on the final examination. Information for Students in MATH 141 2010 01 15 Please do not ask the tutors to provide information as to which textbook sections should be emphasized. Unless you are informed otherwise by the instructors in the lecture sections or published notes — printed, or mounted on the Web — you should assume that all materials listed are included in the syllabus. You are not expected to be able to reproduce proofs of the theorems in the textbook. However, you could be expected to solve problems in which there might be unspecific real variables, rather than specific numbers, and which problems might look like textbook theorems.24 1.8 Preparation and Workload 1.8.1 Prerequisites. It is your responsibility as a student to verify that you have the necessary prerequisite. It would be foolish25 to attempt to take the course without it. Students who obtained only a grade of C in MATH 139 or MATH 140 would be advised to make a special effort to reinforce their foundations in differential calculus; if weakness in MATH 139 or MATH 140 was a consequence of poor preparation for that course, it is not too late to strengthen those foundations as well.26 The fact that MINERVA may permit you to register does not relieve you from the responsibility to observe university regulations concerning prerequisites, and exposes you to the risk of failure in a course for which you are nor properly prepared; students with an F in MATH 139 or MATH 140 could have their registration in MATH 141 annulled. The regulations are in place to protect you! 1.8.2 Calculators The use of calculators is not permitted in either quizzes or the examination in this course. Students whose previous mathematics courses have been calculator-oriented would be advised to make particular efforts to avoid the use of a calculator in solving problems in this course, in order to develop a minimal facility in manual calculation. This means that you are urged to do all arithmetic by hand. Students who use calculators when they answer their WeBWorK problems are undermining the usefulness of the programme to themselves: learn to use the built-in calculation capabilities that are present in WeBWorK. 24 The intention is that you should be learning how to solve problems, but should not have to memorize whole proofs from the textbook. 25 and contrary to McGill regulations 26 The reality of inflated grading at McGill or at your previous institution must not be overlooked: it could happen that students who obtained a grade higher than C in the prerequisite course do not have adequate skills to succeed in MATH 141! The onus is on you to seek help and to take remedial actions where necessary. Information for Students in MATH 141 2010 01 16 1.8.3 Self-Supervision This is not a high-school course, and McGill is not a high school. The monitoring of your progress before the final examination is largely your own responsibility. Students must not assume that they will be exposed in lectures and tutorials to detailed model solutions for every type of Calculus 2 problem. It is essential that you supplement these classes with serious work on your own, carefully reading the textbook and solving problems therein. While the tutors and instructors are available to help you, they cannot do so unless and until you identify the need for help. WeBWorK and quizzes are designed to assist you in doing this. If you encounter difficulties, take them to the tutors during one of their many office hours: you may attend the office hours of any tutor in the course, and are not restricted to those of the tutor of the tutorial in which you are registered. Time Demands of your Other Courses. Be sure to budget enough time to attend lectures and tutorials, for private study, and for the solution of many problems. Don’t be tempted to divert calculus study time to courses which offer instant gratification. While the significance of the tutorial quizzes in the computation of your grade is minimal, these are important learning experiences, and can assist you in gauging your progress in the course. This is not a course that can be crammed for: you must work steadily through the term if you wish to develop the facilities needed for a strong performance on the final examination. Lecture Times, and Preparation for the Lectures The lecture sections in MATH 141 2010 01 meet at the times that have been made available to us: early in the morning, or late in the afternoon. While these times may not appeal to you, you should not underestimate the damage you do to your expectations in the course by missing lectures, either occasionally — when you find it convenient to divert calculus time to other purposes — or systematically. To extract maximum benefit from the lectures, you should peruse the scheduled material before coming to class, trying some of the textbook problems; your instructors invite you to draw their attention to specific difficulties that you encounter before class in the textbook — it may be possible to respond to these difficulties during the lecture. Working Problems on Your Own. An effective way to master the calculus is through working large numbers of problems from the textbook. Your textbook was selected partly because of the availability of an excellent Student Solutions Manual [9]; this manual has brief but complete solutions to most of the odd-numbered exercises in the textbook. The skills you acquire in solving textbook problems could have much more influence on your final grade than either WeBWorK or the quizzes. Information for Students in MATH 141 2010 01 17 When to do the WeBWorK assignment. I recommend that you defer working WeBWorK problems until you have tried some of the easier odd-numbered problems in the textbook. For these you (should) have the Student Solutions Manual to help you check your work. Once you know that you have the basic concepts mastered, then is a good time to start working WeBWorK problems. But these should be done first from a printed copy of your assignment — not worked during real time online. The real uses of WeBWorK and the quizzes. Students often misunderstand the true significance of WeBWorK assignments and the quizzes. While both contribute to your grade, they can help you estimate the quality of your progress in the course. Quizzes are administered under examination conditions, so poor performance or non-performance on quizzes can provide an indicator of your expectations at the final examination; take proper remedial action if you are obtaining low grades on quizzes27 . Since WeBWorK is not completed under examination conditions, the grades you obtain may not be a good indicator of your expectations on the examination; if you require many attempts before being able to solve a problem on WeBWorK, you should use that information to direct you to areas requiring extra study: the WeBWorK grades themselves have little predictive use, unless they are unusually low. However, while both WeBWorK and the quizzes have a role to play in learning the calculus, neither is as important as reading your textbook, working problems yourself, and attending and listening at lectures and tutorials. What to strive for on WeBWorK assignments. Since the practice assignments give you ample opportunity to experiment, your success rate on the assignments “that count” should be close to 100%. If you are needing more than 2 attempts to solve a WeBWorK problem, then you are probably not ready to work the assignment. In order to be able to solve a WeBWorK problem successfully on the first attempt you will need to check your work, and this is a skill that you will need on the final examination, and in the advanced studies or the real world where you may eventually be applying the calculus. 1.8.4 Escape Routes At any time, even after the last date for dropping the course, students who are experiencing medical or personal difficulties should not hesitate to consult their advisors or the Student Affairs office of their faculty. Don’t allow yourself to be overwhelmed by such problems; the University has resource persons who may be able to help you. 27 The worst action is to miss the quizzes, and thereby block out an unwelcome message. Information for Students in MATH 141 2010 01 18 1.8.5 Terminology Do not be surprised if your instructors and tutors use different terminology from what you have heard in your previous calculus course, particularly if that course was at a high school. Sometimes the differences are purely due to different traditions in the professions. “Negative x” Your instructors and tutors will often read a formula −x as minus x, not as negative x. To a mathematician the term negative refers to real numbers which are not squares, i.e. which are less than 0, and −x can be positive if x itself is negative. However, mathematicians will sometimes refer to the operation of changing a sign as the replacement of x by “its negative”; this is not entirely consistent with the usual practice, but is an “abuse of language” that has crept into the professional jargon. Inverse trigonometric functions A formula like sin−1 x will be read as the inverse sine of x — never as “sine to the minus 1” or “sine to the negative 1”. However, if we write sinn x, where n is a positive integer, it will always mean (sin x)n . These conventions apply to any of the functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolic functions, which we have met on general functions, so a formula like f 2 (x) does not have an obvious meaning, and we will avoid writing it when f is other than a trigonometric or hyperbolic function. Logarithms Mathematicians these days rarely use logarithms to base 10. If you were taught to interpret log x as being the logarithm to base 10, you should now forget that — although it could be the labelling convention of your calculator. Most often, if your instructor speaks of a logarithm, and writes log x, he will be referring to the base e, i.e. to loge ; that is, he is referring to the function that calculus books call ln. When a logarithm to some other base is intended, it will either be denoted by an explicit subscript, as log2 , or some comment will be made at the beginning of the discussion, as “all logarithms in this discussion are to the base 2”. Your instructors try to think like mathematicians even when lecturing to their classes, and so we use the language and terminology we use when talking to each other. 1.9 Communication with Instructors and TA’s 1. E-mail messages to your instructor or your TA should be sent to the addresses shown in Table 1.2 and Table 2. Please show your full name and/or student number, so that we can clearly identify you. 2. The only messages sent through WeBWorK should be those generated by the Feedback facility: this means a message that refers to a specific problem on a specific WeBWorK Information for Students in MATH 141 2010 01 19 assignment, generated by clicking on the Feedback button while you are working that problem, and after you have entered your proposed answer(s) into the answer box(es).28 3. Please do not send instructors messages using the Mail facility of myCourses. This facility is difficult for instructors to use, since it is not integrated with the other mail services. We normally disable myCourses mail for that reason. If you had a need to send a message while you are connected to myCourses, just open another window and send a message with your regular e-mail client. 4. Keep your e-mail address up to date Both myCourses and WeBWorK contain an email address where we may assume you can be reached. If you prefer to use another e-mail address, the most convenient way is to forward your mail from your student mailbox, leaving the recorded addresses in these two systems unchanged. 1.10 Commercial tutorial and exam preparation services We do not endorse any commercial tutorial service, nor any service that claims to prepare students to write examinations. We have no way of evaluating the quality of any such operations, nor whether they are conforming to the University’s general practices. Caveat emptor! 28 This facility should be used sparingly; you should not expect instant response, so questions sent close to the due time on the due date will not likely receive a reply before the assignment becomes due. Information for Students in MATH 141 2010 01 20 1.11 Special Office Hours and Tutorials The following chart will show any special activities that are scheduled during the term. This table was last updated on April 09, 2010. Review Tutorial Polar Coordinates Sequences + Series TA/Instructor Dr. Y. Zhao J. Feys UPDATED TO April 17, 2010 location ARTS 145 BURN 1B45 Date 13 April 16 April Time 15:05–16:55 18:05–19:55 Information for Students in MATH 141 2010 01 21 2 Draft Solutions to Quiz Q1 For each of the days of the week a sample quiz is given below. Our policy is that, to earn part marks on any separately numbered part of any question a student must have, in the opinion of the grader, earned at least half of the marks available. 2.1 Instructions to Students 1. Show all your work. Marks may not be given for answers not supported by a full solution. For future reference, the form of your solutions should be similar to those shown in the textbook or Student Solutions Manual for similar problems. 2. In your folded answer sheet you must enclose this question sheet: it will be returned with your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.) All submissions should carry your name and student number. 3. Time = 20 minutes. 4. No calculators are permitted. 2.2 Monday Versions 1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component Z3 rectangles hang by their upper right-hand corner from the graph) to compute (10x − 0 2) dx . No other method will be accepted! Solution: If we divide the interval [0, 3] into n subintervals of equal lengths ∆x = 3n , the right end-points of these intervals are ∆x, 2∆x, . . . , n∆x, so the definite integral is equal to the following limit of a (right) Riemann sum: n n n X X X 2 (10i∆x − 2) ∆x = lim 10(∆x) lim i − 2∆x 1 n→∞ i=1 n→∞ i=1 2 i=1 3 3 n(n + 1) −2· ·n = lim 10 · 2 · n→∞ n 2 n ! n(n + 1) n = lim 45 · −6· 2 n→∞ n n ! ! 1 = lim 45 · 1 · 1 + − 6 = 39 . n→∞ n ! Information for Students in MATH 141 2010 01 22 It would not have been acceptable to evaluate this integral using the Fundamental Theorem, but that theorem could be used by a student to verify her work: Z3 h i3 (10x − 2) dx = 5x2 − 2x = 5 · 32 − 2 · 3 = 45 − 6 = 39. 0 0 2. [10 MARKS] Compute π Z3 (a) (sec x)(7 sec x + 3 tan x) dx . 0 Z1 (b) √ x · 5x2 − 5x − 4 dx . 0 Solution: For these problems the use of the Fundamental Theorem was not excluded. (a) π Z3 π Z3 7 sec2 x + 3 sec x · tan x dx (sec x)(7 sec x + 3 tan x) dx = 0 0 π = [7 tan x + 3 sec x]03 π π = 7 tan + 3 sec − (7 tan 0 + 3 sec 0) 3 √ 3 √ = 7 3 + 3 · 2 − 7 · 0 − 3 · 1 = 7 3 + 3. (b) Z1 0 √ Z 1 5 3 1 x · 5x − 5x − 4 dx = 5x 2 − 5x 2 − 4x 2 2 0 " 2 7 2 5 2 3 = 5 · · x2 − 5 · · x2 − 4 · · x2 7 5 3 8 68 10 −2− =− . = 7 3 21 #1 0 Information for Students in MATH 141 2010 01 23 2.3 Tuesday Versions 1. [10 MARKS] Use a left Riemann sum (i.e. a Riemann sum where the component rectZ2 angles hang by their upper left-hand corner from the graph) to compute (3x + 2) dx . 0 No other method will be accepted! Solution: If we divide the interval [0, 2] into n subintervals of equal lengths ∆x = 2n , the left end-points of these intervals are 0, ∆x, 2∆x, . . . , (n − 1)∆x, so the definite integral is equal to the following limit of a (left) Riemann sum: n n n X X X 2 (3(i − 1)∆x + 2) · ∆x = lim 3(∆x) (i − 1) + 2∆x 1 lim n→∞ n→∞ i=1 ! ! 1 lim 6 1 − + 4 = 6 + 4 = 10 . n→∞ n i=1 12 (n − 1)n 4 = lim 2 · + n→∞ n 2 n ! i=1 It would not have been acceptable to evaluate this integral using the Fundamental Theorem, but that theorem could be used by a student to verify her work: Z2 " 1 (3x + 2) dx = 3 · · x2 + 2x 2 0 #3 = 0 3 · 4 + 2 · 2 = 6 + 4 = 10. 2 2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find the √ x Z cosh t derivative of the function f (x) = dt. Fully simplify your answer! t3 2 √ Solution: Define u = x. Then d dx Z √ 2 x Z u cosh t d du cosh t dt = dt · by the Chain Rule t3 du 2 t3 dt cosh u du · by the Fundamental Theorem = u3 √ dt cosh x 1 1 = dt · x− 2 3 2 x2√ cosh x . = 2x2 Information for Students in MATH 141 2010 01 24 2.4 Wednesday Versions 1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component Z0 rectangles hang by their upper right-hand corner from the graph) to compute 2x2 dx . −2 No other method will be accepted! Solution: If we divide the interval [−2, 0] into n subintervals of equal lengths ∆x = 0−(−2) = 2n , the right end-points of these intervals are −2 + ∆x, −2 + 2∆x, . . . , −2 + n∆x, n so the definite integral is equal to the following limit of a (right) Riemann sum: lim n→∞ n X 2 (−2 + i∆x)2 ∆x = 2 lim 4∆x − 4i(∆x)2 + i2 (∆x)3 n→∞ i=1 n n n X X X i2 i + (∆x)3 = 2 lim 4∆x 1 − 4(∆x)2 n→∞ i=1 i=1 i=1 !2 !3 2 2 2 n(n + 1) n(n + 1)(2n + 1) = 2 lim 4 · · n − 4 · · + · n→∞ n n 2 n 6 ! ! !! 4 1 1 1 + · 1+ 2+ = 2 lim 8 − 8 1 + n→∞ n 3 n n 16 16 = 16 − 16 + = . 3 3 It would not have been acceptable to evaluate this integral using the Fundamental Theorem, but that theorem could be used by a student to verify her work: Z0 " 1 2x dx = 2 x3 3 #0 2 −2 = −2 16 2 3 0 − (−2)3 = . 3 3 2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find the Ze4x p derivative of the function (ln t)2 + 7 dt. Fully simplify your answer! e2x Solution: Define v = e4x and u = e2x . We begin by dividing the interval of integration at some convenient point, which in this proof may not even lie within the interval, and then reverse the limits of integration in the first summand: Z1 p Ze4x p Ze4x p (ln t)2 + 7 dt = (ln t)2 + 7 dt + (ln t)2 + 7 dt e2x e2x 1 Information for Students in MATH 141 2010 01 Z e2x = − p 25 Z (ln t)2 e4x + 7 dt + 1 p (ln t)2 + 7 dt 1 Now we apply the Chain Rule in differentiating each of the summand integrals: d dx Z e4x p (ln t)2 e2x Z e4x p d + 7 dt + (ln t)2 + 7 dt dx 1 1 Z 2x Z e4x p du dv d e p d (ln t)2 + 7 dt · (ln t)2 + 7 dt · − + du 1 dx dv 1 dx p p − (ln e2x )2 + 7 · 2e2x + (ln e4x )2 + 7 · 4e4x p p − (2x)2 + 7 · 2e2x + (4x)2 + 7 · 4e4x √ √ − 4x2 + 7 · 2e2x + 16x2 + 7 · 4e4x . d + 7 dt = − dx = = = = Z e2x p (ln t)2 2.5 Thursday Versions 1. [10 MARKS] Use a left Riemann sum (i.e. a Riemann sum where the component rectZ2 x2 + 1 dx. angles hang by their upper left-hand corner from the graph) to compute −1 No other method will be accepted! Solution: If we divide the interval [−1, 2] into n subintervals of equal lengths ∆x = 2−(−1) = 3n , the left end-points of these intervals are −1, −1 + ∆x, −1 + 2∆x, . . . , −1 + (n − n 1)∆x, so the definite integral is equal to the following limit of a (left) Riemann sum: lim n→∞ = lim n→∞ n X (−1 + (i − 1)∆x)2 + 1 ∆x i=1 n X 2 − 2(i − 1)∆x + (i − 1)2 (∆x)2 · ∆x i=1 n n n X X X = lim 2∆x 1 − 2(∆x)2 (i − 1) + (∆x)3 (i − 1)2 n→∞ i=1 i=1 i=1 !2 !3 3 3 (n − 1)n 3 (n − 1)n(2(n − 1) + 1) = lim 2 · · n − 2 · · + · n→∞ n n 2 n 6 ! ! !! 1 9 1 1 = lim 6 − 9 1 − + 1− 2− = 6 − 9 + 9 = 6. n→∞ n 2 n n It would not have been acceptable to evaluate this integral using the Fundamental Theo- Information for Students in MATH 141 2010 01 26 rem, but that theorem could be used by a student to verify her work: Z2 " 1 x + 1 dx = 2 x3 + x 3 2 −1 #2 −1 ! ! 8 1 + 2 − − − 1 = 6. = 3 3 2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all Z9 ln t dt. critical numbers of the function ex 2 +7x+12 Solution: We first determine the derivative, using the Fundamental Theorem and the du 2 2 Chain Rule. Define u = e x +7x+12 . Then = (2x + 7)e x +7x+12 . dx x2 +7x+12 e Z Z9 d d ln t dt = ln t dt − dx dx ex 2 +7x+12 9 x2 +7x+12 e Z du d ln t dt · = − du dx 9 2 2 = − ln e x +7x+12 · (2x + 7)e x +7x+12 2 = − x2 + 7x + 12 (2x + 7)e x +7x+12 2 = −(x + 3)(x + 4)(2x + 7)e x +7x+12 . This derivative has an exponential factor which cannot equal 0. The function is differentiable for all x, and the derivative is 0 when x = −4, − 72 , −3, so the latter 3 points are the critical numbers of the function. 2.6 Friday Versions 1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component rectZ2 angles hang by their upper right-hand corner from the graph) to compute x2 − 3 dx. No other method will be accepted! 0 Solution: If we divide the interval [0, 2] into n subintervals of equal lengths ∆x = 2n , the right end-points of these intervals are ∆x, 2∆x, . . . , n∆x, so the definite integral is equal Information for Students in MATH 141 2010 01 27 to the following limit of a (right) Riemann sum: n n n X X X 3 4 3 lim (i∆x) − 3 ∆x = lim (∆x) i − 3∆x 1 n→∞ n→∞ i=1 i=1 i=1 !4 2 2 n (n + 1)2 2 = lim · − 3 · · n n→∞ n 4 n !2 1 − 2 = 4 − 6 = −2 . = lim 4 1 + n→∞ n It would not have been acceptable to evaluate this integral using the Fundamental Theorem, but that theorem could be used by a student to verify her work: Z2 0 " #2 1 4 x − 3 dx = · x − 3x = 4 − 6 = −2. 4 0 3 2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all Z4 critical numbers of the function f (x) = sinh t dt. x2 −4x−5 du = 2x − 4. Then dx Z x2 −4x−5 Z 4 d d sinh t dt = sinh t dt − dx x2 −4x−5 dx 4 Z x2 −4x−5 du d sinh t dt · = − du 4 dx = − sinh x2 − 4x − 5 · (2x − 4) . Solution: Define u = x2 − 4x − 5, so The sinh function is 0 if and only if its argument is 0, here if and only if (x − 5)(x + 1) = x2 − 4x − 5 = 0, i.e., if and only if x = 5 or x = −1. Thus the derivative — which is defined for all x — is 0 when x = −1, 2, 5; these are the critical numbers. Information for Students in MATH 141 2010 01 28 3 Draft Solutions to Quiz Q2 For each of the days of the week a sample quiz is given below. Our policy is that, to earn part marks on any separately numbered part of any question a student must have, in the opinion of the grader, earned at least half of the marks available. 3.1 Instructions to Students 1. Show all your work! To be awarded partial marks on a part of a question a student’s answer for that part must be deemed to be more than 50% correct. For future reference, the form of your solutions should be similar to those shown in the textbook or Student Solutions Manual for similar problems. 2. In your folded answer sheet you must enclose this question sheet: it will be returned with your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.) All submissions should carry your name and student number. 3. Time = 25 minutes. 4. No calculators are permitted. Information for Students in MATH 141 2010 01 29 3.2 Monday Versions 1. [10 MARKS] Use appropriate substitutions to compute Z x3 + 3 (a) dx x4 + 12x − 2 R et dt (Hint: Start with the substitution u = et .) (b) et − 8e−t Solution: (a) In general it’s a long procedure to integrate rational function (=ratios of polynomials). However, in this case, one can observe that the numerator is precisely 41 of the 4 derivative of the denominator. So, ifwe use a substitution u = x + 12x − 2, we 3 3 have du = 4x + 12 dx = 4 x + 3x dx. Thus Z x3 + 3 dx = x4 + 12x − 2 = Z 1 du u 4 1 1 ln u + C = ln x4 + 12x − 2 + C . 4 4 (b) The integrand is a rational function, but the variable is an exponential. While we could simplify the integrand before proceeding with a substitution, it appears promising to choose a substitution like u = et . Differentiation yields du = et dt. So we have Z Z et du dt = 8 et − 8e−t u− u Z u = du . u2 − 8 Now we have a rational function in u, which can be simplified by a second substitution like v = u2 − 8, where dv = 2u du. We obtain Z Z ln u2 − 8 √ dv ln v u du = = + C = + C = ln e2t − 8 + C . u2 − 8 2v 2 2 2. [10 MARKS] Let R be the region bounded by the lines x = 1, y = ln 5, and and the curve y = ln x. Compute the volume of the solid of revolution obtained by revolving R about the x-axis. You may either use the method of cross-sections (or washer method) or the method of cylindrical shells. Information for Students in MATH 141 2010 01 30 Solution: 1. (Using Washers=Cross-sections): Z 5 Volume = π (ln 5)2 − (ln x)2 dx . 1 2 ln x We can integrate by parts, taking u = (ln 5)2 − (ln x)2 , dv = dx, so du = − dx, x v = x. We obtain ! Z 5 h i Z 5 2 ln x 2 2 2 2 5 π (ln 5) − (ln x) dx = π x (ln 5) − (ln x) − x· − dx 1 x 1 1 h i5 = π x (ln 5)2 − (ln x)2 + 2(x ln x − x) 1 2 = π (0 + 10 ln 5 − 10) − (ln 5) + 0 − 2 = π 10 ln 5 − (ln 5)2 − 8 . Where the integral of ln x was required, the student could either have quoted it from memory, or could have integrated it by parts as is done in the textbook. 2. (Using Cylindrical Shells): For this purpose we will need the equation of the curve y = ln x in the form x = ey , in order to calculate the horizontal length of the shell of Zln 5 radius y to be e x − 1. Thus the volume is 2π y (ey − 1) dy. To evaluate this integral 0 we may again use integration by parts, here taking U = y, dV = ey − 1, so dU = dy, V = ey − y. ! Z ln 5 Z ln 5 y ln 5 y y (e − y) dy 2π y (e − 1) dy = 2π y (e − y) 0 − 0 0 " !#ln 5 y2 y y = 2π y (e − y) − e − 2 0 ! 2 (ln 5) −4 . = 2π 5 ln 5 − 2 Information for Students in MATH 141 2010 01 31 3.3 Most Tuesday Versions 1. [10 MARKS] Use appropriate substitutions to compute Z e5t + 5 (a) dt e5t + 25t Z √ x3 x2 − 1 dx (Hint: Start with the substitution u = x2 − 1.) (b) Solution: (a) Try the substitution u = e5t + 25t. Then du = 5e5t + 25 dt = 5 e5t + 5 dt. Hence Z Z e5t + 5 1 du 1 dt = = ln u + C e5t + 25t 5 u 5 1 ln e5t + 25t + C . = 5 (b) The hint suggests using the substitution u = x2 − 1 ⇒ du = 2x dx. Then Z Z √ √ du 3 2 x x − 1 dx = (u + 1) u · 2 1 52 1 23 ·u + u +C = 5 3 1 1 2 5 3 = (x − 1) 2 + (x2 − 1) 2 + C. 5 3 (There are variations of this substitution that could have been used. For example, √ 2 we could try the substitution u = x − 1, so u2 = x2 − 1, x2 = u2 + 1, u du = x dx. Then Z Z √ 3 2 x x − 1 dx = (u2 + 1)u · u du Z = u4 + u2 du u5 u3 + +C 5 3 5 1 3 1 2 = x − 1 2 + x2 − 1 2 + C . 5 3 = 2. [10 MARKS] Let R be the region in the first quadrant bounded by the lines y = 0, y = 5x, 6 and the curve y = − 1. Compute the volume of the solid of revolution obtained by x Information for Students in MATH 141 2010 01 32 revolving R about the x-axis. You may either use the method of cross-sections (or disk method) or the method of cylindrical shells. 6 Solution: To determine the point of intersection of the lines y = 5x y = − 1 we solve x the equations simultaneously: eliminating y between the equations yields 5x2 + x−6 = 0, i.e., (5x + 6)(x − 1) = 0, so the curves intersect in points with x = − 56 or x = 1. But x = − 56 corresponds to an intersection point in the 3rd quadrant: the only intersection point in the first quadrant is (x, y) = (1, 5). (a) (Using Washers=Cross-sections): While we can integrate using either of the methods we know, using “washers” will require two separate integrals, since the description of the outer radius of the washer changes at x = 1. !2 Z 1 Z 6 6 2 Volume = π(5x) dx + π − 1 dx x 0 1 " " #1 #6 25 3 36 = π x + π − − 12 ln x + x 3 x 0 1 ! 130 = π − 12 ln 6 . 3 (b) (Using Cylindrical Shells): The right boundary of the region is the curve whose 6 equation can be reformulated as x = . The length of the cylinder with radius y y+1 ! Z5 6 6 y y is, therefore, − . The volume is, therefore, 2π y − dy . Because y+1 5 y+1 5 0 we haven’t yet studied integration of rational functions in general, the easiest way to integrate this now is to use a substitution like u = y + 1 to make the denominator into powers of the variable: Z5 2π 0 ! ! Z 6 6 6 u−1 y y − dy = 2π (u − 1) − du y+1 5 u 5 1 ! Z 2π 6 30 2 = du −u + 2u + 29 − 5 1 u #6 " 2π u3 2 = − + u + 29u − 30 ln |u| 5 5 1 ! 130 = π − 12 ln 6 . 3 Information for Students in MATH 141 2010 01 33 3.4 Most Wednesday Versions 1. [10 MARKS] ! x x 8 (a) Compute the average value of f (x) = cos · sin on the interval 0, · π . 16 16 3 Z 1 (b) Use an appropriate substitution to compute the integral dx . (Hint: 5x e + e−5x Start with the substitution u = e5x .) 3 Solution: Z x x cos · sin dx 16 16 (a) The average is 0 . The integral in the numerator can be Z 8π3 1 dx 0 evaluated by a substitution u = cos 16x , which implies that 8π 3 3 du = − sin x 1 · dx . 16 16 Hence Z 8π 3 0 Z x 3 x · sin dx = cos 16 16 1 Hence the average is 7 4 8 3 ·π−0 √ 3 2 = i √3 7 4 2 3 −16u du = − 4u = . 1 4 21 . 32π (b) Try the substitution u = e5x ; du = u · 5 · dx ⇒ dx = Z 1 dx = 5x e + e−5x Z 1 du . Then 5u du · 1 5u u+ u Z 1 du = 2 5 u +1 1 1 arctan u + C = arctan e5x + C . = 5 5 Information for Students in MATH 141 2010 01 34 ! 6 2. [10 MARKS] Use integration by parts to compute the integral arctan dx. x Solution: We know that the integration of arctan x can be accomplished by integration 6 by parts, so we can try the same tactic here, with u = arctan x , and dv = dx. Then v = x, and −6 1 −6 · dx . du = 2 · 2 dx = x 36 + x2 1+ 6 Z x Hence Z ! ! Z 6 6 x arctan dx = arctan ·x+6 dx . x x 36 + x2 The new integrand can be integrated by using a substitution like w = 36 + x2 , where dw dw = 2x dx , so x dx = : 2 ! ! Z Z 6 6 1 arctan dx = arctan ·x+6 dw x x 2w ! 6 = arctan · x + 3 · ln |w| + C x ! 6 = arctan · x + 3 · ln 36 + x2 + C . x Information for Students in MATH 141 2010 01 3.5 Thursday Versions 1. [10 MARKS] Compute the integral 35 Z cos(5t) · cosh t dt. Solution: We can integrate this product by two successive integrations by parts, judiciously chosen; care is needed in the second application, since a poor decision could reverse the effect of the first integration. We can begin with u = cos 5t, dv = cosh t dt, which imply that du = −5 sin 5t, v = sinh t. Then Z Z cos(5t) · cosh t dt = (cos 5t) · sinh t − (−5 sin 5t) sinh t dt Z = (cos 5t) · sinh t + 5 (sin 5t) sinh t dt . In a second application we can take U = sin 5t, dV = sinh t dt, which imply that dU = 5 cos 5t dt, V = cosh t. Then Z cos(5t) · cosh t dt Z = (cos 5t) · sinh t + 5 (sin 5t) sinh t dt ! Z = (cos 5t) · sinh t + 5 (sin 5t) cosh t − 5(cos 5t) cosh t Z = (cos 5t) · sinh t + 5(sin 5t) cosh t) − 25 (cos 5t) cosh t dt While we haven’t completed the solution yet, we have expressed the original integral in terms of itself; if we shift the integral from the right side of the equation to the left, and combine the 25 copies with the one, we obtain Z (1 + 25) cos(5t) · cosh t dt = (cos 5t) · sinh t + 5(sin 5t) cosh t + C , which we can solve by division by 26, to obtain Z 1 5 cos(5t) · cosh t dt = (cos 5t) · sinh t + (sin 5t) cosh t + C . 26 26 (The name of the constant was changed, since this constant is different from the precedC ing: of course, we could also have written .) 26 2. [10 MARKS] Use a suitable substitution and then integration by parts to compute the Z integral x ln(x+5) dx. (Hint: Start with the substitution u = x+5, then use integration by parts.) Information for Students in MATH 141 2010 01 36 Solution: One can start by defining a substitution u = x + 5, but the substitution could be deferred, or even entirely avoided. For example, just take u = ln(x + 5), dv = x dx, so du = Then Z dx , x+5 v= x2 . 2 x2 x ln(x + 5) dx = (ln(x + 5)) · − 2 Z x2 dx · . 2 x+5 The last integral could be integrated by simply using long division, where x2 = (x + 5)(x − 5) + 25, so Z Z Z 1 x2 x−5 25 1 x2 5x 25 = dx + dx = − + ln |x + 5| + C . (1) 2 x+5 2 2 x+5 4 2 2 Hence Z x ln(x + 5) dx = x2 x2 5x 25 · ln(x + 5)) − + − ln |x + 5| + C1 . 2 4 2 2 The long division step is one of the standard operations we will see when we study the integration of general rational functions. But one could have used, in the second integral in equation (1), a substitution u = x + 5, to obtain Z Z x2 (u − 5)2 dx = du 2(x + 5) 2u ! Z u 25 = −5+ du 2 2u u2 25 (x + 5)2 25 = − 5u + ln |u| + C = − 5(x + 5) + ln |x + 5| + C2 . 4 2 4 2 The substitution could also have been effected at the beginning, and it was this sequence that was contemplated in the problem: Z Z x ln(x + 5) dx = (u − 5)(ln u) du . Now we can integrate by parts, taking U = ln u, dV = (u − 5) du, so that dU = V= Z u2 − 5u, 2 x ln(x + 5) dx = Z (u − 5)(ln u) du du , u Information for Students in MATH 141 2010 01 ! Z ! u2 u2 du = (ln u) − 5u − − 5u · 2 2 u ! Z u2 u = (ln u) − 5u − − 5 du 2 2 ! u2 u2 = (ln u) − 5u − + 5u + C 2 4 ! (x + 5)2 (x + 5)2 = (ln(x + 5)) − 5(x + 5) − + 5(x + 5) + C3 . 2 4 37 Information for Students in MATH 141 2010 01 38 3.6 Friday Versions 1. [10 MARKS] Let R be the region bounded by the lines y = 0, y = ln 7, x = 0, and the curve y = ln x. Compute the volume of the solid obtained by revolving R about the x-axis. [Hint: Sketch the region; then choose the appropriate method.] Solution: (a) (Using Cylindrical Shells): We will need to find the height of the cylindrical shell of radius y, and this will require transforming the equation y = ln x into x = ey . Zln 7 y y The distance from x = 0 to x = e is then e − 0, so the volume is 2πy · ey dy , 0 which we will integrate by parts, taking u = y, dv = ey dy, so that du = dy, v = ey : Zln 7 y 2πy · e dy = 2π 0 7 yey ln 0 Z ln 7 − 2π ey dy 0 7 y ln 7 = 2π yey ln 0 − 2π [e ]0 = 2π((ln 7) · 7 − 0) − 2π(7 − 1) = 2π(7 ln 7 − 6) . (b) (Using Washers): One drawback of this method is that there will be two kinds of washers: For 0 ≤ x ≤ 1 the washers have no hole in the middle, and the radius is a constant, ln 7; but for 1 ≤ x ≤ 7 there will be a hole whose radius is ln x. The area is, therefore, a sum Z 7 Z 1 2 2 π (ln 7)2 − (ln x)2 dx π (ln 7) − 0 dx + 0 1 Z 7 Z 7 2 2 = π (ln 7) − 0 dx − π (ln x)2 dx 0 1 Z 7 Z 7 = π(ln 7)2 dx − π (ln x)2 dx 0 1 Z 7 = 7(ln 7)2 − π (ln x)2 dx . 1 To integrate (ln x)2 we could apply integration by parts twice. Start by taking u = ln x , v = x, and (ln x)2 , v0 = 1, so that u0 = 2 x Z 7 Z 7 h i7 ln x 2 2 · x dx (ln x) dx = (ln x) · x − 2 1 x 1 1 Information for Students in MATH 141 2010 01 = h 39 i7 2 Z 7 (ln x) · x − 2 ln x dx 1 1 h i7 = (ln x)2 · x − 2[x · ln x − x]71 1 2 = [7(ln 7) − 0] + 2[(7 ln 7 − 7) − (0 − 1)] which leads to the same result as found earlier. Z9 2. [10 MARKS] Use an appropriate substitution to compute the integral (Hint: Start with the substitution u = 6 + √ 0 x.) Solution: One substitution that would simplify this integral is u = 6 + u − 6 ⇒ x = (u − 6)2 ⇒ dx = 2(u − 6) du. Thus Z9 0 1 √ dx = 6+ x √ 1 √ dx . 6+ x x ⇒ √ 6+ 9 Z 6 Z 9 = 2 6 2(u − 6) du u ! 6 1− du u = 2[u − 6 ln u]96 = 2(9 − 6 ln 9 − 6 + 6 ln 6) = 6 + 12 ln 2 . 3 √ x = Information for Students in MATH 141 2010 01 201 4 References 4.1 Stewart Calculus Series [1] J. Stewart, Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-01169-X. [2] J. Stewart, Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-01166-5. [3] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-01240-8. [4] J. Stewart, Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008); bundled with Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-42966-X. [5] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-01239-4. [6] J. Stewart, Multivariable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-?????-?. [7] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39330-6. [8] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39321-7. [9] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39333-0. [10] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003); bundled with Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-17-6425411. [11] J. Stewart, Single Variable Essential Calculus (Early Transcendentals). Thomson * Brooks/Cole (2006). Thomson * Brooks/Cole (2003). ISBN 0-495-10957-6. Information for Students in MATH 141 2010 01 202 [12] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003); bundled with Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-53410307-3. [13] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39331-4. [14] Video Outline for Stewart’s Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39325-X. 17 VCR tapes. [15] Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 5th Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39326-8. [16] H. Keynes, J. Stewart, D. Clegg, Tools for Enriching Calculus, CD to accompany [7] and [8]. Thomson * Brooks/Cole (2003). ISBN 0-534-39731-X. [17] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN 0-534-35563-3. [18] J. Stewart, Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN 0-534-36298-2. [19] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Single Variable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN 0-534-36301-6. [20] J. Stewart, L. Redlin, S. Watson, Precalculus: Mathematics for Calculus, Enhanced Review Edition. Thomson * Brooks/Cole. (2006). ISBN: 0-495-39276-6. [21] J. Stewart, Trigonometry for Calculus. Thomson * Brooks/Cole. ISBN: 0-17-641227-1. 4.2 Other Calculus Textbooks 4.2.1 R. A. Adams [22] R. A. Adams, Calculus, Single Variable, Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79805-0. [23] R. A. Adams, Calculus of Several Variables, Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79802-6. Information for Students in MATH 141 2010 01 203 [24] R. A. Adams, Calculus: A Complete Course, Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79131-5. [25] R. A. Adams, Student Solution Manual for Adams’, Calculus: A Complete Course, Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79803-4. [26] R. A. Adams, Calculus: A Complete Course, with Solution Manual, Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-131-30565-4. [27] R. A. Adams, Calculus: A Complete Course Manual, Sixth Edition. Addison, Wesley, Longman, Toronto (2006). ISBN 0-321-27000-2. 4.2.2 Larson, Hostetler, et al. [28] Calculus Instructional DVD Program, for use with (inter alia) Larson/Hostetler/Edwards, Calculus of a Single Variable: Early Transcendental Functions, Third Edition [29]. Houghton Mifflin (2003). ISBN 0-618-25097-2. [29] R. Larson, R. P. Hostetler, B. H. Edwards, D. E. Heyd, Calculus, Early Transcendental Functions, Third Edition. Houghton Mifflin Company, Boston (2003). ISBN 0-61822307-X. 4.2.3 Edwards and Penney [30] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus, Early Transcendentals, Sixth Edition. Prentice Hall, Englewood Cliffs, NJ (2002). ISBN 0-13-041407-7. [31] C. H. Edwards, Jr., and D. E. Penney, Calculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-793076-3. [32] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-079875-4. [33] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-793092-5. [34] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Single Variable Calculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095247-1. Information for Students in MATH 141 2010 01 204 4.2.4 Others, not “Early Transcendentals” [35] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge University Press (1967). [36] H. S. Hall, S. R. Knight, Elementary Trigonometry, Fourth Edition. Macmillan and Company, London (1905). [37] S. L. Salas, E. Hille, G. J. Etgen, Calculus, One and Several Variables, 10th Edition. John Wiley & Sons, Inc. (2007). ISBN 0471-69804-0. 4.3 Other References [38] G. N. Berman, A Problem Book in Mathematical Analysis. Mir Publishers, Moscow, (1975) 1977 [39] D. Ebersole, D. Schattschneider, A. Sevilla, K. Somers, A Companion to Calculus. Brooks/Cole (1995). ISBN 0-534-26592-8. [40] McGill Undergraduate Programs Calendar 2009/2010. Also accessible http://coursecalendar.mcgill.ca/ug200910/wwhelp/wwhimpl/js/html/wwhelp.htm at Information for Students in Lecture Section 1 of MATH 141 2010 01 1001 A Timetable for Lecture Section 001 of MATH 141 2010 01 Distribution Date: Monday, January 04th, 2010 (Last updated on March 29th, 2010. Subject to further correction and change.) Section numbers refer to the text-book. MONDAY 04 11 18 25 01 08 15 22 WEDNESDAY JANUARY FRIDAY 06 §5.1, §5.2 08 §5.3 Tutorials begin week of January 11th, 2010 §5.3, §5.4 13 §5.4, §5.5 15 §5.5 Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010 §6.1 Q1 20 §6.2 Q1 22 §6.3 Q1 WeBWorK Assignment A1 due Jan. 24, 2010 Deadline for withdrawal with fee refund = Jan. 24, 2010 Verification Period: January 25 – 29, 2010 §6.5 A1 27 §7.1 29 §7.1, §7.2 FEBRUARY §7.2 03 §7.3 05 §7.3 WeBWorK Assignment A2 due Feb. 07, 2010 §7.4 Q2 A2 10 §7.4 Q2 12 §7.5, §7.8 Q2 Deadline for web withdrawal (with W) from course via MINERVA = Feb. 14, 2010 §7.8 17 §8.1, §8.2 19 §8.2 Study Break: February 21 – 27, 2010 No lectures, no regular office hours, no regular tutorials! WeBWorK Assignment A3 due Feb. 21, 2010 NO LECTURE, NO TU- 24 NO LECTURE, NO TU- 26 NO LECTURE, NO TU§5.1, §5.2 TORIALS A3 Notation: An Qn X TORIALS TORIALS = Regular WeBWorK Assignment An due about 23:30 hours on the Sunday preceding this Monday = Quiz Qn will be administered at the tutorials this week. = reserved for eXpansion or review Information for Students in Lecture Section 1 of MATH 141 2010 01 1002 Section numbers refer to the text-book. MONDAY 01 08 15 22 29 05 12 WEDNESDAY MARCH FRIDAY 03 §10.2 05 §§10.2–§10.4 WeBWorK Assignment A4 due Mar. 07, 2010 §10.4 A4 Q3 10 §10.4 Q3 12 §10.4, §11.1 Q3 §11.2, §11.3 17 §11.3 19 §11.3 WeBWorK Assignment A5 due Monday, Mar. 21, 2010 §11.4 A5 Q4 24 §11.5 Q4 26 §11.6 Q4 §11.6, §11.7 31 §11.7, X APRIL 02 NO LECTURE Q4 This week’s tutorials are the last (except for Monday tutorials). WeBWorK Assignment A6 due Apr. 04, 2010 NO LECTURE 07 X 09 X X 14 X §10.1,§10.2 Notation: An Qn X = Regular WeBWorK Assignment An due about 23:30 hours on the Sunday preceding this Monday = Quiz Qn will be administered at the tutorials this week. = reserved for eXpansion or review UPDATED TO April 17, 2010 Information for Students in Lecture Section 002 of MATH 141 2010 01 2001 B Timetable for Lecture Section 002 of MATH 141 2009 01 Distribution Date: Monday, January 04th, 2010 (Subject to further correction and change.) Section numbers refer to the text-book. MONDAY 04 11 18 25 01 08 15 22 WEDNESDAY JANUARY FRIDAY 06 §5.3 08 §5.4 Tutorials begin week of January 11th, 2010 §5.4, §5.5 13 §5.5 15 §6.1 Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010 §6.2 Q1 20 §6.2, §6.3 Q1 22 §6.3 Q1 WeBWorK Assignment A1 due Jan. 24, 2010 Deadline for withdrawal with fee refund = Jan. 24, 2010 Verification Period: January 25 – 29, 2010 §6.5 A1 27 §6.5, §7.1 29 §7.1, §7.2 FEBRUARY §7.2 03 §7.2, §7.3 05 §7.3 WeBWorK Assignment A2 due Feb. 07, 2010 §7.3, §7.4 Q2 A2 10 §7.4 Q2 12 §7.5, §7.8 Q2 Deadline for web withdrawal (with W) from course via MINERVA = Feb. 14, 2010 §7.8 17 §8.1, §8.2 19 §8.2 Study Break: February 21 – 27, 2010 No lectures, no regular office hours, no regular tutorials! WeBWorK Assignment A3 due Feb. 21, 2010 NO LECTURE, NO TU- 24 NO LECTURE, NO TU- 26 NO LECTURE, NO TU§5.1, §5.2 TORIALS A3 Notation: An Qn X TORIALS TORIALS = Regular WeBWorK Assignment An due about 23:30 hours on the Sunday preceding this Monday = Quiz Qn will be administered at the tutorials this week. = reserved for eXpansion or review UPDATED TO April 17, 2010 Information for Students in Lecture Section 002 of MATH 141 2010 01 2002 Section numbers refer to the text-book. MONDAY 01 08 15 22 29 05 12 WEDNESDAY MARCH FRIDAY 03 §10.3 05 §10.3, §10.4 WeBWorK Assignment A4 due Mar. 07, 2010 §10.4 A4 Q3 10 §11.1, §11.2 Q3 12 §11.2 Q3 §11.3 17 §11.4 19 §11.4, §11.5 WeBWorK Assignment A5 due Sunday, Mar. 28, 2010 §11.5 A5 Q4 24 §11.6 Q4 26 §11.6 Q4 §11.6 31 §11.7, X APRIL 02 NO LECTURE Q4 This week’s tutorials are the last (except for Monday tutorials). WeBWorK Assignment A6 due Apr. 04, 2010 NO LECTURE 07 X 09 X X 14 X §10.1,§10.2 Notation: An Qn X = Regular WeBWorK Assignment An due about 23:30 hours on the Sunday preceding this Monday = Quiz Qn will be administered at the tutorials this week. = reserved for eXpansion or review Information for Students in Lecture Section 1 of MATH 141 2010 01 3001 C Supplementary Notes for Students in Section 001 of MATH 141 2010 01 C.1 Lecture style in Lecture Section 001 Lecture content. The timetable on pages 1001, 1002 will show you approximately what I plan to discuss at each lecture. I suggest that you look through the material in advance. If you have time to try some of the exercises, and find some that cause you difficulty, you are welcome to bring them to my attention; perhaps I may be able to work some of these examples into the lecture. What goes on the chalkboard? — Should I take notes? I believe strongly that students should not sit in the lecture feverishly copying notes for fear of missing some essential topic; in this course most of what you need to know is contained in the textbook. You should take notes, but you should be trying to think at the same time. The chalkboard will be used for • statement/illustration of specific definitions and theorems • sketching solutions to problems, or classes of problems • a scratchpad Some of this material will be useful to you in learning the material in the course. Even when the material on the board is equivalent to something in your textbook, the act of writing may help you remember it. But much of the material will be restatements of your textbook, so you should normally not panic if you miss something. Graphs My emphasis is on qualitative properties of the graphs of functions, but not on the production of extremely precise graphs. You can expect to see me draw on the chalkboard sketches that are extremely crude approximations of functions, sometimes even caricatures of the true graph. Mathematicians do not base proofs on sketches of graphs — the role of a sketch is usually only to assist the reader to visualize the verbal or symbolic reasoning which accompanies it. Sometimes a graph is used help one discover a phenomenon, but the result would not be acceptable to a mathematician unless it could be proved in a non-graphical way.29 These supplementary notes Section and paragraph headings follow the order of topics in the textbook. While some of the comments or explanations may be helpful in understanding the book, the notes are not required reading for examination purposes. Sometimes it may happen that the discussion of a topic or an exercise evolves during the lecture into one which requires 29 This is why I usually avoid problems in the textbook that appear to be drawing inferences from graphs. Information for Students in Lecture Section 1 of MATH 141 2010 01 3002 more detail than is practical to write on the chalkboard. In such cases you may be referred at the following lecture to supplementary material that will be contained in the notes placed on the Web. Such evolutions are spontaneous and not planned, and cannot be announced in advance. Timing and corrections The notes will usually not be posted until after the lecture. While I do try to check the notes before posting them, there will inevitably be errors: if you see something that doesn’t look right, please ask. The notes will be progressively corrected as misprints and other errors are discovered. C.2 Supplementary Notes for the Lecture of January 04th, 2010 Release Date: January 04th, 2010, subject to revision Textbook Chapter 5. INTEGRALS. (There will be examples, etc., in these notes that were not discussed specifically at the lecture, because of time constraints.) C.2.1 §5.1 Areas and Distances. When, in [1, §2.1], the textbook was motivating the differential calculus, it presented two applications: “The Tangent Problem”, which was geometric; and “The Velocity Problem”, which was physical. Now, in motivating the integral calculus, the author presents two analogous problems: “The Area Problem”, which is geometric; and “The Distance Problem”, which is physical. The Area Problem. The textbook discusses approximation of the area under the graph of y = f (x) between x = a and x = b — more precisely, the area between the graph, the x-axis, and the vertical lines x = a and x = b, as a limit of a sum of areas of narrow vertical rectangles. The approximation is first motivated with simple functions where the area can be bounded above and below by easily computable sums, which together converge to the same value as their number approaches ∞ and their width approaches 0. You should become comfortable using the “sigma notation”, where, for integers n1 and n2 with n1 ≤ n2 , we write n2 X f (i) i=n1 to mean the sum f (n1 ) + f (n1 + 1) + f (n1 + 2) + . . . + f (n2 ) . The textbook then proposes the following definition of the area between the graph of a function, two vertical lines, and the x-axis: Information for Students in Lecture Section 1 of MATH 141 2010 01 3003 Definition C.1 (Preliminary) Let f be a function which is continuous on the interval [a, b]. The area of the region bounded by the graph of f , the x-axis, and the vertical lines x = a and x = b (where a ≤ b) is the limit of a sum of the areas of rectangles “hanging” from the graph, as follows: subdivide the interval [a, b] into n smaller intervals by points x1 , x2 , . . . , xn−1 , where x1 < x2 < . . . < xn−1 , and, for convenience, we define x0 = a, xn = b; select points xi∗ n X ∗ f xi∗ (xi − xi−1 ). (i = 1, . . . , n) such that xi ≤ xi+1 ≤ xi+1 , and consider the sum Rn = i=1 (This definition is “preliminary” in that we haven’t yet argued that such a limit need exist. This is not the type of limit — of a function of one variable — studied in MATH 140. We also need to clarify what restrictions hold for the points x1 , x2 , . . . xn−1 , and how we select the points x1∗ , x2∗ , . . . xn∗ .) Example C.1 (cf. [1, Example 5.1.2, p. 357]) Let a and b be non-negative real numbers, and consider the area under the parabola y = x2 between the vertical lines x = a, x = b, and above the line y = 0. 1. Let’s first consider the special case where the left side of the region is along the y=axis, i.e., where a = 0. From this special case we will be able to derive the general solution. b−0 2. Divide the interval [0, b] into n intervals of equal width ; thus the points x1 , x2 , . . . n b−0 are chosen to be xi = · i (i = 1, 2, . . . , n). For the sample points xi∗ , let’s consider n the points at the right end of each of the subintervals: so xi∗ = xi (i = 1, 2, . . . , n). Then !2 ! !! n X b−0 b−0 b−0 Rn = ·i · ·i − · (i − 1) (2) n n n i=1 b . n The first factor is squared because we are evaluating the function f (x) = x2 at the point b−0 · i. In order to evaluate this sum we need to know the sum of the squares of the n first n positive integers (=natural numbers). If you didn’t learn this in high school, here it is: n X n(n + 1)(2n + 1) . (3) i2 = 6 i=1 where the second factor is simply the common length of all the subintervals, i.e., We won’t be able to prove this formula here: a proof requires use of a tool like Mathematical Induction, which we are omitting from the syllabus. Applying (3) to (2) yields ! 3 1 n(n + 1)(2n + 1) 2+ + 2 n n 6 3 3 =b · . Rn = b · 3 6 n Information for Students in Lecture Section 1 of MATH 141 2010 01 As n is permitted to become arbitrarily large positively, Rn → 3004 b3 . 3 a3 3. If the region starts at the origin and extends to the line x = a, then the area will be . 3 So, if the region we wish to study starts at the line x = a, and ends at the line x = b, b3 − a3 where b ≥ a, we need only subtract one of these areas from the other, obtaining . 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3005 C.3 Supplementary Notes for the Lecture of January 06th, 2010 Release Date: January 06th, 2010, subject to revision C.3.1 §5.1 Areas and Distances (conclusion). The Distance Problem. Here the textbook considers what appears to be a different type of problem, and shows that the solution is the same type of sum met in the Area Problem above. In this case the problem is to determine the distance travelled by a particle moving so that its velocity at time t is a given function f (t). It will be seen that the distance can again be interpreted as a limit of a sum — the same sum that would be seen if we attempted to determine the area in the xt-plane under the graph x = f (t). dx If a particle is known to be moving along the x-axis at a velocity of v(t) = x0 (t) = (t) dt cm/s, how much distance is traversed between time t = a and time t = b? If, by distance, we mean displacement or “signed distance”, where movement to the right counts positively, and to the left negatively30 , then the distance is x(b) − x(a); we shall see that this can be interpreted d as the area under the graph y = x(t) between x = a and x = b, which we will be denoting dt Zb d by x(t) dt . When it is intended to consider all motion as non-negative — the way the dt a odometer of an automobile measures distance, then we would want to find the area under the graph not of the velocity, v(t) = x0 (t), but of the speed, |v(t)| = |x0 (t)|, and the distance travelled would be Z b Z b d 0 x(t) dt . |v (t)| dt = a dt a But in practice the word distance is often used with either meaning, so care is required. I have shown that the area under the graph of the velocity represents the directed distance travelled by the moving particle. But we expressed the velocity as the derivative of the displacement of the particle relative to some fixed origin; and the distance travelled can be expressed as the difference between two values of the displacement. This is a special case of the Fundamental Theorem of the (Integral) Calculus, which will be introduced below. 5.1 Exercises 30 Note that, with this definition, a particle that moves around and then returns to the same point will have travelled a distance of 0. Information for Students in Lecture Section 1 of MATH 141 2010 01 3006 [1, Exercise 20, p. 365] Determine a region whose area is equal to the limit lim n→∞ n X 2 i=1 2i 5+ n n !10 . Do not evaluate the limit at this time. Solution: Take the widths of the approximating rectangles to be constant, 4x = 2 ; n then n of these constant widths would span a distance of length 2. The limit can be n P seen to be lim 4x · (5 + i4x)10 . The rectangles could be interpreted, for example, n→∞ i=1 as “hanging by their upper right hand corners” from the curve y = (5 + x)10 above 11 11 the interval 0 ≤ x ≤ 2. (Later in the course we shall see that the area is (5+2)11 −5 = 1,928,498,618 .) 11 In the lecture I mentioned that the area of this region could be represented in other ways, for example, by hanging the component rectangular elements by their left upper corner !10 n X 2 2(i − 1) from the graph. In that case the sum could have been written as lim 5+ ; n→∞ n n i=1 !10 n−1 X 2 2i 5+ . or, alternatively, as lim n→∞ n n i=0 The original limit sum, or the limits of either of the latter sums, could also be interpreted as being the area under the graph of f (x) = x10 between x = 5 and x = 7. The transition from this interpretation to the earlier one is represented by moving the location of the y-axis — without changing the area of the region. [1, Exercise 21, p. 365] Determine a region whose area is equal to the limit n X iπ π tan . n→∞ 4n 4n i=1 lim Do not evaluate the limit at this time. Solution: You should read the discussion of this problem in your Student Solution Manual. What is different from the preceding example to this one, is that we don’t have any straightforward algebraic way of evaluating the limit of the sum of rectangles obtained to approximate this area. Thus there is going to be something new in our theory if we will be able to determine this area exactly. And, in fact, we shall be able to evaluate this Information for Students in Lecture Section 1 of MATH 141 2010 01 3007 area exactly! The upper boundary of the region in this case could be the graph of the function y = tan x, and the other boundaries of the region could be, in addition to the π x-axis, the vertical lines x = 0 and x = . 4 C.3.2 §5.2 The Definite Integral The formal definition of the integral involves a number of technical difficulties which I shall not consider in detail in this course. You should read the definition the textbook gives of the integral [1, p. 366], but you are not going to be asked to work with it in full generality; in fact the definition given in the textbook is simpler than the definition that is normally used for the Riemann Integral. We would need to appeal to this definition if we wished to formally prove all the properties that the author is going to ascribe to the integral; but we shall not attempt to provide such proofs. The usual definition of the integral would permit the widths of the subintervals, here denoted by ∆x, to be different: ∆x1 for the first subinterval, ∆x2 for the second, etc., and would then require that the largest of them should approach zero. This technicality is needed for general functions, but will not be discussed further in this course. Read the book and be sure you know the definitions of each of the following terms: • sample points • definite integral of f from a to b • integral sign • integrand • limits of integration • lower limit, upper limit • Riemann sum Where we take, as the sample points in the subintervals, global maximum points for the function on the subintervals, we have what is called the upper Riemann sum; analogously, we may speak of the lower Riemann sum. To prove the existence of the definite integral we would want to show that the difference between these sums approaches 0 in the limit. This can be shown to be the case in particular when the function is continuous everywhere, and it is even true in certain more general situations. In this course we will normally be taking the functions to be either continuous or, more generally, “piecewise continuous”; that is, we will permit functions which can be obtained by “gluing” together functions which are continuous over adjacent intervals. As long as there are only a finite number of such components, it can be shown that the Information for Students in Lecture Section 1 of MATH 141 2010 01 3008 integral exists; it doesn’t matter if the function is discontinuous at the finite number of locations were the functions are “glued”. But some of the properties we will be using will apply only to continuous functions, and we may have to break a problem up into parts in order to solve it. More about this later. Evaluating Integrals. Among the integrals discussed in this subsection are several that require the following formulæ for sums of powers of the natural numbers: n X i=1 n X i=1 n X i1 = n(n + 1) 2 (4) i2 = n(n + 1)(2n + 1) 6 (5) i3 = n2 (n + 1)2 4 (6) i=1 to which we could add the following trivial result31 : n X i0 = n . (7) i=1 Formulæ 4, 5, 6 are proved in [1, Appendix E], but you are not expected to have read those proofs. The proofs given are by “Mathematical Induction”.32 Asking the Right Question. The fact that the formulæ for the sums of powers do not appear to follow any general pattern is not because there is no pattern, but simply that we are “asking the wrong question”. If, instead, we had asked for the sums of what are called falling factorials, i.e., products of an integer with successive integers immediately less than it, we would obtain the following, much prettier results. You do not need to remember these formulæ: n X i0 = i=1 n X i=1 31 i = n 1 (8) (n + 1)n 2 (9) A definition that the product of an empty set of numbers is equal to 1 is consistent with the definition of multiplication of real numbers. 32 Mathematical Induction was not an examination topic in MATH 140 2009 09; in the present course you are not expected to know how to apply Mathematical Induction, but interested students are urged to read about it in the textbook [1, pp. 77]. Information for Students in Lecture Section 1 of MATH 141 2010 01 n X i(i − 1) = i=1 n X i(i − 1)(i − 2) = i=1 3009 (n + 1)n(n − 1) 3 (10) (n + 1)n(n − 1)(n − 2) 4 (11) A glance at these formulæ, which are certainly “prettier” than the formulæ for the sums of the powers, shows that the first one, (8) doesn’t look as though it fits. Here again, that is because we are again “asking the wrong question”. Let’s formulate the results slightly differently, including the term i = 0 in each of the sums; only in the case of the 0th powers does this make any difference, since 00 is defined to be 1: n X i0 = i=0 n X i = i=0 n X i(i − 1) = i=0 n X i(i − 1)(i − 2) = i=0 n+1 1 (12) (n + 1)n 2 (13) (n + 1)n(n − 1) 3 (14) (n + 1)n(n − 1)(n − 2) . 4 (15) Now we can see much further; we can even conjecture that there is a general result that encompasses all of these particular cases: n X i(i − 1)(i − 2) · . . . · (i − r + 1) = i=0 (n + 1)n(n − 1)(n − 2) · . . . · (n − r + 1) . r (16) And finally, of what use are these formulæ if we need the sums of the powers of the integers, not the sums of “falling factorials”. Any power of n can be expressed in terms of “falling factorials”, for example n1 = n n2 = n(n − 1) + n n3 = n(n − 1)(n − 2) + 3n(n − 1) + n , so property (16) can provide all the sums we need. The purpose of this parenthetical discussion is to illustrate that the main challenge in proofs by induction is making the right guess, rather than in the details of the proof, which may be routine. Linearity of the summation operator. The textbook discusses some properties of the “sigma” notation; these could be called the linearity properties of the operator Σ, and are all special Information for Students in Lecture Section 1 of MATH 141 2010 01 3010 cases of the following: k+` X (rai + sbi ) = r i=k k+` X ai + s i=k k+` X bi . i=k I may have more to say about the sigma notation after I discuss [1, §5.5], where we shall encounter properties of the integral that have analogues for sums. For the present let it be k+` P noted that the symbol i in ai is not a “free” variable, in that you cannot assign any values to i=k it: it performs a function in the symbol, but that function would be performed equally well if k+` k+` k+` P P P we replaced i by any other symbol that is not already in use, e.g., au , aλ , a♥ . u=k λ=k ♥=k The Midpoint Rule. The “Midpoint Rule” is an approximation formula for definite integrals. Use of an approximation formula entails a willingness to accept an error in the calculation. Mathematicians normally expect to see an estimate of how good or how bad an approximation can be before recommending their use. A partial justification of the Midpoint Rule is contained in [1, §7.7], a section that is to be omitted from the syllabus. For that reason you are asked to omit this subsection: you will not be expected to know anything about the Midpoint Rule. Properties of the Definite Integral – Linearity Properties. erties of the Definite Integral, proving some of them. The textbook lists many prop- Zb 1. c dx = c(b − a) a Zb 2. Zb [ f (x) + g(x)] dx = a a Zb [ f (x) − g(x)] dx = a a f (x) dx Zb 4. g(x) dx Zb c f (x) dx = c a f (x) dx + a Zb 3. Zb Zb f (x) dx − a g(x) dx a for any real numbers a, b, c, and any continuous functions f, g. Information for Students in Lecture Section 1 of MATH 141 2010 01 3011 Some of these properties can be derived from others, or can be combined into a more general formula. So, for example we can prove that Z b Z b Z b (r · f (x) + s · g(x)) dx = r f (x) dx + s g(x) dx (17) a a a Z c Z b Z c f (x) dx = f (x) dx + f (x) dx (18) a a b for any real numbers r, s, a, b, c, and these two equations are equivalent to the properties that the textbook numbers ##2, 3, 4, etc. [1, p. 387]: Z a f (x) dx = 0 (19) a Z b Z b Z b ( f (x) ± g(x)) dx = f (x) dx ± g(x) dx (20) a a a Z b Z b (21) c f (x) dx = c f (x) dx a a The first property in the textbook list, Z b c dx = c(b − a) (22) a states, for c ≥ 0 and b ≥ a, that the area of a rectangle of width b − a and height c is c(b − a). Note that all of these properties hold for constants a, b, c that are positive or or negative! Here one must be careful in interpreting areas, since, in the definite integral, areas are signed — they are either positive or negative: we associate the positive sign to areas under a graph above the x-axis, where the lower limit of the integral is not greater than the upper limit. When the curve is below the x-axis, or the lower limit of the integral is not greater than the upper limit, the area is negative. (This is the case for part of [1, Exercise 22, p. 377] which is solved below: there the portion of the area that was below the x-axis cancelled part of the area above the x-axis; while the net result we obtained — 21 — was positive, it was not equal to the total of the magnitudes of the two areas above and below the x-axis, but√was equal to their difference. The graph√of the integrand crosses the x-axis at the points √ ± 6 − 1. The region under the interval [1, 6−1] can be shown to have (negative) area −4 6+ 283 ; while the region √ √ + 4 over the interval [ 6 − 1, 4] can be shown to have (positive) area 35 6.) 3 Properties of the Definite Integral – Additivity of the Interval. A second type or property listed states, in principle, that the area under a curve is the sum of the areas under any two parts into which the curve can be decomposed: Z c Z b Z c (23) f (x) dx = f (x) dx + f (x) dx a a b Information for Students in Lecture Section 1 of MATH 141 2010 01 3012 for any real numbers a, b, c; here again, the constants are not necessarily positive, so one can interpret the point c as lying outside of the interval [a, b] when a ≤ b. This property implies another property listed on [1, p. 373]: Z b Z a f (x) dx = − f (x) dx (24) a b Information for Students in Lecture Section 1 of MATH 141 2010 01 3013 C.4 Supplementary Notes for the Lecture of January 08th, 2010 Release Date: Friday, January 08th, 2010, subject to correction C.4.1 Summary of the last lectures 1. The objective was to define what we mean by the (signed) area under the graph of a function; or, equivalently, the (signed) distance travelled by a moving particle, giving its instantaneous velocity. Zb 2. For a function f continuous on an interval [a, b] I defined the definite integral n P f (x) dx a lim f (xi∗ )∆xi , called a Riemann sum, and mentioned i=1 max ∆xi → 0 n→∞ that this definition and introduction was simplified for your first encounter, and somewhat lacking in rigour. The treatment of the textbook, which is restricted to continuous functions, does not require the possibility that the width ∆xi of the ith interval be possibly different from the that of other subintervals; thus we will be considering only one width, and denote it simply by ∆x. as a kind of limit 3. I mentioned that it can be shown that the sum must have the same limit for all subdivisions of [a, b] into subintervals [a = x0 , x1 ], [x1 , x2 ], . . ., [xi−1 , xi ], [xi , xi+1 ], . . ., [xn−1 , xn = b], and all choices of xi∗ in the ith subinterval (i = 1, . . . , n). 4. I illustrated computations with Riemann sums using the function f (x) = x2 over the interval [a, b]; the variation I considered had the rectangles “hanging” from the graph of f by their upper right corners, but I suggest that students should rework the example with the rectangles hanging by their upper left corners. In my notes, but not discussed in class, was the example of a similar computation for the function f (x) = x2 + 2x − 5 over the interval [1, 4]; for part of the interval f (x) < 0, and the value of the integral includes a cancellation of “negative” and “positive” contributions to area under the graph of f . 5. I reminded you of the formulæ for summing the 0th, 1st, 2nd, and 3rd powers of the first n natural numbers, and indicated their use in the preceding computation. 6. I indicated that computation of the integral in this way will normally not be necessary, as we will be meeting a theorem today which will enable much practical computation for many functions. However, students are still expected to be able to carry out the calculation of the value of a definite integral using Riemann sums. Information for Students in Lecture Section 1 of MATH 141 2010 01 3014 7. I stated basic properties of the definite integral, most of which can be derived from (17), (26), (23): Z b Z b Z b (r · f (x) + s · g(x)) dx = r f (x) dx + s g(x) dx a a a Z c Z b Z c f (x) dx = f (x) dx + f (x) dx a C.4.2 a b §5.2 The Definite Integral (conclusion) Properties of the Definite Integral – Comparison Properties. The textbook lists 3 properties, which are interrelated — each of them can be used to prove the other 2. Z b f (x) ≥ 0 for a ≤ x ≤ b ⇒ f (x) dx ≥ 0 (25) a Z b Z b f (x) ≥ g(x) for a ≤ x ≤ b ⇒ f (x) dx ≥ g(x) dx (26) a a Z b m ≤ f (x) ≤ M for a ≤ x ≤ b ⇒ m(b − a) ≤ f (x) dx ≤ M(b − a) . (27) a Z b While the integral f (x) dx has, thus far, been defined only for a function f which is cona tinuous on [a, b], we will eventually permit generalizations to that definition. Under those generalizations, properties (25), (26), (27) will continue to hold wherever they “make sense”33 . Some worked examples based on problems in an earlier edition of your textbook. Example C.2 ([7, Exercise 18, p. 391]) “Express the limit as a definite integral on the given n X e xi interval: lim ∆x, on the interval [1, 5].” n→∞ 1 + xi i=1 Solution: This problem is not stated in perfect mathematical language, but we know what the textbook means. We are to consider the interval 1 ≤ x ≤ 5 to be subdivided into n subintervals e xi . Then we are to interpret 1+x as f (xi∗ ), the value of a function of equal length ∆x, so ∆x = 5−1 n i ∗ at a point xi chosen in the ith subinterval, so 1 + (i − 1)∆x ≤ xi∗ ≤ 1 + i∆x . 33 We will even have generalizations permitting infinite values for the integral, and the properties will hold there, provided we don’t have to work with “values” like ∞ − ∞ or 0 × ∞. Information for Students in Lecture Section 1 of MATH 141 2010 01 3015 ex Since the function f (x) = is continuous, by virtue of [1, Theorems 7,9, pp. 124–125], 1+x the limit must exist: Z 5 x n X e xi e lim ∆x = dx . n→∞ 1 + x 1 + x i 1 i=1 (It happens that this integral is one that we will be unable to evaluate exactly, but will only be able to approximate.) Example C.3 ([7, Exercise 42, p. 392] “Evaluate Solution: If in (18) x2 cos x dx.” 1 Z Z c Z b f (x) dx = f (x) dx a we set c = a and b = a, we obtain Z a Z f (x) dx = a b Z a a f (x) dx + a Z c f (x) dx + a from which it follows that R1 f (x) dx a a f (x) dx = 0 . a The present integral is of this type — the limits are equal. This result could also have been inferred from (27), by taking b = a, i.e. in Z b m ≤ f (x) ≤ M for a ≤ x ≤ b ⇒ m(b − a) ≤ f (x) dx ≤ M(b − a) Za a ⇒ 0=m·0≤ f (x) d f ≤ M · 0 = 0 , a implying that the integral is equal to 0. Later in the course we will have a method for evaluating an integral of the form Rb x2 cos x dx, a where a, b are any real numbers. That method34 is not required for this “easy” problem. Example C.4 ([7, Exercise 54, p. 392]) Use the properties of integrals to verify the inequalities, without evaluating the integral: Z π2 π π ≤ sin x dx ≤ . π 6 3 6 34 Integration “by parts” Information for Students in Lecture Section 1 of MATH 141 2010 01 3016 Solution: The integrand, sin x, is an increasing function when x is in the 1st quadrant; this π π implies that, for ≤ x ≤ , 6 2 1 π π = sin ≤ sin x ≤ sin = 1 . 2 6 2 π π π − = . The value 2 6 3 π π of the integral is then bounded by the areas of two rectangles on the base , : the lower 6 2 bound is given by the rectangle whose height is the minimum value of the function, the value at x = π6 ; the upper bound is given by the rectangle whose height is the maximum value, attained at x = π2 : Z π2 1 π π · ≤ sin x dx ≤ 1 · . π 2 3 3 6 The length of the interval over which the integral is being evaluated is The exact value of the integral will eventually be seen to be √ 3 . 2 Example C.5 ([7, Exercise 58, p. 392]) “Use (27) to estimate the value of the integral Z 2 (x3 − 3x + 3) dx .” 0 Solution: On the real line the given function has critical numbers at ±1; of these only x = +1 is in the interval of the integral. By the 2nd Derivative Test x = 1 is a local minimum: the function value there is 1. At the end-points of the interval of integration, 0 and 2, the function has values 3 and 5. We conclude that, on the given interval, the function values are bounded as follows: 1 ≤ x3 − 3x + 3 ≤ 5 . The length of the interval is 2 − 0 = 2. The value of the integral is, therefore, bounded between 2 and 10. (Eventually we shall be able to evaluate this integral exactly, and shall be able to ! 24 3 2 show that its value is − · 2 + 3 · 2 = 4 − 6 + 6 = 4. 4 2 5.2 Exercises [1, Exercise 22, p. 771] “Use the form of the definition of the integral...to evaluate the integral R4 2 (x + 2x − 5) dx.” 1 Solution: Here we are finding the area under the graph of the polynomial x2 + 2x − 5, a continuous function, on the interval 1 ≤ x ≤ 4, i.e., the area of the region bounded by the graph on top, the lines x = 1 and x = 4 on the two vertical sides, and the x-axis on the Information for Students in Lecture Section 1 of MATH 141 2010 01 3017 , and the end-points of the bottom. We divide the interval into n parts of length 4x = 4−1 n 3 ith subinterval are xi−1 = 1+(i−1)∆x = 1+(i−1)· n on the left and xi = 1+i·4x = 1+i· 3n on the right. If we choose the sample points to be the right end-points of the subintervals, i.e., to find the limit of the right Riemann sum, we find that Z 4 (x2 + 2x − 5) dx 1 = lim n→∞ n X (xi2 + 2xi − 5)∆x i=1 !2 ! n X 3 3i 3i 1 + = lim +2 1+ − 5 n→∞ n n n i=1 ! ! ! n X 3i 9i2 3 3i 1+2· + 2 +2 1+ = lim −5 n→∞ n n n n i=1n n n n n n X X X X X 3 X 6 9 6 2 1+ 1 i+ 2 i +2 i−5 = lim 1 + n→∞ n i=1 n i=1 n i=1 n i=1 i=1 i=1 n n n n n n X X 3 X 6X 9 X 2 6X 1+ 1 i+ 2 i +2 i−5 = lim 1 + n→∞ n i=1 n i=1 n i=1 n i=1 i=1 i=1 n n n X 9 X 2 3 6+6X i+ 2 i = lim (1 + 2 − 5) 1+ n→∞ n n n i=1 i=1 i=1 ! 12 n(n + 1) 9 n(n + 1)(2n + 1) 3 = lim −2n + · + 2· n→∞ n 2 n 6 n !! 9 3 3 = lim −2n + 6(n + 1) + 3n + + n→∞ 2 2n n = 21 If we choose the sample points to be the left end-points, i.e., to find the limit of the left Riemann sum, we obtain a very similar sum. We can write the sum, like the preceding, as a sum over the index i ranging from i = 1 to i = n, but where the function is evaluated at the left end-points: Z 4 (x2 + 2x − 5) dx 1 = n X 2 (xi−1 + 2xi−1 − 5)∆x lim n→∞ i=1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3018 !2 ! n X 3 3(i − 1) 3(i − 1) 1 + = lim +2 1+ − 5 n→∞ n n n i=1 ! ! ! n X 3(i − 1) 3 3(i − 1) 9(i − 1)2 + 2 1 + = lim 1+2· + − 5 2 n→∞ n n n n i=1 n n n n n n X X X X X 3 X 9 6 6 2 (i − 1) + 2 (i − 1) − 5 1 = lim 1 + (i − 1) + 2 1+ n→∞ n i=1 n i=1 n n i=1 i=1 i=1 i=1 At this point we can proceed in a variety of ways. One is to observe that the first term in the sums of powers is 0, so that we are summing only n − 1 non-zero powers. A more formal way to to that is to define j=i−1 (28) and to rewrite each of the sums as a sum over j, replacing the value i = 1 by j = 0 and the value i = n by j = n − 1. We then obtain Z 4 (x2 + 2x − 5) dx 1 n−1 n−1 n−1 n−1 n−1 n−1 X X X X X X 3 6 9 6 1+ 1 j+ 2 j2 + 2 j−5 = lim 1 + n→∞ n j=0 n j=0 n j=0 n j=0 j=0 j=0 n−1 n−1 n−1 X X X 3 9 6 + 6 2 j+ 2 j = lim (1 + 2 − 5) 1+ n→∞ n j=0 n j=0 n j=0 ! 12 (n − 1)(n) 9 (n − 1)n(2n − 1) 3 = lim −2n + · + 2· n→∞ n 2 n 6 n !! 9 3 3 = lim −2n + 6(n − 1) + 3n − + n→∞ 2 2n n = 21 again. The definition of j in (28) is analogous to the changes of variables that we will be making in definite integrals in [1, §5.5]: there, as here, we have to change the limits of the integral to correspond to the new values of the variable of integration. What we have here is an example of the “finite difference calculus”, where there are results similar to those that we will be developing in the “infinitesimal calculus”. [1, Exercise 40, p. 378] “Evaluate the integral by interpreting it in terms of areas: R10 |x−5| dx.” 0 Solution: (While we could evaluate this integral using Riemann sums, the intention is that the student interpret the area under the curve in terms of familiar geometric objects, and use known formulæ to determine the value.) Information for Students in Lecture Section 1 of MATH 141 2010 01 3019 The portion of the integral from 0 to 5 is the area of a right-angled triangle whose hypotenuse is on the line y = −(x − 5), with base of length 5, and height 5, so its area is 25 52 = . The portion of the integral from 5 to 10 is the mirror image of the triangle 2 2 described above, this time with hypotenuse along the line y = x − 5; its area is the same as the previous one, so the value of the integral is 25. C.4.3 §5.3 The Fundamental Theorem of Calculus In a number of areas of mathematics there are theorems that have acquired the name “The Fundamental Theorem of...”. The present section is devoted to such a theorem, also known as “The Fundamental Theorem of (the) Integral Calculus”, one part of which relates the value of a definite integral to antiderivatives of its integrand, and provides a method for evaluating such integrals without the need for computing limits of complicated sums. (The formulation of the theorem as being divided into two parts is not completely standard.) Differentiation and Integration as Inverse Processes. Theorem C.6 (The Fundamental Theorem of Calculus) If f is continuous on [a, b], then ! Z x d 1. f (t) dt = f (x). dx a Z b 0 2. F = f ⇒ f (x) dx = F(b) − F(a) . a Definition C.2 We may represent a difference F(b) − F(a) by [F(x)]ba or even more briefly by F(x)]ba if the latter expression is unambiguous. I sometimes use a notation which is not standard, but is unambiguous, and write the preceding difference as F(x)] x=b x=a Information for Students in Lecture Section 1 of MATH 141 2010 01 3020 Example C.7 Following my discussion of [1, §5.2] in these notes there is35 a solution of Z4 [1, Exercise 22, p. 377], in which the integral x2 + 2x − 5 dx is evaluated “from first 1 principles”, proving that its value is 21. Let’s now verify that result using the Fundamental x3 Theorem. One antiderivative of x2 + 2x − 5 is + x2 − 5x. The value of the integral is, 3 therefore, #4 ! ! " 3 64 1 x 2 + x − 5x = + 16 − 20 − + 1 − 5 = 21 . 3 3 3 1 Example C.8 ([7, Exercise 12, p. 402]) “Use Z Part 1 of the Fundamental Theorem of Calculus 10 to find the derivative of the function F(x) = tan θ dθ.” x Rx Solution: First observe that F(x) = − 10 tan θ dθ. Having written the integral in the form to which the Fundamental Theorem applies, i.e., with the variable in the upper limit, we may apply that theorem: the derivative is minus the value of the integrand, tan θ, evaluated where θ = x, i.e., − tan x. (Eventually we will see that F(x) = ln cos x − ln cos(10); again students may then verify that the Fundamental Theorem is giving us the correct derivative.) 35 on pages 3016–3018 Information for Students in Lecture Section 1 of MATH 141 2010 01 3021 C.5 Supplementary Notes for the Lecture of January 11th, 2010 Release Date: Monday, January 11th, 2010 updated on 12 January; subject to further updates and corrections C.5.1 §5.3 The Fundamental Theorem of Calculus (conclusion) Example C.9 ([7, Exercise 10, p. 402]) “Use 1 of the Fundamental Theorem of Calculus Z Part u 1 dx.” to find the derivative of the function g(u) = 2 3 x+ x 1 Solution: The derivative is the value of the integrand, evaluated at the upper limit of the x + x2 1 integral, i.e., where x = u: g0 (u) = . u + u2 u+1 4 − ln ; students may then verify that the (Eventually we will see that g(u) = ln u 3 Fundamental Theorem is giving us the correct derivative.) 5.3 Exercises Z 5x [1, Exercise 54, p. 389] “Find the derivative of the function y = cos(u2 ) du.” cos x Solution: Two observations are necessary: • The Fundamental Theorem is concerned with an integral whose upper limit is variable; if we wish to apply that theorem here, we shall need to transform the problem to one where only the upper limit of the integral(s) is variable. • The Fundamental Theorem is concerned with an integral whose upper limit is the independent variable under consideration; should we wish to permit the upper limit to vary in a more complicated way, we will need to apply the Chain Rule. 1. We shall transform the given integral into a sum of two where the lower limit of each is constant. We do this by splitting the interval of integration, [cos x, 5x] into two parts at a “convenient” point. It is not even necessary that the point we choose be inside the interval, since the property we are applying, (18) on page 3011 of these notes, does not require that fact. I will choose the constant 0 to be the point where the splitting occurs: Z 5x Z 0 Z 5x 2 2 cos(u ) du = cos(u ) du + cos(u2 ) du cos x cos x 0 Z cos x Z 5x 2 = − cos(u ) du + cos(u2 ) du 0 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 Hence d dx Z 5x d cos(u ) du = − dx cos x Z cos x 2 0 d cos(u ) du + dx 3022 Z 5x 2 cos(u2 ) du (29) 0 2. Each of these derivatives will be computed using the Chain Rule, with the intermediate variable being the function appearing as the variable upper limit. In the first case, if we take the intermediate variable to be, say z = cos x, we have Z cos x Z z d d dz 2 cos(u ) du = cos(u2 ) du · dx 0 dz 0 dx dz = cos(z2 ) · dx 2 = cos(cos x) · (− sin x) For the second integral we take the intermediate variable to be w = 5x. Here Z 5x Z w d d dw 2 cos(u ) du = − cos(u2 ) du · dx 0 dw 0 dx dw = cos(w2 ) · dx 2 = cos(25x ) · 5 Combining the two results gives Z 5x Z cos x Z 5x d d d 2 2 cos(u ) du = − cos(u ) du + cos(u2 ) du dx cos x dx 0 dx 0 2 = − cos(cos x) · (− sin x) + cos(25x2 ) · 5 = cos(cos2 x) · sin x + 5 cos(25x2 ) Z x 1 [1, Exercise 58, p. 389] “Find the interval on which the curve y = dt is concave 2 0 1+t+t upward.” 1 −1 − 2x Solution: By the Fundamental Theorem, y0 (x) = ; hence y00 (x) = . 2 1+x+x (1 + x + x2 )2 The denominator of the second derivative is a non-zero square, so it is always positive; 1 the function will be positive whenever −1 − 2x > 0, i.e., whenever x < − : this is where 2 the graph is concave upward. (Eventually we will be able to evaluate the integral explicitly, showing that the curve is ! 2x + 1 π 2 . y = √ arctan √ − 6 3 3 Students may differentiate to check that I have found the correct second derivative.) Information for Students in Lecture Section 1 of MATH 141 2010 01 Zb Za x [1, Exercise 74, p. 390] Suppose that 3023 e x dx . Express b in terms of a. e dx = 3 0 0 Solution: Substituting the values of the given integrals, from Part 2 of the Fundamental Theorem, yields e x ]b0 = 3 e x ]a3 , which implies that eb − e0 = 3 ea − e0 ⇒ eb = 3ea − 2 ⇒ b = ln (3ea − 2) . Problems Plus [1, Exercise 12, p. 413] Find d2 dx2 Z x 0 Z sin t ! √ 1 + u4 du dt . 1 Solution: d2 dx2 C.5.2 Z x 0 Z sin t 1 ! √ 4 1 + u du dt = ! ! Z x Z sin t √ d d 4 1 + u du dt dx dx 0 1 ! Z sin x √ d = 1 + u4 du dx 1 p = 1 + sin4 x · cos x . §5.4 Indefinite Integrals and the “Net Change” Theorem Indefinite Integrals The traditional symbol for a “general” antiderivative F(x) of a function Z f (x) (i.e., some function with the property that F 0 (x) = f (x)) is f (x) dx, which is called the indefinite integral of f (x). Since two antiderivatives differ by a constant (by a corollary to the Mean Value Theorem), we usually write statements in the form Z f (x) dx = F(x) + C where F(x) is one specific antiderivative, and C is a constant of integration, intended to range over all real numbers. Once a particular antiderivative F has been chosen, the particular real Information for Students in Lecture Section 1 of MATH 141 2010 01 3024 number C that applies in a particular situation has to be determined from additional information that is usually available in the problem at hand. Much of this course will be concerned with methods for finding indefinite integrals. While the finding of the indefinite integral may be a difficult problem, the verification that a function F that is claimed to be an antiderivative of f is not, since all that needs to be done is to differentiate F and to check whether the derivative is f . (In principle, if a function F has the property that F 0 = f , then F is an antiderivative of f : thus it is possible to find an antiderivative by guessing, or by simply copying the answer from the back of the textbook or from your neighbour’s work. The intention in the course is that you should normally be expected to be able to show a systematic way of determining an antiderivative; there will be a very few special situations where you will be presented with an antiderivative without a convincing way of finding it.) Z Note the same symbol, a stylized letter S — called the integral sign — is used for both the indefinite integral and the definite integral; while they are related by the Fundamental Theorem, they are different operations. Even though the two operations are different, they share some similar properties. For example, parallel to property (17) of the definite integral on page 3011 of these notes, we can also prove that Z Z Z (r f (x) + sg(x)) dx = r f (x) dx + s g(x) dx , (30) where r, s are any constants. Equations like the preceding, in which an indefinite integral appears on both sides of the equal sign, are normally written without any constant of integration. The “Net Change” Theorem This is simply the author’s name for the second part of the Fundamental Theorem. It is not a term in standard usage, and I am not likely to use it. As a beginning in the development of general techniques for determining indefinite integrals, we can reformulate results that we developed for derivatives. For example, since we know that d sin x = cos x , dx we can reformulate this result as Z cos x dx = sin x + C . Some reformulations require minor changes, e.g., division by an appropriate constant. From the result that d n x = n · xn−1 when n , 0 dx we can conclude that Z 1 xn−1 dx = xn + C when n , 0 n Information for Students in Lecture Section 1 of MATH 141 2010 01 3025 Function f (x) g(x) f (x) + g(x) One antiderivative F(x) G(x) F(x) + G(x) xn+1 xn (n , −1) n+1 1 ln |x| x ex ex cos x sin x sin x − cos x 2 sec x tan x sec x tan x sec x csc x cot x − csc x 1 arctan x 1 + x2 1 −arccot x 1 + x2 1 arcsin x √ 1 − x2 1 − arccos x √ 1 − x2 Table 4: Some Antiderivatives or, after the substitution of m + 1 for n, Z 1 m+1 xm dx = x +C m+1 when m , −1 . Thus Table C.5.2 of antiderivatives on page 3025 of these notes, which I included last semester in my notes in MATH 140 2009 09, can now be recast in the form of Table C.5.2 on page 3026 below. Applications While we have been using the 2nd part of the Fundamental Theorem to express the value of a definite integral in terms of the “net change” in the antiderivative, we can also apply the result in the opposite way: that the net change can be found by evaluating the integral. This is the spirit of the motivation that the author called “The Distance Problem” (see above, page 3005, or the solution below, on page 3029 of [1, Exercise 44, p. 397]]). We will see another example below in the solution of [1, Exercise 62, p. 398]. Information for Students in Lecture Section 1 of MATH 141 2010 01 Z Z [r f (x) + sg(x)] dx Z xn dx Z e x dx Z sin x dx Z sec2 x dx Z sec x tan x dx Z 1 1 + x2 Z 1 √ 1 − x2 = r Z f (x) dx + s xn+1 = +C n+1 = ex + C = − cos x + C = tan x + C = sec x + C = arctan x + C = arcsin x + C 3026 Z g(x) dx k dx = kx + C Z (n , −1) 1 dx = ln |x| + C Z x ax a x dx = + C (a > 0) ln a Z cos x dx = sin x + C Z csc2 x dx = − cot x + C Z csc x cot x = − csc x + C Z 1 = −arccot x + C 1 + x2 Z 1 dx = − arccos x + C √ 1 − x2 Table 5: Very Short Table of Indefinite Integrals Information for Students in Lecture Section 1 of MATH 141 2010 01 3027 Example C.10 ([7, Exercise 10, p. 411]) “Find the general indefinite integral of ! Z 1 2 x +1+ 2 dx .” x +1 Solution: Break the integrand into two parts: the summand at the end is recognizable as the derivative of arctan x; the two terms at the beginning are multiples of powers of x, and we have observed earlier how to integrate them. Thus ! Z Z Z 1 1 2 2 x + 1 dx + x +1+ 2 dx = dx 2 x +1 x +1 1 3 1 = x + x + arctan x + C 3 1 where the letter C represents the constant of integration. Even though there are two indefinite integrals on the right side of the equation, only one constant is needed: if we were to include two constants, as +C1 + C2 , we would not gain any more freedom. Example C.11 ([7, Exercise 11, p. 411]) Find the general indefinite integral of Z √ 2 2 − x dx . that the antiderivative of Solution: One might be tempted, at first, to consider the possibility √ the given 2nd power is a multiple of the 3rd power of 2 − x; unfortunately that temptation will have to be resisted, as the resulting functions will have the correct derivative. The √ not 1 2 simplest approach is to expand the square: since (2 − x) = 4 − 4x 2 + x, Z Z √ 2 √ 2 − x dx = 4 − 4 x + x dx 2 3 1 = 4x − 4 · x 2 + x2 + C 3 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3028 C.6 Supplementary Notes for the Lecture of January 13th, 2010 Release Date: Wednesday, January 13th, 2010, subject to correction C.6.1 §5.4 Indefinite Integrals and the “Net Change” Theorem (conclusion) 5.4 Exercises [1, Exercise 34, p. 397] “Evaluate the integral Z 9 3x − 2 √ dx.” x 1 Solution: Simplify the integrand by dividing the denominator into the two summands of the numerator: ! Z 9 Z 9 √ 3x − 2 2 3 x − √ dx √ dx = x x 1 1 " #9 2 32 1 = 3 · · x − 2 · 2 · x2 3 1 h 3 √ i9 = 2x 2 − 4 x 1 = (2 · 27 − 4 · 3) − (2 · 1 − 4 · 1) = (54 − 12) − (2 − 4) = 44 . [1, Exercise 38, p. 397] “Evaluate the integral Z π3 sin θ + sin θ · tan2 θ dθ .” sec2 θ 0 Solution: When the integrand involves trigonometric functions, one may have to apply a familiar identity to simplify the integration. There is often more than one way to do this. In the present example sin θ + sin θ · tan2 θ = sin θ · cos2 θ + sin3 θ sec2 θ = sin θ · (cos2 θ + sin2 θ) = sin θ so Z π 3 0 sin θ + sin θ · tan2 θ dθ = sec2 θ Z π 3 sin θ dθ 0 1 1 π + cos 0 = − + 1 = . 3 2 2 Eventually you will know how to evaluate the parts of this integral separately, but the present solution is faster than evaluating and adding the parts. π = [− cos θ]03 = − cos Information for Students in Lecture Section 1 of MATH 141 2010 01 3029 3π Z2 [1, Exercise 44, p. 397] Evaluate the integral | sin x| dx. 0 Solution: The integrand is continuous, so we know the integral exists. However, it is not convenient to work with an antiderivative of | sin x |. So we split the interval of h i 3π integration into the parts [0, π] and π, 2 where the integrand is respectively positive and negative, and thereby avoid working with the absolute value. For the integrands, respectively sin x and − sin x, we know antiderivatives − cos x and + cos x, so we may apply the Fundamental Theorem to each part separately: Z 3π 2 Z | sin x| dx = 0 Z π 3π 2 sin x dx + 0 (− sin x) dx π 3π = [− cos x]π0 + [cos x]π2 = (−(−1) + 1) + (0 − (−1)) = 3. [1, Exercise 58, p. 398] The velocity function is v(t) = t2 − 2t − 8 for a particle moving along a line. Find (a) the displacement; and (b) the distance travelled by the particle during the time interval 1 ≤ t ≤ 6. Solution: Denote the position of the particle at time t by x(t); then v(t) = an antiderivative of v. d x(t), dt so x is (a) The displacement is simply the difference between initial and final positions of the moving particle; it is equal to the area under the graph of the velocity function between the appropriate times. Z 6 6 displacement = x(6) − x(1) = [x(t)]1 = v(t) dt 1 Z 6 = (t2 − 2t − 8) dt 1 #6 " 13 2 t − t − 8t = 3 1 ! 1 10 = (72 − 36 − 48) − −1−8 =− 3 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3030 (b) The distance travelled is equal to the area under the graph of the speed function between the appropriate times; the speed is the magnitude of the velocity. Z 6 Z 6 2 t − 2t − 8 dt = distance travelled = |(t + 2)(t − 4)| dt . 1 1 The function (t + 2)(t − 4) changes sign at t = −2, which is outside the interval of integration, and again at t = 4, which is inside the interval of integration. We can break the integral up into two parts at x = 4, and then we can express each of the parts without using absolute value symbols: Z 6 Z 4 Z 6 |(t + 2)(t − 4)| dt = |(t + 2)(t − 4)| dt + |(t + 2)(t − 4)| dt 1 1 4 Z 4 Z 6 = − (t + 2)(t − 4) dt + (t + 2)(t − 4) dt 1 4 Z 4 Z 6 2 = − (t − 2t − 8) dt + (t2 − 2t − 8) dt 1 4 " #4 " #6 13 2 13 2 = − t − t − 8t + t − t − 8t 3 3 4 " # 1 44 98 = −[−18] + = 3 3 Note that the factorization of the quadratic was needed in order to determine where the split the interval of integration, but it did not help in the actually integration operation, and had to be reversed at that stage. [1, Exercise 62, p. 398] “Water flows from the bottom of a storage tank at a rate of r(t) = 200 − 4t litres/min, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 10 minutes.” Solution: Let’s denote by W(t) the amount of water (measured in litres) that has left the tank by time t. We are told that d W(t) = r(t) = 200 − 4t dt Then Z 10 W(10) − W(0) = Z 0 10 = (0 ≤ t ≤ 50). dW dt dt h i10 (200 − 4t) dt = 200t − 2t2 0 = (200 · 10 − 2 · 102 ) − 0 = 1800 , 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3031 so the total amount of water leaving the tank in the first 10 minutes is 1800 litres. Example C.12 ([7, Exercise 26, p. 402]) “Use Part 2 of the Fundamental Theorem of Calculus Z3 to evaluate the integral, or explain why it does not exist: x−5 dx.” −2 Solution: The integrand is not defined at the point 0 in the interval [−2, 3]. This means that we cannot even speak of the integral at this time. (Later we will generalize our definition of integral to permit us to consider certain “improper” integrals where there is an infinite discontinuity. But even that generalization will not apply to this problem, although it is premature to consider it here. Look at this problem again when we study [1, §7.8].) The theorem is not applicable because the integrand is not defined at one point in the domain, and the discontinuity is neither removable nor a jump discontinuity: a removable discontinuity would have no effect at all, and a jump discontinuity could be accommodated by the method of Example C.14 above, i.e., by splitting the integral into two parts at the jump discontinuity. ( x if −π ≤ x ≤ 0 . Use Part 2 of Example C.13 ([7, Exercise 42, p. 403]) Let f (x) = sin x if 0 < x ≤ π Zπ the Fundamental Theorem of Calculus to evaluate the integral f (x) dx, or explain why it −π does not exist. Solution: The function f is defined piecewise, by gluing together one function defined on [−π, 0] and another defined on (0, +π]; however, it is continuous, as lim− x = 0 = lim+ sin x, x→0 x→0 and f is defined at x = 0 and equal to the common value of the two one-sided limits. We know 1 that one antiderivative of x is x2 ; and that one antiderivative of sin x is − cos x. Consider the 2 ( x2 if −π ≤ x ≤ 0 2 function F(x) = . This function is evidently differentiable at all 1 − cos x if 0 < x ≤ π points except possibly 0; and F is an antiderivative of f on the interval −π < x < +π, with the possible exception of the point x = 0. At x = 0 the two one-sided limits of difference quotients are x2 3 −0 x x 1 = lim+ = · 0 = 0 x→0 3 3 F(x) − F(0) (1 − cos x) − 0 lim = lim+ x→0+ x→0 x x 1 − (1 − 2 sin2 2x ) − 0 = lim+ x→0 x F(x) − F(0) lim− = lim− x→0 x→0 x Information for Students in Lecture Section 1 of MATH 141 2010 01 = lim+ sin 2x x 2 sin 2x lim x x→0+ 2 x · sin 2 x→0 = 3032 ! · lim+ sin x→0 x = 1 · 0 = 0, 2 so the function is also differentiable at x = 0, where the derivative is equal to 0, i.e., to f (0); thus F is an antiderivative of f on the interval (−π, +π). Thus we can apply the Fundamental Theorem: ! Zπ (−π)2 π2 f (x) dx = F(π) − F(−π) = (1 − cos π) − =2− . 2 2 −π But finding the antiderivative was a complicated computation, and rendered the problem more difficult than necessary. Instead, one should proceed as follows, applying (23) in these notes, page 3011: Zπ Z0 Zπ f (x) dx = −π f (x) dx + −π 0 Z0 Zπ = x dx + −π " f (x) dx # 2 0 sin x dx 0 x 2 + [− cos x]π0 −π ! (−π)2 = 0− + (− cos π + cos 0) 2 π2 π2 = − + (−(−1) + 1) = 2 − . 2 2 = The lesson to be learned from this example is that there are often advantages to splitting up an integral, even if it is theoretically possible to evaluate it without doing so. Integration of “piecewise continuous” functions. The definition [1, Definition 2, p. 366] of a definite integral given by the textbook, is more restrictive than necessary. This definition requires that the integrand be continuous throughout the interval of integration. In fact, the definition can be weakened to apply to a broader class of functions. While we don’t require full generality in this course, we do wish to be able to apply the theory to functions that have isolated “jump” discontinuities; (we can handle removable discontinuities just be “removing” them, i.e., by extending the function to the appropriate value at the points missing in the original definition). If a function f is continuous on an interval [a, c], except for a point b in (a, c) Information for Students in Lecture Section 1 of MATH 141 2010 01 3033 where lim− f (x) and lim+ f (x) both exist, but are not equal, we will define x→b x→b Z Z c f (x) dx = a Z b c f (x) dx + a f (x) dx (31) b that is, we will define the integral to be such as to satisfy the additivity property (23) we saw earlier in these notes, on page 3011. Note, however, that we cannot apply to the whole interval Rb the Fundamental Theorem to evaluate integral f (x) dx if f has a discontinuity at a point c a such that a < c < b. Example C.14 Suppose that ( f (x) = Z −1 if x ≤ 0 1 if x > 0 2 Evaluate f (x) dx. −1 Solution: We cannot apply the Fundamental Theorem to the entire interval [−1, 2], as the integrand is discontinuous at point 0. So we split the integral at the point x = 0, which is the point of discontinuity. The two integrals we obtain now satisfy the conditions of the Fundamental Theorem: Z 2 Z 0 Z 2 f (x) dx = f (x) dx + f (x) dx −1 −1 0 Z 0 Z 2 = (−1) dx + 1 dx −1 = C.6.2 [−x]0−1 0 + [x]20 = (−0 + (−1)) + (2 − 0) = 1. §5.5 The Substitution Rule The “Substitution Rule” is a reformulation, in terms of the integral, of the Chain Rule. Theorem C.15 Let u = g(x) and f (x) be respectively differentiable and continuous on a given interval; then Z Z Z 0 0 ( f ◦ g)(x) · g (x) dx = f (g(x)) · g (x) dx = f (u) du . In applying this theorem we usually begin with a “complicated” integral, whose form we try to interpret like the left side of the above equation, and try to find an appropriate “substitution” of the form u = g(x) which will transform the integral into one whose integrand is one that we Information for Students in Lecture Section 1 of MATH 141 2010 01 3034 are able to integrate. In practice one works with the differentials dx and du in a “mechanical” way that can be justified by the theorem. Proof: Let F and g be differentiable. Then F 0 (g(x)) · g0 (x) = d F(g(x)) . dx Integrating with respect to x, we have Z Z d 0 0 F (g(x)) · g (x) dx = F(g(x)) dx = F(g(x)) + C. dx If we define u = g(x), f = F 0 , then Z Z Z d 0 f (g(x)) · g (x) dx = F(g(x)) dx = F(u) + C = f (u) du . dx In practice this “substitution” is often applied by “mechanically”, substituting functions and differentials, and such operations can be shown to be fully justifiable. The general idea in looking for substitutions is to try to reduce the complication of the original indefinite integral. This is a subjective term, and different users may find a variety of distinct substitutions which they will find helpful in evaluating an indefinite integral. Finding the appropriate substitution is one step in solving a problem. In the first problems in the list of exercises the author suggests a substitution which will be helpful; eventually students are expected to find an appropriate substitution on their own — often there are several possible choices. You should be experimenting with different substitutions and getting to know the types of problems each of them is useful in solving. Example C.16 Earlier, in Example C.11 on page 3027 of these Znotes, I considered [7, Exercise √ 2 11, p. 411], which was concerned with the indefinite integral of 2 − x dx. We evaluated this integral by expanding the square and then integrating Z the powers of x separately. Could √ 10000 we use the same methods for the indefinite integral 2− x dx? Solution: While we could expand the integrand in this case too, the result would have 10,001 terms, each of which would have to be integrated. The result we would obtain would not be very useful. Consider the following alternative approach, that Z could also have been used when the exponent was 2. We have here an integral of the form f (x) = x10000 , and g(x) = 2 − √ x. Define u = g(x) = 2 − √ ( f ◦ g)(x) · g0 (x) dx, where 1 x, so that du = − √ dx, so 2 x √ dx = −2 x du = 2(u − 2)du . Information for Students in Lecture Section 1 of MATH 141 2010 01 Z Z √ 10000 2− x dx = 2 u10000 (u − 2) du Z = 2 u10001 − 2u10000 du 2 4 u10002 − u10001 + C 10002 10001 √ √ 2 4 = (2 − x)10002 − (2 − x)10001 + C 10002 ! √ 10001 √ 10001 2 x 4 = (2 − x) − − +C 10002 (10001)(10002) = 3035 Information for Students in Lecture Section 1 of MATH 141 2010 01 3036 C.7 Supplementary Notes for the Lecture of January 15th, 2010 Release Date: Friday, January 15th, 2010 (subject to revision) C.7.1 §5.5 The Substitution Rule (conclusion) Review The Substitution Rule (cf., p. 3033) states that Let u = g(x) and f (x) be respectively differentiable and continuous on a given interval; then Z Z Z 0 0 ( f ◦ g)(x) · g (x) dx = f (g(x)) · g (x) dx = f (u) du In C.16, discussed at the end of the last lecture, I showed how we could evaluate Z Example √ 10000 √ 2− x dx using the substitution u = g(x) = 2 − x, obtaining √ dx = −2 x du = 2(u − 2)du Z Z √ 10000 ⇒ 2− x dx = 2 u10000 (u − 2) du Z = 2 u10001 − 2u10000 du 2 4 u10002 − u10001 + C 10002 10001 √ √ 2 4 = (2 − x)10002 − (2 − x)10001 + C 10002 ! √ 10001 √ 10001 2 x 4 = (2 − x) − − +C 10002 (10001)(10002) R Example C.17 [1, Example 6, p. 403] “Calculate tan x dx.” Solution: This problem does not, at first, appear to be a candidate for a substitution; but, by sin x expressing the tangent as , the textbook shows that it can be simplified by treating cos x cos x as the new variable. That is, by defining u = cos x, which implies that du = − sin x dx, we can evaluate the integral as follows: Z Z sin x dx tan x dx = Z cos x d(cos x) = − cos x = Information for Students in Lecture Section 1 of MATH 141 2010 01 3037 Z Z du which has the form − , which we should recognize as − d(ln |u|). Thus one subu R stitution that is indicated is u = cos x: the integral becomes − d(ln |u|) = − ln |u| + C = − ln | cos x| + C. The integral could also be expressed as ln | sec x| + C or as ln |k sec x|, where C or k are constants of integration. Three other indefinite integrals can be evaluated in similar ways; perhaps you should remember36 them together:37 Z tan x dx = ln | sec x | + C = − ln | cos x | + C (32) Z cot x dx = − ln | csc x | + C = ln | sin x | + C (33) Z tanh x dx = ln cosh x + C (34) Z coth x dx = ln | sinh x | + C (35) Z Example C.18 ([7, Exercise 2, p. 420]) “Evaluate the integral x(4 + x2 )10 dx by making the substitution u = 4 + x2 .” Solution: First we should observe that we could solve this problem without any substitution at all. We could expand the 10th power of the binomial into a polynomial of degree 20, multiply each of the terms by an additional x, and then integrate the resulting sum of 11 terms one by one. The resulting polynomial would be the correct solution. But what would you do it the exponent were not 10, but 10000? Applying the method of substitution, we set du u = g(x) = 4 + x2 , f (u) = u10 , so = g0 (x) = 2x. Then dx Z Z 10 10 1 d 2 x 4+x dx = 4 + x2 · 4 + x2 dx 2 dx Z 11 1 d = 4 + x2 dx 2 dx 11 1 = 4 + x2 + C . 22 In practice we often operate mechanically with differentials; from u = 4 + x2 we have du = 2x dx, so Z Z Z 2 10 10 x 4+x dx = xu dx = u10 · x dx Z 1 1 = u10 · du = u11 + C . 2 22 36 37 This doesn’t mean to memorize them — just to remember the “trick” needed to evaluate them. Why are absolute signs missing in one of these cases? Information for Students in Lecture Section 1 of MATH 141 2010 01 3038 But it would be bad form to stop here, since our answer has been expressed in terms of u, not the original variable x; so we continue = 11 1 4 + x2 + C . 22 Example C.19 ([7, Exercise 30, p. 421]) Evaluate the indefinite integral Z ax + b dx . √ ax2 + 2bx + c Solution: Later in the term we will consider the integration of any function of the form kx + ` , √ ax2 + 2bx + c but the present problem concerns a special case which is easy to handle. (Note that the textbook should have insisted that not all of a, b, c can be 0, as then the ratio is not defined.) Again, we begin by attempting to simplify the integrand. It would appear that the quadratic function whose square root appears in the denominator is the feature we should attempt to simplify. One way to do this is to define u = ax2 + 2bx + c. This implies that du = 2(ax + b) dx, so we obtain Z Z 1 du dx = √ √ 2 u ax2 + 2bx + c 1 = u2 + C 1 = (ax2 + 2bx + c) 2 + C . ax + b 1 The problem could also be solved by the substitution v = (ax2 + 2bx + c) 2 . Example C.20 ([7, Exercise 38, p. 421], slightly changed) Evaluate the indefinite integral Z √ 7 x3 + 1 · x5 dx . Solution: The method I propose to use here is based on the fact that d(xn ) = nxn−1 dx for any integer n. When an integrand is expressible in terms of a power of the variable, short by just 1, then this type of substitution is often a good way to begin simplifying it, although further Information for Students in Lecture Section 1 of MATH 141 2010 01 3039 steps might be needed. So, in this case, I begin with u = x3 , which implies that du = 3x2 dx, i.e., that x2 dx = 13 du. I then obtain Z √ Z √ 7 1 7 5 3 x + 1 · x dx = u + 1 · u du 3 after which I consider further simplification. At√this point I would like to simplify the 7th root. 7 I can do this by either setting v = u+1, or w = u + 1. In the first case I would obtain dv = du, while, in the second, u + 1 = w7 , so du = 7w6 dw. Z √ Z √7 1 1 7 u + 1 · u du = v · (v − 1) dv 3 3Z 8 1 1 = v 7 − v 7 dv 3 " # 1 7 157 7 87 = v − v +C 3 15 8 " # 1 7 7 15 8 7 7 = (u + 1) − (u + 1) + C 3 15 8 " # 1 7 3 7 15 8 3 = (x + 1) 7 − (x + 1) 7 + C 3 15 8 Z p Z 1 1 7 or (u + 1) · u du = w · w7 − 1 · 7w6 dw 3 3Z 7 = w14 − w7 dw 3 " # 7 1 15 1 8 = w − w +C 3 15 8 7 7 15 8 8 = (u + 1) 7 − (u + 1) 7 (u + 1) 7 + C 45 24 7 3 7 15 8 8 = (x + 1) 7 − (x3 + 1) 7 (x3 + 1) 7 + C . 45 24 Definite Integrals Thus far I have been applying the Substitution Rule to indefinite integrals. Substitution may be applied to a definite integral Z b f (g(x)) g0 (x) dx a in two ways: • Apply the Substitution Rule to the corresponding indefinite integral, Z f (g(x))g0 (x) dx ; Information for Students in Lecture Section 1 of MATH 141 2010 01 3040 then apply the second part of the Fundamental Theorem to the resulting antiderivative; or • A variant of the Substitution Rule can be formulated specifically for definite integrals. Using the same functions as described earlier, it is Z b Z g(b) 0 f (g(x))g (x) dx = f (u) du . a g(a) That is, change the limits of the new integral to the values that u has when x = a and x = b. These two methods are equivalent — use whichever you prefer. Symmetry The textbook reviews the definitions of even and odd functions, and considers their definite integrals over a finite interval centred at the origin. The author shows that the definite integral of an even function f over the interval −a ≤ x ≤ +a will be twice the integral over the interval 0 ≤ x ≤ a; and the integral of an odd function over the same interval will be 0, because the integral to the left of 0 will cancel the integral to the right of 0. The proofs are simple applications of the Substitution Rule. Z π6 Example C.21 Evaluate the definite integral tan2 θ dθ. − π6 Solution: Normally we will evaluate an integral of this type by replacing the integrand, tan2 θ, by sec2 θ − 1. In this case we can also use the symmetry of the integrand and the interval of integration to further simplify the calculations: Z π 6 − π6 Z 2 tan θ dθ = π 6 − π6 Z = 2 sec2 θ − 1 dθ π 6 sec2 θ − 1 dθ 0 since the integrand is an even function (Prove this!) π = 2 [tan θ − θ]06 ! 1 π = 2 √ − . 3 6 5.5 Exercises Z [1, Exercise 28, p. 407] “Evaluate the indefinite integral tan−1 x dx.” 1 + x2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3041 Solution: In looking for a substitution, our general intention is to try to simplify the integrand. In the present integrand, we might be expected to consider the arctangent factor the most complicated, so I will try to simplify by the substitution u = tan−1 x. 1 dx, which is also present in the integral. The integral This implies that du = 1 + x2 becomes Z Z tan−1 x 1 2 (tan−1 x)2 dx = u du = u + C = +C. 1 + x2 2 2 We can verify our work by differentiating the function we claim to be an antiderivative: ! d 1 −1 2 1 1 tan−1 x tan x = · 2 tan−1 x · = . dx 2 2 1 + x2 1 + x2 Z [1, Exercise 24, p. 407] Evaluate the indefinite integral (1 + tan θ)5 sec2 θ dθ . Solution: In this case one should observe that there is a factor sec2 θ, which we recognize as the derivative of tan θ. A first simplification would be obtained by the substitution u = tan θ since then du = sec2 θ dθ, or dθ = cos2 θ du. The integral transforms to Z Z 5 2 (1 + tan θ) sec θ dθ = (1 + u)5 sec2 θ cos2 θ du Z = (1 + u)5 du . One can observe that the integral here will be a constant multiple of (1 + u)6 , and then replace u by tan θ. If you don’t make the observation, a second substitution would be in order. I would observe that, in the last mentioned integral, the “most complicated” factor is 1 + u; so a substitution v = 1 + u could be attempted. This yields Z ⇒ dv = 0 Z + du (1 + u)5 du = v5 dv 1 6 ·v +C 6 1 = · (1 + u)6 + C 6 1 = · (1 + tan θ)6 + C . 6 = Information for Students in Lecture Section 1 of MATH 141 2010 01 3042 Z sin x dx.” 1 + cos2 x Solution: First we see the cosine in the denominator; then we see in the numerator sin x dx,Z which is −d(cos x). This suggests a substitution u = cos x. The integral be−du comes du = − arctan u + C = − arctan(cos x) + C. 1 + u2 Z x [1, Exercise 42, p. 407] “Evaluate the indefinite integral dx.” 1 + x4 Solution: (I am going to begin the solution by making a poor choice for a substitution, a choice that I will then “fix” by following it with a second substitution. Then I will observe what would have been a better first choice.) Examining the integral we see that the powers of x present are x1 in the numerator, and x4 in the denominator. A first idea might be to try u = x4 . That would yield Z Z x 1 1 dx = du √ 1 + x4 4 u(1 + u) [1, Exercise 36, p. 407] “Evaluate the indefinite integral which looks more√complicated than before. However, we could then undertake a second substitution, v = u, which would correspond to an original substitution of v = x2 ; this would produce Z Z x 1 1 dx = dv 4 1+x 2 1 + v2 1 1 = arctan v + C = arctan x2 + C 2 2 An experienced student would have been able to see the substitution v = x2 immediately: a substitution u = x2 is indicated when the entire integrand can be expressed in simple terms of x2 , with the exception of one extra single power of x that is “left over”. Z π6 [1, Exercise 57, p. 407] Evaluate the definite integral tan3 θ dθ. − π6 Solution: Eventually you will be able to integrate the cube of the tangent; but, at this point, you may not be able to do that. However, you can observe that the integrand is an odd function of θ. (Why?) Hence the integral from − π6 to 0 will be equal in magnitude but opposite in sign to the integral from 0 to π6 ; so the given integral is 0. Z π6 [1, Exercise 57, p. 407] “Evaluate the definite integral tan3 θ dθ.” − π6 Solution: Eventually you will be able to integrate the cube of the tangent; but, at this point, you may not be able to do that. However, you can observe that the integrand is an Information for Students in Lecture Section 1 of MATH 141 2010 01 3043 odd function of θ. (Why?) Hence the integral from − π6 to 0 will be equal in magnitude but opposite in sign to the integral from 0 to π6 ; so the given integral is 0. Z 2 √ [1, Exercise 65, p. 407] “Evaluate the definite integral x x − 1 dx.” 1 Solution: This integral can be simplified by defining u = x − 1, so du = dx.38 Then Z √ Z √ x x − 1 dx = (u + 1) u du Z 3 1 = u 2 + u 2 dx 2 52 2 23 u + u +C 5 3 2 2 5 3 = (x − 1) 2 + (x − 1) 2 + C 5 3 = Taking the differences of the values of this antiderivative (i.e., with any specific value for C, e.g., with C = 0) at the limits yields " = = #2 2 2 5 3 (x − 1) 2 + (x − 1) 2 5 3 !1 ! 2 2 2 2 5 3 5 3 (2 − 1) 2 + (2 − 1) 2 − (1 − 1) 2 + (1 − 1) 2 5 3 5 3 ! 2 2 16 + − (0 + 0) = 5 3 15 Alternatively, the Definite Integral version of the Substitution Rule could have been used, to obtain #2−1 ! ! " 2 52 2 32 2 52 2 23 2 52 2 32 u + u = 1 + 1 − 0 + 0 etc. 5 3 1−1 5 3 5 3 The Logarithm Defined as an Integral (material in §5.6 of the 5th Edition) 39 While it is theoretical, the discussion below is an essential part of the course; its purpose is to substantiate some of the discussions in [1, §§1.5, 1.6] by replacing the weakest parts of the definitions given at that point in the textbook. This is one of several possible ways of treating exponentials and √ An even better substitution would be v = x − 1, which would imply that v2 = x − 1, so dx = 2v dv, and the R1 integral would be equal to (v2 + 1)v · 2v dv etc. 38 39 0 The textbook material has been moved in the 6th edition to [1, Appendix G, pp. A48-A55] Information for Students in Lecture Section 1 of MATH 141 2010 01 3044 logarithms “properly”; the treatment in [1, §§1.5, 1.6] was necessarily inadequate, as there was a perceived urgency to make logarithms and exponentials available to students who didn’t have the background for a more substantial treatment. The Natural Logarithm Previously we had “defined” the logarithm to be the inverse of the exponential function, whose definition was, at best, intuitive. Now we sketch very briefly a more rigorous definition of both functions, beginning with the logarithm. This theory has been delayed until now because parts of the proofs require the concept of substitution in a definite integral. Henceforth we take the following as our primary definition: Definition C.3 Let x be any positive real number. Define Z x 1 ln x = dt 1 t 1 i.e., it is the area under the right branch of the hyperbola y = , between t = 1 and t = x. t Which Definition is “Correct”? The definitions in [1, §§1.5, 1.6] were all we had available at that time to follow the objective of the textbook to introduce the exponential and logarithm functions “early”. We had to rely largely on intuition in deriving properties of these functions. Now that we have the integral available, we can return and replace those inadequate definitions by some that are more rigorous. Even the present definitions have some issues, since we haven’t rigorously proved all the properties that we are using for the definite integral. While most mathematicians today would probably follow the present development — of introducing the logarithm first, as a definite integral, and the exponential as its inverse — there are other possible ways of defining the functions. Yet another way will be available to you if you follow the theory of infinite series beyond what we will be doing in [1, Chapter 11]. Properties of the Logarithm diately obtain the fact that Theorem C.22 From the first part of the Fundamental Theorem we imme- d 1 ln x = . dx x We can also prove the following, using basic properties of the integral: Theorem C.23 Let x, y be positive real variables. Then 1. ln(xy) = ln x + ln y (36) Information for Students in Lecture Section 1 of MATH 141 2010 01 3045 2. ln 1 = 0 3. ln 4. 1 = − ln x x d 1 ln |x| = dx x 5. lim ln x = ∞ x→∞ 6. lim+ ln x = −∞ x→0 While I do not suggest you memorize the proofs, except possibly that of the first part below, they provide simple exercises on various aspects of the integral. Proof: 1. This is a simple application of properties of the definite integral, and in the Substitution Rule. Z xy 1 ln xy = dt t 1 Z x Z xy 1 1 = dt + dt t 1 t x by decomposing the interval [1, xy] into [1, x] and [x, xy] Z x Z y 1 1 = dt + x du 1 t 1 xu using the substitution t = xu Z x Z y 1 1 = dt + du = ln x + ln y 1 t 1 u Z1 2. ln 1 = 1 dt = 0 since the integration is over an interval of length 0. t 1 1 1 3. Using a substitution u = , where dt = − 2 du, we obtain t u Z 1x 1 1 ln = dt x 1 t ! Z x 1 = u − 2 du u 1 Z x 1 du = − ln x = − 1 u Information for Students in Lecture Section 1 of MATH 141 2010 01 4. Recall that ( d +1 if |x| = −1 if dx It is convenient to write these values in the form |x| d |x| = x dx undefined 3046 x>0 . x<0 if x,0 if x=0 . Hence, by the Chain Rule and the Fundamental Theorem, 1 |x| 1 d (ln |x|) = · = dx |x| x x when x , 0. 5. We will reconsider this proof when we study “harmonic series” in [1, §11.2, Example 7, Z2n 1 dt. We pp. 691-692]. Consider the Riemann sum obtained for the definite integral t 1 will hang the rectangles by their upper right-hand corners, so that the Riemann sum is clearly less than the integral. Then the area under the curve is greater than the sum 1 1 1 1 + + + ... + n . 2 3 4 2 Let’s collect these terms into groups ending with the reciprocal of each power of 2. We obtain 1 1 = 2 2 1 1 1 1 1 + > + = 3 4 4 4 2 1 1 1 1 1 1 1 1 1 + + + > + + + = 5 6 7 8 8 8 8 8 2 ... ... 1 1 1 1 1 + n−1 + . . . + n > 2n−1 · n = n−1 2 2 +1 2 2 2 n We see that the area of the rectangles under the curve is 2 . As we allow n → ∞ this lower bound for the area approaches ∞, implying that the full area surely → ∞. 1 6. Let’s make the substitution y = , so that, as x → 0+ , y → +∞. Then x 1 lim+ ln x = lim ln = − lim ln y = −∞ . y→+∞ y→+∞ x→0 y Information for Students in Lecture Section 1 of MATH 141 2010 01 3047 7. I presented the preceding proof because it is classical. The definition of the logarithm as a definition integral permits a much simpler proof of the same result. Since 1t ≤ 1 for Zx Zx 1 t ≥ 1, dt ≥ 1 dt = x − 1 → ∞ as x → ∞. t 1 1 The construction began with a formal definition for the (natural) logarithm. At this point we can define the (natural) exponential function as the inverse of the logarithm. We begin by calling the function exp(x); only after we prove that it behaves the way we expect a power to behave, do we revert to the familiar notation. The Natural Exponential Function We could now justify the various equations that we used intuitively in the “early transcendentals” treatment in [1, §§1.5, 1.6]. As mentioned, we begin by showing that ln, as defined above, is invertible. Then we temporarily call the inverse function, i.e., ln−1 x, exp(x). Thus we have, Theorem C.24 exp x = y exp(ln x) ln(exp x) exp(x + y) ⇔ = = = ln y = x x x (exp x) · (exp y) exp x exp(x − y) = exp y y (exp x) = exp(xy) (37) (38) (39) (40) (41) (42) following which, knowing that the exponent rules are satisfied, we define Definition C.4 e = exp 1 (43) and change our notation by recognizing that exp x = e x . (44) We also show that Theorem C.25 d (exp x) = exp x . dx Using standard properties of the definite integral we can also show that (45) Theorem C.26 lim exp x = 0 (46) lim exp x = ∞ (47) x→−∞ x→∞ Information for Students in Lecture Section 1 of MATH 141 2010 01 3048 General Exponential Functions The preceding constructions have been for the natural logarithm and the natural exponential. We may now extend these definitions to exponentials to any positive base a , 1. Definition C.5 For any positive real number a and any real number x we define a x = e x ln a . (48) This leads to the exponent rules for general (positive) bases: a x+y = a x · ay ax a x−y = y a (a x )y = a xy (ab) x = a x · b x and to the derivative property d x (a ) = a x · (ln a) dx (49) (50) (51) (52) (53) General Logarithmic Functions Finally the inverse function of a x is named loga , and its familiar properties are developed, e.g., Theorem C.27 loga x = y ⇔ ay = x d 1 loga x = dx x ln a The Number e Expressed as a Limit 1 Theorem C.28 Let a > 0. Then lim (1 + ax) x = ea x→0 Proof: lim (1 + ax) x→0 1 x 1 ln(1+ax) x = lim e x→0 since exponential and logarithm are inverses 1 = lim e x ·ln(1+ax) x→0 by the exponent rules ln(1+ax)−ln(1+0·x) ln(1+ax) x = lim e x = lim e x→0 x→0 (54) (55) Information for Students in Lecture Section 1 of MATH 141 2010 01 3049 ln(1 + ax) − ln(1 + 0 · x) x = e x→0 by the continuity of the exponential function d ln(1 + ax) x=0 = e dx lim by definition of the derivative at x = 0 a = e 1 + ax x=0 by the Chain Rule a = e C.7.2 5 Review True-False Quiz Students tend to avoid the True-False questions because these are unlikely to appear on quizzes or examinations. It’s true that I would not wish to have a simple TrueFalse question in any test, since there would be a 50% chance of a correct guess. However, the True-False questions in your textbook are much more difficult than that, as they ask “Determine whether the statement is true of false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.” This requirement of a proof or a counterexample makes these problems very challenging. The odd-numbered problems are solved in your Student Solutions Manual [3]. I consider some of the even-numbered problems below. [1, True-False Exercise 2, p. 409] If f and g are continuous on [a, b], then Z b Z b Z b [ f (x) · g(x)] dx = f (x) dx · g(x) dx a a a Solution: This statement is FALSE. That is, it is not true for all constants a, b and for all functions f, g continuous on [a, b]. There will, of course, be functions and intervals for which the statement is true, but the generalized statement, quantified for all functions and intervals is not true. While you would have know to suspect this after we study [1, §7.1], but you should be suspecting it already, and be able to construct a simple counterexample. Here is one: Define f (x) = g(x) = x. Then Z Z b b f (x) dx = a Z a x2 g(x) dx = 2 Z b b [ f (x) · g(x)] dx = a a #b = a x3 x dx = 3 b2 − a2 . 2 #b 2 = a b3 − a3 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3050 b2 − a2 b2 − a2 b3 − a3 While the polynomials · and look different, that does not consti2 2 3 tute a counterexample, as it could happen that we have two algebraic functions that are equal by virtue of some identity that we don’t happen to recognize at the moment. So, in order to complete the counterexample, we need to find specific values of a and b that will entail that Z b Z b Z b f (x) dx · g(x) dx , [ f (x) · g(x)] dx , a i.e., that a a b2 − a2 b2 − a2 b3 − a3 · , . 2 2 3 4 One such example is a = 0, and b having any value except 0, : 3 b2 b2 b4 b3 · = , ; 2 2 4 3 for example, take b = 1. But the counterexample given is far from the simplest! For example, take f (x) = g(x) = 1 for all x. Then the left side of the claimed equation is equal to b − a, while the right side is (b − a)2 , which will be different from b − a except where b − a = 1 or b = a. [1, True-False Exercise 4, p. 409] This is very similar to [1, True-False Exercise 2, p. 409]; you should have no trouble constructing a counterexample. [1, True-False Exercise 8, p. 409] If f and g are differentiable, and f (x) ≥ g(x) for a < x < b, then f 0 (x) ≥ g0 (x) for a < x < b. Solution: This statement is false. Think geometrically. The condition that f (x) ≥ g(x) tells us that the graph of f is above the graph of g; the condition that f 0 (x) ≥ g0 (x) tells us that the graph of f is steeper than the graph of g. So we can construct a counterexample by finding, for a convenient function g, a function that is larger, but whose graph is not so steep. For example, take g(x) = 0 for all x; now all we need is a positive function f whose graph is sloping downward; for example, f (x) = e−x will work, with any interval a ≤ x ≤ b. A simpler counterexample is f (x) = a + b − x, g(x) = 0. [1, True-False Exercise 14, p. 409] All continuous functions have antiderivatives. Solution: This problem is interesting, as it follows [1, True-False Exercise 13, p. 409], which states that “All continuous functions have derivatives.” That preceding statement is false, and you should be able to give counterexamples (e.g., |x| is continuous, but lacks Information for Students in Lecture Section 1 of MATH 141 2010 01 3051 a derivative at x = 0). The present statement is TRUE, by virtue of the first part of the Fundamental Theorem. If f is continuous on an interval [a, b], then Z x d f (x) = f (t) dt . dx a Exercises !9 !9 !9 n 9 1 1 2 3 [1, Exercise 70, p. 411] Evaluate lim + + + ... + n→∞ n n n n n Solution: The form of this limit should suggest that it is a Riemann sum. There will be more than one way to interpret this limit as such, but I will choose the interpretation that is, in some sense, obvious. First, we see a factor 1n that is tempting to interpret as a common subinterval length ∆x; so let’s do that: consider the number of subintervals to be n, and the length of the interval to be 1 — it’s convenient to take a = 0 and b = 1, i 9 but these are not the only possible choices. Then we can interpret as f (xi∗ ); the n most convenient choice is to think of f (x) = x9 , and xi∗ chosen as the right end-point of the subinterval [xi−1 , xi ]. Thus we see that the limit can be interpreted as the limit of an upper Riemann sum associated with the integral Z 1 x9 dx 0 x10 and its value is the value of the integral, i.e., 10 #1 = 0 1 . 10 Information for Students in Lecture Section 1 of MATH 141 2010 01 3052 C.8 Supplementary Notes for the Lecture of January 18th, 2010 Release Date: Monday, January 18th, 2010 (subject to revision) Textbook Chapter 6. APPLICATIONS OF INTEGRATION. C.8.1 §6.1 Areas between Curves If, for a ≤ x ≤ b, continuous functions f and g have the property that f (x) ≥ g(x) — i.e., if the graph of f is not lower than the graph of g for that interval, then the area bounded by the graphs and the vertical lines x = a and x = b is given by the integral Z b ( f (x) − g(x)) dx . a Before working some examples, I make several observations: • When g is the 0 function — i.e., when the lower R b boundary of the region is the x-axis — this formula reduces to the definite integral a f (x) dx. • We can find the areas of certain regions by decomposing them into parts that can be described as above. Sometimes there is more than one “natural” way to decompose the region, but all decompositions should yield the same area. • An analogous formula can be applied if we consider the “region” bounded by graphs of the form x = f (y), x = g(y), for a ≤ y ≤ b. In that case we say that we are integrating with respect to y or integrating along the y-axis. • Often we wish to find the area between two graphs determined by points where they intersect. In these cases the vertical sides of the region have zero length. • In solving problems it is useful to make a sketch showing an “element of area” — a thin rectangle whose width is shown as dx or ∆x in the case of integration with respect to x, and analogously for the case of integration with respect to y. It is difficult for me to include such sketches in these notes, but they will be shown on the chalkboard. The sketch is not a formal part of the solution, but you are likely to find it helps you formulate your solution. • Where two graphs cross in several places, so that the area between them is in several pieces, be sure you understand what you intend if you write an integral that extends over several pieces: if the curves interchange positions, with the upper one becoming the lower, then the signs of the areas will change, and some of the areas will cancel. If this is not what you intend, you must either write the integrand with absolute signs, as | f (x)−g(x)|, or, equivalently, find the area of each of the pieces, and add their magnitudes so as to prevent cancellation. Information for Students in Lecture Section 1 of MATH 141 2010 01 3053 Example C.29 ([7, Exercise 8, p. 442]; see Figure 1 on page 3053) To find the area of the 1.0 0.75 0.5 0.25 0.0 −1.0 −0.5 0.0 0.5 1.0 x Figure 1: The region(s) bounded by y = x2 and y = x4 region bounded by the curves y = x2 and y = x4 . Solution: Note that the text-book used the word “region” in a general way. Not all authors use the word in this way. For this problem there are two connected regions bounded by the curves, and the intention of the textbook was that you find the areas of both of them together. To determine where the curves intersect, we solve their equations, obtaining (x, y) = (0, 0) and (x, y) = (±1, 1). Note that it is not enough to guess the coordinates of these points of intersection from the graph: you must determine the coordinates by rigorous solution of the equations! Information for Students in Lecture Section 1 of MATH 141 2010 01 3054 First solution: Integration with respect to x. 2 Z 1 2 x −x 0 4 " x3 x5 dx = 2 − 3 5 #1 = 0 2 2 4 − = . 3 5 15 Second solution: Integration with respect to y. I find the area of the region in the first quad√ √ rant. The equations of the curves are, respectively, x = y and x = 4 y. The area is Z 1 0 1 4 y −y 1 2 " 4 54 2 32 dy = ·y − ·y 5 3 #1 = 0 4 2 2 − = . 5 3 15 Now double this area. Note that the order of the curves in this integral is the reverse of that used when we integrated with respect to x, because we now order them by their relative distance from the y-axis. While the limits of the two integrals look as though they are the same, the limits in the first case refer to the extreme values of x, while the second refer to the extreme values of y. Example C.30 ([7, Exercise 22, p. 442]) (see Figure 2 on page 3055) To find the area of the region bounded by the curves y = sin x and y = sin 2x between the vertical lines x = 0 and x = π2 . Solution: To determine the points of intersection of the curves, we solve y = sin x with y = sin 2x, and solve sin x − sin 2x = 0. Since sin 2x = 2(sin x) · (cos x), the intersections must satisfy sin x · (1 − 2 cos x) = 0: either 2 cos x = 1 or we must be unable to divide by sin x, because sin x = 0 . In the interval 0 ≤ x ≤ π2 , we must therefore have either x = π3 (to make cos x = 12 ) or x = 0 √ π (to make sin x = 0). This yields the points of intersection (0, 0), and , 3 . The 3 2 regions to be considered therefore consists of a lune-shaped region with one end at the origin √ √ and the other at π3 , 23 ; and a second region which begins at π3 , 23 and ends at the vertical line x = π2 . We will find the areas of each of the regions and add them, since that appears to be what the textbook expects. (The wording of the question is not totally unambiguous; some readers might be justified in assuming that the author intended signed areas to be used, rather than the absolute values, as I am taking.) Information for Students in Lecture Section 1 of MATH 141 2010 01 3055 1.0 0.75 0.5 0.25 0.0 0.0 0.25 0.5 0.75 1.0 1.25 1.5 Figure 2: The region(s) bounded by y = sin x, y = sin 2x between x = 0 and x = For the left-most region the curve y = sin 2x is above y = sin x, so the area will be Z π 3 0 " # π3 1 (sin 2x − sin x) dx = − cos 2x + cos x 2 " # ! # "0 1 1 1 1 1 = − · − + − − +1 = 2 2 2 2 4 For the right region the orders of the curves are reversed, and the area is Z π 2 π 3 " (sin x − sin 2x) dx = 1 − cos x + cos 2x 2 # π2 π 3 π 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 " = Thus the total area is 1 4 + 1 4 3056 # " !# 1 1 1 1 1 −0 + · (−1) − − + − = 2 2 2 2 4 = 21 . 6.1 Exercises [1, Exercise 18, p. 420] (see Figure 3 on page 3056) To find the area of the region bounded 10.0 7.5 5.0 2.5 0.0 −3 −2 −1 0 1 2 3 −2.5 Figure 3: The region(s) bounded by y = 8 − x2 , y = x2 between x = ±3 by the curves y = 8 − x2 and y = x2 between the vertical lines x = −3 and x = 3. Solution: Let’s first determine where the curves intersect. Solving the equations shows that the intersections occur when x = ±2, y = 4, i.e., in the points (±2, 4). But we are asked to find the area from x = −3 to x = 3. Thus this region has three parts: Information for Students in Lecture Section 1 of MATH 141 2010 01 3057 • −3 ≤ x ≤ −2, where 8 − x2 ≤ x2 ; • −2 ≤ x ≤ 2, where 8 − x2 ≥ x2 ; • 2 ≤ x ≤ 3, where 8 − x2 ≤ x2 . A naive attack gives the following: Z 3 Area = |(8 − x2 ) − x2 | dx −3 Z −2 Z 2 Z 3 2 2 2 2 = |(8 − x ) − x | dx + |(8 − x ) − x | dx + |(8 − x2 ) − x2 | dx −3 −2 2 Z −2 Z 2 = − (8 − x2 ) − x2 dx + (8 − x2 ) − x2 dx −3 −2 Z 3 − (8 − x2 ) − x2 dx 2 Z −2 Z 2 Z 3 2 2 = − (8 − 2x ) dx + (8 − 2x ) dx − (8 − 2x2 ) dx −3 −2 2 " #−2 " #2 #3 " 2 3 2 3 2 3 = − 8x − x + 8x − x + 8x − x 3 −3 3 −2 3 2 " ! # " ! !# 16 16 16 = − −16 + − (−24 + 18) + 16 − − −16 + 3 3 3 " !# 16 − (24 − 18) − 16 − 3 92 = 3 This attack is naive in that we have not taken advantage of symmetry. Since the integrand is an even function, and the interval of integration is symmetric about x = 0, we can find the integral from 0 to 3 and double it: that means evaluating 2 integrals instead of 3. The areas could also be found by integrating with respect to y, where we would have to sum three pieces (or two if we make use of symmetry). For example, the middle region could be described as being double p the following region: bounded by the y-axis, the √ curve x = y, and the curve x = 8 − y. This approach is cumbersome, as the region √ has to be broken up further into two pieces: one bounded by the y-axis, x = y, and p y = 4, and the other bounded by they-axis, y = 4, and x = y − 8. [1, Exercise 32, p. 420] “Evaluate the integral and interpret it as the area of a region: Z 4 √ x + 2 − x dx .” 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3058 Solution: (see Figure 2 on page 3058) We can interpret this as the area of the region 4 3 2 1 0 −2 −1 0 1 Figure 4: The region(s) bounded by y = 2 √ 3 4 x + 2, y = x between x = 0 and x = 4 √ bounded by the curve y = x + 2, the line y = x, and the vertical lines x = 0 and x = 4. If we square the former equation, we obtain y2 = x+2, which can be seen to be a parabola that is symmetric about the x-axis, with its vertex on the line x = −2 and opening to the right. Squaring the equation caused the new equation to include the lower branch of this parabola; the original equation represents only the upper branch. The upper branch of the parabola and the line y = x meet both in the origin and in the point (2, 2). (The other point of intersection of the parabola and the line does not lie on the upper branch of the parabola, and so is extraneous to this discussion.) Information for Students in Lecture Section 1 of MATH 141 2010 01 3059 The simplest way of evaluating this integral is to break it up into two parts: Z 4 √ Z 2√ Z 4 √ x + 2 − x dx + x − x + 2 dx x + 2 − x dx = 0 0 2 2 4 3 3 (x + 2) 2 x2 (x + 2) 2 x2 = − + − + 3 3 2 2 2 2 # "0 #2 " # " # " 2 32 2 32 2 32 2 32 = ·4 −2 − ·2 −0 + − ·6 +8 − − ·4 +2 3 3 3 3 16 2 3 2 3 16 = − 2 − · 2 2 − · 6 2 + 18 + −2 3 3 3 3 √ 44 4 √ 2−4 6 = − 3 3 [1, Exercise 50, p. 421] 1. “Find the number a such that the line x = a bisects the area 1 under the curve y = 2 , 1 ≤ x ≤ 4 x 2. “Find the number b such that the line y = b bisects the area in part (a).” Solution: 1. We solve for a the equation Z a 1 1 dx = x2 Z 4 a 1 dx x2 to obtain a = 85 . It follows that the area of half of the region is the common value h ia of the two integrals, 1x = 83 . 1 2. The right side of the region has an irregular boundary: the upper part has equation y = x12 , i.e., x = √1y ; the lower part is on the line x = 4. The two parts of the 1 boundary meet in the point 4, 16 . Thus the area of the rectangle bounded by the 1 3 lines x = 1, x = 4, y = 0, y = 16 is 16 . As this is less than half the total, we know 1 that b > 16 . This fact is important, as it tells us that we can represent the upper half of the area — the portion above the line y = b, by the integral ! Z 1 1 √ − 1 dy , y b where the −1 in the integrand represents the lower boundary of the region — now viewed as being “under” x = √1y and “over” x = 1. Setting this integral equal to 38 Information for Students in Lecture Section 1 of MATH 141 2010 01 3060 h √ i1 and solving, we obtain 2 y − y = 38 , which evaluates to b √ 3 1−2 b+b= 8 √ 3 ⇔ ( b − 1)2 = r8 √ 3 ⇔ b=1± 8 r 11 3 ⇔ b= ± 8 2 One of the values we obtain is greater than 1, which contradicts our assumption that 0 ≤ b ≤ 1. The other gives the bisecting line r 3 11 y= − ≈ 0.150255129. 8 2 The value we rejected has a geometric significance: it is the height of a horizontal line at which the area bounded by the curve y = x12 , the line x = 1 and that line is equal to 38 . 1 And what would have happened if we failed to observe that b ≥ 16 ? We would have obtained an equation for b that could not be solved; it would be equivalent to 5 1 requiring that 85 − 3b = 16 when 0 ≤ b ≤ 16 , which are contradictory statements. √ The area which I found to equal 1 − 2 b + b by integrating with respect to y could 1 also be evaluated by integrating with respect to x: here the line y = b meets y = x2 in the point √1b , b , so the area is Z √1 b 1 ! √ √ 1 − b dx = b − 2 b + 1 = ( b − 1)2 , 2 x which we must again equate to 3 8 and solve. Information for Students in Lecture Section 1 of MATH 141 2010 01 3061 C.9 Supplementary Notes for the Lecture of January 20th, 2010 Release Date: Wednesday, January 20th, 2010 (subject to revision) C.9.1 §6.2 Volumes Just as we defined the area of a region as the limit of a sum of narrow rectangles, we can define volumes as limits of sums of thin elements; these elements can be assembled in various ways. In this course you will see volumes expressed either as sums of thin slices with planar sides (called laminæ), and — for solids with rotational symmetry (called solids of revolution) — as sums of thin shells with cylindrical sides. Some problems will lend themselves only to one type of dissection; where a problem can be approached in more than one way, it is instructive to try it in several ways, in order to verify your answer and also to gain experience in choosing the method that is more efficient for different types of problems. In this section we will be considering dissections into thin slices; where the slice has the shape of a disk with a concentric disk cut from its centre, the author uses the term washer40 . Example C.31 1. (see Figure 1 on page 3053) ([7, Exercise 2, p. 452]) “Find the volume 7 2.5 6 2 5 1.5 4 3 1 2 0.5 1 0 0 0.20.40.60.8 1 1. 0 0 0.5 1 1.5 2 4. Figure 5: Regions for Example C.31 of the solid obtained by rotating the region bounded by the curves y = e x , y = 0, x = 0, x = 1 about the x-axis.” We will first solve the problem as stated, and then consider some related problems obtained by changing some of the data. 40 Students whose native language is not English sometimes cannot understand why the author will use the name of a household appliance here; this is another meaning of the English word washer, which refers to a thin disk with a hole in the middle Information for Students in Lecture Section 1 of MATH 141 2010 01 3062 Solution: (Remember to come back to this problem when we study [1, §6.3], to solve it using the methods of that section.) We will decompose the solid into laminæ that are thin disks. The “washers” will be obtained by rotating about the x-axis the element that we would have used for the area if we had found the area by integrating with respect to x. For the disk whose faces are centred at points (x, 0) and (x + ∆x, 0), the volume obtained by hanging the element from its upper left corner on the curve is πy2 ∆x = πe2x ∆x. By the same reasoning that led us to express areas as definite integrals, we have Z 1 Volume = πe2x dx 0 π 2x 1 π = ·e = · (e2 − 1) 2 2 0 2. Now let us change the problem, asking that the solid be rotated about the y-axis. Here the description of the cross sections will depend on the height of the cross section. We will be expressing everything in terms of y, not x, and integrating “with respect to y”. For y ≤ 1 (the height where the curve cuts the y-axis), the cross sections are disks of radius 1, and the volume of that part of the solid is Z 1 π12 dy = πy]10 = π. 0 For y ≥ 1 the cross sections are “washers”, with outer radius 1 and inner radius x; to express this in terms of y we need to rewrite the equation of the curve y = e x in an equivalent way that expresses x in terms of y, as x = ln y. The washers at height y have volume π(12 − x2 )∆y = π(1 − (ln y)2 )∆y so this part of the solid of revolution has volume Z e π (1 − (ln y)2 ) dy . 1 where the upper limit of integration is the ordinate of the point where the line x = 1 meets the curve y = e x . You are not quite ready to complete this integration. You can R1 make a substitution like x = ln y, under which the integral transforms to π 0 (1−x2 )e x dx, and you know how to integrate part of this integral, as Z 1 π 1e x dx = πe x ]10 = π(e − 1) , 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3063 R1 but you are not ready to integrate π x2 e x dx. We will, in [1, §7.1], develop a method 0 R to determine that x2 e x dx = (x2 − 2x + 2)e x + C (which is obvious by differentiation). With that we know that Volume = π[e x − (x2 − 2x + 2)e x ]10 = π[(−x2 + 2x − 1)e x ]10 = π . 3. As another variant on this problem, consider the following: “Find the volume of the solid obtained by rotating the region bounded by the curves y = e x , y = 0, x = 0, x = 1 about the line y = −3.” The analysis is similar to what we did originally, except that the laminæ now have a hole of radius 3 in the middle, and the outer radius is 3 units larger. This leads to an integral Z 1 Z 1 2 x 2 Volume = π (e + 3) − 3 dx = π e2x + 6e x dx 0 0 " #1 1 = π e2x + 6e x 2 !0 !# " 1 π 2 1 2 e + 6e − +6 = e + 12e − 13 = π 2 2 2 4. Finally, suppose that the right boundary of the region generating the solid by revolution changed from x = 1 to y = e2 (x − 1). This new right boundary also passes through (1, 0), but meets the curve in the point (x, y) = 2, e2 . The volume will be Z 1 Z 2 2x x 2 2 2 π (e ) − e (x − 1) dx Z 2 Z 2 π 2 2x 4 e −1 +π e dx − πe (x − 1)2 dx = 2 1 1 " 2x #2 " #2 e π 2 4 1 3 = e −1 +π − πe (x − 1) 2 2 1 3 1 ! ! e4 − 1 e4 e4 1 = π −π =π − 2 3 6 2 πe dx + 0 1 Example C.32 ([7, Exercise 8, p. 452]) Find the volume of the solid obtained by rotating the region bounded by the curves y = sec x, y = 1, x = −1, x = 1 about the x-axis. Solution: The lines x = ±1 intersect y = sec x in the respective points (1, ± sec 1). The area of the washer centred on the x-axis between cross sections x = X and x = X+4X is approximately π(sec2 X − 12 ) · 4X. The volume of revolution will be Z 1 π sec2 x − 1 · dx = π[tan x − x]1−1 = 2π(tan 1 − 1). −1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3064 (Had the integral involved arctan 1, you would have been expected to simplify it further; but you cannot evaluate tan 1 without calculators or techniques that you will not meet until Calculus 3.) Example C.33 ([7, Exercise 56, p. 454]) Here is a problem where the solid is not generated by revolving a plane region about an axis. “Find the volume of the solid S: the base of S is the parabolic region {(x, y)|x2 ≤ y ≤ 1} ; cross-sections perpendicular to the y-axis are equilateral triangles.” √ Solution: The cross-section of the base at level y has ends with coordinates (± y, y), so the √ √ √ √ √ length of the base is 2 y, and the area of the triangular cross-section is 12 ·2 y· 3 y = 3·y. Integrating along the y axis we find that √ 1 Z 1√ 3 2 Area = 3y dy = y 2 0 0 √ √ 3 2 3 = (1 − 02 ) = 2 2 Example C.34 ([7, Exercise 15, p. 458]) (Where this problem appeared in the cited textbook, students were asked to solve it using the method of “shells”. Let’s see if we can solve it using the method of “washers”.) To find the volume of the solid of revolution generated by revolving the region bounded by y = x2 , y = 0, x = 1, x = 2 about the axis x = 1. Solution: The washers generated by elements of area parallel to the x-axis will have two kinds of descriptions, depending on whether their cross sections are above or below the point where x = 1 meets y = x2 , i.e. (1, 1): below y = 1 the cross sections are rectangles of width 2 − 1 = 1; above y = 1 the cross sections are rectangles obtained from such a rectangle as below y = 1 by cutting away a rectangle of length x2 starting at the left end. But, in evaluating the limit of the sum of the volumes of these washers, we shall be integrating with respect to y, so we need to express the dimensions and position of the rectangular element in terms of y. The √ equation of the right branch of the parabola y = x2 is x = y. The hole in the washer has √ radius x − 1 = y − 1; the washer will have area √ √ π(12 − ( y − 1)2 ) = π(2 y − y) . The line x = 2 meets the parabola in the point (2, 4), so we shall integrate for y ranging from 0 to 4: from 0 to 1 using the constant integrand π12 , and from 1 to 4 using the integrand √ π(2 y − y). The volume of revolution is " #4 Z 1 Z 4 1 4 32 y2 √ 2 π1 dy + π(2 y − y) dy = πy 0 + π y − 3 2 1 0 1 # " " # 4 1 4 17π = π+π ·8−8 −π ·1− = 3 3 2 6 Information for Students in Lecture Section 1 of MATH 141 2010 01 3065 which is the same result given in the textbook for the solution that could be obtained in the next section using the method of cylindrical shells. 6.2 Exercises [1, Exercise 12, p. 430] Find the volume of the solid obtained by rotating the region bounded by the curves y = e−x , y = 1, x = 2 about the line y = 2. Solution: Through the point (x, 2) on the line y = 2 the cross section will be an annulus (ring) with outer dimension 2−e−x and inner dimension 2−1 = 1; the annulus is bounded by concentric circles centred at the point (x, 2). The volume will be Z 2 −2 2 2 π 2−e − π1 dx 0 Z 2 = π 3 − 4e−x + e−2x dx . 0 The first summand of the integrand has antiderivative 3x; the second summand can be integrated by using a substitution u = −x, which leads to an antiderivative +4e−x ; the last summand can be integrated by using a substitution u = −2x, which leads to an antiderivative − 21 · e−2x . Putting these three components together we find the value of the integral to be " #2 ! ! 1 −2x 4 1 1 −x π 3x + 4e − e = π 6+ 2 − 4 −π 0+4− 2 e 2e 2 0 ! 5 4 1 = π + 2− 4 . 2 e 2e [1, Exercise 39, p. 431] The textbook asks you to use a Computer Algebra System to find the volume of the solid generated by revolving about the line y = −1 the region bounded by y = sin2 x, y = 0, for 0 ≤ x ≤ π. No Computer Algebra System is needed, although we haven’t yet seen how to integrate this. Here is the full solution. The cross sections are washers with centre on the line y = −1, outer radius ending on the curve y = sin2 x, and inner radius ending on the line y = 0 (the x-axis). The volume is thus Z π 0 π 2 2 2 (sin x − (−1)) − 1 !2 1 − cos 2x dx = π + 1 − 1 dx 2 0 Z π π = (3 − cos 2x)2 − 4 dx 4 0 Z π π = 5 − 6 cos 2x + cos2 2x dx 4 0 Z π Information for Students in Lecture Section 1 of MATH 141 2010 01 = = = 3066 ! Z π π 1 + cos 4x 5 − 6 cos 2x + dx 4 0 2 Z π π (11 − 12 cos 2x + cos 4x) dx 8 0 " #π π π sin 4x 11π2 = · 11π = 11x − 6 sin 2x + 8 4 0 8 8 [1, Exercise 44, p. 431] Describe the solid whose volume is represented by the integral Z π 2 π [(1 + cos x)2 − 1] dx . 0 Solution: The dx tells us that the integration is along the x-axis. That is, the planes of the washers are perpendicular to the x-axis. The integrand is the difference of 2 squares, multiplied by π. We may interpret this as the area of a washer whose outer radius is 1 + cos x, and whose inner radius is 1. If we interpret 1 + cos x as cos x − (−1), we can interpret this as the radius of a disk whose centre is at the point (x, y) = (x, −1), generated by a radius extending from that centre to the point (x, cos x) on the graph of the cosine function. The subtracted term −π12 can be interpreted as the area of a disk whose centre is at the same point, but its radius extends from that point (x, −1) to the point (x, 0) above it on the x-axis. Thus the integral represents the solid of revolution about the line y = −1 of the region bounded by the graph of the cosine function, and the x-axis, between x = 0 and x = π2 . (This integral is not difficult to evaluate. We will see in [1, Chapter 7] that, if we replace cos2 x by 12 (1 + cos 2x), the value of the integral is " #π x sin 2x 2 π2 π 2 sin x + + = 2π + . 2 4 0 4 This is not the only way of interpreting the integral. We could also reason that it represents the volume of a region rotated about the x-axis, bounded by the y-axis, the line y = 1, and the graph of y = cos x + 1.) [1, Exercise 49, p. 431] Find the volume of a right circular cone with height h and base radius r. Solution: This is a standard problem that every student should be able to work. A cone is a surface generated by joining to all points curve in a plane a fixed point (called the apex outside of the plane. A cone is circular if the base curve is a circle. It is right circular if the apex is located on the normal to the plane of the curve through the centre of that base circle. Information for Students in Lecture Section 1 of MATH 141 2010 01 3067 It is convenient to set up coordinate axes so that the cone is generated by a right angled x y triangle with height h and base r, i.e., with hypotenuse along the line + = 1, which r h triangle is to be rotated about the y-axis. The horizontal rectangular elements of area, with height dy and length x are rotated about the y-axis to generatedisk-shaped laminæ. y Expressing x as a function of y for the hypotenuse, we have x = r 1 − , so the crossh y 2 2 sectional area at height y is πr 1 − . Hence the volume must be h Z h y 2 πr2 1 − dy . h 0 We know how to evaluate an integral of this type by expanding the square. But it is y 1 easier to make a change of variable: u = 1 − , so du = − dy, dy = −h du and h h " 3 #0 Z 0 u 1 2 2 2 Volume = − πr u h du = πr h = (πr2 h) . 3 1 3 1 (cf. [9, p. 47]). If you remember this as 13 π × the area of the base, you will know this as a special case of a general theorem. In fact, it is not hard to show that the area is not affected by the fact that the cone is a right cone: even if the apex were moved to a location not over the centre of the circular base, the area would not change. Moreover, it can be shown that even the fact that the base is circular is not relevant! The volume of any “cone” is as shown: 1 × π × Area of base × height . 3 For example, the case of a square base is discussed in [1, Example 8, p. 429]. [1, Exercise 55, p. 432] Find the volume of the solid S which is a tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm. Solution: I find it convenient to locate the three perpendicular sides along the coordinate axes, with one vertex at the origin. So I locate the vertices of the tetrahedron at (0, 0, 0), (3, 0, 0), (0, 4, 0), (0, 0, 5). This tetrahedron is just a pyramid or cone on a triangular base; thus we know by the theory of [1, Example 8, p. 429] that the volume will be 13 × 12 × 3 × 4 × 5 = 10. But I will pretend we don’t know that. Consider cross sections perpendicular to the z-axis. These are triangles whose x-dimension at height z will be 35 (5 − z) (by similar triangles in the xz-plane); and whose y=dimension Information for Students in Lecture Section 1 of MATH 141 2010 01 3068 at height z will be 45 (5 − z), (again by similar triangles). The area of the triangle will be 1 (5 − z)2 , so the volume will be 2 Z 5 0 " #5 1 12 6 1 2 2 3 · · (5 − z) dz = − (5 − z) = − 0 − 53 = 10 . 2 25 25 3 25 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3069 C.10 Supplementary Notes for the Lecture of January 22nd, 2010 Release Date: Friday, January 23nd, 2010, subject to revision C.10.1 §6.3 Volumes by Cylindrical Shells The integrand when using cylindrical shells The volume of a right circular cylinder (i.e., with a disk as base, axis perpendicular to the base) is — as you can easily prove using washers or otherwise — the product of the area its base and its height. If we consider a hollowed out cylinder of radius r2 , in which an inner cylinder of radius r1 is removed, then the volume will be, if the height is h, πr22 h − πr12 h = π r22 − r12 h . Let’s assume that r2 = r1 + 4r, and expand this product: π r22 − r12 h = π (r1 + 4r)2 − r12 h = π r12 + 24r · r1 + (4r)2 − r12 h = 2πh4r · r1 + πh(4r)2 Both of these products approach zero as we allow 4r → 0. If we wish to determine the volume of a solid of revolution by decomposing it into cylindrical shells about the axis of revolution, it would appear that we should add elements of volume 2πh4r · r1 + πh(4r)2 and then permit the number of shells to approach infinity, and the width ∆r → 0. It can be shown that, in any such limiting process, the sum of the terms of type πh(4r)2 approaches 0; that is, not only do the individual terms πh(∆r)2 approach 0, as we permit 4r → 0, but even the sum of these terms, increasing arbitrarily in number as 4r → 0, also approaches 0. Thus the R volume sought can be viewed as a Riemann sum, leading to a definite integral of the form 2πr1 h dr. Thus, to find the volume of a solid that we can decompose into elements which are cylindrical shells, we need only consider an integral related to terms of the first type. The integrand can be interpreted as the product of 2πr1 and h, i.e., as the area of a rectangle obtained by “cutting open” the inner surface of the cylindrical shell and “unrolling” it; then the 4r can be interpreted as the thickness of a thin rectangular lamina based on that rectangle. Of course, when you attempt to do that, you find that there will be a small error in that product, but, as I have mentioned, it can be shown that the totality of these errors approaches 0 as we replace the sum of volumes of elements by the definite integral. All that remains to be done is to express h in terms of the radius, and to determine the appropriate limits for integration. If you choose to approach these problems by substituting in formulæ, you are urged to remember how to generalize to situations where the axis of circular symmetry is parallel to Information for Students in Lecture Section 1 of MATH 141 2010 01 3070 but different from one of the coordinate axes. I do not recommend memorizing these formulæ. Among the following examples are some that were not discussed in the lecture. One of these, which you should certainly read shows how to find the volume of a sphere; instead of this example, I worked a more difficult — but much more instructive — example which finds the volume of a torus. 6.3 Exercises [1, Exercise 14, p. 436] (see Figure 6 on page 3070) “Use the method of cylindrical shells to 4 3 2 1 0 -2 -1 0 1 2 3 4 -1 Figure 6: The region(s) bounded by x + y = 3 and x = 4 − (y − 1)2 find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. Sketch the region and a typical shell: x + y = 3, x = 4 − (y − 1)2 .” Information for Students in Lecture Section 1 of MATH 141 2010 01 3071 Solution: The parabola meets the line in the points (0, 3) and (3, 0), on the coordinate axes. 1. First we follow the instructions, using the method of cylindrical shells. The elements of area that generate the shells will be narrow horizontal rectangles; at height y the rectangle has width (4 − (y − 1)2 ) − (3 − y), obtained by expressing the equations in the form x =function of y. The circumference of the base of the generated cylinder is then 2πy, and the volume is Z 3 2π (4 − (y − 1)2 ) − (3 − y) y dy 0 Z 3 = 2π 3y2 − y3 dy 0 " #3 27π y4 3 = . = 2π y − 4 0 2 2. Now let us compute the volume using washers. The equation of the line may be rewritten as y = 3 − x; but the √ parabola now splits into 2 curves — the√ upper branch has equation y = 1 + 4 − x, and the lower has equation y = 1 − 4 − x. The description of the element of area that generates the washer will change at x = 3.√ To the left of x = 3 the element of area at horizontal position x has height (1 + 4 − x) − (3 − x). We need to compute the area of the annulus the element generates. For that purpose the length of the element is not enough, as we need to know the distance√from the axis about which it is revolving. The outer radius of the washer is 1 + 4 − x, and the inner radius is 3 − x, so the area of the annulus is the difference between the areas of two disks: √ π(1 + 4 − x)2 − π(3 − x)2 and the volume of the solid generated up to x = 3 is Z 3 √ (1 + 4 − x)2 − π(3 − x)2 dx . π 0 √ The elements of area to the right of x = 3 have inner radius 1 − 4 − x, and outer √ radius 1 + 4 − x, so the area of the annular cross-section is 2 2 √ √ √ π 1 − 4 − x − π 1 + 4 − x = 4π 4 − x , R4 √ and the volume is 4π 3 4 − x dx. The total volume by the method of washers is Z 3 Z 4√ √ 2 2 π 4 − x dx (1 + 4 − x) − π(3 − x) dx + 4π 0 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3072 Z 3 Z 4√ √ = π −(x − 1)(x − 4) + 2 4 − x dx + 4π 4 − x dx 0 3 Z 1 Z 0 √ √ 2 = π (u − 3u + 2 u) (−1) du + π 4 u(−1) du 4 1 using the substitution u = 4 − x, du = −x, x = 4 − u 27π = as before. 2 [1, Exercise 20, p. 437] “Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. Sketch the region and a typical shell: y = x2 , x = y2 , about y = −1.” Solution: 1. At height y the cylinder is generated by an element of area whose horizontal di√ mension is y − y2 , and whose vertical dimension is ∆y; the circumference of the circle generate by a point at one end of this element under revolution is 2π(y−(−1)) (since the radius is the distance between a point (x, y) and the point (x, −1) below it), so the volume is Z 1 √ y − y2 2π(y + 1) dy 0 Z 1 3 √ y 2 − y3 + y − y2 dy = 2π 0 " #1 2 52 1 4 2 23 1 3 = 2π y − y + y − y 5 4 3 3 0 29 = π. 30 2. It wasn’t asked for in the problem, but let’s find the same volume by the√method of √ washers. The external radius of the vertical washer at x is x − (−1) = x + 1; the internal radius of that washer is x2 − (−1) = x2 + 1. The area of the cross section will be π times the difference of the squares of the radii, i.e., √ π ( x + 1)2 − (x2 + 1)2 . The integral from x = 0 to x = 1 is again equal to 29 π. 30 [1, Exercise 29, p. 437] The following integral represents the volume of a solid; describe the solid: Z 3 0 2πx5 dx . Information for Students in Lecture Section 1 of MATH 141 2010 01 3073 Solution: Since the problem appears in this section, the textbook expects you to interpret the integral as the result of application of the method of cylindrical shells. If we interpret 2πx5 · ∆x as (2πx) · x4 · ∆x, we see that it is the volume of the solid generated by rotating about the y-axis the region bounded by the x-axis and the curve y = x4 from the point (0, 0) to the line x = 3. But the author should not have used the definite article the, since the integral can be interpreted in other ways. For example, we can interpret it as resulting from application √ 2 of the method of washers also. This time we interpret 2πx5 · ∆x as π 2x5 · ∆x: it is the area of the solid obtained by rotating about the x-axis the region bounded by that √ axis and the curve y = 2x5 between x = 0 and x = 3. π Z4 [1, Exercise 32, p. 437] The integral solid. Describe the solid. 2π(π − x)(cos x − sin x) dx represents the volume of a 0 Solution: The same integral could easily represent more than one solid. One interpretation is the following: The solid is a solid of revolution around the vertical line x = π, generated by revolving the region bounded by the y-axis and the graphs y = cos x and y = sin x up to the point where they intersect, (x, y) = π4 , π4 . But the region could be deformed vertically without changing the volume. For example, the region could be taken to be bounded by the lines y = 0, x = 0, ! π π π −x+x −x−x 1 2 2 y = cos x − sin x = sin · cos = √ cos − x 2 2 4 2 from x = 0 to x = π4 . [1, Exercise 43, p. 437] Use cylindrical shells to find the volume of a sphere of radius r. Solution: Here is a case where the solid is prescribed, but not the way of generating it. A sphere can be generated by the rotation of a half-disk around its diameter. I chose to take 2 2 2 the diameter p as the y-axis, and the equation of its boundary as x +y = r , more precisely, as x = + r2 − y2 where −r ≤ y ≤ r. The length of a vertical rectangular √ element of 2 2 area a√distance of x from √ the axis is the distance between the points (x, r − x ) and (x, − r2 − x2 ), i.e., 2 r2 − x2 . This is the height of the cylindrical generated when the element is rotated about the axis. The base of the shell is at a distance of x from the axis, so it generates a circular band (annulus) of radius 2x, hence of circumference 2πx. Here I have not been precise about whether this is the inner or outer circumference of the annulus, since, in the limit, these distinctions have no effect on the calculations. The thickness of the annulus is represented by the differential, dx, in the integral, and we Information for Students in Lecture Section 1 of MATH 141 2010 01 obtain a volume of Z r 0 3074 i √ 4π h 2 4πr3 3 r 2πx(2 r2 − x2 ) dx = − (r − x2 ) 2 = . 0 3 3 (56) Of course, this volume could also be evaluated by the method of washers, see [1, Example 1, pp. 423-424]. If you found the number of variables in equation (56), you could amplify the notation, and write Z x=r i √ 4πr3 4π h 2 3 x=r 2πx(2 r2 − x2 ) dx = − = (r − x2 ) 2 for all r . (57) x=0 3 3 x=0 [1, Exercise 44, p. 437] Use cylindrical shells to find the volume of the solid torus generated by rotating a disk of radius r around a line located a distance R from its centre. Solution: We can take the disk to be bounded by (x − R)2 + y2 = r2 , and the axis of rotational symmetry of the torus (doughnut) to be the y-axis. The boundary of the disk p 2 ispthe graphs of 2 functions, y = ± r − (x − R)2 , so the height of the element of area is 2 r2 − (x − R)2 . The volume is Z R+r Z r p √ 2 2 2πx · 2 r − (x − R) dx = 2π (u + R)2 r2 − u2 du R−r −r the substitution uZ = x − R Z under r √ r √ = 4π u r2 − u2 du + 4πR r2 − u2 du −r −r • Here the first integral can be seen to be the area under the graph of an odd function from −r to the " symmetrically#rlocated value +r, so the volume is 0; it may also be 4π 3 integrated as · (r2 − u2 ) 2 = 0. 3 −r This integral could also be evaluated “na¨ıvely” (as I did in the lectures), by using one of several possible substitutions, or by observation, since the integrand is ev! 1 3 3 idently the derivative of · − · (r2 − u2 ) 2 , which, when evaluated between −r 2 2 and +r gives a difference of 0. √ • The second integral is 2πR times the area under the curve y = r2 − u2 from u = −r to u = r, which can be seen to be the area of a half disk of radius r, which we know πr2 . Thus the volume is 2π2 Rr2 . to be 2 [1, Exercise 45, p. 437] I began the lecture by finding the volume of a right circular cone (already considered in [1, Exercise 49, p. 431], solved in the notes of the previous lecture. A solution to this variation of the problem can be found in the Student Solutions Manual [3]). Information for Students in Lecture Section 1 of MATH 141 2010 01 C.10.2 3075 §6.4 Work This section has been omitted from the syllabus because it involves physical concepts that some students from outside of the Faculties of Science and Engineering might not be prepared for. If you are a Science or Engineering student, you are urged to peruse the section and try the problems. Your instructors and TA’s will be happy to help you with any difficulties. Information for Students in Lecture Section 1 of MATH 141 2010 01 3076 C.11 Supplementary Notes for the Lecture of January 25th, 2010 Release Date: Monday, January 25th, 2010, subject to further revision Review of preceding two lectures We have studied how to use a definite integral to determine volumes of solids, by expressing the solid as the limit of a union of thin layers: in [1, §6.2] as the union of thin laminæ with planar sides and, in [1, §6.3], for solids of revolution, as the union of thin circular cylindrical shells. Where the solid is obtained by revolving a plane region about a line in that plane the solid can be called a solid of revolution about that line, and the methods of both sections are, in principle, applicable. I do not recommend attacking these problems by substitution in formulæ, as there are many variants to be considered, and the blind use of formulæ often leads to disaster when the wrong formula is applied, or the right formula is applied incorrectly. I list some formulæ below just to summarize the nature of the results we have found, with the intention that, in each case, you decompose the solid in one of the methods that is applicable and set up the integral by carefully examining the decomposition. 1. When a solid is decomposed into thin laminæ between parallel planes which are perZb pendicular to the x-axis: the volume is expressible in the form A(x) dx, where the a decomposition ranges between values x = a and x = b, and A(u) is an expression for the cross-sectional area at x = u. (If the decomposition is along the y-axis, then the limits of the integral will be the limiting values of y, and we will usually express the integral in terms of y; of course, the x or y that appears in the integral is irrelevant, since these are bound or “dummy” variables, that are part of our notation, and don’t actually affect the numerical value of the integral.) 2. When a decomposition as in item 1 above has circular symmetry as a solid of revolution about a line x = c of a region in the xy-plane, the laminæ can be interpreted as being generated by revolving a thin rectangle, and will consist of disks, possibly with a hole in the middle — the textbook calls them “washers”. The integrand will be an expression of the form π (r2 (x))2 − π (r1 (x))2 , where r2 and r1 are the outer and inner radius of the disk, determined by the distances of the two ends of the thin rectangle from the point of intersection of the extended rectangle with the axis of rotational symmetry of the solid. Here we need to be prepared to work with either x or y, and possibly to consider rotation about a line parallel to but distinct from the coordinate axes. 3. In the case of decomposition into cylindrical shells of a solid with rotational symmetry about an axis, the cylindrical shells can also be interpreted as being generated by rotating a thin rectangle about the axis — but here the long dimension of the rectangle is parallel to the axis. The integrand may also be interpreted as the area of a “flattened” cylindrical Information for Students in Lecture Section 1 of MATH 141 2010 01 3077 shell, obtained by slitting the shell open and unrolling it; one dimension will be the circumference of the circle generated by revolving any point of the cylinder around the axis of symmetry — this factor will be of the form 2πx when we are at distance x from the axis of rotational symmetry; the other dimension will be the height of the cylindrical shell. Again, we need to adjust this formula according to the orientation of the axis of rotational symmetry, and according as the axis is or is not a coordinate axis. Before proceeding with the topics scheduled for today, I reminded students of the ease in which we can determine the volume of a sphere of radius r. A solution using cylindrical shells ([1, Exercise 43, p. 437]) can be found beginning on page 3073 of these notes. We can find the volume using washers by taking the equation of circle to be x2 + y2 = r2 , and revolving the upper semi-disk around the x-axis, obtaining " #r ! ! Z x3 r3 4 r3 r 2 2 2 3 3 r 2π(r − x ) dx = 2π r x − = 2π r − − 2π −r + = · πr3 . 3 −r 3 3 3 − C.11.1 §6.5 Average value of a function In this section the textbook defines what is meant by the term average of a continuous function or a function “pieced” together from continuous functions over an interval a ≤ x ≤ b. The definition is a generalization of the definition familiar to you of the average of a finite number of numbers y1 , y2 , . . . , yn , i.e., n P yi i=1 average = . n If, in the finite case just mentioned, we treat each of yi as the height of a rectangle of unit width, situated so that its base is placed on the x-axis with its corners at (i − 1, 0) and (i, 0), and with n X height yi , then the sum yi is the area under the graph of the function f defined by i=1 y1 if 0 ≤ x ≤ 1 y2 if 1 < x ≤ 2 f (x) = ... yn if n − 1 < x ≤ n For any function that is piecewise continuous on the interval a ≤ x ≤ b, we define Z b 1 average of f over [a, b] = f (x) dx , b−a a Rb f (x) dx = aR b 1 dx a Information for Students in Lecture Section 1 of MATH 141 2010 01 3078 and so our definition is consistent with the earlier definition when the function is defined at a finite number of points x1 , . . . , xn : we are simply extending the definition of that function by extending the value at any integer point to the entire interval of unit length to the left of the point. The Mean Value Theorem for Integrals If f Ris continuous on [a, b], the Mean Value Thex orem may be applied to the function g(x) = a f (t) dt, which we know from the Fundamental Theorem to be differentiable. It asserts the existence of a point c in (a, b) such that g(b) − g(a) = g0 (c), i.e., such that b−a Z b f (t) dt = g(b) − g(a) = f (c) · (b − a) , a Rb or f (x) dx f (c) = aR b a 1 dx Rb = a f (x) dx b−a . [1, Exercise 23, p. 445]. As mentioned in connection with the Mean Value Theorem in Math 140, the spirit of this theorem is in the existence of the point c, not in the specific values that c takes. The theorem is not constructive: it proves the existence without telling you how to find the point. (A constructive proof of a theorem proves existence by providing a way of finding the point; the proof we have given for the present theorem is not constructive.) Example C.35 [1, Exercise Z 13, p. 445] The textbook asks you to prove that, if f is any contin3 uous function for which f (x) dx = 8, then f takes on the value 4 somewhere in the interval 1 1 ≤ x ≤ 3. The MVT for Integrals tells you that there is a point c such that 1 ≤ x ≤ 3 and Z 3 1 f (x) dx = f (c) 3−1 1 and the left side of this equation is exactly 82 = 4. What if we were to ask the same question with 4 replaced by another real number, e.g., 4.001? The answer would now be negative: as a counterexample41 take the constant function f (x) = 4. This function is continuous and has the desired area over the interval 1 ≤ x ≤ 3; but it never assumes the value 4.001. 41 an example to disprove a general statement: since the statement refers to all functions with a given property, we can disprove it by exhibiting just one example. Information for Students in Lecture Section 1 of MATH 141 2010 01 3079 Average velocity Suppose that the position at time t of a particle moving along the x-axis is f (t). In [1, §3.7, p. 221] the textbook has defined the Average Velocity of the particle over a time interval a ≤ t ≤ b to be ∆x f (b) − f (a) = . ∆t b−a You may recall being told by your instructor that the use of the word average there was also a generalization of the traditional meaning recalled above for the average of a finite set of numbers. You can now see that, since Rb d f (t) dt f (b) − f (a) dt = a , b−a b−a that use of the word is consistent with the use we have defined here. In other words, the Average Velocity is, in fact, the average of the velocities. So the earlier use of the word “average” was, though premature, consistent with the generalization that was planned. Previously “Average Velocity” was a two-word name for a concept, and you would not be justified in treating the first word as a modifier of the second; now you may interpret that as a conventional use of language, where average is an adjective. When mathematicians name concepts we try to make the nomenclature intuitive, and consistent with earlier usage. 6.5 Exercises √ [1, Exercise 4, p. 445] Find the average value of the function g(x) = x2 1 + x3 on the interval [0, 2]. Solution: 1 average = 2−0 Z 2 √ x2 1 + x3 dx 0 We apply the substitution u = x3 , so du = 3x2 dx and Z 23 √ 1 1 average = 1 + u · du 2−0 0 3 #8 i 1 2 1h 3 3 3 = · (1 + u) 2 = 9 2 − 1 2 6 3 9 0 26 = 9 √ (We could also use the substitution u = 1 + x3 .) Had the problem asked us to “sketch a rectangle whose area is the same as the area under the graph of g,” we could take a rectangle based on the interval 0 ≤ x ≤ 2, with height 269 .) Information for Students in Lecture Section 1 of MATH 141 2010 01 3080 Example C.36 (taken from a problem book for students in Russian technical universities) [38, Problem 1647, p. 126] Suppose that a trough has a parabolic cross-section with equation of the form y = K x2 , and measures 1 meter across the top and is 1.5 meters deep. Determine the average depth. Solution: From the data we know the cross section passes through the point 12 , 23 , so 32 = 2 K 23 , and K = 6. The depth of the trough at position x is 32 − 6x2 , so R 12 3 2 − 6x dx 1 −2 2 average depth = 1 − − 12 2 ! Z 12 3 2 = 2 − 6x dx 2 0 by symmetry, since the integrand is an even function " # 12 3x 3 = 2 − 2x =1 2 0 so the average depth is 1 meter, i.e., two-thirds of the way to the bottom of the trough. 6 Review [1, Exercise 32, p. 447] “Let R1 be the region bounded by y = x2 , y = 0, and x = b, where b > 0. Let R2 be the region bounded by y = x2 , x = 0, and y = b2 , to the right of the line x = 0. 1. “Is there a value of b such that R1 and R2 have the same area? 2. “Is there a value of b such that R1 sweeps out the same volume when rotated about the x-axis and the y-axis? 3. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated about the x-axis? 4. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated about the y-axis?” Solution: Note that the statement of the problem in the textbook is ambiguous, as the description of R2 applies to one region to the right of the y-axis and another to the left; for that reason I have added the italicized words. Now the two regions combine to form the rectangle {(x, y)| 0 ≤ x ≤ b, 0 ≤ y ≤ b2 }. 1. We are asked to investigate solutions of the equation Z b Z b b3 2b3 2 = , x dx = (b2 − x2 ) dx ⇔ 3 3 0 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3081 which has no positive solution. (We could have evaluated either or both of these areas by integration along the y-axis, writing the equation of the right branch of the √ parabola as x = y. With both integrations along that axis the equation would be Z b2 Z b2 √ √ (b − y) dy = y dy . 0 0 2. I will find the volume about the x-axis using washers, and the volume about the y-axis using the same elements of area, using cylinders. Equating the two areas I obtain the equation " 5 #b " 4 #b Z b Z b x x 2 2 2 π x dx = 2πx · x dx ⇔ π = 2π 5 0 4 0 0 0 which is equivalent to b4 (2b − 5) = 0 and has one positive solution, b = 52 . (This problem could be solved by evaluating either of the integrals in the “other” way.) 3. We have determined the volume swept out by rotating R1 about the x-axis to be πb5 . If we compute the volume obtained by rotating R2 about the x-axis, we find it 5 to be " #b Z b x5 4π 5 2 2 2 2 4 dx = π b x − π b − x = b using washers, and 5 0 5 0 Z 2 5 ib2 4π 5 √ y dy = 2π · y 2 = b using cylindrical shells. 0 5 5 0 Equating the volume obtained by rotating R1 about the same axis to this, we obtain 5 πb5 = 4πb , which has no positive solution. 5 5 b2 2πy · 4. The volume obtained by rotating R1 about the y-axis has been determined above to be π2 b4 . The volume obtained by rotating R2 about the y-axis is Z b2 Z √ b2 2 π( y) dy = π 0 0 " y2 y dy = π 2 #b2 = π 4 b 2 = π 4 b 2 0 using washers, or Z b 2 2πx b − x 0 2 " b2 x 2 x 4 − dx = 2π 2 4 #b 0 using shells. Here we see that the two volumes of revolution are equal for all values of the parameter b. Information for Students in Lecture Section 1 of MATH 141 2010 01 3082 C.12 Supplementary Notes for the Lecture of January 27th, 2010 Release Date: Wednesday, January 27th, 2010, subject to further revision Textbook Chapter 7. TECHNIQUES OF INTEGRATION. C.12.1 §7.1 Integration by Parts Earlier we developed the Substitution Rule for evaluating integrals, from the Chain Rule for differentiation. Now we will develop another procedure for evaluation of integrals, called Integration by Parts, based on the Product Rule for differentiation. As with the Substitution Rule, this rule will be applicable to both definite and indefinite integrals. If does not affect the (“independent”) variable and so there will be no change to limits in definite integrals. Starting from the Product Rule, d d d [ f (x) · g(x)] = f (x) · g(x) + f (x) · g(x) , dx dx dx we integrate all 3 members with respect to x: # # Z Z " Z " d d d [ f (x) · g(x)] dx = f (x) · g(x) dx + f (x) · g(x) dx dx dx dx and observe that the integral on the left is simply f (x) · g(x) + C. Moving the terms around gives the Rule of Integration by Parts: # # Z " Z " d d f (x) · g(x) = f (x) · g(x) dx + f (x) · g(x) dx dx dx ⇔ # # Z " Z " d d f (x) · g(x) dx = f (x) · g(x) − f (x) · g(x) dx , dx dx # # Z " Z " d d f (x) · g(x) dx = f (x) · g(x) − f (x) · g(x) dx , dx dx Z Z or, compactly, f dg = f g − g d f . Traditionally we often name the functions u and v, or variations of these symbols42 — since the solution of a specific problem may require multiple applications of integration by parts. These equations are always true for differentiable functions, but we shall be applying them when they tend to replace a difficult integral by one that is “easier” to evaluate. Usually the 42 like u1 , v1 , U, V, u˜ , v˜ , . . . Information for Students in Lecture Section 1 of MATH 141 2010 01 3083 applications will be such that the integrand admits a “natural” factorization into two factors, where either one of them becomes “simpler” under differentiation, or one becomes “simpler” under integration. I begin with an example where the use of Integration by Parts is obviously indicated. Z Example C.37 To integrate xe x dx. Solution: If we factorize the integrand into u = x, v0 = e x , then u is “simplified” by differentiation, since u0 = 1, while v0 is not “complicated” by differentiation, where v = e x . We obtain Z Z x x xe dx = xe − 1e x dx = xe x − e x + C . Example C.38 Sometimes the factorization of the integral is less than obvious. Consider the problem of integrating ln x (cf. [1, Example 2, p. 454]). The “factorization” we choose is u = ln x, dv = dx . (58) By (58), dx du = , and x v = x where we have chose one convenient antiderivative. Z Z dx ln x dx = (ln x)x − x· x = (ln x)x − x + C . This is a derivation you should remember, as we often need an antiderivative of a logarithm.43 Example C.39 Similar to the preceding example is the integration of arctan x (cf. [1, Example 5, p. 456]). Here again the function does not admit a well defined factorization, but we can try (58) and obtain Z Z x dx . arctan x dx = x · arctan x − 1 + x2 43 Perhaps one should remember it in the “more general” form Z ln |x| dx = x ln |x| − x + C . Information for Students in Lecture Section 1 of MATH 141 2010 01 3084 Now apply a substitution to the remaining integral, either v = x2 ⇒ dv = 2x dx or w = 1 + x2 ⇒ dw = 2x dx Hence Z Z 1 dw w = x · arctan x − ln |w| + C = x · arctan x − ln |1 + x2 | + C . arctan x dx = x · arctan x − Of course, the absolute signs are not needed here because the argument of the logarithm function is evidently non-negative. There are several “standard” situations where we need to use integration by parts. Suppose we need to integrate a function of one of the following forms, where P(x) is some polynomial: P(x) · sin x, P(x) · cos x, P(x) · e x , P(x) · cosh x, P(x) · sinh x. In each of these cases the derivative of the polynomial is “simpler” (here meaning “of lower degree”), while the integral of the other factor is “not more complicated”. Repeated applications cause the polynomial to disappear, leaving only an integral involving the second factor. Here again one should remember the derivation, but not memorize the formulæ, since they can be easily reconstructed, and there are too many variations to memorize. The rule of integration by parts need not be used in isolation: it may be necessary to precede or follow its use by substitutions, and several applications of integration by parts could be needed to complete the solution to a problem. Two applications of Integration by Parts? We saw in connection with substitutions that we might need to use the procedure more than once. Can that occur with Integration by Parts? It can always occur, but, if not done carefully, the second iteration will reverse the action of the first, and return us to the integral that we began with. Where the purpose of using Integration by Parts is to truly simplify the integrand, then it is unlikely you will make this tactical error. But consider the first example in the next paragraph. Solving an equation to evaluate an indefinite integral Sometimes the application of integration by parts does not appear to make any progress, but a second or more applications may eventually produce a constraint on the integral, which enables one to evaluate it. This idea will be extended later. Information for Students in Lecture Section 1 of MATH 141 2010 01 3085 Z Example C.40 To evaluate e x cos x dx. Solution: Set u = e x , dv = cos x, so that du = e x dx, v = sin x. Then Z Z x x e cos x dx = e sin x − e x sin x dx . The new integral is of similar difficulty to the old one. We apply integration by parts again: U = e x , dV = sin x, dU = e x dx, V = − cos x: Z Z x x e cos x dx = e sin x − e x sin x dx ! Z x x x = e sin x − −e cos x + e cos dx Z x x = e sin x + e cos x − e x cos dx Could this be an instance of the pitfall that was described in the preceding paragraph? Fortunately not. The same integral appears on both sides of the equation, but with different coefficients. If we move the integral from the right side to the left, we obtain Z 2 e x cos x dx = e x sin x + e x cos x dx + C (59) which implies that Z e x cos x dx = 1 x (e sin x + e x cos x) + C 2 (60) Several comments are appropriate: 1. If we had taken the second application of Integration by Parts as U = sin x, dV = e x dx, dU = cos x dx, V = e x , then we would have obtained Z Z x x e cos x dx = e sin x − e x sin x dx ! Z x x x = e sin x − e sin x − e cos dx Z Z x x x = e sin x − e sin x + e cos dx = e x cos x dx , which is a tautology. The statement obtained would not be incorrect, but we would have wasted our time to obtain an end result that could have been stated immediately, and which does not get us any closer to solving the problem at hand. Information for Students in Lecture Section 1 of MATH 141 2010 01 3086 2. Why did the constant of integration appear in equations (59), (60) but not in the preceding equation? The preceding equation was of the form Z Z f (x) dx = g(x) + h(x) dx . When both sides of an equation contain an indefinite integral, one understands that each side is a set of functions which differ by a constant; no more generality is achieved if we add the notational comment that one may add a constant to one side. But, when one side would consist of a single function — here it is e x sin x + e x cos x — the inclusion of a constant changes the meaning from one specific function to all functions obtained from it by adding any real number. 3. When we integrate the dv term, we appear to be selecting a specific antiderivative. Is this restrictive? Should we be including a constant of integration? Try to convince yourself that he selection of a specific antiderivative is not at all restrictive; that is, if you do include a constant of integration, the changes to the formula will cancel each other out. For that reason you should always choose the simplest antiderivative of v that is convenient. Reduction Formulæ Since integration by parts is helpful when the integrand is a product of two functions, one of which does not become “more complicated” under differentiation, and the other of which does not become “more complicated” under integration, this method should be able to assist in the integration of the product of a polynomial and a sine, cosine, exponential, or hyperbolic function. But what happens if we have to integrate other products of these functions? In some cases the differentiations and integrations do not produce major changes of simplicity, but they can lead to information that enables us to determine the integral, in the way in which the preceding example was solved. This method is particularly important when we wish to obtain an algorithm for evaluating certain general classes of integrals. Sometime we will need to use the methods of the preceding paragraph in this connection, and sometimes not. Z Example C.41 Find a general formula for evaluating I(n) = xn e−x dx where n is a positive integer. Solution: Let u = xn , dv = e−x dx, so du = nxn−1 dx, v = −e−x . Then Z Z n −x n −x x e dx = −x e + n xn−1 e−x dx . (61) or I(n) = −xn e−x + n · I(n − 1) . For any specific value of n, this formula may be used to evaluate Information for Students in Lecture Section 1 of MATH 141 2010 01 3087 Z x5 e−x dx, we have the integral recursively. For example, if we need to know Z x5 e−x dx = I(5) = = = = = = = −x5 e−x + 5 · I(4) −x5 e−x + 5(−x4 e−x + 4 · I(3)) −x5 e−x + 5(−x4 e−x + 4(−x3 e−x + 3 · I(2))) −x5 e−x + 5(−x4 e−x + 4(−x3 e−x + 3(−x2 e−x + 2 · I(1)))) −x5 e−x + 5(−x4 e−x + 4(−x3 e−x + 3(−x2 e−x + 2(−x1 e−x + 1 · I(0))))) −x5 e−x − 5x4 e−x − 5 · 4x3 e−x − 5 · 4 · 3x2 e−x − 5 · 4 · 3 · 2xe−x − 5!e−x + C − x5 + 5x4 + 5 · 4x3 + 5 · 4 · 3x2 + 5 · 4 · 3 · 2x1 + 5 · 4 · 3 · 2 · 1x0 e−x + C Example C.42 [1, Exercise 50, p. 458] Let n be an integer greater than 1. Find a procedure — i.e., a reduction formula — that can be used to evaluate secn x. Solution: Z Z n sec x dx = secn−2 x · sec2 x dx Applying integration by parts with u = secn−2 x, dv = sec2 x dx, we set du = (n − 2) secn−3 x · sec x tan x dx = (n − 2) secn−2 x · tan x dx, We obtain Z v = tan x . Z n n−2 sec x dx = sec tan2 x · secn−2 x dx x · tan x − (n − 2) Z n−2 = sec n−2 = sec x · tan x − (n − 2) x · tan x − (n − 2) (sec2 x − 1) · secn−2 x dx Z secn x − secn−2 x dx which may be solved for the desired indefinite integral. First we move all copies of the same integral to one side of the equation: Z Z n n−2 (n − 1) sec x dx = sec x · tan x dx + (n − 2) secn−2 x dx . Dividing by n − 1 yields the reduction formula Z Z 1 n−2 n n−2 sec x dx = sec x · tan x + secn−2 x dx . n−1 n−1 (62) Information for Students in Lecture Section 1 of MATH 141 2010 01 3088 Since we know the integrals of the 0th and 2nd powers of the secant we can now find the integral of any even positive power. For the odd positive powers we may reduce the problem to the integration of sec x, which has not been achieved yet. Example Z C.43 ([7, Exercise 36, p. 480]) Follow a substitution by integration by parts to inte2 grate x5 e x dx. Solution: The obvious substitution is u = x2 , so du = 2x dx. Then Z Z 1 5 x2 x e dx = u2 eu du 2 to which we apply integration by parts with U = u2 , dV = eu du, dU = 2u du, V = eu ! Z 1 2 u = u e − 2 u du 2 Z 1 2 u = u e − ueu du 2 to which we apply integration by parts with u = u, dv = eu du, du = du, v = eu ! Z 1 2 u u u = u e − ue − e du 2 1 2 u = u e − (ueu − eu ) + C 2 ! 1 2 = u − u + 1 eu + C 2 ! 1 4 2 2 = x − x + 1 ex + C 2 Your solution is incomplete unless you express the integral in terms of the original variable. (This section will be discussed further at the beginning of the next lecture.) Information for Students in Lecture Section 1 of MATH 141 2010 01 3089 C.13 Supplementary Notes for the Lecture of January 29th, 2010 Release Date: Friday, January 27th, 2010, revised 01 February, 2010, subject to further revision C.13.1 §7.1 Integration by Parts (conclusion) Recapitulation In the last lecture I introduced “Integration by Parts” — an integration technique related to the Product Rule of differentiation. I discussed the routine types of applications we will see, as well as application to the integration of ln x and arctan x. I ended with an application which required two successive applications of the technique, followed by the solution of an equation. 7.1 Exercises [1, Exercise 10, p. 457] “Evaluate the integral Z arcsin x dx.” Solution: Since you probably don’t know an antiderivative of the inverse sine, but do know its derivative, we can try integration by parts with u = arcsin x, dv = dx, so 1 du = √ dx and v = x. (I say try because not every attempt to apply one of the 1 − x2 integration rules will be successful: the rules are valid, and do convert the given integral into another; but if you are unable to evaluate the new integral, you haven’t fully solved the problem, and sometimes you may have made it even more difficult to solve.) Z Z x arcsin x dx = x arcsin x − dx √ 1 − x2 to which we apply the substitution w = 1 − x2 , so dw = −2x dx ! Z 1 1 = x arcsin x − √ − dw w 2 1 √ = x arcsin x + · 2 w + C 2√ = x arcsin x + 1 − x2 + C √ Other substitutions that could have been used are w = x2 , w = 1 − x2 . Z 1 r3 [1, Exercise 30, p. 458] Several methods suggest themselves for evaluating dr. √ 0 4 + r2 One solution using integration by parts can be based on √ r u = r2 , dv = √ dr ⇒ du = 2r dr, v = 4 + r2 . 4 + r2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3090 Hence Z 1 0 r3 √ 4 + r2 Z 1 √ i1 h √ s 2 dr = r 4 + r − 2 r 4 + r2 dr 0 " 0 h √ i1 1 = r s 4 + r2 − 2 · 0 2 " 2 # 1 r −8 √ 2 = · r +4 3 0 √ 16 − 7 5 = 3 2˙ 3 (4 + r2 ) 2 3 #1 0 which is approximately 0.115841, where the second integral was found by observation again. Some students may have difficulty observing the integral of dv, but a substitution could make that phase easier. √ For a solution that does not use integration by parts, try the substitution u = 4 + r2 . Then u · du = r dr, and r2 = u2 − 4. Z 1 0 r3 Z dr = √ 4 + r2 √ 5 (u2 − 4) du 2 # √5 u3 = − 4u 3 √2 16 − 7 5 = 3 " as before. Alternatively, one can use the substitution v = r2 + 4, so dv = 2r dr. Then Z 5 Z 1 v − 4 dv r3 dr = √ √ v 2 4 0 4 + r2 # " √ 5 1 32 ·v −4 v = 3 √ 4 ! √ 5 5 8 = − 4 5 − −8 , 3 3 which is the same value as obtained earlier twice. Information for Students in Lecture Section 1 of MATH 141 2010 01 [1, Exercise 44, p. 458] 1. Prove the reduction formula Z n−1 1 n−1 n cos x dx = cos x · sin x + cosn−2 x dx . n n R 2. Use Part 1 to evaluate cos2 x dx . R 3. Use Parts 1, 2 to evaluate cos4 x ds. 3091 Z (63) Solution: First observe that the textbook has overlooked the restriction that n ≥ 1; the equation is not meaningful when n = 0. We do know that, when n = 1, Z 1 cos1 x dx = sin x + 0. (64) 1 1. For n ≥ 2, we may take u = cosn−1 x, dv = cos x; then du = (n − 1) cosn−2 x · (− sin x) dx Z by the Chain Rule, and v = sin x. Let’s denote cosn x dx by In . Then a first application of Integration by Parts yields Z Z n n−1 cos x dx = cos x · sin x + (n − 1) cosn−2 x · sin2 x dx Z n−1 = cos x · sin x + (n − 1) cosn−2 x · (1 − cos2 x) dx = cosn−1 x · sin x + (n − 1) (In−2 − In ) . Collecting all terms in In to the left side of the equation, and dividing by n − 1, we obtain 1 n−1 In = · cosn−1 x · sin x + In−2 n n for n ≥ 2, as desired. 2. When n = 2, the reduction formula reduces to Z 1 1 I2 = cos x · sin x + dx 2 2 x 1 = cos x · sin x + + C. 2 2 3. When n = 4, a second application gives 1 3 I4 = cos3 x · sin x + I2 4 4 ! 1 3 x 1 3 = cos x · sin x + cos x · sin x + +C 4 4 2 2 3 3 1 · cos3 x sin x + · cos x sin x + · x + C = 4 8 8 Information for Students in Lecture Section 1 of MATH 141 2010 01 C.13.2 3092 §7.2 Trigonometric Integrals This section is concerned with integrating functions that can be expressed simply in terms of trigonometric functions. The techniques rely on heavy use of familiar trigonometric identities. In particular, integration of the following types of functions is considered: • products of non-negative powers of sin x and cos x • products of non-negative powers of tan x and sec x The integration of other functions that can be reduced to functions of these two types is also considered. Strategies are developed for products of these types. However, students may well be able to use techniques already seen to integrate certain integrals of these types in ways other than those suggested here. To reiterate: YOU MAY BE ABLE TO INTEGRATE CERTAIN FUNCTIONS OF THE TYPES LISTED BY USING OTHER METHODS. The objective in all of these procedures is to “simplify” the integration; where the function is a product of trigonometric functions, this “simplification” is usually measured by a reduction in the total degree of the product, i.e., in the total number of trigonometric factors. Since the total is finite, repeated applications of such procedures will eventually result in successful integration. R R Two ways of integrating sin2 x dx and cos2 x dx: One way of evaluating these integrals is to use one of the double angle formulæ from trigonometry: sin 2θ = 2 sin θ · cos θ cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ (65) (66) which follow from the formulæ for the sine and cosine of a sum. Two formulæ involving cos 2θ may be rewritten as 1 (1 − cos 2θ) 2 1 cos2 θ = (1 + cos 2θ) 2 sin2 θ = From these we obtain Z 2 sin x dx = = Z cos2 x dx = = Z 1 (1 − cos 2x) dx 2 ! 1 1 x − sin 2x + C 2 2 Z 1 (1 + cos 2x) dx 2 ! 1 1 x + sin 2x + C 2 2 (67) (68) Information for Students in Lecture Section 1 of MATH 141 2010 01 3093 R As we saw earlier in connection with cos2 x dx, another way to evaluate these integrals is through integration by parts. See [1, Exercises 43(a), 44(a)(b) p. 458] which describe how to use the reduction formula in [1, Example 6, p. 457] and an analogue for cosines for these purposes. Z Strategy for evaluating sin m x · cos n x dx 0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing an integral that may be evaluated immediately, the last 2 steps will lead to a simplification in the integrand, after which the procedure is begun again. 1. If n is odd, use the identity cos2 x = 1 − sin2 x to convert all but one of the cosine factors into a function of sines. Then apply the substitution u = sin x with du = cos x dx. 2. If m is odd, proceed analogously to the preceding: use the same identity to convert all but one of the sine factors into a function of cosines. Then apply the substitution u = cos x, with du = − sin x dx. 3. If both m and n are even, use the identities cos2 x = 12 (1+cos 2x) and sin2 x = 21 (1−cos 2x) to express the integrand as a sum of products of sines and cosines of 2x. The degrees of these terms will be less than the degree of the integrand we started with, so that we have simplified the problem, and can repeat the procedure until we have completed the integration. 4. When both m and n are odd, other identities may also be used to simplify the integrand; for example, we can use sin 2x = 2 sin x · cos x combined with the two double angle formulæ mentioned immediately above. Z Example C.44 To determine sin4 x · cos2 x dx. Solution: Z !2 1 − cos 2x 1 + cos 2x sin x · cos x dx = · dx 2 2 ! Z 1 − cos 2x − cos2 2x + cos3 2x dx = 8 Z 1 1 1 + cos 4x = ·x− sin 2x − dx 8 16 16 Z d 1 1 − sin2 2x · sin(2x) dx + 16 dx Z 4 2 etc. (Use the substitution u = sin 2x to evaluate the last integral.) Information for Students in Lecture Section 1 of MATH 141 2010 01 (to be continued) 3094 Information for Students in Lecture Section 1 of MATH 141 2010 01 3095 C.14 Supplementary Notes for the Lecture of February 01st, 2010 Release Date: Monday, February 01st, 2010, subject to revision C.14.1 §7.2 Trigonometric Integrals (conclusion) Z Strategy for evaluating sin m x · cos n x dx (Repetition from last lecture). 0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing an integral that may be evaluated immediately, the last 2 steps will lead to a simplification in the integrand, after which the procedure is begun again. 1. If n is odd, use the identity cos2 x = 1 − sin2 x to convert all but one of the cosine factors into a function of sines. Then apply the substitution u = sin x with du = cos x dx. 2. If m is odd, proceed analogously to the preceding: use the same identity to convert all but one of the sine factors into a function of cosines. Then apply the substitution u = cos x, with du = − sin x dx. 3. If both m and n are even, use the identities cos2 x = 12 (1+cos 2x) and sin2 x = 21 (1−cos 2x) to express the integrand as a sum of products of sines and cosines of 2x. The degrees of these terms will be less than the degree of the integrand we started with, so that we have simplified the problem, and can repeat the procedure until we have completed the integration. 4. When both m and n are odd, other identities may also be used to simplify the integrand; for example, we can use sin 2x = 2 sin x · cos x combined with the two double angle formulæ mentioned immediately above. Z Example C.45 To determine sin4 x · cos2 x dx. Solution: !2 1 − cos 2x 1 + cos 2x sin x · cos x dx = · dx 2 2 ! Z 1 − cos 2x − cos2 2x + cos3 2x = dx 8 Z 1 1 + cos 4x 1 ·x− sin 2x − dx = 8 16 16 Z d 1 + 1 − sin2 2x · sin(2x) dx 16 dx etc. (Use the substitution u = sin 2x to evaluate the last integral.) Z Z 4 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3096 Z tan m x·sec n x dx. Strategy for evaluating Earlier I discussed an algorithm for integrating products of non-negative integer powers of sines and cosines. I now consider another important class of products of trigonometric functions. 1. The basis of the solution I propose for most of these cases is to use the facts that d(tan x) = sec2 x dx d(sec x) = sec x tan x dx sec2 x = tan2 x + 1 2. If n is even and n ≥ 2, then we can replace the factor secn x by sec2 x n−2 n−2 d d 2 tan x = tan2 x + 1 2 · tan x dx dx which, when multiplied by tanm x — where m has any value — yields an integral that is simplified by the substitution u = tan x. Z 3. If n = 0 we have tanm x dx: (a) When m = 0, the solution is R dx = x + C. R (b) When m = 1, the function integrates as tan x = ln | sec x| + C = − ln | cos x| + C. (c) When m ≥ 2, one may detach 2 powers of the tangent from the others, replacing them by sec2 x − 1, thereby reducing the problem Z to the integration of a lower power of the tangent, and an integral of the form tanm−2 x · sec2 x dx, which can be integrated following a substitution u = tan x. Henceforth we may assume that n is odd. 4. If m is odd, we may detach one power of tan x from tanm x and one power of sec x from secn x, and write the integrand as which is equal to tan2 x m−1 2 · secn−1 x · d sec x dx m−1 d sec2 x − 1 2 · secn−1 x · sec x dx and can be integrated after a substitution u = sec x. Information for Students in Lecture Section 1 of MATH 141 2010 01 3097 5. In the only remaining cases m is even and n is odd. One method would be to transform the entire integrand into powers of sec x and then to use the reduction formula [1, Exercise 50, p. 458] to reduce everything to the problem of integrating sec x. Other reductions are possible: for example, into a function expressible in terms of cos x, with one factor cos x left over — this would permit a substitution u = cos x that converts the problem to the type we shall meet in [1, §7.3]; alternatively, one may express the integrand as a sum of powers of tan x, and develop a reduction formula for them (cf. [1, endpapers, item 75 p. 9.]). Z The integral tan x dx. As observed above, Z tan x dx = − ln | cos x| + C = ln | sec x| + C . Z The integral sec x dx. The textbook observes that Z sec x dx = ln | sec x + tan x| + C . Do not forget the absolute value signs when you quote these results, although you may expect that in many of the problems you consider, where sec x + tan x is positive, omission of the signs may not produce any visible error. While it is possible to prove the validity of the preceding equation simply by differentiation of the alleged antiderivative, a more direct proof requires some ingenuity.44 Z Strategy for evaluating cot m x · csc n x dx. Analogues of the preceding techniques can simplify integrals of these types. 44 One way to derive this result is to observe that sec x = = = cos x 1 = cos x 1 − sin2 x cos x (1 − sin x)(1 + sin x) cos x 1 cos x + 2 1 − sin x 1 + sin x using ideas of partial fractions that we shall be meeting in [1, §7.4]. Each of these summands can be integrated as a logarithm (See [33, pp. 505-506 ].) Information for Students in Lecture Section 1 of MATH 141 2010 01 3098 “Other” trigonometric identities. By adding and subtracting the expansions of sin(A ± B) and cos(A ± B), one may obtain the following identities 1 (sin(A − B) + sin(A + B)) sin A · cos B = (69) 2 1 (cos(A − B) − cos(A + B)) sin A · sin B = (70) 2 1 (cos(A − B) + cos(A + B)) cos A · cos B = (71) 2 These identities permit the derivation of another class of useful identities. If we replace A − B V +U V −U and A + B respectively by U and V — equivalently, if we replace A by and B by , 2 2 we obtain V +U V −U sin U + sin V = sin · cos (72) 2 2 −V + U V +U sin U − sin V = sin · cos (73) 2 2 V +U V −U cos U + cos V = cos (74) · cos 2 2 V +U V −U cos U − cos V = sin (75) · sin 2 2 Z Example C.46 Integrate (sin 50x · cos 12x) dx. Solution: (One possible solution) Z Z 1 (sin 38x + sin 62x) dx (sin 50x · cos 12x) dx = 2 1 1 1 1 = − · · cos 38x − · · cos 62x + C 2 38 2 62 Example C.47 ([7, Exercise 62, p. 488]) Find the volume obtained by rotating the region bounded by the curves y = cos x, x = 0, y = 0, x = π2 about the axis y = 1. Solution: I give a solution, by washers. (The problem can be solved by cylindrical shells also, but that method is much more difficult.) Z π2 Volume = 12 − (1 − cos x)2 dx 0 Z π2 (2 cos x − cos2 x) dx = " 0 x sin 2x = 2 sin x − − 2 4 π = π 2− . 4 # π2 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3099 7.2 Exercises π Z3 tan5 x sec6 x dx. [1, Exercise 30, p. 466] To evaluate 0 Solution: This integral can be evaluated in at least two different ways. 1. Since the exponent of the secant is even, we can use a substitution u = tan x, so du = sec2 x dx. π Z3 Z 5 tan 6 tan x sec x dx = π 3 2 u5 u2 + 1 du tan 0 0 √ 3 Z = Z 0 √ 3 = " 2 u5 u2 + 1 du u9 + 2u7 + u5 0 10 8 6 2 du # √3 u u u + + 10 4 6 0 243 81 27 981 = + + = 10 4 6 20 = 2. Alternatively, since the exponent of the tangent is odd, we can use a substitution v = sec x, so dv = sec x · tan x dx: π Z3 Z 5 2 6 tan x sec x dx = (v2 − 1)2 v5 dv 1 0 " v10 v8 v6 = − + 10 4 6 981 = 20 Z #2 1 dx . cos x − 1 Solution: This integral is not of any of the forms shown in the chapter, so some ingenuity is needed. I give more than one solution. [1, Exercise 48, p. 466] To evaluate Information for Students in Lecture Section 1 of MATH 141 2010 01 3100 1. First solution, using double angle formula. When you see cos x − 1, that should suggest the identity cos 2θ = 1 − 2 sin2 θ. Applying that identity here, with θ = 2x , yields Z Z dx dx = cos x − 1 −2 sin2 2x Z 1 x x = − csc2 dx = cot + C 2 2 2 2. Second solution, using the identity sin2 x + cos2 x = 1. The idea here resembles the rationalization of fractions involving square roots, seen earlier. ! Z Z dx cos x + 1 1 = · dx cos x − 1 cos x − 1 cos x + 1 Z cos x + 1 = dx 2 − sin x Z Z cos x = − dx − csc2 x dx sin2 x The first integral may be evaluated in several ways, for example by using the substitution u = sin x, so du = − cos x dx. That integral becomes Z Z cos x du dx = u2 sin2 x 1 = − + C1 = − csc x + C1 . u The second integral can be seen immediately to be − cot x + C2 . The two together give us Z dx = csc x + cot x + C . cos x − 1 But, are the two answers equal? This can be seen through trigonometric identities. For example 2 cos2 2x 1 + cos x x csc x + cot x = = x x = cot sin x 2 sin 2 · cos 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3101 C.15 Supplementary Notes for the Lecture of February 03rd, 2010 Release Date: Wednesday, February 03rd, 2010, subject to revision C.15.1 §7.3 Trigonometric Substitution In this section we describe a type of substitutions which simplify certain commonly met integrals. We approach these substitutions in the reverse direction from that used earlier. Whereas earlier we investigated substitutions of the form u = g(x), this time we will usually formulate our substitutions first in the form x = h(u); that is, we will look for a substitution that will simplify the integrand, and then try to implement it. We usually proceed “mechanically” in these problems; but, in principle, we are postulating the existence of an inverse function, and should be checking that there really is an inverse whenever we use the method. I will try to go through some of those steps in the first examples we consider, but, in practice, we often become careless and don’t check everything unless something indicates a problem. This is unwise; the reason that it “works” is that we usually confine the substitutions to certain well understood pairs of functions/inverses, where all of the snags have already been worked out. Z 2 √3 x3 Example C.48 [1, Exercise 4, p. 472] To evaluate dx. √ 0 16 − x2 Solution: √A first look at the integrand suggests that the complication comes from the expression 16 − x2 in the denominator. We could try to simplify by giving this a new name, √ u = 16 − x2 . That leads to x du = − √ dx ⇒ x dx = −u du 16 − x2 and so the integral becomes Z 2 4 " u3 (u − 16) du = − 16u 3 #2 2 = 4 40 . 3 But we would like to illustrate the notion of trigonometric substitution here (in a prob√ 2 lem where it isn’t really needed!) The component 16 − x suggests that we might wish to interpret the problem geometrically, with this component arising from an application of Pythagoras’s Theorem to a right-angled triangle; equivalently, from the identity that sin2 θ + cos2 θ = 1 . To do this, we can first divide out the factor 16, which, when it leaves the square root, will reappear outside as 4: r r x 2 2 √ x =4 1− . 16 − x2 = 4 1 − 16 4 Information for Students in Lecture Section 1 of MATH 141 2010 01 3102 This suggests a substitution x = sin v 4 or, equivalently, x = 4 sin v x that will make into a sine or a cosine — either one will work. When we express the substi4 tution that way we are really working with the inverse — to conform with our earlier theory about substitutions we should be expressing the new variable in terms of the old; so we should be starting with x v = arcsin . 4 √ √ x 3 The range of values of x that interest us is 0 ≤ x ≤ 2 3, equivalently 0 ≤ ≤ , and 4 2 we know that the inverse sine function is defined over this domain. So, beginning with the substitution above, we obtain dv = 1 1 dx , 2 · dx = √ 4 16 − x2 1 − 4x and the integral transforms as follows: √ 2 3 Z Z x3 dx = √ 16 − x2 0 arcsin arcsin 04 π 3 √ 2 3 4 64 sin3 v dv Z = 64 sin3 v dv 0 which we proceed to evaluate using the methods of the preceding section: Z π 3 64 0 Z π 3 3 sin v dv = 64 " sin v 1 − cos2 v dv 0 1 = 64 − cos v + cos3 v 3 # π3 = 0 40 3 In practice this method works smoothly, but one must occasionally be careful about the evaluation of the inverse function, remembering our original definitions of the restricted interval The method is indicated whenever we see expressions like √ √ where the inverse was taken. 2 2 1 − x or, more generally, a − x2 , since we can transform the latter into the former by division by a positive real number. The inverse cosine could be used instead of the inverse sine, and the results would be no more difficult. Information for Students in Lecture Section 1 of MATH 141 2010 01 3103 Z √ x2 − 4 dx. x4 Solution: We need a substitution that will convert x2 to 4 times the square of a secant.45 One way to achieve this is to make u have the property that Example C.49 ([7, Exercise 8, p. 494]) Evaluate x = 2 sec u ; so the actual substitution will be x u = arcsec , 2 which implies that dx = 2 sec u tan u du . √ √ Under the substitution what happens to x2 − 4? It becomes 4 tan2 u, i.e., 2| tan u|. Do we need the absolute signs? Recall that the inverse secant takes its values in the two intervals 0 ≤ u < π2 and π ≤ u ≤ 3π . In these two intervals the tangent is always positive, so the absolute 2 signs may be dropped. Z √ 2 Z x −4 2 tan u dx = · 2 sec u · tan u du 4 4u x 16 sec Z 1 = sin2 u · cos u du 4 1 = sin3 u + C 12 We can’t leave the answer in this form, as it must be expressed in terms of the original variable x. Since sin u = tan u · cos u = = tan u sec u √ 1 x2 − 4 2 x 2 , 3 1 1 (x2 − 4) 2 sin3 u = · 12 12 x3 Hence 45 Z √ 3 (x2 − 4) 2 x2 − 4 dx = +C. x4 12x3 Alternatively, we could make x2 four times the square of a hyperbolic cosine. Information for Students in Lecture Section 1 of MATH 141 2010 01 3104 Example C.50 I have avoided computing the area of a disk until now. It is trivial with a 2 2 2 trigonometric substitution. If the disk is the √ set of points (x, y) such that x + y ≤ R , then it is the region bounded by the 2 graphs y = ± R2 − x2 . Its area is Z R√ Z R √ √ 2 2 2 2 R − x − − R − x dx = 2 R2 − x2 dx −R −R Z R √ = 4 R2 − x2 dx , 0 since the integrand is even, and the interval of integration is symmetric around 0. The substitution u = cos−1 Rx for 0 ≤ x ≤ R implies that cos u = Rx or x = R cos u, where the interval of integration is now from u = cos−1 0 = π2 to u = cos−1 1 = 0; dx = −R sin u du; √ R2 − x2 = R| sin u| = R sin u, since the sine is positive in this interval. The integral transforms to Z R√ Z 0 2 2 4 R − x dx = 4 |R sin u|(−R sin u) du π 2 0 = −4R Z 0 2 π 2 sin2 u du since the sine is positive for 0 ≤ u ≤ " #0 2u 2 u = −4R − sin 2 4 π 2 π 2 = −4R 0 − = πR2 4 π 2 If we had not used the evenness of the integrand to reduce the original problem to integrating over the interval 0 ≤ x ≤ R, there would have been a serious difficulty. That is because the inverse cosine function takes its values between 0 and π. If we had naively carried through the substitution over the entire interval −R ≤ x ≤ R, we would have obtained Z Rp Z −π 2 2 2 4 R − x dx = 4 |R sin u|(−R sin u) du −R Z = 4 Z = 4 π 2 Z 0 − π2 |R sin u|(−R sin u) du + 4 π 2 |R sin u|(−R sin u) du 0 Z 0 π 2 = −4R2 (R sin u)(−R sin u) du + 4 Z 0 π 2 Z sin2 u du + 4R2 0 − π2 (−R sin u)(−R sin u) du 0 − π2 sin2 u du Information for Students in Lecture Section 1 of MATH 141 2010 01 Z = −4R2 0 − π2 Z sin2 u du + 4R2 π 2 sin2 u du . 0 But note that the integrand is an even function, so the two integrals would cancel, and the answer would be 0. This is clearly incorrect, but what went wrong? The error was in attempting to replace x by R cos u: u is not uniquely defined for − π2 ≤ x ≤ π2 ! We could, though, have used a substitution u = sin−1 Rx over the full interval, and the correct answer would have been obtained. We would have defined u = arcsin Rx , so sin u = Rx , cos u du = R1 · dx. Z R 4 p Z R2 − x2 dx = 4 −R π 2 − π2 (R cos u)R cos u du Z π 2 1 + cos 2u du 2 − π2 " #π sin wu 2 2 = 2R u + 2 −π 2π π 2 + 0 − − + 0 = πR2 . = 2R 2 2 = 4R 2 3105 Information for Students in Lecture Section 1 of MATH 141 2010 01 3106 C.16 Supplementary Notes for the Lecture of February 05th, 2010 Release Date: Friday, February 05th, 2010 C.16.1 §7.3 Trigonometric Substitution (conclusion) Table of trigonometric substitutions given in the textbook: Expression √ a2 − x2 √ a2 − x2 √ a2 + x2 √ x 2 − a2 I can expand the table of substitutions of this type Inverse Substitution Substitution Identity θ = arcsin ax (−a ≤ x ≤ +a) x = a sin θ − π2 ≤ θ ≤ π2 1 − sin2 θ = cos2 θ θ = arccos ax (−a ≤ x ≤ +a) x = a cos θ (0 ≤ θ ≤ π) 1 − cos2 θ = sin2 θ θ = arctan ax (−∞ < x < +∞) x = a tan θ − π2 ≤ θ ≤ π2 1 + tan2 θ = sec2 θ θ = arcsec ax (−∞ < x ≤ −a or a ≤ x < ∞) x = a sec θ π ≤ θ < 3π or 2 π 0≤θ< 2 sec2 θ − 1 = tan2 θ I have shown both sine and cosine versions of the first substitution, and could similarly have produced a cotangent version of the tangent substitution, and a cosecant version of the secant substitution; in practice the latter two variants are not used frequently, and usually offer no advantages over the substitutions shown. Another trigonometric substitution will be discussed in the next section. It is used not to simplify a square root, but to simplify denominator terms of the form x2 + a2 . 7.3 Exercises Z 1 dt. √ t2 − 6t + 13 Solution: I shall complete the square of the quadratic polynomial in the denominator in order that, after a first substitution, this integral will be of a type that we recognize. Since ! t − 3 2 2 2 + 1 , t − 6t + 13 = (t − 3) + 4 = 4 2 [1, Exercise 24, p. 472] Evaluate Information for Students in Lecture Section 1 of MATH 141 2010 01 a first substitution u = 3107 t−3 , which implies that dt = 2 du, could be applied: 2 Z Z dt du = . √ √ t2 − 6t + 13 u2 + 1 Now I take u = tan θ, i.e., θ = arctan u. Thus − π2 < θ < + π2 . Z Z sec2 θ = dθ √ | sec θ| u2 + 1 Z = | sec θ| dθ Z = sec θ dθ du since − π π <θ< 2 2 = ln | sec θ + tan θ| + C 12 2 = ln ± 1 + tan θ + tan θ + C where the + sign is taken since |θ| < π2 and the cosine and secant are positive in Quadrants ##1,4, 1 2 2 = ln 1 + u + u + C !2 12 ! t − 3 t − 3 + = ln 1 + +C 2 2 12 2 = ln t − 6t + 13 + (t − 3) + (C − ln 2) And I could rename the constant with a single letter, e.g., K = C − ln 2. Hyperbolic substitutions It is possible to achieve the same sorts of simplifications by using inverse hyperbolic functions. Since we have spent little time in becoming comfortable with the hyperbolic functions, I will not discuss these substitutions in general, but may apply them in specific cases. Example C.51 This is [1, Exercise 31, p. 472]. 1. Use trigonometric substitution to show that Z √ dx = ln x + x2 + a2 + C √ x2 + a2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3108 2. Use hyperbolic substitution x = a sinh t to show that Z dx −1 x = sinh +C √ a x 2 + a2 Solution: 1. We can simplify the surd by making x = a tan u. But our theory of substitutions requires that we express u as a function of x; so we confine ourselves, for example, to − π2 < x < π2 , and define u = arctan ax on that interval. Then du = = ⇒ dx = Z √ dx x2 + a2 = = dx x2 a2 · 1 a dx = 2 a x + a2 1+ a dx sec2 u 1 sec2 u du a Z sec2 u du | sec u| Z sec u du = ln | sec u + tan u| + C since the secant is positive in the given interval r x x 2 + 1 + C = ln + a a √ 2 2 = ln x + x + a − ln a + C √ = ln x + x2 + a2 + C 0 where we absorb the subtracted logarithm of a constant into a new constant of integration. 2. The hyperbolic function sinh is invertible, since its derivative is positive always. If we wish to have x = a sinh u, we can define u = sinh−1 ax . Taking differentials gives r p √ x2 dx = a cosh u du = a 1 + sinh2 u du = a + 1 du = x2 + a2 du 2 a Hence Z Z dx = du = u + C √ x 2 + a2 x = arcsinh + C . a Information for Students in Lecture Section 1 of MATH 141 2010 01 Z Example C.52 ([7, Exercise 26, p. 494]) Evaluate Solution: Completion of the square yields 3109 x2 dx. √ 4x − x2 !2 x − 2 . 4x − x = −(x − 4x) = 4 − (x − 4x + 4) = 4 − (x − 2) = 2 1 − 2 2 Hence 2 Z √ x2 4x − x2 2 Z dx = 2 x2 2 dx 2 1 − x−2 2 q Z = 2 (2u + 2)2 2 du √ 2 1 − u2 x−2 under substitution u = 2 Z = 4 sin2 v + 2 sin v + 1 dv under substitution v = arcsin u = arcsin Z 1 − cos 2v = 4 dv − 8 cos v + 4v 2 = 6v − sin 2v − 8 cos v + C = 6v − 2(4 + sin v) cos v + C x−2 2 Now the arcsine function takes values between − π2 and π2 , in which interval the cosine is positive; hence p √ 1√ cos v = + 1 − sin2 v = 1 − u2 = 4x − x2 . 2 We conclude that Z x2 x−2 x+6 √ dx = 6 arcsin − · 4x − x2 + C . √ 2 2 2 4x − x Z 1 Example C.53 ([7, Exercise 28, p. 494]) Evaluate 5 dx. 5 − 4x − x2 2 Solution: As in the preceding example, I shall complete the square of the quadratic polynomial in the denominator in order that, after a first substitution, this integral will be of a type that we recognize. Since !2 x + 2 2 2 2 , 5 − 4x − x = −(x + 4x − 5) = 9 − (x + 2) = 9 1 − 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 a first substitution u = 3110 x+2 , which implies that dx = 3 du, could be applied: 3 Z Z 1 1 du dx = . 5 5 81 (1 − u2 ) 2 5 − 4x − x2 2 Now we can apply a second substitution u = sin φ — actually it is φ = arcsin u – where du dφ = 1 . We obtain 1 − u2 2 Z Z du dφ 1 1 = 5 81 81 cos4 φ (1 − u2 ) 2 Z 1 = sec4 φ dφ 81 Z 1 tan2 φ + 1 sec2 φ dφ = 81 " # 1 1 3 = tan φ + tan φ + C . 81 3 The function φ was defined to be an arcsine, so its values are in the interval − π2 ≤ φ ≤ π2 , in which the cosine is positive. Hence tan φ = = sin φ = cos φ q u = √ 1 − u2 + 1 − sin2 φ x+2 3 1− x+2 3 q sin φ 2 = √ x+2 9 − 4x − x2 " # 1 1 3 (x + 2)3 x+2 and tan φ + tan φ = + 3 1 81 3 243 5 − 4x − x2 2 81 5 − 4x − x2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3111 C.17 Supplementary Notes for the Lecture of February 08th, 2010 Release Date: Monday, February 08th, 2010 C.17.1 §7.4 Integration of Rational Functions by Partial Fractions In this section we shall see that an entire class of functions can be integrated by a systematic algebraic decomposition procedure, followed by specific methods for the various components into which we decompose the functions. An example to illustrate the general procedure Before describing the general procedure let us consider some examples that will make it easier to comprehend. Z 1 [1, Exercise 14, p. 482] Evaluate the integral dx. (x + a)(x + b) Solution: There will be two quite different solutions, depending on whether a = b. Case a = b: This can be integrated directly by observation, or by using the substitution u = x + a. Z 1 1 dx = − +C. 2 (x + a) x+a Case a , b: We try to decompose the integrand into the sum of fractions whose denominators are, respectively, x + a, and x + b. The degrees of the numerators must be less than these polynomials of degree 1, so they have to have degree 0, i.e., they have to be constants. So suppose that 1 α β = + . (x + a)(x + b) x + a x + b It can be shown algebraically that such a decomposition is always possible; all that is missing is to know the values of the constants α and β. The last equation, after multiplication of both sides by (x + a)(x + b), yields 1 = α · (x + b) + β · (x + a) . Here are 2 ways to find α and β: 1. Express both sides of the equation as polynomials in x, and equate the coefficients of corresponding powers of x. The left side is 1x0 + 0x1 . Thus degree 0: 1 = α · b + β · a degree 1: 0 = α + β Information for Students in Lecture Section 1 of MATH 141 2010 01 3112 Solving these equations gives 1 b−a 1 β= a−b α= 2. This equation must be true for all values of x. Give x “convenient” values to obtain equations in α and β, and solve those equations. Two “convenient” values are x = −a and x = −b. They give, respectively, the following equations: 1 b−a 1 1 = β · (0) + β · (−b + a) ⇒ β = a−b 1 = α · (−a + b) + β · 0 ⇒ α = We may now complete the integration: ! Z Z 1 1 1 1 dx = − + dx (x + a)(x + b) a−b x+a x+b 1 = (− ln(x + a) + ln(x + b)) + C a−b 1 |x + b| = ln +C a − b |x + a| 1 x + b +C = ln a − b x + a This family of functions could be written in other ways. For example, since C · (a − b) = ln eC·(a−b) , we could call eC·(a−b) K, and write the family as ! 1 x + b 1 x + b + ln K = ln ln K · a−b x + a a−b x + a where K serves as the constant of integration. Polynomials Recall that a polynomial [1, p. 28] is a function of the form P(x) = an xn + an−1 xn−1 + · · · + a1 x1 + a0 x0 where a0 , . . . , an are real numbers, called the coefficients of the polynomial. Except for the zero polynomial, which is the constant function 0, all polynomials will have a largest integer m such that am , 0; m is called the degree of the (non-zero) polynomial, and am is called the leading coefficient; a0 is called the constant term. We usually write x0 simply as 1, and x1 simply as x. Information for Students in Lecture Section 1 of MATH 141 2010 01 Rational Functions 3113 A rational function is a ratio of polynomials of the form A(x) B(x) where A and B are polynomial functions. The function will have as discontinuities the roots of B, if there are any. The goal of partial fraction decompositions In a partial fraction decomposition we express a ratio of polynomials as a sum of “partial” fractions — fractions that have special properties. In these “partial” fractions the denominator polynomials will always be powers of one polynomial that is irreducible, i.e., it cannot be factored further (unless we move beyond the real number system to, for example, the complex number system — with which you are not expected to be familiar). It is a theorem of algebra that Theorem C.54 If a non-zero real polynomial is irreducible then it must be of one of the following forms: 1. a non-zero constant 2. a polynomial of degree 1, of the form ax + b, where a , 0. 3. a polynomial of degree 2 without real roots, i.e., of the form ax2 + bx + c where a , 0 and b2 < 4ac. The only type of ratio to which we apply this decomposition is one where the degree of the numerator is strictly less than the degree of the denominator; if the rational function that we start with does not have this property, then we will have to preprocess it to obtain a ratio of this type; after the procedure of partial fraction decomposition is applied, the resulting “partial” fractions will have the same property — that the degree of their numerator will always be less than that of the denominator. We will then show that we are able to integrate all partial A(x) fractions of these types, and so we will be able to integrate all rational polynomials . B(x) The procedure we shall develop depends on having such a factorization of the denominator of the given ratio into irreducible polynomials. In this course we shall not be concerned with finding the factorization itself, beyond knowing • that P(a) = 0 ⇒ x − a divides P(x) — the so-called “Factor Theorem”; • how to factorize quadratic polynomials, both by using the quadratic formula, and by “completing the square”; • that, for any positive integer n, an − bn = (a − b)(an−1 + an−2 b + . . . + abn−2 + bn−1 ) Information for Students in Lecture Section 1 of MATH 141 2010 01 3114 • that, for any positive integer n, a2n+1 + b2n+1 = (a + b)(a2n − a2n−1 b + . . . − ab2n−1 + b2n ). Sometimes a factor x − a may divide P(x), so that P(x) = (x − a)Q(x), and again x − a may also divide Q(x), and possibly divide even more quotients. We may then speak of the multiplicity of the factor x − a of P(x), and say that a is a root of P(x) of that multiplicity. “Long division” of polynomials. You should know from high school how to divide a polynomial B(x) into a polynomial A(x) and obtain a quotient polynomial and a remainder polynomial. The procedure is similar to long division of integers, and you will be reminded very briefly of it at the lecture. (You will see an example of long division of polynomials in the left margin of [1, p. 474], but the way you learned to write such a calculation may be slightly different from that used in the textbook.) The first step. The procedure for integrating obtaining a quotient and a remainder: A(x) begins with the division of B(x) into A(x), B(x) A(x) = Q(x) · B(x) + R(x) where the remainder has degree less than that of the divisor B(x). This is an essential step; if it is omitted, it may cause the subsequent steps to fail. However, it is not necessary if the degree of B is greater than the degree of A, since, in that case, the quotient will be 0 and the remainder will be A(x). An example to illustrate the general procedure. Z 1 [1, Exercise 14, p. 482] Evaluate the integral dx. (x + a)(x + b) Solution: I discussed an example of this type at the last lecture; the solution (when a , b), shown on pages 3111 through 3112 of these notes, may be written as ! Z Z 1 1 1 1 dx = + − (x + a)(x + b) a−b x+a x+b 1 (− ln(x + a) + ln(x + b)) + C = a−b 1 |x + b| = ln +C a − b |x + a| 1 x + b +C ln = a − b x + a Information for Students in Lecture Section 1 of MATH 141 2010 01 3115 Suppose that, instead of the given integral, we wished to integrate Z 3 x + (a + b)x2 + abx + 1 dx . (x + a)(x + b) This function cannot be expanded into partial fractions until it is arranged that the degree of the numerator — presently 3 — be less than the degree of the denominator — 2. If we divide the denominator into the numerator, we find that x3 + (a + b)x2 + abx + 1 = x · (x + a)(x + b) + 1 . ! Z 1 Hence the integral may be expressed as x+ dx, and its value will be (x + a)(x + b) 1 x + b x2 +C + ln 2 a − b x + a What would have happened if we had attempted to expand this function into partial fractions? No such decomposition can exist. Using the first method (equating coefficients of corresponding powers) we would have obtained 4 equations which would overdetermine the constants α and β, and which would be inconsistent — there would be no solution. But, if we used the second method, and didn’t take enough equations, we might not notice that there was an inconsistency, and the alleged partial fraction would be incorrect. The general procedure. The general procedure has many facets, and some will not be explicitly discussed in the lecture; you are expected to work many problems in this section of the textbook, not only problems discussed in the lecture or appearing on WeBWorK or the quizzes. Remember the steps we need to follow: 1. The first step in any of these problems is to ensure that the degree of the numerator must be less than that of the denominator. If it is not, you must divide the denominator into the numerator, obtaining a quotient and a remainder. The quotient integrates as a polynomial, and we are left with the ratio of the remainder to the denominator, for which the methods we are discussing will enable a complete solution. 2. Factorize the denominator into a product of polynomials of degrees 1 and 2 which are irreducible, i.e., which do not factorize further into lower degree polynomials. The quadratic factors will be those having no real roots. Group factors which are exactly the same together, so that your denominator is a product of powers of distinct, irreducible polynomials. 3. Then the fraction must be decomposed into partial fractions using methods we have illustrated before, and continue in this lecture. Information for Students in Lecture Section 1 of MATH 141 2010 01 3116 Z x2 − 2x − 1 dx. (x − 1)2 (x2 + 1) Solution: The denominator has two distinct irreducible factors: x − 1, of degree 1 and multiplicity 2, and x2 + 1, an irreducible quadratic factor of degree 2 and multiplicity 1. The degree of the numerator is less than 4, which is the degree of the denominator, so there is no need for any long division. One type of partial fraction decomposition is of the form Example C.55 [1, Exercise 28, p. 482] Evaluate the integral x2 − 2x − 1 αx + β γx + δ = + (x − 1)2 (x2 + 1) (x − 1)2 x2 + 1 in which we take the most general numerator in each case, of degree less than the degree of the denominator. This is not the most useful type of partial fraction decomposition, but we will carry this step out and then improve on it. Multiplying both sides by (x − 1)2 (x2 + 1), we obtain a polynomial equation x2 − 2x − 1 = (αx + β)(x2 + 1) + (γx + δ)(x2 − 2x + 1) ⇔ x2 − 2x − 1 = (α + γ)x3 + (β − 2γ + δ)x2 + (α + γ − 2δ)x + (β + δ) Equating coefficients of corresponding powers of x yields equations corresponding to the terms of degrees 3, 2, 1, 0: α+γ β − 2γ + δ α + γ − 2δ β+δ = = = = 0 1 −2 −1 which we proceed to solve, obtaining (α, β, γ, δ) = (1, −2, −1, 1) , so the decomposition is x−2 −x + 1 x2 − 2x − 1 = + 2 . 2 2 2 (x − 1) (x + 1) (x − 1) x +1 (to be continued) Information for Students in Lecture Section 1 of MATH 141 2010 01 3117 C.18 Supplementary Notes for the Lecture of February 10th, 2010 Release Date: Wednesday, February 10th, 2010 C.18.1 §7.4 Integration of Rational Functions by Partial Fractions (conclusion) The general procedure (continued). Example C.56 (continued from Example2010:31 on page 3116 of these notes, cf. [1, Exercise 28, p. 482]) We shall see that we will be able to integrate the second summand immediately, by breaking it into two parts. We can integrate the first summand if we first apply the substitution u = x − 1: ! Z Z Z x−2 u−1 1 1 dx = du = − du (x − 1)2 u2 u u2 1 = ln |u| + + C u 1 = ln |x − 1| + +C x−1 In practice we anticipate the results of this substitution by refining the partial fraction decomposition: in place of a summand of the form an−1 xn−1 + an−2 xn−2 + . . . + a0 (x − a)n we repeatedly divide x − a into the numerator, so that we can express the numerator as a sum of powers of x − a; then we decompose the fraction and divide excess powers of x − a, to obtain a decomposition of the form bn bn−1 b1 + ... + n n−1 (x − a) (x − a) x−a in which we can integrate each of the summands at sight. This decomposition can be accomplished as a second phase of partial fraction decomposition, or immediately, by assuming a decomposition of the form ζ η γx + δ x2 − 2x − 1 = + + 2 2 2 2 (x − 1) (x + 1) (x − 1) x−1 x +1 which leads to the polynomial identity x2 − 2x − 1 = ζ(x2 + 1) + η(x − 1)(x2 + 1) + (γx + δ)(x − 1)2 ⇔ x2 − 2x − 1 = (η + γ)x3 + (ζ − η − 2γ + δ)x2 + (η + γ − 2δ)x + (ζ − η + δ) Information for Students in Lecture Section 1 of MATH 141 2010 01 3118 which we proceed to solve, obtaining (ζ, η, γ, δ) = (−1, 1, −1, 1) , Z = = = = x2 − 2x − 1 dx (x − 1)2 (x2 + 1) Z Z Z Z 1 −x 1 −1 dx + dx + dx + dx (x − 1)2 (x − 1)1 x2 + 1 x2 + 1 1 1 + ln |x − 1| − ln |x2 + 1| + arctan x + C x−1 2 1 1 + ln |x − 1| − ln(x2 + 1) + arctan x + C x−1 2 1 |x − 1| + ln √ + arctan x + C x−1 x2 + 1 7.4 Exercises Z 1 dx . (x + 5)2 (x − 1) Solution: Since the degree of the numerator is 0, and that of the denominator is 2 + 1 = 3 > 0, we can skip the long division step. We have to decompose the fraction into a sum of partial fractions, which we may take to be of the form [1, Example 19, p. 482] Evaluate the integral 1 A B C = + + . 2 2 1 (x + 5) (x − 1) (x + 5) (x + 5) (x − 1)1 Multiplying both sides by (x + 5)2 (x − 1), we obtain 1 = A(x − 1) + B(x − 1)(x + 5) + C(x + 5)2 . (76) 0x2 + 0x1 + 1x0 = A(x − 1) + B(x2 + 4x − 5) + C(x2 + 10x + 25) . (77) i.e., One way to obtain the values of A, B, C is simply to equate coefficients of like powers of x: 0 = B+C 0 = A + 4B + 10C 1 = −A − 5B + 25C 1 1 which we may solve to show that (A, B, C) = − 61 , − 36 , 36 . Information for Students in Lecture Section 1 of MATH 141 2010 01 3119 Another method of solution is to assign to x “convenient” values of x, and thereby obtain more convenient equations to solve. Two values which re “convenient” are x = 1 and x = −5, and they yield from equation (76) the following equations: x = 1 ⇒ 1 = 36C and x = −5 ⇒ 1 = −6A 1 implying that A = − 16 and C = 36 . To obtain the value of B we would need a third equation. This could be obtained by equating coefficients, as in the earlier method, or by simply choosing another value; e.g., x = −1 ⇒ 1 = −2A − 8B + 16C = 2 16 − 8B + 6 36 which yields the same values as before. Integration is straightforward: Z Z 1 1 − 36 − 16 1 36 dx = + + dx 2 2 (x + 5) (x − 1) (x + 5) x+5 x−1 1 1 1 1 · − ln |x + 5| + ln |x − 1| + K . = 6 x + 5 36 36 This indefinite integral could be further simplified, e.g., by combining the two logarithmic terms into the logarithm of a quotient of polynomials of degree 1. Repeated irreducible quadratic factors. x4 + 1 dx. x(x2 + 1)2 Solution: What distinguishes this example from those studied earlier is the multiplicity of the irreducible quadratic factor x2 + 1 in the denominator; this is the first case we have seen where that multiplicity exceeds 1. Before I begin the application of the methods of this section, I will apply a substitution that could simplify this problem. By setting u = x2 , I find that du = 2x dx, x2 = u − 1, so Z Z x4 + 1 1 u2 + 1 dx = du . x(x2 + 1)2 2 u(u + 1)2 Example C.57 [7, Exercise 38, p. 504] To integrate We can seek a partial fraction decomposition: u2 + 1 A B C = + + 2 2 u(u + 1) u (u + 1) u+1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3120 which implies identity of the numerators after the right side is taken to a common denominator: u2 + 1 = A(u + 1)2 + Bu + Cu(u + 1) in which I will take several “convenient” values of u to obtain equations that can be solved for the coefficients; u = −1 ⇒ 2 = 0 − B + 0 ⇒ B = −2 u=0 ⇒ 1=A u = 1 ⇒ 2 = 4A + B + 2C ⇒ C = 0 Hence Z ! Z x4 + 1 1 1 2 dx = − du x(x2 + 1)2 2 u (u + 1)2 1 1 = · ln |u| + +K 2 u+1 1 = ln |x| + 2 +K. x +1 The preceding was fortuitous, a consequence of the fact that a substitution was possible. Now let’s use the problem as an example for the implementation of partial fraction decomposition. We proceed analogously to the last phase of the preceding example, always taking the numerator to be the most general polynomial whose degree is less than the degree of the irreducible factor in the denominator, but now taking separate summands for the powers of that irreducible factor. Thus we assume a decomposition of the form x4 + 1 α βx + γ δx + η = + 2 + 2 2 2 2 x(x + 1) x x +1 x +1 and multiply through by the denominator on the left, to obtain the polynomial identity 2 x4 + 1 = α x2 + 1 + (βx + γ)x + (δx + η) x2 + 1 x in which the identification of coefficients of like powers of x leads to the 5 equations α+δ η 2α + β + δ γ+η α = = = = = 1 0 0 0 1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3121 implying that (α, β, γ, δ, η) = (1, −2, 0, 0, 0) . x4 + 1 1 −2x = + . 2 2 2 x(x + 1) x x +1 2 Z Z Z x4 + 1 1 −2x dx = dx + dx 2 2 x(x + 1) x x2 + 1 2 1 = ln |x| + 2 +K, x +1 the same solution as was foruitously found earlier by using a substitution. Z constant , where n is x2 + 1 n x2 + 1 an integer greater than 1, we can begin the integration by a substitution u = arctan x, which 1 implies that du = 2 dx, and x +1 Z Z 1 1 du n dx = 2n−2 2 sec u x +1 Z = cos2n−2 u du !n−1 Z 1 + cos 2u = du , 2 1 n dx. Should one of the partial fractions be of the form which, in principle, we know how to integrate. Z Example C.58 To evaluate the indefinite integral, Let x = tan u, i.e., u = arctan x. Then du = Z Z 1 dx = x2 + 1 2 Z dx . 1 + x2 1 x2 + 1 2 dx. 1 du sec2 u cos2 u du = Z 1 + cos 2u du 2 ! 1 u sin 2u = + +C 2 2 4 = Information for Students in Lecture Section 1 of MATH 141 2010 01 arctan x 2 arctan x = 2 arctan x = 2 arctan x = 2 = 3122 sin u · cos u +C 2 tan u + +C 2 sec2 u x + +C 2 2 tan u + 1 x + +C 2 2 x +1 + The presence of the arctangent function in the integral indicates that this could not have been evaluated without some step equivalent to this use of the arctangent function in a substitution. Rationalizing Substitutions. In some integrals in which the integrand is not originally a rational function, it can be transformed into a rational integrand by an appropriate substitution. Example C.59 ([7, Exercise 40, p. 504]) “Make Z a substitution to express the integrand as a 1 rational function, and then evaluate the integral dx.” √ x − √x + 2 Solution: One substitution that suggests itself is u = x + 2. Under this substitution, u2 = x + 2, dx = 2u du, Z Z 2u 1 dx = du √ (u − 2)(u + 1) x− x+2 ! Z 2 2 1 = + du 3 u−2 u+1 2 (2 ln |u − 2| + ln |u + 1|) + C = 3 2 √ 2 √ = ln x + 2 − 2 x + 2 + 1 + C 3 2 √ 4 √ = ln x + 2 − 2 + ln x + 2 + 1 + C , 3 3 √ where I removed the absolute signs from the second term because x + 2 + 1 must be positive. The arctangent substitution needs to be used when the quadratic, irreducible factors contain a first degree term. Z dx . Example C.60 To integrate 2 4x + 4x + 3 Solution: Begin by dividing out the coefficient of the 2nd degree term, then completing the square: ! !2 !2 3 1 1 1 2 2 4x + 4x + 3 = 4 x + x + = 4 x + + = 2 + 4 x + . 4 2 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3123 1 It is clear that we can partially simplify the integral by a substitution of the form u = x + , 2 Z du where du = dx. The integral becomes . In order to interpret the integral as an 2 + 4u2 arctangent, we need to make some type of scale change. for example, we could interpret it as Z Z Z du du du 1 1 = = . √ 2 2 2 + 4u 2 1 + 2u 2 1 + ( 2 · u)2 √ √ Thus an appropriate original substitution would be v = 2 · x + 12 , where dv = 2 dx, and Z dx 1 = 2 4x + 4x + 3 4 = = = = Z dx x2 + x + 43 Z 1 dv √ v2 +1 4 2 Z 2 2 dv 1 √ 2 v +1 2 2 1 √ arctan v + C 2 2 ! √ 1 1 √ arctan 2x + √ + C ; 2 2 2 you should verify the correctness of this integration by differentiation. Information for Students in Lecture Section 1 of MATH 141 2010 01 3124 C.19 Supplementary Notes for the Lecture of February 12th, 2010 Release Date: Friday, February 12th, 2010 C.19.1 §7.5 Strategy for Integration The textbook suggests a strategy for solving integration problems: 1. Simplify the integrand if possible. 2. Look for an obvious substitution. 3. Classify the integrand according to its form. 4. Try again. The preceding instructions are vague, depend on your experience and your intuition, and are occasionally not appropriate, as sometimes the best way to attach a problem will be obscured by these methods. So be prepared to try again. As for experience, you need to work many problems to acquire it. Table of integration formulæ Many of these formulæ are just recasts of familiar differentiation formulæ that you already know. But you should remember the integrals of tan x, cot x, sec x, csc x, even though you may know how to derive some of them. Can We Integrate All Continuous Functions? Read this subsection. For most of the functions you will meet in this course it will be possible to integrate them. If that is not the case, it could be that we are asking you indirectly to do something else than to integrate. You are not expected to be able to detect which functions cannot be integrated in terms of elementary functions: this is a difficult problem even for a trained mathematician. “You may be assured, though, that the integrals in the...exercises are all elementary functions.” 7.5 Exercises Z x ln x dx. √ x2 − 1 Solution: A first impression is that the most complicated part of the integrand is the logarithm, and I would like to dispose of it. One way to do that would be through a substitution, but that could well render the denominator rather complicated. I propose to try integration by parts, and to assign u and v so that the ln x term is part of u. But how much of the integrand should be taken for u? I observe that if I take only the ln x term, it leaves a function that is easy to integrate, so it is an ideal first step: [1, Exercise 58, p. 489] Evaluate u = ln x ⇒ du = dx x Information for Students in Lecture Section 1 of MATH 141 2010 01 3125 x dx x2 − 1 But what is v? If you don’t see, use the substitution U = x2 : Z Z x 1 1 dx = dU . √ √ 2 2 U −1 x −1 dv = √ You should be able to evaluate this last integral by sight; but, if you can’t, try the substitution V = U − 1: Z Z √ √ 1 1 1 1 dU = v= √ √ dV = V + C = x2 − 1 + C . 2 2 V U −1 Hence Z √ 2 √ x ln x x −1 2 dx = (ln x) x − 1 − dx . √ x x2 − 1 The problem isn’t solved yet, but at least the logarithm is gone. Now I propose to try a trigonometric substitution to simplify the square root: Z w = arcsecx ⇒ x2 − 1 = tan2 w, Z √ Z x2 − 1 dx = x Z = Z = x dx = sec2 w tan w dw : | tan w| · sec w tan w dw sec w tan2 w dw (sec2 w − 1) dw = tan w − w + C √ sec2 w − 1 − arcsec x + C = √ = x2 − 1 − arcsec x + C from which we may conclude that Z √ √ x ln x dx = (ln x) x2 − 1 − x2 − 1 + arcsec x + C1 √ x2 − 1 This was not the only way to solve this problem, and it may not have been the best! But I have written down the way I solved it first. One shorter, but less intuitive way would √ x be to use the substitution u = x2 − 1, which implies that du = √ dx. Then x2 − 1 Z Z Z √ x ln x 1 ln(u2 + 1) du . dx = ln u2 + 1 du = √ 2 2 x −1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3126 Does this look simpler to you? It can be integrated by parts by taking u = ln(u2 + 1), du dv = du, so du = 2 2 u +1 Z Z 1 1 u2 2 2 ln(u + 1) du = u ln(u + 1) − du . 2 2 u2 + 1 The point is that the last integrand is rational, and we know how to integrate all such functions! (Indeed, division of the denominator into the numerator yields a quotient of 1 and a remainder of -1, so the integral can be evaluated as Z Z Z u2 1 du = 1 du − du u2 + 1 u2 + 1 = u − arctan u + C √ √ = x2 + 1 − arctan x2 + 1 + C Another strategy would be to use the same trigonometric inverse substitution immediately: w = arcsec x ⇒ dx = sec w tan w dw ⇒ Z Z x ln x du = sec2 w ln sec w dw . √ x2 − 1 In the last integral you should recognize ln sec w as being an antiderivative of tan w. This suggests using integration by parts with u˜ = ln sec w and d˜v = sec2 w dw: du˜ = tan w dw, v˜ = tan w. Z Z 2 sec w ln sec w dw = (ln sec w)(tan w) − tan2 w dw Z = (ln sec w)(tan w) − (sec2 w − 1) dw = (ln sec w)(tan w) − tan w + w + C √ √ = (ln x) x2 − 1 − x2 − 1 + arcsec x + C In several places I have casually suppressed absolute value signs; these steps can be justified, and are consequences of the way in which we defined the inverse functions. Z 2π √ Exercise C.1 To integrate 1 + sin x dx. 0 Solution: This is an interesting integral which, at first glance, does not appear to fit into any of the families of integrals we have been studying. However, it can easily be seen that, except for the determination of a sign, the integral is not very difficult to evaluate. But the sign question is delicate, and even the previous edition of a well known text book overlooked√this difficulty. In the Student Solution Manual to that textbook the answer was given to be −2 1 − sin t + C, and the following hint was given for integration: Information for Students in Lecture Section 1 of MATH 141 2010 01 “Multiply numerator and denominator of the integrand by √ 3127 1 − sin x.” The hint is a good one, and does, indeed, lead to one way of solving the problem. Unfortunately, the answer that was given in that textbook was correct only within certain intervals. Solution following the suggestion Z √ s √ Z Z √ 1 − sin x 1 − sin2 x dx = 1 + sin x dx = 1 + sin x · √ dx 1 − sin x 1 − sin x Z r Z cos2 x | cos x| = dx = dx √ 1 − sin x 1 − sin x Had the absolute signs not been present, we could make the substitution u = sin x, subject to which we would have du = cos x · dx Z Z cos x 1 dx = du √ √ 1−u 1 − sin x √ = −2 1 − u + C √ = −2 1 − sin x + C √ Unfortunately, this function has derivative 1 + sin x only when cos x is non-negative, i.e., only when 4n−1 · π ≤ x ≤ 4n+1 · π, where n is any integer. For example, the value of the definite 2 2 Z 2π √ h √ i2π integral 1 + sin x dx is certainly not equal to −2 1 − sin x = 0 , since the integrand 0 0 √ is positive for most x, so the area under the curve y = 1 + sin x must be positive; the correct value is h i π2 h √ i 3π2 h √ i2π √ −2 1 − sin x + 2 1 − sin x π + −2 1 − sin x 3π 0 2 2 h √ i 3π2 h √ i π2 = 2 1 − sin x π + −2 1 − sin x −π 2 √ 2 √ √ = 2 2+2 2=4 2,0 Other ways to find the indefinite integral While we cannot remove the sign difficulties in this problem, we can show that the problem does, in fact, lend itself to a more systematic integration — i.e., the hint given above is not really necessary. One way to see this is to remember that we have a trigonometric identity that expresses 1 + cos θ as a square. But, as the trigonometric function given here is a sine, and not a cosine, one must first arrange for the Information for Students in Lecture Section 1 of MATH 141 2010 01 3128 presence of a cosine. One way is as follows: Z √ Z r π 1 + sin x dx = 1 + cos − x dx 2 Z r π x = − dx 2 cos2 4 2 π x √ Z = 2 cos − dx 4 2 Another approach, suggested by a student in this course several years ago, is to observe that r √ x x x x 1 + sin x = sin2 + cos2 + 2 sin · cos 2 2 2 2 x x = cos + sin . 2 2 To integrate this we need to determine the sign of the inside the absolute signs. This xfunction √ π can be done by observing that it is equal to 2 sin + , essentially the same function as 2 4 determined just above. A more systematic approach would have been to attempt to simplify the original integral by the substitution u = sin x, which would imply that du = cos x dx, so Z √ Z √ 1+u du 1 + sin x dx = cos x Z √ Z 1+u 1 = ± du = ± du √ √ 1−u 1 − u2 √ √ = ∓2 1 − u + C = ∓2 1 − sin x + C where the sign still depends upon the interval in which x is located. √ What, then, is one antiderivative of 1 + sin x? We have found that, if we confine π < x < 4n+3 π, any antiderivative has the form ourselves to one interval of the form 4n+1 2 2 √ −2 1 − sin x + C; and, if we confine ourselves to one interval of the form 4n+3 π < x < 2 √ 4n+5 π, any antiderivative has the form 2 1 − sin x + C. By choosing the constants to make 2 the function continuous (indeed, differentiable) we can patch such subfunctions together to form an antiderivative which is valid over an extended domain. The function f defined by the Information for Students in Lecture Section 1 of MATH 141 2010 01 3129 following table is one such antiderivative: x f (x) ··· · ·√· √ −3 −1 π ≤ x < 2 π −4√ 2 + 2√ 1 − sin x 2 −1 π ≤ x < 12 π 0 √2 − 2 √1 − sin x 2 1 π ≤ x < 23 π 0 √2 + 2 √1 − sin x 2 5 3 π ≤ x < 2π 4 √2 − 2 √1 − sin x 2 5 7 π ≤ x < 2π 4 √2 + 2 √1 − sin x 2 9 7 π ≤ x < 2π 8 √2 − 2 √1 − sin x 2 9 11 π ≤ x < 2 π 8 √2 + 2 √1 − sin x 2 11 π ≤ x < 13 π 12 2 − 2 1 − sin x 2 2 ··· ··· Z 2π √ We can verify that the value of the definite integral 1 + sin x dx is [ f (x)]2π 0 = f (2π) − 0 √ √ √ √ √ f (0) = 4 2 − 2 1 − sin 2π − 0 2 − 2 1 − sin 0 = 4 2. C.19.2 §7.6 Integration Using Tables and Computer Algebra Systems (OMIT) This section “is not examination material, but students are to try to solve the problems manually.” C.19.3 §7.7 Approximate Integration (OMIT) This section is not part of the syllabus of this course. Information for Students in Lecture Section 1 of MATH 141 2010 01 3130 C.20 Supplementary Notes for the Lecture of February 15th, 2010 Release Date: Monday, February 15th, 2010 C.20.1 §7.8 Improper Integrals Piecewise continuous integrands Consider a function f that is continuous everywhere in an interval [a, b], including continuity from the right at b and from the left at a. For such functions we have developed the theory of the definite integral, and the Fundamental Theorem applies. Now suppose that there is a point c such that a < c < b, where f has a jump discontinuity: lim f (x) and lim+ f (x) both exist, but are different. It is possible to define the integral of f as x→c x→c− follows: Z Z Z b c f (x) dx = a b f (x) dx + a f (x) dx , c and it can be shown that the familiar properties of the definite integral hold here, with the exception of the Fundamental Theorem. We can proceed in the usual way with such integrals, until we need to actually evaluate them; then we must split them up at the jump discontinuity before we attempt to apply the Fundamental Theorem. More generally, this definition “works” for functions with many jump discontinuities.46 47 Other types of generalizations In this section we wish to consider other types of generalizations related to a condition of the original definition of the definite integral that fails to be satisfied. We will follow the terminology of the textbook, calling two types of “impropriety” Type 1 and Type48 2, but students should be aware that these terms are not in universal use. The general spirit of these definitions, and of the preceding generalization to jump discontinuities, is that the familiar properties of integrals proved for continuous functions should hold for these broader classes wherever they make sense. We shall still have to keep away from any situation that might lead us to attempt to, for example, add +∞ to −∞, and any other undefinable operations. Remember that, in this theory, you must rely on the definition, and not attempt to write down what “makes sense”. The restrictions in some of these definitions are needed to avoid paradoxes elsewhere. 46 We will not explore here what limits — if any — exist on the number of discontinuities. We have already seen that there may be other situations where the splitting of an integral into pieces for different parts of the domain may simplify the integration; but, where there is a jump discontinuity, the splitting is not by choice, it is by our definition of the meaning of the generalized symbol. 48 Note that the terms Type 1 and Type 2 refer to a type of “impropriety” — not to a type of integral. One improper integral could contain multiple instances of Type 2, and as many as 2 instances of an impropriety of Type 1. 47 Information for Students in Lecture Section 1 of MATH 141 2010 01 3131 Type 1: Infinite Intervals Our original definition of the definite integral was given for a finite interval. If we wish to speak of an integral where either or both limits are infinite, we need to define what these are to mean. The definition we give is one that is consistent with the definition for finite intervals, and preserves those properties of the integral that are meaningful when the limits before infinite. I repeat the boxed definition on [1, p. 509]: Z∞ Zt f (x) dx = lim f (x) dx provided the integral on the right exists for Definition C.6 1. t→∞ a a all t ≥ a, and the limit exists as a finite number. The “improper” integral on the left is then said to converge or to be convergent. If the limit does not exist, the improper integral is divergent. 2. An analogous definition holds when the upper limit is −∞, or when the lower limit is either of ±∞. Read the textbook. 3. When both limits of the definite integral are infinite, we define the value to be the sum of two integrals obtained by splitting the domain. It can be shown that it doesn’t matter where the line is split, the following definition will always give the same value: Z∞ Za Z∞ f (x) dx = f (x) dx + f (x) dx . −∞ −∞ a Of course, the integrand f must have the entire real line R as its domain! Z∞ 4. It is ESSENTIAL to understand that, in the definition of , TWO limits have to be de−∞ Zt termined independently. IT IS NOT CORRECT TO CONSIDER ONLY lim f (x) dx. t→∞ −t In this case one has to add two finite numbers; if either of the limits does not exist as a finite number — and that includes being infinite — the sum is not finite, and the improper integral does not exist. Type 2: Discontinuous Integrands I have observed above how to cope with a finite, jump discontinuity in the integrand. Here we are interested in other types of discontinuity, in particular, discontinuities where the function has a vertical asymptote. If the discontinuity occurs at the left end-point of the interval, then the value of the “improper” integral is defined by Zb Zb f (x) dx = lim+ f (x) dx . t→a a t Information for Students in Lecture Section 1 of MATH 141 2010 01 3132 If the discontinuity is at the right end-point b, then the value of the “improper” integral is defined by Zb Zt f (x) dx = lim− f (x) dx . t→b a a And, if the discontinuity occurs at a point c between a and b, then the definition is based on splitting the integral into two parts: Zb Zc Zb f (x) dx = f (x) dx + f (x) dx , a a c where both of the summands on the right are defined to be limits as above, and the two limits have to be evaluated independently. It is important to understand that in this case it is not acceptable that the two limits be linked so that one may affect the other. Z2 1 Example C.61 The improper integral dx does not converge, since at least one of the x−1 0 limits Zt lim− t→1 0 1 dx , x−1 Z2 lim+ t→1 t 1 dx x−1 does not exist; indeed, neither of them exists! We say that this improper integral diverges. Without this severe definition we might find that some of the properties we wish the integral to possess might not be present. Z ∞ 1 Example C.62 ([1, Example 4, p. 511]) “For what values of p is the integral dx conxp 1 vergent?” Solution: Z ∞ Z t 1 1 dx = lim dx t→∞ 1 x p xp 1 lim [ln x]t1 when p = 1 t→∞ h i = t 1 lim x1−p when p , 1 t→∞ 1 1− p lim [ln t − 0] = ∞ when p = 1 t→∞ " # 1 lim 1 − 1 = 1 when p > 1 = t→∞ t p−1 1 − p p − 1 h i 1 lim t1−p − 1 = ∞ when p < 1 1 − p t→∞ Information for Students in Lecture Section 1 of MATH 141 2010 01 3133 This result will be needed in Chapter 11 in connection with the “p-series test” [1, p. 700]. In 1 the special case p = 1, it states that the area under the hyperbola y = from x = 1 indefinitely x to the right is unbounded. By symmetry, that region should have the same area as the area between x = 0 and x = 1 under the same curve, and over the line y = 1. That area is the value of the improper integral Z 1 Z 1 1 1 dx = lim+ dx t→0 0 x t x = lim+ [ln x]1t t→0 = − lim+ ln t = +∞ t→0 Example C.63 Consider the improper integrals Z +∞ Z +∞ 1 x dx , dx . 2 2 −∞ x + 1 −∞ x + 1 In both of these cases we must split the interval of integration at a finite point, and consider two improper integrals of Type 1 independently. In the first case an antiderivative is arctan x, so Z +∞ Z 0 Z +∞ 1 1 1 dx = dx + dx 2 2 x2 + 1 −∞ x + 1 −∞ x + 1 0 Z 0 Z b 1 1 = lim dx + lim dx a→−∞ a x2 + 1 b→+∞ 0 x2 + 1 = lim (0 − arctan a) + lim (arctan b − 0) a→−∞ b→+∞ π π = − − + = π, 2 2 and the improper integral is convergent. (We could have split the interval at any convenient point other than 0.) But, in the second case, we have Z +∞ Z 0 Z +∞ x x x dx = dx + dx 2 2 2 x +1 −∞ x + 1 −∞ x + 1 0 Z 0 Z b x x = lim dx + lim dx a→−∞ a x2 + 1 b→+∞ 0 x2 + 1 ! ! ln(x2 + 1) ln(x2 + 1) = lim 0 − + lim −0 . a→−∞ b→+∞ 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3134 Here we find that each of the limits is infinite: the integral is said to be divergent. In assigning a meaning to a doubly infinite integral we do not permit +∞ and −∞ to be approached at the same rate. That is, it is not permitted to interpret this integral as being equal Za x to lim dx. (Here the finite, symmetric integral is 0, so the limit is 0. But we do not 2 a→∞ x +1 −a give that meaning to the original, improper integral, which we insist diverges. Extended Definition of the Integral The original definition in the textbook — in terms of Riemann sums — was for continuous functions f , where the interval of integration was a finite interval [a.b]. This definition has been extended in several different ways. 1. First extension: to functions with a finite number of finite jump discontinuities. The same notation, viz. Zb f (x) dx a is used; but it now means the sum of integrals over the various (disjoint) subintervals where the function is continuous. Remember that continuity over a closed interval [a, b] means • continuity at every point x such that a < x < b, • continuity from the right at x = a; and • continuity from the left at x = b. (When a = b the integral is defined to be 0, with no continuity restrictions.) There is a need for an explicit definition here, since, where a function f defined on a ≤ x ≤ b has a jump discontinuity at c, where a < c < b, then the function will usually be continuous on one side of c, so, for the subinterval on the other side of c, there is still the need for a limiting operation of some kind, since our original definition did require continuity of the integrand over the full closed interval of the integral. If it happens that the jump discontinuity is removable, then it can simply be ignored.) 2. Type 1 Improper Integral with one limit of integration equal to +∞. We define Z+∞ Zb f (x) dx = lim f (x) dx ; and b→+∞ a a Za Za f (x) dx = +∞ lim f (x) dx . b→+∞ b Information for Students in Lecture Section 1 of MATH 141 2010 01 3135 3. Type 1 Improper Integral with one limit of integration equal to −∞. We define Zb Zb f (x) dx = lim f (x) dx ; and a→−∞ −∞ Z−∞ a Za f (x) dx = f (x) dx . lim a→−∞ b b 4. Type 1 Improper Integral with one limit of integration equal to −∞ and one limit of integration equal to +∞. We define Z+∞ Za Z+∞ f (x) dx = Z −∞ −∞ f (x) dx + −∞ Z a Z −∞ f (x) dx = +∞ f (x) dx and, a f (x) dx + f (x) dx , a +∞ where a is any convenient point chosen to break the line into two semi-infinite rays. In practice one often chooses a = 0, but that is not required. It is not permitted to compute the two limits together — each of them must exist as a finite limit, independent of the other. 5. Type 2 Improper Integral where end-point a is a point of infinite discontinuity. Here we “excise” the point a and define Zb Zb f (x) dx = lim+ f (x) dx . c→a a c 6. Type 2 Improper Integral where end-point b is a point of infinite discontinuity. Here we “excise” the point b and define Zb Zc f (x) dx = lim− f (x) dx . c→b a a 7. Type 2 Improper Integral where a point c such that a < c < b is a point of infinite discontinuity. Here we “excise” the point c and define Zb Zc f (x) dx = a Zb f (x) dx + a f (x) dx . c Information for Students in Lecture Section 1 of MATH 141 2010 01 3136 This means that the integral is the sum of two separate limits, where the “bad” point has been excised at one end of each of the smaller intervals of integration. The two limits must be computed separately, and both of them must exist (as finite limits) for the integral to be said to converge. It is not permitted to compute the two limits at the same time in a symmetric way. With these extended definitions we can show that the familiar rules for definite integrals are still operational, where they make sense. Because the definitions for improper integrals involve limits, we use the terminology of convergence and divergence, the same terminology we could use in connection with other limits, and which we will see again when we study infinite sequences and series. Comparison Test for Improper Integrals The definitions I have sketched concern limits of the values of certain integrals. Where we are unable to evaluate certain integrals directly, we can still justify a comparison theorem similar to that we saw in connection with finite integrals: Theorem C.64 (Comparison “Test” for Improper Integrals) Suppose that f and g are continuous functions such that, on the interval a ≤ x ≤ ∞, 0 ≤ g(x) ≤ f (x). Then Z∞ Z∞ g(x) dx converges f (x) dx converges ⇒ a a Z∞ Z∞ diverges ⇒ g(x) dx a f (x) dx diverges a Ze3 Example C.65 Evaluate the definite integral 1 dx . Show that the integral is a conver√ x ln x gent, improper integral, and find its value. Solution: The integrand is not defined at x = 1, since one factor in the denominator is expressed in terms of ln x. For x > 1 the substitution u = ln x is valid. Hence Ze3 1 dx = lim+ √ t→1 x ln x Ze3 t dx √ x ln x Z3 1 √ du t→1 u ln t h √ i3 = lim+ 2 u ln t t→1 h √ h√ i √ i √ √ = lim+ 2 3 − 2 ln t = 2 3 − 2 lim+ ln t = 2 3 , = lim+ t→1 t→1 Information for Students in Lecture Section 1 of MATH 141 2010 01 since, as t → 1+ , ln t → 0, so right at t = 1. 3137 √ √ √ ln t → 0 = 0, by the continuity of the function t from the 7.8 Exercises Z∞ [1, Exercise 18, p. 515] Determine whether the integral dz is convergent or diz2 + 3z + 2 0 vergent. Evaluate it if it is convergent. Solution: Using standard methods of Partial Fractions we can show that ! Z Z dz 1 1 = − dz z2 + 3z + 2 z+1 z+2 = ln |z + 1| − ln |z + 2| + C z + 1 +C. = ln z + 2 Hence Z∞ 0 " # z + 1 a dz = lim ln a→∞ z2 + 3z + 2 z + 2 0 1 + 1a 1 = lim ln − ln a→∞ 2 1+ 2 a 1 = ln 1 − ln = ln 2 2 ∞ Z dz as the difference of the improper Note that we could not have expressed 2 z + 3z + 2 Z∞ integrals 0 dz and z+1 Z∞ 0 dz , as both of these improper integrals are divergent and the z+2 0 difference of their “values” is not defined. [7, Exercise 52, p. 516] “Use the Comparison Theorem to determine whether the integral Z∞ x dx is convergent or divergent.” √ 1 + x6 1 Solution: At first glance, this integral suggests a substitution u = x2 . While that would simplify its form, it would not enable us to integrate it immediately, and it is not necessary, since we can prove the convergence without this step. A simpler attack is to Information for Students in Lecture Section 1 of MATH 141 2010 01 observe that 1 + x6 > x6 , so, for positive x, Zt consider the limit lim t→∞ 1 3138 √ 1 1 1 + x6 > x3 , √ < 3 . Hence we can 1 + x6 x " #t x 1 1 dx = lim − = 1 − lim 1 = 1. The convergence of the 3 t→∞ t t→∞ x x 1 larger, improper integral implies the convergence of the given one. [1, Exercise 75, p. 517] “Show that Z∞ x2 e −x2 1 dx = 2 Z∞ 2 e−x dx .” 0 0 Solution: This is an interesting question, because we cannot express an antiderivative 2 of e−x in terms of elementary functions. So, at first glance, one wonders how it will be possible to work with these integrals. Before doing that, I must prove that the integrals are convergent — otherwise we don’t have any right to include them as numbers in an equation. I note that x ≥ 1 ⇒ −x2 ≤ −x (multiplying the inequality by a negative number) 2 ⇒ e−x ≤ e−x (exponential function is increasing) Zt Zt 1 1 −x2 e−x dx = − t e dx ≤ ⇒ for t ≥ 1 e e 1 1 1 Hence, as t → ∞, the integral on the right approaches , i.e., the improper integral e Z∞ 1 e−x dx = . e 1 Z∞ 2 e−x dx is also con- By the [1, Comparison Theorem, p. 514], the improper integral 1 1 vergent, and is less than . But we were considering the integral from 0, not from 1! e The integral from 0 to 1 can be bounded in another way, since the reasoning given above is valid only for x ≥ 1. For example, 2 −x2 < 0 ⇒ e−x < e0 = 1 Z1 Z1 2 ⇒ e−x dx < 1 dx = 1 0 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 Hence Z1 Z∞ e −x2 Z∞ −x2 e dx = 2 e−x dx < 1 + dx + 1 . e 1 0 0 3139 This is not the exact value of the integral; in fact, it can be shown that √ Z∞ −x2 e π , 2 dx = 0 but you are not expected to know this fact, nor how to prove it. 2 Now to prove the desired equality. Let us apply integration by parts with dv = xe−x dx 2 and u = x, so v = − 12 e−x , and du = dx: Zt 2 −x2 xe 1 2 dx = − xe−x 2 0 = By l’Hospital’s Rule, t 1 + 2 2 2et #t 1 + 2 0 Zt 2 e−x dx 0 Zt 2 e−x dx. 0 t 1 = lim = 0. 2 t→∞ et t→∞ 2tet2 lim Hence Z∞ Zt 2 −x2 xe 0 dx = lim t→∞ 2 x2 e−x dx 0 Z 1 2 = lim 0t e−x dx t→∞ 2 y Z∞ 1 2 = e−x dx . 2 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3140 C.21 Supplementary Notes for the Lecture of February 17th, 2010 Release Date: Wednesday, February 17th, 2010 Textbook Chapter 8. FURTHER APPLICATIONS OF INTEGRATION. C.21.1 §8.1 Arc Length Just as with the earlier concepts of area, volume, and average, we are faced first with adopting a definition that appears to have the properties that we associate with the concept, and, at the same time, is workable in practice. The length of an arc will be defined to be the limit — if there is a limit — of the sum of the lengths of the sides of an approximating polygon formed by choosing points closer and closer together on the curve, and joining them by line segments. Note that we haven’t even defined what we mean in general by a curve, so the definition we give will apply at first only to the graph of a function. Suppose that we wish to find the length of the arc of the graph of y = f (x) between the points (a, f (a)) and (b, f (b)). We can subdivide the interval [a, b] on the x-axis by intermediate vertices, so that we have a sequence a = x0 , x1 , x2 , . . . , xn = b of points on the x-axis. If we define ∆xi−1 = xi − xi−1 , and ∆ f (xi−1 ) = ∆yi−1 = yi − yi−1 = f (xi ) − f (xi−1 ) then the distance between successive points Pi−1 = (xi−1 , f (xi−1 )) and Pi = (xi , f (xi )) is p p |Pi−1 Pi | = (xi − xi−1 )2 + (yi − yi−1 )2 = (∆xi )2 + (∆yi )2 , which square root can be expressed as either of the following: s s !2 !2 ∆ f (xi ) ∆xi 1+ · ∆xi = 1 + · ∆yi . ∆xi ∆yi Note that the orientation of the increments in x and y is not relevant, as the increments appear in these formulæ only as magnitudes. When we pass to the limit, as the “mesh” of points selected on the x-axis become closer and closer together, the first of these expressions, gives rise in the limit to the following integral representing the length of the arc: Zb p a 1 + ( f 0 (x))2 dx. Information for Students in Lecture Section 1 of MATH 141 2010 01 3141 If the curve is given by an equation in the form x = g(y), then we find the arc length from the point (g(k), k) to (g(`), `) to be Z` p 1 + (g0 (y))2 dy. k When the function whose graph is y = f (x) is invertible, both formulæ are applicable, and they give the same length.49 Evaluation of these integrals can often require an approximation method, as the integrands tend to be of types for which a function expressible in terms of elementary functions is unavailable. For that reason the problems that one meets in calculus books are often confined to a small set of functions for which antiderivatives can be found. The Arc Length Function. If we fix a point on a curve, we can then define a function that expresses distance along the curve from the fixed point. This distance is expressed as an integral with a variable upper limit, and is signed, so that, in effect, we have parameterized the curve with a variable — usually denoted by the symbol s — uniquely denoting the position of a point on a path along the curve. This practice differs from that employed when we evaluate the length of the arc between two points, where only the magnitude is of interest. I have written “...on a path along the curve” rather than “...on the curve”, because we shall be generalizing these ideas in [1, Chapter 10], to consider curves that are not the graphs of functions; in those generalizations a curve may cross itself, and the same point could be traversed more than once by a point whose path we are studying: in that case it is the length of the path that will be given by the arc length function. Example C.66 Circumference of a circle. What is the circumference of the circle x2 +y2 = R2 (where R > 0)? Solution: Since the equation given is not the graph of a function (because the curve crosses some vertical lines more than once), let’s find the length of the upper arc x = −R to r from √ x 2 2 2 x = +R, and double it. This is given by the function f (x) = R − x = R 1 − . R x R y0 = − q 1 + y0 2 = 1− 49 2 1 − Rx 1 2 x R Passage between the two forms can be seen to result from the change of variable given by y = f (x). Information for Students in Lecture Section 1 of MATH 141 2010 01 ZR Circumference = 2 1 q 1− −R ZR = 4 1 q 1− 0 3142 2 dx x R 2 dx x R since the integrand is even, and the interval symmetric around 0 Zt 1 = 4 lim− q 2 dx t→R 1 − Rx 0 since the integral has a Type 2 impropriety at x = R t ZR = 4R lim− t→R 0 1 du √ 1 − u2 under the substitution u = Za = 4R lim− a→1 0 x R 1 du √ 1 − u2 under the substitution a = arcsin Z a = 4R lim− a→1 t R cos v dv | cos v| arcsin 0 under the substitution v = arcsin u arcsin Z a = 4R lim− 1 dv a→1 0 since cos v is non-negative for − π2 ≤ v ≤ π 2 a = 4R lim− [v]arcsin 0 a→1 = 4R arcsin 1 by continuity of arcsin π = 4R · = 2πR . 2 Of course, we didn’t need to apply this last substitution v = arcsin u because we know two Information for Students in Lecture Section 1 of MATH 141 2010 01 antiderivatives of √ 1 1 − u2 3143 : t ZR 4R lim− t→R 0 t 1 du = 4R lim− [arcsin u]0R √ t→R 1 − u2 t = 4R lim− arcsin − 0 t→R R = 4R [arcsin 1 − 0] by continuity of arcsin π = 4R − 0 = 2πR . 2 x2 ln x Example C.67 ([7, Exercise 8, p. 552]) Find the length of the arc y = − , (2 ≤ x ≤ 4). 2 4 Solution: Note the way in which the information is presented: we need a description of the underlying curve, here given by an equation, and specifications of the portion of the curve whose length is to be determined, here given by an interval 2 ≤ x ≤ 4 or in the alternative notation x ∈ [2, 4]. Only the absolute value of the length is of interest, so we need not be careful about which of the end-points 2, 4 is in which limit of the integral; alternatively, it is the absolute value of the integral that we seek. x2 ln x dy 1 − ⇒ = x− 2 4 dx 4x s ! p 1 1 ⇒ 1 + (y0 )2 = 1 + x2 − + 2 16x2 r p 1 1 1 0 2 2 1 + (y ) = x + + = x + ⇒ 2 16x2 4x ! 1 1 Hence the arc length between x = 2 and x = 4 (where the function x + is equal to x + 4x 4x is y= ! Z4 Z4 1 x + 1 dx = dx x+ 4x 4x 2 2 " 2 #4 x ln x = + 2 4 2 √4 ln 2 ln 2 2 ln 2 −2− =6+ = 6 + ln 2 . = 8+ 4 4 4 Information for Students in Lecture Section 1 of MATH 141 2010 01 3144 Example C.68 ([7, Exercise 12, p. 552]) Find the length of the curve y = ln x for 1 ≤ x ≤ Solution: Let’s attack this problem by integrating along either axis. √ 3. Integrating along the x-axis: y = ln x ⇒ dy 1 = dx x s dy 1+ dx ⇒ Hence r !2 = 1+ 1 x2 √ Z3r arc length = 1+ 1 dx . x2 1 To complete the integration, one approach is to try the substitution u = simplify the integral. Then √ x2 + 1 to x dx = u du Z2 u2 1 1 + 2 dx = du x u2 − 1 √ √ Z3r 1 2 Z2 1 + = √ 2 1 2 u−1 − 1 2 u+1 du r 2 u − 1 = u + ln u + 1 √2 s√ √ 1 2−1 = 2 − 2 + ln √ − ln √ 3 2+1 √ ln 3 1 = 2− 2− − ln √ 2 2+1 √ √ ln 3 + ln( 2 + 1) , etc. = 2− 2− 2 Integrating along the y-axis: Here the curve can be rewritten as x = ey , whose length is to be found for 0 ≤ y ≤ ln23 . ln 3 Z2 √ 1 + e2y dy arc length = 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z2 = √ 2 v2 dv v2 − 1 under substitution v = Z2 1 1 2 1 + = − 2 dv v−1 v+1 √ = = = √ 1 + e2y 2 #2 1 v − 1 v + ln 2 v + 1 √2 √ √ ln 3 1 2−1 2− 2− − ln √ 2 2 2+1 √ √ √ ln 3 1 2 − 1 2 − 1 2− 2− − ln √ · √ 2 2 2+1 2−1 2 √ ln 3 1 √ 2− 2− − ln 2 − 1 2 2 " = 3145 = 2− √ 2− √ ln 3 + ln( 2 + 1) . 2 Example C.69 ([7, Exercise 14, p. 552]) Find the length of the curve y2 = 4x, 0 ≤ y ≤ 2. 2 Solution: Since x = y4 , dx = 2y , and the arc length is the integral dy Z2 r π 1+ 0 y2 4 Z4 sec v · 2 sec2 v dv dy = 0 under the substitution y = 2 tan v, i.e., v = arctan 2y π = [tan v · sec v + ln | sec v + tan v|]04 √ √ = 2 + ln 2 + 1 . (I have applied the reduction formula for §7.1].) R sec3 v dv determined in [1, p. 458, Exercise 50, 8.1 Exercises [1, Exercise 12, p. 530] Find the length of the curve y = ln(cos x) for 0 ≤ x ≤ π . 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3146 q 0 Solution: y = − tan x ⇒ 1 + (y0 )2 = | sec x|. π Z3 Arc length = | sec x| dx 0 π Z3 = sec x dx 0 π = [ln | sec x + tan x|]03 √ √ = ln |2 + 3| − ln |1 + 0| = ln(2 + 3) . C.21.2 §8.2 Area of a Surface of Revolution I develop a formula for the area of a surface of revolution by observing that an element of arc of length ∆s will, when rotated about an axis whose distance from the element is r, generate an element of surface whose area is approximately 2πr · ∆s. Remembering that s s !2 !2 ∆ f (x) ∆x ∆s = 1 + · ∆x = 1 + · ∆y . ∆x ∆y we can integrate with respect to either x or y as the conditions of the problem demand (provided the derivative exists). (to be continued) Information for Students in Lecture Section 1 of MATH 141 2010 01 3147 C.22 Supplementary Notes for the Lecture of February 19th, 2010 Release Date: Friday, February 19th, 2010; corrected 10 March, 2010 subject to further correction C.22.1 §8.2 Area of a Surface of Revolution (conclusion) 8.2 Exercises [1, Exercise 12, p. 537] Find the area of the surface obtained by rotating the curve x = 1 + 2y2 (1 ≤ y ≤ 2) about the x-axis. Solution: dx x = 1 + 2y2 ⇒ = 4y dy s !2 p dx = ⇒ 1+ 1 + (4y)2 dy Z2 p ⇒ Surface Area = 2π y 1 + (4y)2 dy 1 π 3 2 = 1 + 16y2 2 24 1 √ π √ = · 65 65 − 17 17 24 [1, Exercise 22, p. 537] ...Find the area of the surface obtained by rotating the curve y = √ 2 x + 1 (0 ≤ x ≤ 3) about the x-axis. Solution: ⇒ Area of revolution √ x2 + 1 x ⇒ y0 = √ 2 x +1 √ q 1 + 2x2 2 ⇒ 1 + (y0 ) = √ 1 + x2 √ Z3 √ 1 + 2x2 = dx 2π 1 + x2 · √ 1 + x2 y= 0 Z3 √ = 2π 1 + 2x2 dx 0 UPDATED TO April 17, 2010 Information for Students in Lecture Section 1 of MATH 141 2010 01 2π √ 2 = 3148 √ arctan Z3 2 sec3 θ dθ 0 √ under the substitution θ = arctan(x 2) √ arctan Z3 2 √ π π 3 2 = √ sec θ dθ + √ [tan θ · sec θ]arctan 0 2 2 0 by the reduction formula [1, Exercise 50, p. 458] √ π 3 2 = √ [tan θ · sec θ + ln | sec θ + tan θ|]arctan 0 "2 # √ √ √ 1 = π 3 19 + √ ln( 19 + 3 2) . 2 Note that the textbook suggested the use of either a table of integrals or a computer algebra system, but that neither was needed, as the solution of this problem is well within the abilities of a student in this course (if she has time to do the calculations). [1, Exercise 24, p. 537] Find the area of the surface obtained by rotating the curve y = ln(x+1) (0 ≤ x ≤ 1) about the y-axis. Solution: dx x+1 y = ln(x + 1) ⇒ dy = ⇒ Area of surface q 1 + (y0 )2 = Z1 s = 2π x 1 + 0 1+ (u − 1) 1 (x + 1)2 1 dx (x + 1)2 r Z2 = 2π s 1+ 1 du u2 1 under the substitution u = x + 1 Z2 √ Z2 √ 1 + u2 1 + u2 du − 2π du = 2π u Z2 1 √ 1 + u2 du = 1 arctan Z 2 sec3 θ dθ π 4 1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3149 under the substitution θ = arctan u 1 2 [tan θ · sec θ + ln | sec θ + tan θ|]arctan = π 4 2 by the reduction formula [1, Exercise 50, p. 458] √ i √ √ 1 h √ = 2 5 + ln( 5 + 2) − 2 + ln( 2 + 1) 2 arctan Z2 √ Z 2 1 + u2 (sec θ · tan θ + csc θ) dθ du = u π 4 1 under the substitution θ = arctan u 2 = [sec θ + ln | csc θ − cot θ|]arctan π 4 √ √ √ 5 − 1 √ − 2 + ln( 2 − 1) = 5 + ln 2 etc. Here again the integral does not require special software or tables, just patience. [1, Exercise 4, p. 562] 1. The curve y = x2 , (0 ≤ x ≤ 1), is rotated about the y-axis. Find the area of the resulting surface. 2. Find the area of the surface obtained by rotating the curve in part 1 about the x-axis. Solution: 1. Since dy = 2x, dx Z Area 1 = 2π " √ x 1 + 4x2 dx 0 3 2 1 = 2π · · 1 + 4x2 2 3 8 π 32 = 5 −1 . 6 Z1 2. Here Area = 2π #1 0 √ x2 1 + 4x2 dx . To simplify this integral I begin with the sub- 0 stitution 2x = tan θ, i.e., θ = arctan 2x. Then 2 dx = sec2 θ dθ, so arctan Z 2 Area = 2π 0 sec2 θ tan2 θ · sec θ · dθ 4 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 π = 4 = π 4 3150 arctan Z 2 tan2 θ · sec3 θ dθ 0 arctan Z 2 sec5 θ − sec3 θ dθ . 0 Earlier, in Example C.42 on 3087 I had solved [1, Exercise 50, p. 458], proving a reduction formula (Equation (62))for integrals of positive integer powers of the secant: Z Z 1 n−2 n n−2 sec x dx = sec x · tan x + secn−2 x dx . n−1 n−1 From this formula we see that Z Z 1 1 3 sec θ dθ = sec θ · tan θ + sec θ dθ 2 2 1 1 = sec θ · tan θ + ln | sec θ + tan θ| + C 2 2Z Z 1 3 5 3 sec θ dθ = sec · tan θ + sec3 θ dθ 4 4 ! 1 3 1 1 3 sec · tan θ + sec θ · tan θ + ln | sec θ + tan θ| + C = 4 4 2 2 1 3 3 = sec3 · tan θ + sec θ · tan θ + ln | sec θ + tan θ| + C 4 8 8 Note that the value of these antiderivatives (with C = 0) at θ = 0 is 0, and that √ sec arctan 2 = 5. Hence " #arctan 2 π 1 1 1 3 Area = sec θ · tan θ − sec θ · tan θ − ln | sec θ + tan θ| 4 4 8 8 0 √ π 3 1 2 · 5 2 · 2 − 5 2 · 2 − ln 5 + 2 = 32 √ 9 √ ln 5 + 2 . = π 5− 16 32 C.22.2 §8.3 Applications to Physics and Engineering (OMIT) Omit this section. (But read it if you are majoring in Physics or Engineering!) Information for Students in Lecture Section 1 of MATH 141 2010 01 C.22.3 §8.4 Applications to Economics and Biology (OMIT) Omit this section. (But read it if you are majoring into economics or biology.) C.22.4 §8.5 Probability (OMIT) Omit this section. Textbook Chapter 9. DIFFERENTIAL EQUATIONS. (OMIT) No parts of this chapter are included in the syllabus. 3151 Information for Students in Lecture Section 1 of MATH 141 2010 01 3152 C.23 Supplementary Notes for the Lecture of March 01st, 2010 Release Date: Monday, March 01st, 2010 Important Announcement In earlier versions of the timetable for Section 1 of MATH 140 2010 01, the times shown for Quizzes Q3 , and Q4 were not correct. Quiz Q3 will be administered during the week March 08-12, and Quiz Q3 will be administered during the week March 22–26. The timetable has now been corrected. Textbook Chapter 10. PARAMETRIC EQUATIONS AND POLAR COORDINATES. C.23.1 §10.1 Curves Defined by Parametric Equations A parameter is just a variable. When we call a variable by this term we usually are thinking of a function or set of functions involving the variable as representing a family of objects. In this first contact, the family will be a set of points. We will be taking the variable to be a real variable, and so it is natural to consider not only the family of points on the graphs of a function, but also the way in which the points are generated by the assignment of real numbers to the parameter as representing a point moving along the curve. We can name the parameter with any available symbol. Often we use the letter t; and a common use for this representation is to treat t as time, so that the curve can be thought of as the trajectory of (i.e., the path traced out by) a moving point. If we adopt this point of view, and if the parameter values are chosen from an interval a ≤ t ≤ b, then we can speak of the curve (x(t), y(t)), and can think of (x(a), y(a)) and (x(b), y(b)) as, respectively, the initial and terminal points. Curves given parametrically in this way need not be graphs of functions: a curve may cross vertical lines more than once, and may even cross itself, possibly more than once. Graphing Devices This subsection may be omitted, as, in this course, we shall not be concerned with the use of graphing devices. Can the graph of a function, given non-parametrically, be expressed in parametric form? The curve y = f (x) can be expressed in parametric form as, for example, x = t, y = f (t). But there are infinitely other ways of expressing it parametrically, for example x = t3 , y = f (t3 ), Information for Students in Lecture Section 1 of MATH 141 2010 01 3153 but not necessarily x = t2 , y = f (t2 ), since the latter would include only the points with non-negative abscissæ. Parametric vs. nonparametric representation of a curve When we represent a curve in parametric form, the parametrization sometimes contains information beyond what is available in a non-parametric representation. Often we can see a dynamic way of actually tracing out the curve by allowing the parameter to range through its domain. So, for example, the curve x = cos t y = sin t can be thought of as being traced out by a point that moves around the unit circle centred at the origin, starting at the point (1, 0) at time t = 0, counterclockwise at a rate of 1 radian per unit time as t increases (and clockwise at a rate of 1 radian per unit time as t decreases). If we compare the curve with x = cos 2t y = sin 2t we see that both trace out the same circle, covering the curve infinitely often, but the second curve moves twice as fast. If we wanted to eliminate the multiple covering, we could include the inequalities 0 ≤ t < 0 in the first case, or 0 ≤ t < π2 in the second. The standard way to transform from parametric to non-parametric equations is to eliminate the parameter “between” the equations, which can be interpreted as solving one equation for the parameter, and substituting that value into the second equation. However, this operation sometimes “loses information”. “Cartesian” representation of curves Where the textbook speaks of a Cartesian representation, I prefer to speak of a non-parametric representation. All the representations for curves considered in these two sections are Cartesian, since they all refer to the system of representation associated with Ren´e Descartes. In the following two sections we will be considering curves represented in another way — in the so called polar representation. Example C.70 [7, Exercise 13, p. 656] asks you to find a Cartesian equation of the curve x = cos2 θ, y = sin2 θ, − π2 < θ < π2 . We can eliminate θ by solving one equation for θ and substituting into the second; or, more elegantly, by adding the two equations together, obtaining an equation that does not involve θ: x+y=1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3154 But this equation does not convey all of the information we started with. One type of information that has been lost is the fact that both x and y, being squares, are non-negative. Thus it is not the whole line x + y = 1 that is equivalent, but only the line segment joining the points (0, 1) and (1, 0). This feature is essential, since the doubly-infinite line is not the curve represented by the parametric equations. Another type of information that is contained in the parametrization is the way in which this line segment is traversed. The point is moving back and forth between the point (1, 0) and the point (0, 1), covering the open segment in a parameter interval of length π4 . Distance is not covered at a constant speed — the point moves fastest in the middle of the segment; but, so far, we are not interested in how fast the point is moving. There is more than one correct way to describe this curve, but giving the equation x + y = 1 is not enough. Example C.71 [7, Exercise 17, p. 656] (a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases: x = cosh t y = sinh t For every value of t a point with these coordinates can be seen to lie on the curve x2 − y2 = 1, which is a hyperbola with two branches, one opening to the right, and passing through the point (1, 0), the other opening to the left, and passing through the point (−1, 0); the curves are both asymptotic to the lines y = ±x. But, since the hyperbolic cosine is positive, the parametrization applies only to the right branch of the hyperbola: the curve comes in from −∞ from below, passes through (1, 0) at t = 0, and the moves off along the upper half as t → +∞. So one way to describe the curve non-parametrically is x2 − y2 = 1 x ≥ 0. This curve may be parameterized in other ways. For example, we could represent it as x = sec t y = tan t (cf. [7, Exercise 14, p. 656]) but this time we need to restrict the values the parameter may take, for example by π π − <θ< . 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3155 The Cycloid In this subsection the textbook describes the construction of this interesting curve. You are not expected to remember specific properties of this curve, nor its history. However, you should be able to work with this curve if the parametric equations are given to you, in the same way as you would be expected to be able to work with any reasonable curve given parametrically. The particular parametrization developed in the textbook is x = r(θ − sin θ), y = r(1 − cos θ). Families of Parametric Curves Here the author considers a family of curves given parametrically, in which the various members of the family all have a similar equation obtained by assigning different values to a variable in the parametric equations. That variable is also called a parameter, and the family of curves could be called a parametric family of curves represented by parametric equations. Of course, the parameter that represents the family of curves is not the same as the parameter that represents the family of points on any specific curve. For example, x2 + y2 = a2 could be considered a parametric family of curves (circles) given in non-parametric form; we could represent the same family in parametric form by x = a cos t y = a sin t where the parameter t represents position on a specific curve, and the parameter a represents the different curves in the family. We could also interpret the equation by switching the roles of the parameters: the curve x = a cos t y = a sin t (78) with t constant and a variable represents the points on the line through the origin inclined to the positive x-axis by an angle of t radians. This point of view will become important in [1, §10.3]. Here again, you are expected to be able to work with reasonable families of parametric curves, but not to know specific properties of those families (with the exception of the obvious parametric equations for familiar curves, like the circle). Example C.72 Equation of a line in the plane. The line in the plane through the point (x0 , y0 ) and with slope m has non-parametric equation y = y0 + m(x − x0 ). It can be represented parametrically in infinitely many ways. If we choose to relate the parameter to distance along the line, one can show that the following equations represent the line: x = x0 + t y = y0 + mt m Check that the line segment joining any two points on this line has slope = m. (Take two 1 points with coordinates (x0 + a, y0 + ma) and (x0 + b, y0 + mb).) Information for Students in Lecture Section 1 of MATH 141 2010 01 C.23.2 3156 §10.2 Calculus with Parametric Curves Tangents If a curve is given parametrically by (x, y) = (x(t), y(t)), then (subject to certain conditions) we can differentiate y with respect to x by passing through an intermediate variable t. Recalling that dx dt dx · = = 1, dt dx dx we have dy dt dy = · dx dt dx dy 1 = · dt dx dt dy in terms of the derivatives of the functions f and dx d2 y g. This calculation can be extended to the second derivative 2 , although the expressions are dx not as pretty: ! d2 y d dy = dx2 dx dx ! d dy dt · = dt dx dx dy d dt 1 = · dt dx dx dt dt d2 y dx d2 x dy dt2 · dt − dt2 · dt 1 = · dx !2 dx dt dt d2 y dx d2 x dy · − 2 · 2 dt = dt dt !dt 3 dx dt so we can determine the first derivative Rather than substituting in this formula, memorized, you are advised to be able to carry out this computation for a specific parameterized curve. It enables us to study the concavity of Information for Students in Lecture Section 1 of MATH 141 2010 01 3157 a parameterized curve, and to apply the 2nd derivative test if necessary in an optimization problem. At this point in the lecture I began to discuss the curve which is featured in [1, Exercise 6, p. 636]. The discussion will be continued at the next lecture. Information for Students in Lecture Section 1 of MATH 141 2010 01 3158 C.24 Supplementary Notes for the Lecture of March 03rd, 2010 Release Date: Wednesday, March 03rd, 2010 C.24.1 §10.2 Calculus with Parametric Curves (continued) Example C.73 (cf. [1, Exercise 6, p. 636]) (see Figure 7 on page 3158) Let’s investigate the 1 0.5 -1.5 -1 -0.5 0 0.5 1 1.5 0 -0.5 -1 -1.5 -2 Figure 7: The curve x = cos θ + sin 2θ, y = sin θ + cos 2θ curve given parametrically by x = cos θ + sin 2θ y = sin θ + cos 2θ (79) (80) Information for Students in Lecture Section 1 of MATH 141 2010 01 3159 Differentiating the parameterizing functions, we obtain dx = − sin θ + 2 cos 2θ dθ = −4 sin2 θ − sin θ + 2 ! 1 1 = −4 sin2 θ + sin θ − 4 2 !2 1 33 = −4 sin θ + − 8 64 √ √ 1 + 33 1 − 33 = −4 sin θ + sin θ + 8 8 dy = cos θ − 2 sin 2θ dθ ! 1 = −4 cos θ sin θ − 4 The textbook asks us to find an equation for the tangent to the curve at the point with parameter ! dx dy value θ = 0. We find that, when θ = 0, , = (2, 1), so the slope of the tangent at the dθ dθ point (x(0), y(0)) = (1, 1) is 12 , and an equation for the tangent is 1 y = 1 + (x − 1) 2 or x − 2y + 1 = 0. We can now use the same curve to illustrate some of the other theory of this section. For dy example, we can determine where the curve is horizontal, by solving the equation = 0, dx dy which implies that = 0. We find that this happens when dθ 1 , 4 which we could proceed to solve. We can also determine where the curve is vertical, by dx = 0; this happens when determining where dθ √ −1 ± 33 sin θ = 8 cos θ = 0 or sin θ = All of these equations could be solved. Since the curve is expressed entirely in terms of sin θ, cos θ, sin 2θ, cos 2θ, and these functions are periodic, repeating themselves after θ passes Information for Students in Lecture Section 1 of MATH 141 2010 01 3160 through an interval of length 2π, we can see the whole curve by confining θ to the interval 1 π 3π when 0 ≤ θ ≤ 2π. In that interval the cosine vanishes when θ = , , and sin θ = 2 2 4 1 1 θ = arcsin and π − arcsin . These are the 4 points where the curve is horizontal; one of 4 4 those point is the origin, because the curve passes through the origin several times, one of those times with a horizontal tangent. The curve has the shape of a “3-leafed rose” or a “trefoil” (3leafed clover) centred at the origin, with one petal bisected by the y-axis. We will see other curves with this shape when we study polar coordinates in [1, §10.3]. For the same curve given parametrically by equations (79), (80), let’s consider the following question, similar to [1, Exercise 25, p. 636]: “Show that the curve..has (several)...tangents at (x, y) = (0, 0), and find their equations.” Solution: If we set the coordinates equal to zero and solve, we can simplify the resulting equations, to obtain: cos θ · (1 + 2 sin θ) = 0 (sin θ − 1) · (1 + 2 sin θ) = 0 . The curve passes through the origin whenever both of the equations are satisfied. This means that either 1 sin θ = − , (81) 2 or both50 of the following equations must hold: cos θ = 0 sin θ = 1 . (82) (83) π For θ between 0 and 2π, these last equations are satisfied when θ = ; equation (81) is satisfied 2 7π 11π when θ = , . The curve, which has the shape of a 3-leafed clover or a “3-leafed rose”, 6 6 passes through the origin 3 times, and we can find the slopes of the tangents in the usual way, dy by taking the ratio dθ . dx dθ 50 Note that the second equation implies the first, but the first does not imply the second; we could thus have shown only the second equation. Information for Students in Lecture Section 1 of MATH 141 2010 01 3161 Areas. Consider the arc of the curve x = f (t), y = g(t) determined by points with parameter value t between α ≤ t ≤ β. Since Z β Z f (β) 0 g(t) · f (t) dt = y · dx , α f (α) the first integral represents the area between the arc and the x-axis. Remember, though, that the curve need no longer be the graph of a function, so it could cross vertical lines more than once. This means that the area could be “folded”, and there might be portions that could be counted negatively and canceling the portions that you are interested in. For that reason you should use this integral only where you are clear about the shape of the region whose area you are finding, and there might be situations where the region should be broken up into parts and the areas of the parts found separately. Arc Length. In a similar way to the preceding, we can argue that the length of the arc α ≤ t ≤ β of the curve x = u(t), y = v(t) between α ≤ t ≤ β is given by the integral s !2 !2 Z β dx dy + dt . dt dt α Note that, when we consider the parametrization x=t y = f (t) (a ≤ x ≤ b) of the graph of the function y = f (x) between the points (a, f (a)) and (b, f (b)), this reduces to the formula we derived earlier. Here again, be careful that you are finding the length of the curve that you intend. In this case there cannot be any cancellation, since the integrand is a square root, which cannot be negative. If you obtain a negative length, it could simply be a consequence of the direction in which you have parameterized the curve, which is harmless; or of an error you have made in you calculations, which is serious. Surface Area. We can also adapt, in the obvious ways, our previous formulæ for area of surfaces of revolution. Example C.74 Let’s determine a formula for the surface area of a doughnut. Suppose that the doughnut is generated by the curve x = R + r cos θ y = 0 + r sin θ Information for Students in Lecture Section 1 of MATH 141 2010 01 3162 where 0 ≤ θ ≤ 2π, where this circle, centred at the point (x, y) = (R, 0), is revolved around the y-axis. Assume R ≥ r. s !2 !2 Z 2π dy dx Area of revolution = 2π(R + r cos θ) + dθ dθ dθ 0 Z 2π √ = 2π(R + r cos θ) r2 dθ 0 2 = 2πr [Rθ + r sin θ]2π 0 = 2πr[R · 2π − R · 0] = 4π rR. Volumes. The textbook appears to say little about this, but here again the earlier formulæ can be adapted, to determine, for example, the volume of revolution generated by a given curve about a given axis. 10.2 Exercises [1, Exercise 54, p. 637] “Find the total length of the astroid x = a cos3 t, y = a sin3 t.” Solution: This curve is generated over an interval of length 2π: we can take 0 ≤ t ≤ 2π. the curve looks like a deformed circle, that has been pinched towards the centre at the points away from where it crosses the coordinate axes. s !2 !2 Z 2π dy dx + dt Total length = dt dt 0 Z 2π r 2 = −3a cos2 · sin t 2 + 3a sin2 t · cos t dt 0 Z 2π = 3|a| · | cos t · sin t| dt 0 By symmetry we can find the length of one quarter of the curve: Z π 2 Total length = 4 3|a| · | cos t · sin t| dt 0 Z π 2 = 4|a| 3 cos t · sin t dt 0 h iπ = 6|a| sin2 t 2 = 6|a| . 0 [1, Exercise 48, p. 637] Find the length of the loop of the curve x = 3t − t3 , y = 3t2 . Information for Students in Lecture Section 1 of MATH 141 2010 01 3163 Solution: Where does this loop begin and end? We need to find all parameter values t1 , t2 where t1 , t2 but 3t1 − t13 = 3t2 − t23 3t12 = 3t22 The equations can be simplified to (t1 − t2 )(3 − t12 − t1 t2 − t22 ) = 0 (t1 − t2 )(t1 + t2 ) = 0 When we know that t1 , t2 , the equations further simplify to the system 3 − t12 − t1 t2 − t22 = 0 t1 + t2 = 0 √ which imply that t1 = −t√2 = ± 3. Thus these two curves intersect in the points that have parameter values ± 3. Since s dx dt !2 dy + dt !2 q = 3 − 3t2 2 + (6t)2 = 3(1 + t2 ) , the length of the arc of the loop is Z √ 3 √ − 3 h i √3 √ 3(1 + t2 ) dt = 2 3t + t3 = 12 3 , 0 since the integrand is an odd function, and the limits of integration are symmetrically located around t = 0. Information for Students in Lecture Section 1 of MATH 141 2010 01 3164 C.25 Supplementary Notes for the Lecture of March 05th, 2010 Release Date: Friday, March 05th, 2010 C.25.1 §10.2 Calculus with Parametric Curves (conclusion) One topic remaining, not discussed in the previous lectures is summarized in the following subsection, which was included earlier in these notes on page 3160: Areas. Consider the arc of the curve x = f (t), y = g(t) determined by points with parameter value t between α ≤ t ≤ β. Since Zβ Zf (β) g(t) · f 0 (t) dt = y · dx , α f (α) the first integral represents the area between the arc and the x-axis. Remember, though, that the curve need no longer be the graph of a function, so it could cross vertical lines more than once. This means that the area could be “folded”, and there might be portions that could be counted negatively and canceling the portions that you are interested in. For that reason you should use this integral only where you are clear about the shape of the region whose area you are finding, and there might be situations where the region should be broken up into parts and the areas of the parts found separately. I illustrate this theory with the following example: 10.2 Exercises (conclusion) [1, Exercise 34, p. 637] “Find the area of the region enclosed by the astroid x = a cos3 θ, y = a sin3 θ.” This is the same curve considered above in [1, Exercise 54, p. 637]. Let’s consider the arch in the first quadrant. The area will be 4 times the area in the first quadrant, which is π π Za Z2 Z2 dx(θ) y dx = y(θ) · dθ = a sin3 θ · −3a cos2 θ · (− sin θ) dθ . dθ 0 0 0 You know one way to evaluate an integral of this type — by replacing sin2 θ and cos2 θ 1 − cos 2θ 1 + cos 2θ respectively by and . A variant of that method is as follows: 2 2 π π Z2 Z2 a sin3 θ · −3a cos2 θ · (− sin θ) dθ = 3a2 a sin4 θ · cos2 θ dθ 0 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3165 π Z2 2 = 3a 1 − cos 2θ sin 2θ · 2 2 !2 dθ 0 π Z2 3a = sin2 2θ − sin2 2θ · cos 2θ dθ 8 2 0 π 3a = 8 2 Z2 ! 1 − cos 4θ 2 − sin 2θ · cos 2θ dθ 2 0 " 3a2 θ sin 4θ sin3 2θ = − − 8 2 8 6 2 3πa = , 32 so the area of the interior of the astroid is Laboratory Project: B´ezier Curves C.25.2 # π2 0 3πa2 . 8 Omit this subsection. §10.3 Polar Coordinates Polar Curves. In the polar coordinate system we locate points in the plane by taking a special point, called the pole, a special half-line or ray which emanates from the pole, called the polar axis and a direction for measuring positive angles — usually taken to be the counter-clockwise direction. Any point can be located if we know its distance from the pole, usually denoted by r, and the angle that the line joining them makes with the polar axis in the positive direction, usually denoted by θ. However, the angle is not unique, since angles of θ and θ + 2nπ will give the same point for any integer n. So here we have one of the essential differences between polar and cartesian coordinates: Theorem C.75 The polar coordinates of a point are never unique. In the case of the pole itself, the angle θ is totally undetermined: once we know that r = 0, any angle θ will give the same point. Convention permitting negative r. It is convenient to broaden the multiplicity of coordinates by permitting the distance from the pole to be negative. We do this by agreeing that (r, θ) and (−r, θ + π) represent the same point. This convention permits for continuous representation of certain curves, but causes complications at various stages: occasionally additional care is required. Information for Students in Lecture Section 1 of MATH 141 2010 01 3166 Relations between polar and cartesian coordinates. In theory we can set up polar and cartesian systems independently in the plane, placing the pole at any convenient place. In practice we often place the pole at the origin of a cartesian system, with the polar axis along the positive x-axis. When the author of the textbook suggests that you are to consider two systems at the same time, and gives no other information, this is what he expects you to do.51 When the polar and cartesian systems are placed in this “standard” way, the following relationships hold: x = r · cos θ y = r · sin θ 2 x + y2 = r2 y = tan θ x Note that, while it is possible to transform from polar to cartesian coordinates without ambiguity, it is not always possible to move painlessly in the other direction; this is because of the non-uniqueness of polar coordinates, about which much more will be said. If you are given an equation in cartesian coordinates, you can transform it to polar by substituting the appropriate formulæ for x and y and simplifying — try [1, Problems 21-26, p. 648]; if you are given the coordinates of a point in polar coordinates, you can transform to cartesian in the same way — try [1, Exercises 3-4, p. 647]. Consequences of Non-uniqueness of Polar Coordinates. This is a difficult topic, and will require considerable practice before you will become comfortable with it! Some of the problems you will have are related to understanding what is meant by an equation for a curve: you need to understand that any curve can be represented in multiple ways, even when we use cartesian coordinates. But, when the representation of the points themselves is not unique, the results can be confusing. We saw with parametric equations that the same point on a curve can appear more than once on a “curve”. In that context there was a “natural” way of tracing out the curve, by following increasing values of the parameter. When we come to study polar coordinates, the situation is much more complicated, because there is no “natural” way of following the generation of the curve, and no one set of coordinates for a point has preferential status with respect to others. Polar curves can be expressed by any relationship between r and θ, although more often than not the equation will be in the form r = f (θ); however, you will see some equations in the form π θ = f (r) — for example, the equation θ = represents the line through the pole inclined at an 4 π angle of to the polar axis. 4 51 However, there is an important application, involving conic sections — ellipses, parabolæ, hyperbolæ, where we place the origin at a different point; this topic is not on the syllabus of the present course, but you may read about it in [1, §10.6]. Information for Students in Lecture Section 1 of MATH 141 2010 01 3167 Symmetry When we studied symmetry in the context of cartesian coordinates, we could have considered [1, pp. 19-20, 308] • reflective symmetry in a vertical line, particularly in the y-axis; • reflective symmetry in a horizontal line, particularly in the x-axis; • rotational symmetry about a point, particularly about an angle of π around the origin (which is equivalent to reflective symmetry in both the x-axis and the y-axis); and • periodicity of a graph under a horizontal shift. These are not the only types of symmetries that a graph can possess, so the study in the textbook is selective, and we haven’t investigated thoroughly what principles were involved. Similarly, when we consider symmetry in the context of polar coordinates, we will not attempt a thorough study, but will consider types of symmetries which the polar system is particularly able to accommodate. Here are the types of symmetry that the textbook mentions: • reflective symmetry in the polar axis — exhibited by the invariance of the equation of a curve under the substitution θ 7→ −θ.52 • rotational symmetry under a rotation through an angle of π around the pole — exhibited by the invariance of the equation under the transformation r 7→ −r or under θ 7→ θ + π π • reflective symmetry under reflection in the line θ = , exhibited by invariance under the 2 transformation θ 7→ π − θ. Again, these are not the full range of symmetries that can occur in the plane. Because of the non-uniqueness of coordinates, curves can have the symmetries listed without exhibiting invariance under the transformations listed. For example, the curve θ = π is the line that extends the polar axis. It certainly has symmetry in the polar axis, but the equation is not unchanged when we replace θ by −θ. You are not expected to be an expert in the subject of symmetries. Graphing Polar Curves with Graphing Devices. course. 52 Omit this section — this is a device-free This description is incomplete. See the discussion below on page 3177 of Example C.78 ([7, Exercise 37, p. 678]). Information for Students in Lecture Section 1 of MATH 141 2010 01 3168 Where is the point with polar coordinates (r, θ)? The ambiguity in polar coordinates is not in locating a point with given coordinates — it is only that all points possess multiple sets of coordinates. If you are given the coordinates (r, θ), you may locate the point by: • First, locating the ray obtained by turning the polar axis (the distinguished ray that emanates from the pole, relative to which we refer all coordinate angles) through an angle of θ in the positive direction; and • then – if r ≥ 0, proceeding along that ray a distance of r; or – if r ≤ 0, proceeding along the extension of the polar axis beyond the pole a distance of −r. Example C.76 ([7, Exercise 2, p. 677]) Plot the point whose polar coordinates are −1, − π2 . Then find two other pairs of polar coordinates of this point, one with r > 0, and one with r < 0. Solution: The second coordinate tells us that the point is on the line inclined to the polar axis at an angle of − π2 radians measured in the positive direction: this takes us to the ray which is obtained by turning the polar axis in the clockwise direction through a right angle. But then the negative first polar coordinate tells us to proceed along the opposite ray for 1 unit. If the polar system is superimposed on a cartesian system in the usual way, the point is the unit point on the positive y-axis. This point has the following other sets of polar coordinates: with positive r: 1, π2 + 2nπ 3π with negative r: −1, −π + 2mπ or −1, + 2`π 2 2 where n, m, and ` are any integers. Example C.77 ([7, Exercise 12, p. 677]) “Sketch the region in the plane consisting of points whose polar coordinates satisfy the conditions −1 ≤ r ≤ 1, π4 ≤ θ ≤ 3π .” 4 Solution: Let’s separate the portion of the region described by positive r, zero r, and negative r: positive r. 0 < r ≤ 1, π4 ≤ θ ≤ 3π describes a sector of the unit disk centred at the pole — all 4 points in the region bounded by two perpendicular radii bisecting the first and second quadrants. The points on the radii and the bounding circle are included. zero r. The region described by r = 0, is irrelevant for the pole. π 4 ≤θ≤ 3π 4 consists of the pole alone: the θ coordinate negative r. −1 ≤ r < 0, π4 ≤ θ ≤ 3π describes the region antipodal to the one described for 4 . positive r — between radii at angles π + π4 and π + 3π 4 Information for Students in Lecture Section 1 of MATH 141 2010 01 3169 10.3 Exercises [1, Exercise 14, p. 648] “Find a formula for the distance between the points with polar coordinates (r1 , θ1 ) and (r2 , θ2 ).” Solution: One way to solve this problem is to transform to cartesian coordinates, then apply the usual distance formula, after which cancellations will occur. A more direct way is to apply the Law of Cosines to a triangle whose vertices are the pole and the two given points. That is easy if we assume that r1 and r2 are non-negative, so they represent the lengths of the sides of the triangle. To solve the problem in this way we need to consider several cases. The result is the same: the distance is q r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) . [1, Exercise 18, p. 648] Identify the curve r = 2 cos θ + 2 sin θ. Solution: The textbook suggests finding a cartesian equation first. This is not always easy, but is not impractical in the present problem. If we multiply the given equation by r, we obtain the equation r2 − 2r cos θ − 2r sin θ = 0 which we see is equivalent to x2 + y2 − 2x − 2y = 0 i.e., to (x − 1)2 + (y − 2)2 = 2 √ which is a circle with radius 2 centred at the point (1, 1). However, we must note that, in multiplying an equation by a factor — equivalently in multiplying the two equations r = 2 cos θ + 2 sin θ r = 0 (84) (85) we were, in the second equation, possibly permitting new points to be included in the “curve”. We must carefully analyze whether the equation r = 0 did that. But we know that r = 0 represents only the pole! And the pole was already on the given curve: it appears there as ! 3π . (r, θ) = 0, 4 So multiplying by r = 0 does not introduce any new points.53 53 But what would happen if you were to multiply an equation like r = 1 by r = 0? Information for Students in Lecture Section 1 of MATH 141 2010 01 3170 [1, Exercise 30, p. 648] Describe the curve with the equation r2 − 3r + 2 = 0. Solution: The equation can be rewritten as (r − 1)(r − 2) = 0 , which is satisfied by all points whose r-coordinate is either 1 or 2, i.e., by all points on either of two concentric circles around the pole. [1, Exercise 33, p. 648] Describe the curve with the equation r = 2(1 − sin θ) (θ ≥ 0). Solution: (see Figure 8 on page 3170) Under the transformation θ 7→ π − θ the equation -2 -1 0 1 2 0 -1 -2 -3 -4 Figure 8: The cardioid with equation r = 2(1 − sin θ) is unchanged, so the curve has symmetry in the y-axis. This is a “heart-shaped” curve, Information for Students in Lecture Section 1 of MATH 141 2010 01 3171 called a cardioid. The textbook describes another example in [1, Example 9, pp. 645646], where the author shows, as we could show here, that the curve is tangent to the y-axis. We call the point of tangency (here, the pole) a cusp. (The condition θ ≥ 0 does not restrict the curve in any way, since the function 2(1 − sin θ) is periodic with period 2π: any part of the curve that might require a negative value θ0 of θ to represent it, can also be represented by the θ0 + 2πn, where n is any integer; taking n sufficiently large will make this value positive.) Information for Students in Lecture Section 1 of MATH 141 2010 01 3172 C.26 Supplementary Notes for the Lecture of March 08th, 2010 Release Date: Monday, March 08th, 2010 C.26.1 §10.3 Polar Coordinates (continued) 10.3 Exercises (conclusion) [1, Exercise 34, p. 648] Describe the curve with equation r = 1 − 3 cos θ. (see Figure 9 on page 3172) 2 1 -4 -3 -2 -1 0 0 -1 -2 Figure 9: The lima¸con r = 1 − 3 cos θ, Solution: It is not practical to represent this curve in cartesian coordinates. We note that, when θ is replaced by −θ, the equation is unchanged. This tells us that the curve is Information for Students in Lecture Section 1 of MATH 141 2010 01 3173 symmetric about the polar axis. Try tracing it out by starting with the point with θ = 0. If we superimpose a cartesian system in the usual way, that point is the point with cartesian coordinates (−2, 0). The curve passes through the pole first when θ = arccos 13 , about 70 degrees. It crosses the y-axis at the point with cartesian coordinates (x, y) = (0, 1), and then moves to its maximum distance of 1 + 3 = 4 from the pole when θ = π. Then it returns, crossing the negative y-axis at (x, y) = (0, −1), and passing again through the pole. The curve is one of the family called lima¸cons, see them sketched in [1, Example 11, p. 647]. Tangents to Polar Curves I discuss how to determine the tangents to curves of the form r = f (θ). The discussion is based on the observation that, if we superimpose polar and cartesian systems in the most usual way, we can express x and y in terms of θ by x = f (θ) · cos θ y = f (θ) · sin θ and thereby interpret θ as a parameter in the parametric representation of a curve. We find that dy dr · sin θ + r · cos θ dy = dθ = dθ . dr dx dx · cos θ − r · sin θ dθ dθ (86) In the case of the cardioid r = 2(1 − sin θ), we find that dy 2 sin θ − 1 = dx 4 cos θ when cos θ , 0. When θ = 0, at the unit point on the initial ray, the tangent has slope − 14 . As θ increases to π6 the tangent becomes vertical; then, as θ > π6 and θ → π2 − , the tangent approaches the vertical, with positive slope. Similarly, as θ → π2 + , the tangent approaches the vertical, with negative slope. We say that this curve has a cusp at the pole. [1, Exercise 35, p. 648] Describe the curve with equation r = θ, with θ ≥ 0. Solution: This curve is a spiral, turning around the pole. (see Figure 10 on page 3174) If, however, we were to ask about the curve with equation r = θ, with θ ≤ 0, it is also a spiral around the pole. (see Figure 11 on page 3175) The superposition of the two curves is shown in Figure 12 on page 3176. Information for Students in Lecture Section 1 of MATH 141 2010 01 30 20 10 -30 -20 -10 0 10 20 30 0 -10 -20 -30 Figure 10: The spiral with equation r = θ, (θ ≥ 0) 3174 Information for Students in Lecture Section 1 of MATH 141 2010 01 3175 30 20 10 -30 -20 -10 0 10 20 30 0 -10 -20 -30 Figure 11: The spiral with equation r = θ, (θ ≤ 0) Information for Students in Lecture Section 1 of MATH 141 2010 01 30 20 10 -30 -20 -10 0 10 20 30 0 -10 -20 -30 Figure 12: The full spiral with equation r = θ, −∞ < θ < +∞ 3176 Information for Students in Lecture Section 1 of MATH 141 2010 01 3177 Example C.78 ([7, Exercises 37, 40, p. 678]) Describe the curves with equations r = sin 2θ, r = sin 5θ. Solution: First let’s consider the curve r = sin 2θ. (see Figure 13 on page 3177) While the 0.6 0.4 0.2 0 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 -0.2 -0.4 -0.6 Figure 13: The “4-leafed rose” with equation r = sin 2θ, equation does not remain unchanged when we replace θ by −θ, it changes to r = − sin 2θ, which contains the same points, since it can be rewritten as −r = sin 2(θ + π), which can be obtained from the original equation by the transformation (r, θ) 7→ (−r, θ + π) . Thus this curve is symmetric about the polar axis. It is also symmetric about the y-axis, since the replacement of θ by π2 − θ leaves the equation unchanged. The curve is a “4-leafed rose”, where each petal is tangent to the cartesian axes, if located in the usual way. Information for Students in Lecture Section 1 of MATH 141 2010 01 3178 But, when the multiplier of θ is an odd integer, the situation changes. (see Figure 14 on page 3178) The curve r = sin 5θ is again a “rose”, passing through the pole every 36 degrees. 0.8 0.4 0 -0.8 -0.4 0 0.4 0.8 -0.4 -0.8 Figure 14: The “5-leafed rose” with equation r = sin 5θ, It is not symmetric about the y-axis, nor about the x-axis. (It is symmetric under reflections in other axes, and also under rotation through certain angles around the pole.) Example C.79 ([7, Exercises 42, p. 678]) Sketch the curve with the polar equation r2 = sin 2θ. Note that there are no points on this curve for π2 < θ < π2 since, in that interval, sin 2θ < 0. The entire curve is traced out for 0 ≤ θ ≤ π2 : every value of θ gives rise to two points on the graph. (see Figure 15 on page 3179) Example C.80 Where is the curve r = sin 2θ horizontal? Information for Students in Lecture Section 1 of MATH 141 2010 01 3179 0.8 0.3 0.2 0.4 0.1 0 -0.8 -0.4 0 0 0.4 0.8 -1 -0.5 0 0.5 1 -0.1 -0.4 -0.2 -0.3 -0.8 Figure 15: The lemniscates r2 = sin 2θ, r2 = cos 2θ Solution: We set dy = 0 in (86): dx dr · sin θ + r · cos θ = 0 dθ ⇔ 2 cos 2θ · sin θ + sin 2θ · cos θ = 0 ⇔ 2(1 − 2 sin2 θ) · sin θ + 2 sin θ · cos2 θ = 0 ⇔ 2 sin θ(2 − 3 sin2 θ) = 0 so the tangents will be horizontal when r sin θ = 0, ± i.e., when r θ = 0, ± arcsin C.26.2 2 , 3 2 , π ± arcsin 3 r 2 . 3 §10.4 Areas and Lengths in Polar Coordinates Areas To find the area bounded by a curve given in polar coordinates we express the area as the limit of a sum of narrow triangles whose bases are along the bounding curve, and whose upper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of the Information for Students in Lecture Section 1 of MATH 141 2010 01 curve r = f (θ) between θ = a and θ = b is then Zb a 1 · ( f (θ))2 dθ . 2 3180 Information for Students in Lecture Section 1 of MATH 141 2010 01 3181 C.27 Supplementary Notes for the Lecture of March 10th, 2010 Release Date: Wednesday, March 10th, 2010 C.27.1 §10.4 Areas and Lengths in Polar Coordinates (continued) Areas To find the area bounded by a curve given in polar coordinates we express the area as the limit of a sum of narrow triangles whose bases are along the bounding curve, and whose upper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of the curve r = f (θ) between θ = a and θ = b is then Zb a 1 · ( f (θ))2 dθ . 2 (This could be shown in two different ways: by treating the element of area as a sector of a circle, or as a narrow, isosceles triangle.) As with all other formulæ involving polar coordinates, one must use this formula with care. Be sure that you know precisely what the region looks like; in case of doubt, break the region up into parts, and find the areas of the parts separately. The limits must be chosen carefully, to be sure that, for example, you are not computing more 1 area than you intend. For example, the curve r = cos θ is a circle of radius 2 centred at the point 21 , 0 , and passing through the pole. The curve is swept out as θ ranges from 0 to π, so the area of the disk is 1 2 Zπ 0 1 cos2 θ dθ = 4 Zπ (1 + cos 2θ) dθ = π . 4 0 Compare this with the area of the disk r = 1, centred at the pole, where the curve is swept out as θ ranges from 0 to 2π; Z2π 1 Area = 1 dθ = π . 2 0 In that case, if we were to stop at π, we would obtain only the area of the upper half-disk. Rather than attempting to memorize rules about limits for integrals, I suggest you carefully analyze each problem individually. Finding the intersections of curves in polar coordinates The textbook [1, p. 651-652] discusses the difficulties that of finding intersections which are a consequence of the multiple sets of coordinates for points. Some textbooks suggest that finding the intersections can only be done visually, and that is not true. In fact, intersections can be found algebraically, but one Information for Students in Lecture Section 1 of MATH 141 2010 01 3182 must be careful and thorough. I attach below a discussion I prepared for a class some years ago, where another textbook had made a false statement that I felt obliged to correct; the error is continued in the current edition of the same textbook. Example C.81 In an example in another textbook [31, Example 8, p. 579], [30, Example 8, p. 635] the objective is to find the points where the curves r = 1 + sin θ r2 = 4 sin θ (87) (88) intersect. It was stated in the textbook solution that only one of the points of intersection can be found algebraically, and that the others can be found only “when the equations are graphed”. We show here all intersection points can be found algebraically! We never resort to calculations on a sketch: all procedures can be justified theoretically — the sketch serves only to help visualize a situation that can be adequately described verbally and/or with mathematical formulæ. (cf. Figure 16, page 3182) Had the curves been given in cartesian coordinates, we could 2 1 –1 –0.5 0 0.5 1 –1 –2 Figure 16: Intersecting polar curves r = 1 + sin θ, r2 = 4 sin θ have found all intersections by solving the equations simultaneously. Why can’t we solve the Information for Students in Lecture Section 1 of MATH 141 2010 01 3183 polar equations in the same way? The difficulty derives from the fact that any point has infinitely many different polar representations. More precisely, a point that can be represented by polar coordinates (r, θ) also has coordinates ((−1)n r, θ + nπ), where n is any integer — positive or negative; moreover, the pole can be represented by (0, θ), where θ is any real number. To determine the points of intersection, one must consider the possibility that the same point appears with different coordinates. Solve the given equations algebraically: By eliminating sin θ between the two equations, we obtain r2 = 4(r−1), which implies that (r−2)2 = 0, so r = 2, and sin θ = 2−1 = 1. Hence π π θ = +2mπ, where m is any integer, and the points of intersection are 2, + 2mπ : but, 2 2 by the convention describedabove, these are representations of the same point, whose π “simplest” representation is 2, . 2 Transform the equations in all possible ways and solve again: Apply to one of the equations the transformation (r, θ) 7→ (−r, θ + π) (89) and solve it with the original form of the other equation. Repeat this process until the equations transform to a pair already solved. Equation (87) transforms to −r = 1 + sin(θ + π) (90) which is equivalent to r = −1 + sin θ (91) √ which√equation we solve with (88). Eliminating sin θ yields r = 2 ± 2 2, so sin θ = 3 ± 2 2. √ The upper sign is inadmissible, as a sine cannot exceed 1 in magnitude. Hence r = 2 − 2 2 and √ sin θ = 3 − 2 2 . (92) √ √ The solutions to (92) are θ = sin−1 3 − 2 2 + 2mπ and θ = π − sin−1 3 − 2 2 + 2mπ; we may take m = 0, as all other values of m give the same two points: √ √ √ √ 2 − 2 2, sin−1 3 − 2 2 and 2 − 2 2, π − sin−1 3 − 2 2 . As the first coordinate in these cases is negative, we could equally well represent the points as √ √ 2 2 − 2, sin−1 3 − 2 2 + π and √ √ 2 2 − 2, − sin−1 3 − 2 2 . A second application of (89), to (90), restores the original equation; hence there are no other intersection points, except possibly the pole. Information for Students in Lecture Section 1 of MATH 141 2010 01 3184 ! 3π Check whether the pole is on both curves: On (87) the pole appears as 0, , etc.; on (88) 2 it appears as (0, 0), etc. Thus the pole is also a point of intersection. The reason we did not find it when we solved pairs of equations is that it appears on the two curves only with different sets of coordinates, no two related by (89). Example C.82 [7, Exercise 29, p., 683] “Find the area of the region that lies inside both curves: r = sin θ, r = cos θ.” Solution: (see Figure 17 on page 3184) The curves are circles through the pole. To see this, 1.0 0.75 0.5 0.25 0.0 −0.5 −0.25 0.0 0.25 0.5 0.75 1.0 −0.25 −0.5 Figure 17: Curves r = sin θ, r = cos θ multiply the first by r = 0 (which, of course, could be bringing the pole into the curve). The resulting equation is r2 = r sin θ, which, in cartesian coordinates, would be x2 + y2 = y, a Information for Students in Lecture Section 1 of MATH 141 2010 01 3185 circle with centre (in cartesian coordinates) 0, 12 and radius 12 . The operation of multiplying by r = 0 did not, however, alter the curve, since the pole was already on the curve, with polar coordinates (r, θ) = (0, 0). In the same way we can show that the second curve is a circle of 1 the same radius centred at the point (r, θ) = 2 , 0 ; polar coordinates for the centre of the first circle are, for example, 12 , π4 . Where do the curves meet? (This was [7, Exercise 37, p. 683], which should have preceded the present problem in the exercises!) We can begin this investigation by solving their equations, which imply that tan θ = 1, so θ = π4 + nπ, where n is any integer; that implies that r = ± √12 : the + sign is associated with the cases where n is even, the − sign with those where n is odd. But (r, θ) = √12 , π4 + 2mπ and (r, θ) = − √12 , π4 + (2m + 1)π are all the same point — having cartesian coordinates 12 , 12 . But the curves also meet at the pole, even though this did not show up when we solved the equations. This is because the pole appears on the first curve with θ = nπ, and on the second curve with θ = (n + 21 )π. The only practical way to determine whether curves intersect at the pole is to examine each of them separately for that point. Could it be that the curves meet in any other points? While a sketch does not suggest that, we should never rely on a sketch! If we transform the first equation under the transformation (r, θ) 7→ (−r, θ + π), we find that the equation does not change (except for a sign change on both sides); the same applies to the second equation. Thus, in the present example, there are no intersections other than the pole in which a point will appear on the two curves with different coordinates. We can comfortably proceed to the integration part of the problem, confident that we have not missed any points of intersection. We can see that the region whose area is sought is symmetric about the line θ = π4 , so it suffices to find half of it and to double it. The lower half of the area is subtended at the pole by the arc 0 ≤ θ ≤ π4 of the circle r = sin θ; hence π Area 1 2 = 2· Z4 sin2 θ dθ 0 1 = 2 π 4 Z (1 − cos 2θ) dθ 0 " 1 sin 2θ = θ− 2 2 # π4 = 0 π 1 − . 8 4 (You could have solved this problem using Cartesian coordinates.) 10.4 Exercises Information for Students in Lecture Section 1 of MATH 141 2010 01 3186 [1, Exercise 22, p. 653] “Find the area enclosed by the loop of the strophoid r = 2 cos θ − sec θ.” Solution: (see Figure 18 on page 3186) (The textbook did not expect a rigorous proof 0.6 0.4 0.2 0 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -0.2 -0.4 -0.6 Figure 18: The strophoid r = 2 cos θ − sec θ that the curve crosses itself only at the pole.) First we observe that the curve is entirely traced out when θ passes through an interval of length 2π; so, without limiting generality, we may confine ourselves to such an interval, say − π2 < θ < π2 , and thereby remove the some possible ambiguity of coordinates. We must exclude the points where sec θ is undefined. For convenience, let’s confine θ to the union of intervals − π π <θ< 2 2 and π 3π <θ< . 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3187 Consider a point on the second branch, say with θ = φ + π, where − π2 < φ < π2 . We find that, for this point, r = 2 cos(φ + π) − sec(φ + π) = −2 cos φ + sec φ = −(2 cos φ − sec φ) So we see that the point is the same as the point (−r, φ). Thus, in order to see the whole curve, it suffices to investigate angles θ between − π2 and π2 ; some of the points will, however, appear with negative r-values. We observe that the replacement of θ by −θ does not change the equation: this tells us that the curve is symmetric about the polar axis and its extension. For the problem to be meaningful, there should be just one loop. The curve passes through the pole when 0 = 2 cos θ−sec θ, equivalently, when cos θ = ± √12 . In the interval to which we have confined θ, this will occur when θ = ± π4 . If we follow the curve as θ comes from − π2 , we find that r is negative, and the point is in the 2nd quadrant. It passes through the pole first when θ = − π4 , and is then in the fourth quadrant, striking the polar axis when θ = 0, at the point (r, θ) = (1, 0) and it then moves into the first quadrant, following the mirror image of the portion in the fourth quadrant, passing through the pole again when θ = π4 , and moving into the 3rd quadrant. If we consider the cartesian coordinates of the curve in parametric form, we have x = 2 cos2 θ − 1 y = sin 2θ − tan θ and see that, as cos θ → 0, x → −1, and y → ±∞, so the curve is asymptotic to the vertical line x = −1. Our present problem is to determine the area of the loop from θ = − π4 to θ = π4 , or, by symmetry, π 2 1 2 π ! Z4 Z4 2 (2 cos θ − sec θ) dθ = 4 cos2 θ + sec2 θ − 4 dθ 0 0 π Z4 = 2 cos 2θ + sec2 θ − 2 dθ 0 π = [sin 2θ + tan θ − 2θ]04 π π = 1+1− −0=2− . 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3188 [1, Exercise 26, p. 653] Find the area of the region that lies inside the curve r = 2 + sin θ (93) r = 3 sin θ . (94) and outside the curve Solution: (see Figure 19 on page 3188) The first step is to determine where the curves 3 2 1 0 -2 -1 0 1 2 -1 Figure 19: Curves r = 2 + sin θ, r = 3 sin θ intersect. We can start by solving equations (93) and (94) algebraically. We find that r = 3 and θ = π2 +2nπ,where n is any integer. The second curve is a circle with centre at the point (r, θ) = 32 , π2 . The first curve appears to be some sort of oval, and touches the Information for Students in Lecture Section 1 of MATH 141 2010 01 3189 circle at its topmost point, which we shall call A, with coordinates (r, θ) = 3, π2 . Could there be any other points of intersection? There don’t appear to be any, from the sketch, but we can resolve this question algebraically. The pole does not lie on first curve, since that would entail that 0 = 2 + sin θ, i.e., that the sine of an angle exceeds 1 in magnitude. Since that is not possible, the pole cannot lie on the curve. If there were to be any other intersections, they would have to have different sets of coordinates on the 2 curves. We investigate what happens to (94) under the transformation (r, θ) 7→ (−r, θ + π) and find that the equation does not change significantly: it becomes −r = sin(θ + π) = − sin θ, which is evidently equivalent to the original equation. Equation (93) changes to −r = 2 + sin(θ + π) = 2 − sin θ, or r = −2 + sin θ. (95) When we solve this equation with (94), we find that r = −3, sin θ = −1: in the interval 3π 0 ≤ θ ≤ 2π we find only the point (r, θ) = −3, 2 , which is the same point as (r, θ) = 3, π2 , so we don’t find any new intersections. Thus there are no other intersections than A. A direct way to solve this problem is to find the area of the region bounded by the outer curve, and then subtract from it the area of the disk inside. The outer curve is traced out as θ ranges from 0 to 2π, so the area of the region it bounds will be 1 2 Z2π Z2π 1 2 (2 + sin θ) dθ = (4 + 4 sin θ + sin2 θ) dθ 2 0 0 ! 1 − cos 2θ 4 + 4 sin θ + dθ 2 0 " #2π 1 9θ 1 9π = − 4 cos θ − sin 2θ = 2 2 4 2 0 1 = 2 Z2π The inner curve bounds an area of 1 2 Zπ 9 9 sin θ dθ = 4 Zπ 2 0 (1 − cos 2θ) dθ 0 #π " 9 9 1 = θ − sin 2θ = · π . 4 2 4 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3190 What would have happened if we had taken the upper limit of integration to be 2π? We would have found twice the area, because the entire circle is swept out as θ ranges over an interval of length π. Thus we see that the area of the region between the curves is 9π 9π 9π − = . 2 4 4 Could we find the area by taking the difference of two squares under the integral sign, as is suggested in [1, Example 2 and Figures 5 and 6, p. 651]? We could do this for the portion of the area above the polar axis and its extension, where we would obtain 1 2 Zπ (2 + sin θ)2 − (3 sin θ)2 dθ 0 1 = 2 Zπ (4 + 4 sin θ − 4(1 − cos 2θ)) dθ 0 1 = [−4 cos θ + 2 sin 2θ]π0 = (2 + 0) − (−2 + 0) = 4 2 For the remainder of the area only curve (93) and the extended polar axis serve as a boundary: 1 2 " Z2π 1 9θ 2 (2 + sin θ) dθ = − 4 cos θ − 2 2 π " 1 (9π − 4 − 0) − = 2 1 sin 2θ 4 #2π π !# 9π 9π − (−4) − 0 = −4 2 2 which, when added to the area of the upper portion, gives the same area as before. What would have happened if we had subtracted the square of 3 sin θ over the full range of the integration? We would have been subtracting the area of the inside disk twice! And what would have happened if we had taken (95) as the equation of the outer curve? If we used it only to find the area of the (larger) region bounded, we would obtain the correct value there. But, if we had taken the difference of squares under the integral sign, we would obtain values for regions that do not correspond to the one whose are we are trying to find. Before finding an area in polar coordinates, (or, indeed, in cartesian coordinates as well), you are urged to make a sketch and study the element of area that you are summing in the limit, to ensure that the integral represents the area that you are seeking. Information for Students in Lecture Section 1 of MATH 141 2010 01 3191 Example C.83 Another example of finding the intersections of curves. Find the intersections of the lima¸con r = 1 − 2 cos θ and the circle r = 1. [29] Solution: (see Figure 20 on page 3191) When we solve the given equations, we find the inter- 1.5 1 0.5 0 -3 -2 -1 0 1 -0.5 -1 -1.5 Figure 20: Intersections of the limac¸on r = 1 − 2 cos θ with the circle r = 1 section points (r, θ) for which cos θ = 0, i.e., 1, ± π2 . But a glance at the graphs shows that there appears to be a third point of intersection. Under the transformation (r, θ) → (−r, θ + π), the given equation for the circle r = 1 does not change in a significant way. However, the equation of the lima¸con changes to r = −1 − 2 cos θ; when we solve this latter equation with the equation r = 1, we find that cos θ = −1, so the point (r, θ) = (1, π) is another point of intersection: i.e., (1, π) on the circle, (−1, 0) on the original equation for the limac¸on. Information for Students in Lecture Section 1 of MATH 141 2010 01 3192 To finish this investigation we should investigate whether the pole is on both curves, since that may not be detected by this method of transforming the equations. We note, however, that the curve r = 1 cannot possibly contain the pole, since all of its points are 1 unit from the pole. Thus the investigation is complete: there are three point of intersection, and we have now found them all. Example C.84 A problem from a recent examination: [12 MARKS] Use polar coordinates — no other method will be accepted — to 1 find the area of the region bounded by the curve r = 2 and the line r = , and cos θ containing the pole. Solution: (see Figure 21 on page 3193) The first step is to determine where the two curves intersect. While the points of intersection can all be found algebraically, that procedure is tedious, so it is best to make a sketch first to see whether the elaborate procedure is required. You should be able to see immediately that the curve r = 2 is a circle of radius 2 centred at the pole. But what about r = sec θ? Since the secant is never less than 1 in magnitude, you know this curve does not pass through the pole. You also know that, as θ ranges between − π2 and π2 , the function ranges from −∞ to +∞ monotonely, so the curve will be closest to the pole when θ = 0 — when it is 1 unit from the pole. As θ increases to + π2 , r increases steadily — monotonely — the curve crossing the circle r = 2 and moving a greater and greater distance from the pole. What happens when r < 0? A point with coordinates (−a, θ), where a > 0 will still be a distance a from the pole! (It’s located on the ray obtained by turning the polar axis through an angle of θ + π.) So, as r → −∞, the curve again crosses each of the circles centred at the pole and moves a greater and distance from the pole. This is all you need to know in order to solve the problem, but you should be able to see that this is a very simple curve: just 1 multiply both sides of r = by cos θ, and you see that the equation is r cos θ = 1, or, in cos θ cartesian coordinates, x = 1 — the curve is a straight line perpendicular to the polar axis! We have seen that there will be just 2 points of intersection of the curves. Solving their equations by eliminating r we obtain cos θ = 12 , so θ = ± π3 : the points of intersection are π (r, θ) = 2, ± . 3 We saw that the element of area in the form of a thin triangle with apex at the pole (or a thin sector of a disk with centre at the pole) has area of the form 12 r2 dθ. In this problem such an element would have to be described in two different ways: π π − ≤ θ ≤ : The area of this triangle is 3 3 π Z3 √ π 1 1 1 3 dθ = [tan θ] 3. π = −3 2 cos2 θ 2 − π3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3193 2 1 0 -2 0 -1 1 2 -1 -2 Figure 21: Intersections of the curve r = sec θ with the circle r = 1 π 5π ≤θ≤ : The area of this portion of the disk is 3 3 5π 1 2 Z3 π 3 5π 8π 1 . 22 dθ = [4θ] π3 = 3 2 3 Thus the area of the region determined by the two curves, and containing the pole, is √ 8π 3+ . 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3194 This problem could also have been solved by subtracting from the area of the disk, π22 , the area given by π Z3 π 1 1 4π √ (22 − sec2 θ) dθ = [4θ − tan θ]−3 π = − 3. 3 2 2 3 − π3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3195 C.28 Supplementary Notes for the Lecture of March 12th, 2010 Release Date: Friday, March 12th, 2010 C.28.1 §10.4 Areas and Lengths in Polar Coordinates (conclusion) Arc Length To develop a formula for the arc length of a curve given in polar coordinates as r = f (θ), we can apply the theory of curves given in parametric form to the curve x = f (θ) · cos θ y = f (θ) · sin θ We find that dx dθ so the length is given by the integral !2 dy + dθ Zb !2 dr = dθ s dr dθ L= a !2 + r2 !2 + r2 dθ where the limits θ = a and θ = b need to be determined from the parametrization of the curve. As usual, you should be careful to determine the appropriate values of θ to define the portion of the curve that interest you. 10.4 Exercises (continued) [1, Exercise 46, p. 654] Find the exact length of the arc of the polar curve r = e2θ from θ = 0 to θ = 2π. Solution: Z2π Length s dr dθ = !2 + r2 dθ 0 Z2π q = 2e2θ 2 + e2θ r2 dθ 0 √ Z 2θ = 5 e dθ 2π √ = 0 5h 2 e2θ i2π 0 = √ 4π 5 e −1 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3196 [1, Exercise 54, p. 654] Graph the curve r = cos2 2θ , and find its length. Solution: The equation can be simplified to read 1 + cos 2 · 2θ r= 2 which we recognize to be a cardioid. As θ ranges over the interval 0 ≤ θ ≤ 2π the entire curve is traced out once. s !2 Z2π dr Length = + r2 dθ dθ 0 1 = 2 = 1 2 1 = 2 Z2π p 0 Z2π 0 Z2π p (− sin θ)2 + (1 + cos θ)2 dθ 2(1 + cos θ) dθ r θ 2 2 cos2 dθ 2 0 2π Z θ cos dθ 2 = 0 θ π θ 2π − 2 sin 2 0 2 π = 2 − (−2) = 4 . = C.28.2 2 sin §10.5 Conic Sections Omit this section. Textbook Chapter 11. INFINITE SEQUENCES AND SERIES. Information for Students in Lecture Section 1 of MATH 141 2010 01 C.28.3 3197 §11.1 Sequences A sequence is an ordered set of objects, usually labelled with either the non-negative integers 0, 1, . . . , n, . . . or the positive integers 1, 2, . . . , n, . . . . In this chapter we shall be interested in sequences of real numbers or of real functions. Technically, such a sequence is a function that maps either the non-negative integers or the positive integers on to the set of real numbers (or to the set of real functions). We usually denote a sequence by a single letter, e.g. a, and show the labelling by either a parenthesized value, as a(n), or a subscripted values, as an ; when we wish to talk about a sequence with specific terms, we may just list the terms with a general term that shows the pattern we are describing, if there is one, as in 1, 1, 2, 3, 5, 8, 13, 21, 24, 55, 89, . . . which is the Fibonacci sequence, in which every entry past the second is the sum of the two entries immediately preceding it. Unless the general term is described unambiguously (as here, where Fn+2 = Fn+1 + Fn for all n ≥ 3), there will be more than one way to generalize a pattern that appears to hold between a few terms at the beginning of the sequence. Notation When a sequence consists of terms a0 , a1 , a2 , . . . we may speak of the sequence {an }, or perhaps the sequence {an }n=0,1,... , or sometimes simply the sequence an . Other notations are also possible, and should be understandable from the context. The limit of a sequence We define lim an = L in a way analogous to the definition of n→∞ lim f (x) = L; another way of writing the same definition is x→∞ an → L as n → ∞. The precise definition is to be found as [1, Definition 2, p. 677], but is not on the syllabus — you are not expected to be able to work with the precise definition of the limit of a sequence. Note that we usually do not write the braces { and } when we speak of the limit of a sequence. Since you bring to this course some understanding of limits of functions, we will occasionally appeal to the following 11.1 Exercises ( ) ln n [1, Exercise 32, p. 685] Determine whether the sequence converges. ln 2n Solution: ln n + ln 2 − ln 2 ln n = ln 2n ln n + ln 2 ln 2 = 1− ln 2n Information for Students in Lecture Section 1 of MATH 141 2010 01 3198 The subtracted fraction has a constant numerator, but the denominator is increasing, so the fraction is decreasing as n increases; hence 1 minus the fraction is increasing; in fact, we can see that the subtracted fraction is approaching 0, so the given fraction is approaching 1. C.28.4 Sketch of Solutions to Problems on the Final Examination in MATH 141 2005 01 (not for discussion at the lecture) The best way to prepare for the examination in MATH 141 2010 01 is to study the textbook, working problems, and verifying your solutions using the Student Solutions Manual: for several reasons I don’t recommend your studying exclusively from old examinations. However the old examinations do have some uses, and I am including one of them here, with solutions, as a timely reminder that less than 1 month remains between now and the final examination in this course. I am hoping to discuss last April’s examination briefly in a final lecture; there will not be time for me to discuss the following examination in the lectures. 1. SHOW ALL YOUR WORK! Z 3 (a) [4 MARKS] Evaluate |x − 1| dx . 0 Solution: Z Z 3 |x − 1| dx = 0 Z 1 3 |x − 1| dx + Z 0 |x − 1| dx 1 1 Z 3 = (1 − x) dx + (x − 1) dx 0 1 " #1 " 2 #3 x2 x = x− + −x 2 0 2 1 ! !! 1 3 1 5 = −0 + − − = 2 2 2 2 Z 5√ d (b) [3 MARKS] Evaluate 4 + t2 dt . dx x Z 5√ Z x√ √ d d Solution: 4 + t2 dt = − 4 + t2 dt = − 4 + x2 . dx x dx 5 Information for Students in Lecture Section 1 of MATH 141 2010 01 3199 Z x2 d (c) [3 MARKS] Evaluate sec t dt . dx 2π Solution: Let u = x2 . By the Chain Rule Z x2 Z u d d du sec t dt = sec t dt · = sec u · 2x = sec(x2 ) · 2x . dx 2π du 2π dx Z √3 (d) [3 MARKS] Evaluate x5 x3 + 1 dx . 1 Solution: Let u = x3 + 1 3 , so u3 = x3 + 1, 3u2 du = 3x2 dx. Z x5 √3 37 43 Z 3 3 7 4 x + 1 x + 1 u u x3 + 1 dx = u3 − 1 ·u·u2 du = − +C = − +C 7 4 7 4 2. SHOW ALL YOUR WORK! For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X 2 − cos n n=2 n 1 2 − cos n Solution: This is a positive series. Since 0 < ≤ , for all n, the terms n n are bounded below by the terms of the harmonic series, a positive series known to diverge. We may apply the Comparison Test to such pairs of series, and may conclude that the given series also diverges. ∞ X n(−3)n (b) [4 MARKS] 4n n=0 Solution: As formulated in your textbook, we may apply the Ratio Test to the sequence of absolute values of ratios of terms to their predecessors. Here (n+1)(−3)n+1 ! 4n+1 1 3 3 lim n(−3)n = lim 1 + · = < 1, n→∞ n 4 4 n→∞ n 4 from which we may conclude that the given series is (absolutely) convergent. Information for Students in Lecture Section 1 of MATH 141 2010 01 (c) [4 MARKS] ∞ X n=2 3200 1 n ln n Solution: Define f (x) = x ln1 x . Then f is a positive, continuous function, taking on the values of the given sequence at the positive integer points. By the Integral Test, the series and the following improper integral will either both converge or both diverge. Z ∞ Z b dx dx = lim = lim [ln ln x]b2 = lim ln ln b − ln ln 2 b→∞ b→∞ b→∞ x ln x 2 2 x ln x which approaches +∞ as n → ∞. Hence the series diverges (to ∞). 3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral or a sum, product, or quotient of several definite integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems: (a) [6 MARKS] The area of the region bounded by the parabola y = x2 , the x-axis, and the tangent to the parabola at the point (1, 1). Solution: The tangent to the parabola at the point x, x2 has slope 2x; at (1, 1) the slope is 2. The equation of the tangent is y = 1 + 2(x − 1) = 2x − 1, which line intercepts the x-axis in the point 12 , 0 . If we evaluate the area by integration with respect to x, we have Z 1 2 0 2 x dx + Z 1 1 2 " 3 # 12 " #1 x (x − 1)3 1 1 x − (2x − 1) dx = + = + 3 0 3 24 24 1 2 2 1 = 12 We can also integrate with respect to y: ! " #1 Z 1 2 32 y2 y 2 1 1 1 y+1 √ − y+ dy = − y + + =− + + = . 2 3 4 2 0 3 4 2 12 0 (Note that you were not expected to evaluate the integrals: I did so in order to use the opportunity of having 2 methods in order to verify my work.) (b) [3 MARKS] The volume of the solid obtained p by rotating about the line y = 4 the region bounded by x = 0 and the curve x = sin y (0 ≤ y ≤ π). Solution: This problem is from your textbook [7, Exercise 25, p. 459], and is solved in the Student Solution Manual, [9, p. 269]; it is one of the problems for which the CD-Roms accompanying the textbook provide extra help. Information for Students in Lecture Section 1 of MATH 141 2010 01 3201 Integrating by the method Shells centred on the line y = 4, we obtain Z π of Cylindrical p the volume to be 2π (4 − y) sin y dy. 0 (c) [3 MARKS] The area of the surface obtained by revolving about the y-axis the curve y = e x , 1 ≤ y ≤ 2. Z ln 2 √ 2πx 1 + e2x dx. InteSolution: Integrating with respect to x gives the integral 0 s Z 2 1 grating with respect to y gives the integral 2π(ln y) 1 + 2 dy. y 1 2x over the interval 0 ≤ x ≤ (1 + x2 )2 2. R 2 2x dx R 2 0 (1+x2 )2 2x Solution: The answer expected was either 12 0 (1+x dx or . The inteR2 2 )2 1 dx 0 gral can be evaluated by using substitutions like u = x2 , or u = x2 + 1. (d) [2 MARKS] The average value of the function 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z 2x3 + 3x2 + 3 dx . x2 + x − 12 Solution: Since the degree of the numerator of the rational function which is the integrand is not less than the degree of the denominator, the first step is to divide denominator into numerator, obtaining 2x3 + 3x2 + 3 23x + 15 = (2x + 1) + 2 . 2 x + x − 12 x + x − 12 R The polynomial quotient may be integrated to (2x + 1) dx = x2 + x + C. The excess must be expanded using the method of Partial Fractions. Assuming an expansion of the form A B 23x + 15 = + , (x + 4)(x − 3) x + 4 x − 3 we multiply both sides by (x+4)(x−3) to obtain the identity 23x+15 = A(x−3)+B(x+4). We may obtain 2 equations for A, B by assigning to x the successive values x = 3 and x = 4, obtaining B = 12 and A = 11. Hence ! Z Z 2x3 + 3x2 + 3 11 12 dx = 2x + 1 + + x2 + x − 12 x+4 x−3 = x2 + x + 11 ln |x + 4| − 12 ln |x − 3| + C , Information for Students in Lecture Section 1 of MATH 141 2010 01 3202 which I then checked by differentiation. 5. SHOW ALL YOUR WORK! (a) [9 MARKS] Use m ≥ 2, Z integration by parts to prove that, for Z 1 m−1 m m−1 cosm−2 x dx cos x dx = cos x · sin x + m m integers Solution: (The instructions asked that the problem be solved using Integration by Parts. Otherwise, the reduction formula could also have been proved by differentiating both sides and showing that the same derivative was obtained.) Let u = cosm−1 x, dv = cos(x) · dx, so du = (m − 1) cosm−2 x · (− sin x) dx and v = sin x. Then Z Z m m−1 cos x dx = cos x · sin x + sin2 x(m − 1) cosm−2 x dx Z m−1 = cos x · sin x + (m − 1) cosm−2 x − cosm x dx Z m Z m−1 ⇒m x dx = cos x · sin x + (m − 1) cosm−2 x dx Z Z 1 m−1 m m−1 ⇒ cos x dx = cos x · sin x + cosm−2 x dx . m m (b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate Z π2 cos6 x dx. 0 Solution: 6 cos x dx = = = = Z 1 5 5 cos x · sin x + cos4 x dx 6 6 ! Z 1 5 1 3 5 3 2 cos x · sin x + cos x · sin x + cos x dx 6 6 4 4 ! ! Z 5 5 1 1 1 5 3 cos x + cos x sin x + cos x · sin x + dx 6 24 8 2 2 ! 1 5 5 5 5 3 cos x + cos x + cos x sin x + x + C . 6 24 16 16 Evaluating between 0 and 5π 5π π we obtain 0 + = . 2 32 32 Information for Students in Lecture Section 1 of MATH 141 2010 01 3203 6. SHOW ALL YOUR WORK! Consider the curve C defined by x = 2 cos t − cos 2t y = 2 sin t − sin 2t . (a) [8 MARKS] Determine the points where the arc of the curve given by 7π π ≤t≤ 4 4 has a vertical tangent. Solution: dx = −2 sin t + 2 sin 2t dt dy = 2 cos t − 2 cos 2t dt Tangents are vertical where dy dx is infinite, i.e., where dx dt = 0 but dy dt , 0. dx = 0 ⇔ −2 sin t + 4 sin t · cos t = 0 dt ! 1 ⇔ sin t cos t − =0 2 1 ⇔ sin t = 0 or cos t = 2 π 5π ⇔ t = π, , 3 3 π 7π for t restricted to lie in the interval ≤ t ≤ . We check the values of dy at these dt 4 4 three points, and find that the respective values are −4, 2, 2, none of which is 0. Hence the tangents will be vertical at the following points: t=π π t= 3 t= 5π 3 (x, y) = (−3, 0) √ 3 3 (x, y) = , 2 2 √ 3 − 3 (x, y) = , − 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3204 (b) [4 MARKS] Determine the length of the arc of the curve given by 0 ≤ t ≤ 2π . Solution: dx dt !2 dy + dt !2 = (−2 sin t + 2 sin 2t)2 + (2 cos t − 2 cos 2t)2 = 4 + 4 − 8(sin t · sin 2t + cos t · cos 2t) = 8 − 8 cos(2t − t) = 8(1 − cos t) . The length of the arc is r √ Z 2π √ √ Z 2π t 8 1 − cos t dt = 8 2 sin2 dt 2 0 0 Z 2π sin t dt = 4 2 0 Z 2π t = 4 sin dt 2 0 t 2π = −8(−1 − 1) = 16 . = −8 cos 2 0 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Determine whether the following integral is convergent; if it is convergent, determine its value: Z 1 dx √ −1 1 − x2 Solution: The integral is improper because the integrand is not defined — becomes infinite — at both ends of the interval of integration. According to the definition, we must integrate between points away from ±1, and allow the limit to be taken independently. The safest way to do that is to split the improper integral into two, and then to evaluate the two of them separately. Z b Z 0 Z 1 dx dx dx = lim+ + lim− √ √ √ a→−1 a −1 1 − x2 1 − x2 b→+1 0 1 − x2 = lim+ (arcsin 0 − arcsin a) + lim− (arcsin b − arcsin 0) a→−1 b→+1 π π − 0 = π. = 0− − + 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3205 (b) [5 MARKS] Determine whether the following series is conditionally convergent, absolutely convergent, or divergent. ∞ X (−1)n n=1 n! nn Solution: I will apply the Ratio Test to the sequence of absolute values of the ratios !−n (−1)n+1 (n+1)! 1 (n + 1)! nn 1 (n+1)n+1 = of terms to their predecessors: → · = 1+ n! n+1 n n! n e (−1) (n + 1) nn as n → ∞. As this limit is less than 1, the given series is absolutely convergent, hence divergent. (c) [3 MARKS] Determine whether the sequence an = ln(n + 1) − ln n is convergent; if it is convergent, carefully determine its limit. Solution: This is not a question about telescoping series! It is a problem about sequences! ! 1 ln(n + 1) − ln n = ln 1 + . As n → ∞, 1 + 1n → 1, so ln 1 + n1 → ln 1 = 0 (by n continuity of the logarithm function from the right at the point 1). 8. SHOW ALL YOUR WORK! [12 MARKS] Find the area of the region bounded by the curves r = 4 + 4 sin θ r sin θ = 3 which does not contain the pole. Solution: The problem is to determine the area cut off from the cardioid r = 4 + 4 sin θ by the line y = 3. First we must determine where the curves cross. Solving the polar equations by eliminating sin θ between them we obtain r2 − 4r − 12 = 0, equivalent to (r − 6)(r + 2) = 0. The values r = 6, −2 yield corresponding values sin θ = 12 , − 32 . Of these the second is impossible, as a sine cannot be less than 1. We conclude that sin θ = 12 and 0 ≤ θ ≤ 2π are θ = π6 , 5π : the curves intersect 6 the values of θ in the interval π 5π in (r, θ) = 6, 6 and (r, θ) = 6, 6 . A naive way to solve the problem is to integrate directly, Z 5π6 1 Area = (4 + 4 sin θ)2 − (3 csc θ)2 dθ 2 π6 ! ! Z 5π6 1 − cos 2θ 1 2 + 2 sin θ + 1 − 9 csc θ dθ 16 = 2 π6 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3206 5π 1 [24θ − 32 cos θ − 4 sin 2θ + 9 cot θ] π6 6 2 √ = 8π + 12 3 . = Another way to find the area is to observe that the triangle with base √ on y = 3√between the points of intersection and its third vertex at the pole has area 12 (6 3)·3 = 9 3. Then Z 5π6 1 (4 + 4 sin θ)2 dθ. one can simply subtract this area from π 2 6 Information for Students in Lecture Section 1 of MATH 141 2010 01 3207 C.29 Supplementary Notes for the Lecture of March 15th, 2010 Release Date: Monday, March 15th, 2010 C.29.1 §11.1 Sequences (conclusion) Theorem C.85 [1, Theorem 3, p. 678] If f is a function, and {an } is such that f (n) = an for all n, and if lim f (x) = L, then lim an = L. x→∞ n→∞ We also generalize our definition of the limit of a sequence to permit limits to be ±∞; the definition is again analogous to the corresponding definition for the meaning of lim f (x) = x→∞ ±∞. A sequence that has limit L is said to converge to L; if there is no limit, the sequence is said to diverge. We do not permit L to be ±∞ in this usage, so a sequence whose limit is ±∞ is said to diverge. Limit Laws We can prove limit laws for sequences analogous to the limit laws we saw in the previous course for functions. I shall not give the details in these notes. Remember the usual restriction that we cannot divide by 0, so there must be a restriction on the quotient law. The limit laws can be extended to infinite limits whenever the operations can be justified; so, we can think of ∞ + ∞ = ∞, but we cannot attach a meaning to ∞ + (−∞), nor of 0 · ∞, as it is not possible to assign to these expressions a meaning that will be consistent with the algebraic operations on real numbers. Increasing and Decreasing Sequences We defined the concepts of increasing and decreasing in connection with real functions of a real variable (cf. [1, p. 20]. The identical definitions apply for sequences, where we consider the domain of the function to be the positive or non-negative integers. When a sequence is either increasing or decreasing we say that it is monotonic or monotone. Sometimes we find it convenient to work with a slightly weaker property than increasing, and may speak of a sequence as being non-decreasing, which permits the function to remain constant for a while. A function is bounded above if there exists a number M which is greater than all values of the function. For example, the sine function is bounded above, since sin x ≤ 1 for all x in its domain. We could also have observed that sin x < 10000, and conclude that such a statement justifies the conclusion that the function is bounded above: we don’t care how high above the graph of the function the bounding line appears, only that such a line exists. But the function tan x is not bounded above. In the same way we can define bounded below and bounded — meaning bounded both above and below. An important theorem we shall need in this chapter is Information for Students in Lecture Section 1 of MATH 141 2010 01 3208 Theorem C.86 1. A sequence that is bounded above and monotonely increasing (or even monotonely non-decreasing) is convergent. 2. A sequence that is bounded below and monotonely decreasing (or even monotonely nonincreasing) is convergent. (Note that the theorem stated here is stronger than [1, Monotonic Sequence Theorem, p., 683].) Example C.87 [1, Exercise 36, p. 685] Determine whether the sequence an = ln(n + 1) − ln n converges. Solution: We might be tempted to analyze a difference by using the difference law for limits. But we find that each of the terms approaches +∞, and we cannot give a meaning to ∞ − ∞. So that approach will not work. However, if we observe that the difference of logarithms is ! n+1 1 ln = ln 1 + n n we can reason as follows. As n → ∞, n1 → 0, 1 + 1n → 1 + 0 = 1. By continuity of the logarithm, the sequence will approach ln 1 = 0. (I am using Theorem C.85 above.) Sequences of powers For a fixed real number a we know the behavior of a function a x as x → ∞, and how it depends on a; this permits us to study the behavior of an when the exponent is restricted to integer values: lim an = +∞ if a > 1 lim an =1 if a = 1 lim an =0 if −1 < a < 1 n→∞ n→∞ n→∞ lim an does not exist if a ≤ −1 n→∞ (Note that, when a limit is ±∞, as in the first case above, we still say that the limit does not exist!) n! Example C.88 Consider the sequence {an }, where an = n . Does it converge? If so, to what n value? Solution: Consider the limit of the ratio of an+1 to an . It is nn 1 1 an+1 (n + 1)! = · = → n an n! (n + 1)n+1 e 1 + 1n as n → ∞. Thus, for sufficiently large n, every term is less than half of the one before it, which 1 of the size; the is positive. Compared to the nth term, the (n + 10)th will be less than 1000 n + 20th will be less than one-millionth the size, etc. The sequence is thus approaching 0. Information for Students in Lecture Section 1 of MATH 141 2010 01 3209 Example C.89 ([7, Exercise 57, p. 711]) Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? an = cos nπ 2 Solution: The values of this sequence are (starting at n = 0) 1, 0, −1, 0, 1, 0, −1, 0, . . .. The sequence is neither increasing nor decreasing, so it is not monotonic, and the [1, Monotonic Sequence Theorem, p. 683] is not applicable. The sequence is bounded. It does not converge — but that is not a consequence of the fact that the sequence does not satisfy the conditions of cos nπ 2 the theorem. (Compare with the sequence which converges to 0.) n 11.1 Exercises (continued) [1, Exercise 52, p. 685] Investigate the convergence or divergence of the sequence an = 1 · 3 · 5 · . . . · (2n − 1) n! Solution: 1 · 3 · 5 · . . . · (2n − 1) n! 2 · 4 · . . . · (2n − 2) ≥ 1· n! 2n−1 = n an = We can determine the limit of this ratio by L’Hospital’s Rule: 2n−1 2n−1 · ln 2 ln 2 n = lim = lim · 2 = +∞ n→∞ n n→∞ n→∞ 2 1 lim since it is a non-zero multiple of the sequence of powers of a constant whose magnitude exceeds 1; thus the original sequence diverges. Example C.90 For each of the following sequences, determine whether it converges; if it does, find the limit. sinh x cosh x r 3 + sin n 2. an = n 1. an = Information for Students in Lecture Section 1 of MATH 141 2010 01 3210 Solution: 1. sinh x e x − e−x = lim x n→∞ cosh x n→∞ e + e−x 1 − e−2x = lim n→∞ 1 + e−2x lim 1 − e−2x n→∞ = lim 1 + e−2x lim n→∞ 1 = =1 1 2. Consider the fraction under the radical sign: the numerator is bounded, while the denominator approaches infinity, so the fraction approaches 0. By a familiar result about the composition of continuous functions, the square root approaches 0. C.29.2 §11.2 Series The sum of the terms in a sequence. Having defined what we mean by convergence of a sequence {an } — analogous to the existence, for a function a(x), of lim a(x) — we proceed x→∞ to apply that concept to generalize the concept of addition. To a mathematician, the operation of addition is a binary operation on numbers, that is, it maps two real numbers on to one real number. This operation has a number of properties that mathematicians take as primitive: • it is commutative: x + y = y + x for all real numbers x and y; • it is associative: (x + y) + z = (x + y) + z for all real numbers x, y, z; • it has an “additive identity”, which we call zero and denote by 0, with the property that x + 0 = x for every real number x; and other properties that we will not discuss here. The second property permits us to talk of more general sums than of just 2 numbers. If we have three real numbers x, y, z, then the second property permits us to define x + y + z to be the common value of (x + y) + z and (x + y) + z; we can generalize this idea to the sum of x1 + x2 + . . . + xn , although we will not consider the details of this definition in this calculus course.54 But the definition we obtain in this way applies only to a finite sequence of numbers. In this section we consider a broader 54 This is a topic that would likely appear in a first Abstract Algebra course, like MATH 235. Information for Students in Lecture Section 1 of MATH 141 2010 01 3211 definition, to permit us to speak of the sum of all the terms in a sequence. In this context we no longer write the sequence with commas, as a0 , a1 , . . . , an , . . . , but, instead, place plus signs between the terms, writing a0 + a1 + . . . + an + . . . , or, more compactly, as ∞ X an , n=0 and we speak of a series, rather than a sequence. Note that the word series is both singular and plural in English: there is no word “serie” in English55 . Our definition proceeds from the partial sums, which we define to be the terms in the sequence a0 , a0 + a1 , a0 + a1 + a2 , ... , a0 + a1 + . . . + an , ... or, more compactly, as 0 X n=0 an , 1 X an , ... , m X an , ... , n=0 n=0 all of which are finite sums, and therefore well defined. (Note that we had to use a different letter — m — for the general term in this sequence, because the letter n was “busy”56 .) We say that ∞ X an = L n=0 if L is the limit of the sequence of partial sums of the series. The series is then said to converge, or to be convergent; if it does not converge, it diverges or is divergent. Here, as with series and functions, we generalize to write that a sum = ∞ or = −∞, but still describe such series as divergent. We will often consider series which we will know to be convergent, without being able to specify the precise value of the limit. “Geometric” series. A geometric series is one, each of whose terms after the first is a constant multiple of its predecessor. We will often represent such series as a + ar + ar2 + . . . + arn + . . . 55 56 But the singular word for “series” in French is s´erie, and the plural is s´eries. Technically, we call n a bound variable in this case. Information for Students in Lecture Section 1 of MATH 141 2010 01 3212 where a , 0 and the “common ratio” is the ratio of each term to its predecessor, here denoted by r. By “standard” methods, usually seen in high school, one can prove that the value of the partial sum of m + 1 terms is m a (1 − rm ) X if r , 1 a + ar + . . . + arm = arn = . 1 − r (m + 1)a if r = 1 n=0 This ratio can be seen to have the following limit properties: a = m 1−r X = +∞ a + ar + . . . + arm + . . . = lim arn = m→∞ = −∞ n=0 diverges if −1 < r < 1 if r ≥ 1 and a > 0 . if r ≥ 1 and a < 0 if r ≤ −1 and a , 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3213 C.30 Supplementary Notes for the Lecture of March 17th, 2010 Release Date: Wednesday, March 17th, 2010 C.30.1 §11.2 Series (conclusion) Divergence of the “harmonic” series. 1+ The harmonic series is 1 1 1 + + ... + + ... . 2 3 n In the next section we will see a proof, using an improper integral, that this series diverges. Here we shall see a simpler proof without integrals. We need only observe the following inequalities: 1 3 1 5 1 2 ≥ 1 2 1 4 + 1 4 > 1 4 + + 16 + 17 + 1 8 > 1 8 + 18 + 18 + = 2 4 = 1 2 1 8 = 4 8 = 1 2 ... 1 1 1 1 1 1 + n−1 + ... + n + n > 2n−1 · n = +1 2 +2 2 −1 2 2 2 Thus we can make the partial sums as large as we like by proceeding out sufficiently far in the series. It follows that ∞ X 1 = +∞ , n n=1 2n−1 so the harmonic series diverges. “Necessary” and “sufficient” conditions. Suppose that A and B are sentences that may be true or false. If A cannot be true except when B is true, we say that the truth of A entails the truth of B, or that A implies B and write this symbolically as A ⇒ B. We call A a sufficient condition for B. An example of such an implication where x and y are members of the “universe” of real numbers is x = y2 − 2y + 4 ⇒ x ≥ 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3214 since y2 − 2y + 4 = (y − 1)2 + 3, and a square cannot be negative. We can interpret the statement A⇒B in another way, by observing that, when A is true, B cannot be false. We can say that B is a necessary condition for A, since A cannot be true unless B is true. Mathematicians usually search for conditions that are both necessary and sufficient, since these can characterize a situation. But often we have to be satisfied with conditions that are either of one type or the other. A “necessary” condition for convergence of a series The implication that interests use here is in the “universe” of series of real numbers. It states that ∞ X n=0 an is convergent ⇒ lim an = 0 . n→∞ This — once it has been proved — is a test that can be applied to a series to see whether the series is convergent. It can never prove that a series is convergent! What it can prove, sometimes, is that a certain series is not convergent, since no series that “fails the test” can be convergent. Many textbooks call this test the “nth term test”; your textbook calls it “‘The’ test for divergence”. This is not a standard term, and you might wish to avoid using it, since a listener who has not read Stewart’s books might not know what you are referring to.57 In practice you should “internalize” this test and always apply it, since you might otherwise waste time trying to prove that a divergent series is convergent. “Telescoping” series. Sometimes the partial sums can be interpreted as sums where there is heavy cancellation of intermediate terms, leaving only a few at each end. An example is ! ∞ ∞ X X 1 1 1 1 = − where the partial sum of the first N terms is simply 1 − . n(n + 1) n=1 n n + 1 N+1 n=1 (cf. [1, Example 6, p. 691]). Changes of variable in a sum. We have already seen the concept of changing a variable in connection with definite integrals. We can carry out similar changes in sums — both finite and infinite, although we will not investigate all the details. So, for example, we can change the 57 You will meet other tests that could be given such a name, although this is the most likely one to distinguish in this way. Stewart’s name for the test may be no more objectionable than the name in general use: it is not good form to use the symbol n in the name of a test, since the test does not depend on giving that particular name to the variable that indexes the members of the sequence; my objections to the name Stewart assigns are that (1) it is not universally accepted; and (2) the definite article suggests uniqueness, and this is not the only test that exists. Information for Students in Lecture Section 1 of MATH 141 2010 01 name of the index of summation, and write m X ar in place of the sum r=0 3215 m X an , by a “change of n=0 variable” given by r = n. More generally, we could define s = n + 4, say, and change into m+4 X m X an n=0 a s−4 . Notice how the “limits of summation” have to be changed in the same way that s=4 we changed the limits of integration in a definite integral. Operations on series. constant: Convergent series may be added term by term, or multiplied by a ∞ X (an + bn ) = n=0 ∞ X an + n=0 ∞ X n=0 can = c ∞ X ∞ X bn n=0 an n=0 But note well: while you may add convergent series term by term, you may not rearrange a series; we shall see later that rearrangement of the terms can result in a series having a different sum, or even in a convergent series being rendered divergent. 11.2 Exercises ∞ X 2 is convergent or diver+ 4n + 3 n=1 gent by expressing the partial sum as a telescoping sum (as in [1, Example 6, p. 691]). If it is convergent, find its sum. [1, Exercise 36, p. 695] Determine whether the series n2 Solution: Since 2 2 1 1 = = − , n2 + 4n + 3 (n + 1)(n + 3) n + 1 n + 3 the Nth partial sum is equal to N X n=1 ! N X 2 1 1 = − n2 + 4n + 3 n+1 n+3 n=1 = = N X n=1 N X n=1 X 1 1 − n + 1 n=1 n + 3 N X 1 1 − n + 1 m=3 m + 1 N+2 Information for Students in Lecture Section 1 of MATH 141 2010 01 = N X n=1 = 2 X n=1 = 2 X n=1 3216 setting m = n + 2 in the second sum N+2 X 1 1 − n + 1 n=3 n + 1 replacing the variable m by n N N N+2 X X X 1 1 1 1 + − − n + 1 n=3 n + 1 n=3 n + 1 n=N+1 n + 1 N+2 X 1 1 − n + 1 n=N+1 n + 1 1 1 1 1 = + − − . 2 3 N+2 N+3 As N → ∞, this partial sum approaches 12 + 13 = 56 . (In this type of problem we don’t expect students to be write the proof out formally in this way; I have included this solution so that you can see how the problem can be solved “properly”, without the use of “dots”. ∞ X n ln [1, Exercise 40, p. 695] Determine whether the series is convergent or divergent n+1 n=1 by expressing sn is a telescoping sum (as in [1, Example 6, p. 691]). If it is convergent, find its sum. Solution: In a similar way to that shown above in the solution to [1, Exercise 36, p. 695], we can show that the partial sum of the first N terms of the series is ln 1 − ln N + 1. As N → ∞ the partial sum approaches −∞: the series diverges. [1, Exercise 44, p. 695] Express the “repeating decimal expansion” 6.254 as a ratio of integers. Solution: By a “repeating decimal expansion”, we intend that the digits under the horizontal line are to be repeated as a subsequence of the expansion indefinitely; thus 6.254 = 6.25454545454... , which we can interpret as the sum of a series 6+ 2 54 54 54 + + + + ... 10 1000 100000 10000000 which can be seen as a constant added to a geometric series: ! !2 !3 2 1 54 1 1 1 + 6+ + + + + . . . 10 1000 100 100 100 Information for Students in Lecture Section 1 of MATH 141 2010 01 whose sum is 6+ 3217 54 1 1 3 344 2 =6+ + + · = . 1 10 1000 1 − 100 5 55 55 (In this way we can show that any “repeating decimal” is a rational number.) [1, Exercise 50, p. 695] Find the values of x for which the series the sum of the series for those values of x. ∞ X (x + 3)n n=0 2n converges. Find !n ∞ X (x + 3) : it can be viewed as a Solution: The given series can be rewritten as 2 n=0 x+3 geometric series with initial term 1, and common ratio . We know that a geometric 2 series converges if and only if its common ratio is less than 1 in magnitude. Thus the x + 3 < 1. Solving this inequality for x we obtain given series converges iff 2 x + 3 < 1 ⇔ −1 < x + 3 < +1 2 2 ⇔ −2 < x + 3 < 2 ⇔ −5 < x < −1 . Thus the given series converges for and only for x ∈ (−5, −1). For x in this interval the sum of the series is first term 1 2 = =− . x+3 1 − common ratio 1 − 2 1+x C.30.2 §11.3 The Integral Test and Estimates of Sums Thus far we have met only one “test” that can be applied to series to investigate whether the series converges. That is the test Stewart calls “The Test for Divergence”, and it gives negative information: it states that, if lim an does not exist, or if the limit does exist but its value is n→∞ ∞ X not 0, then the series an is divergent; this test cannot be used to confirm a suspicion that a n=0 series does converge. This is the only test we shall have that can be applied to “general” series, where the signs of the terms do not follow a specific pattern. The Integral Test. The idea of this test is to interpret the terms of a series ∞ X n=1 an as a sum of areas. We can do this by interpreting an as the area of a rectangle whose width is 1, and Information for Students in Lecture Section 1 of MATH 141 2010 01 3218 whose height is an . The test will be restricted to sequences where the terms are monotonely decreasing. Suppose that we know of a function f defined on the interval [1, ∞), whose graph passes through the left upper end-points of the rectangles. Then the area under the curve will be less than the sum of the areas of the rectangles. If we can show that the area under the curve is infinite, we will be able to conclude that the sum of the series is divergent. In a similar way, if we can pass the graph of a function f through the right upper endpoints of the rectangles, then the area under the curve will exceed the sum of the areas of the rectangles: if the area under the curve is convergent, then the same can be said about the series. In these ways we can shown that the series converges precisely when the integral representing the area under the curve converges! We call this result The “Integral” Test. Information for Students in Lecture Section 1 of MATH 141 2010 01 3219 C.31 Supplementary Notes for the Lecture of March 19th, 2010 Release Date: Friday, March 19th, 2010; corrected on 08 April, 2010; subject to further correction Review of the preceding lecture • Theorem: ∞ X ai converges ⇒ lim an = 0 n→∞ i=0 • The contrapositive of this theorem is “The” Test for Divergence: If lim an either does n→∞ ∞ X not exist, or does exist but does not equal 0, then the series ai diverges. i=0 • The harmonic series diverges — proved by grouping successive terms. • “telescoping” series • evaluating a “repeating” decimal expansion as a rational number • The Integral Test — requiring the association with a positive series of a continuous, decreasing, positive function f taking, at the positive integer points, values equal to the terms of the series C.31.1 §11.3 The Integral Test and Estimates of Sums (conclusion) Theorem C.91 (The Integral Test) If f is a continuous, positive, decreasing58 function defined on the interval [1, ∞), and if the sequence {an } has the property that an = f (n). Then ∞ X n=1 Z∞ an is convergent ⇔ f (x) dx is convergent . 1 The convergence or divergence of both the series and the improper integral do not depend on the lower limits of the sum and the definite integral. (However, when, below, we investigate bounds for the actual value of the sum of the series, then our bounds will depend on the specific values we choose for the limits of sum and integral.) 58 Actually, it suffices for the function to be non-increasing: we can permit the function values to stay at the same level so long as they eventually decrease again and eventually approach 0. Information for Students in Lecture Section 1 of MATH 141 2010 01 3220 Tests of Positive Series for convergence. Most of the tests we shall meet require a specific arrangement of signs for the terms. In fact, all but one of them will require that all the signs are the same. We call such series “positive” series, thinking of all the signs as being +; but similar results hold when all signs are −. The first test of positive series will be discussed in this section. I do not think of this as the most elementary of the tests, and would have preferred to discuss the material of the next sections first; but I will grudgingly follow the order of topics in the textbook. “Ultimate” satisfaction of properties imposed on the terms of a series. In the statement given for the a sum beginning with the 1st term, and compared it with the integral whose lower bound was x = 1. Both of these lower bounds can be replaced by any integers you like, provided the series and function are defined whenever you refer to them. Changing these lower bounds does not change the truth of the logical equivalence. It could, however, change the value of the sum of the series or the definite integral. X 1 Application: the “p-series”. The “p-series” are the series , where p is a positive np constant. The integral test shows that X 1 is convergent if and only if p > 1 . np The case p = 1 is the harmonic series, which we have already shown to be divergent, using a different proof. (This application is sometimes called the “p-series test”.) Since the tests we will meet in [1, §11.4] involve comparing a given series with series that we know to be convergent and divergent, the p-series are particularly important as they provide us with a family of series that can be used for comparison purposes. Estimating the Sum of a Series In the proof of the Integral Test we compared the sum of a series of decreasing positive terms with the area under the graph of a function that is positive and decreasing. This comparison can be refined, and gives rise to inequalities that bound the value of the sum from below and above. P By associating with a decreasing, positive series an a decreasing positive function f , we N X are able to “trap” the value of the partial sums an between the definite integrals giving the n=1 areas for regions respectively contained in and containing the region represented by the partial sum, when interpreted as the area of a region formed by rectangles of width 1 and respective heights 1, 2, . . ., N. From the inequalities Z N N X an ≤ a1 + f (x) dx n=1 1 Information for Students in Lecture Section 1 of MATH 141 2010 01 N X Z 3221 N+1 an ≥ f (x) dx 1 n=1 we Z ∞are able to relate the convergence of the infinite sum to that of the improper integral f (x) dx. 1 Let us, extending the notation used in the enunciation of the theorem, define ∞ X Rn = am ; m=n+1 that is, Rn is the sum of the “tail” or Remainder of the series, starting with the (n + 1)st term. Then Z ∞ Z ∞ f (x) dx ≤ Rn ≤ f (x) dx , n+1 or, alternatively, Z n Z ∞ ∞ f (x) dx ≤ Rn ≤ an+1 + n+1 f (x) dx , n+1 or Z ∞ 0 ≤ Rn − f (x) dx ≤ an+1 . n+1 In particular, when n = 0, this gives the inequalities 0≤ ∞ X n=1 Z ∞ an − f (x) dx ≤ a1 . 1 While, for the purposes of testing convergence, it is sufficient to demand “ultimate” satisfaction of the conditions of the Integral Test, this is not sufficient if we wish to determine bounds for the sum. In that case the estimations discussed above depend on the comparison of the area under a curve and the sum of the areas of the step function that represents the series; these comparisons are valid only when the condition the the function be decreasing is satisfied. It is possible to refine this type of bounding of partial sums and remainders, as in [1, Examples 5,6, pp. 701-702], but we will not go further in this course. ∞ X 1 converges, and we n(n + 1) n=1 found the sum. The integral test could be used to prove that it converges, but the test would not give the exact value of the sum, although it would provide bounds which compare the sum Example C.92 We proved earlier that the telescoping series Information for Students in Lecture Section 1 of MATH 141 2010 01 3222 to the improper integral Z ∞ 1 dx = lim ((ln a − ln 1) − (ln(a + 1) − ln 2)) a→∞ x(x + 1) a = lim ln + ln 2 a→∞ a+1 = ln 1 + ln 2 = 0 + ln 2 = ln 2 , 1 so the sum of the series is bounded between ln 2 = 0.6931471806 and ln 2+ 12 = 1.1931471806, which is a weaker result than the fact we already know that the sum is equal to 1. 11.3 Exercises [1, Exercise 6, p. 703] Use the Integral Test to determine whether the series ∞ X √ n=1 convergent or divergent. 1 n+4 is Solution: We begin by verifying that the Integral Test is applicable to this series: • Since the terms are reciprocals of square roots, they are positive. 1 • The function f (x) = √ is decreasing. This can be shown either by showing x+1 3 that f 0 (x) = − 12 (x + 1)− 2 < 0, or by the following reasoning: x is an increasing function of x x√+ 1 is an increasing function of x √ x + 1 is an increasing function of x since preserves order 1 is a decreasing function of x. √ x+1 ⇒ ⇒ ⇒ Z∞ The Integral Test tells us that the series will converge iff the improper integral √ 1 converges. But Z∞ √ 1 dx x+1 Za = lim √ a→∞ 1 dx x+1 ia h √ = lim 2 x + 1 = +∞ . 1 a→∞ so the series must also diverge. [1, Exercise 20, p. 704] To investigate the convergence of ∞ X n=1 n2 1 . − 4n + 5 dx x+1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3223 Solution: We know that Z Z 1 1 dx = dx = arctan(x − 2) + C , 2 x − 4x + 5 (x − 2)2 + 1 Z ∞ 1 and might be tempted to compare the integral dx with the series. This, 2 x − 4x + 5 1 however, is not directly possible from the Integral Test, because the function f (x) = 1 is decreasing only for x ≥ 2: from x = 1 to x = 2 the function is increasing. 2 x − 4x + 5 ∞ X 1 Thus we can apply the integral test to the series , provided we first verify 2 x − 4x + 5 n=2 that the necessary conditions are satisfied by the function f whose values at the positive integers coincide with the terms of the series: 1. Is f decreasing? d f (x) = dx −(2x − 4) x2 − 4x + 5 −2(x − 2) . (x − 2)2 + 1 2 = The numerator is negative for x > 2; the denominator is a non-zero square, always positive. Thus the ratio is negative for x > 2, and so f is decreasing for x > 2. 2. Is f continuous? f is a rational function, and [1, Theorem 5(b), p. 122] all rational functions are continuous on their domain. (The domain of f is all of R.) Bounding the sum for n ≥ 2 from below yields ∞ X n=2 1 ≥ 2 n − 4n + 5 Z ∞ 2 x2 dx − 4x + 5 = lim arctan(a − 2) − arctan 0 a→∞ π π −0= . = 2 2 Bounding the sum for n ≥ 3 from above yields ∞ X n=3 1 ≤ 2 n − 4n + 5 Z ∞ 2 π = , 2 x2 dx − 4x + 5 Information for Students in Lecture Section 1 of MATH 141 2010 01 3224 which implies that ∞ X n=2 Thus the series 1 1 ≤ 2 + 2 n − 4n + 5 2 − 4(2) + 5 ∞ X n=1 Z 2 ∞ x2 dx π =1+ . − 4x + 5 2 1 1+π converges, and its sum is bounded between and n2 − 4n + 5 2 3+π . (If we were to add more of the terms at the beginning of the series manually, we 2 could determine much “tighter” bounds for the sum of the series.) [1, Exercise 28, p. 704] Find the values of p for which the series ∞ X n=3 vergent. 1 is conn ln n[ln(ln n)] p Solution: The denominators of the fractions being summed are products of n, ln n, and a power of ln ln n, all of which are increasing functions of n. Hence we may apply the 1 Integral Test with the function f (x) = , confident that it is decreasing x(ln x) ([ln(ln x)] p ) and continuous. When x ≥ 3, ln x > 1.098 > 1, so ln ln x > 0, and f (x) > 0. When p = 1, Za dx = [ln ln ln x]a3 = ln ln ln a − ln ln ln 3 → ∞ x ln x[ln(ln x)]1 3 as a → ∞; hence, for the series to converge, we must certainly have p > 1; (any positive power of p less than 1 will yield an integral which is larger than the integral we obtain when p = 1). But, when p > 1, Za 3 " #a dx 1 1 1 1 =− · → p−1 p x ln x[ln(ln x)] p − 1 (ln ln ln x) p − 1 ln ln ln 3 3 as a → ∞; since the improper integral is divergent for 0 < p < 1, the positive series is also divergent. Thus the given series diverges for 0 < p ≤ 1, and converges for p > 1 — the same values for the exponent as for p-series. UPDATED TO April 17, 2010 Information for Students in Lecture Section 1 of MATH 141 2010 01 3225 C.32 Supplementary Notes for the Lecture of March 22nd, 2010 Distribution Date: Monday, March 22nd, 2010, subject to further revision Review of the last lecture • After reviewing the previous lecture, I completed my treatment of the Integral Test. • It time permits, I may return to further discussion of the use of integrals to bound series. • Discussion included [1, Exercise 28, p. 704], to behave “similarly” to p-series. ∞ X n=3 1 , which can be shown n(ln n) (ln ln n) p • There are often many, significantly different choices of series with which to compare. You need to work problems and check your solutions afterward in the manual to develop the necessary experience. C.32.1 §11.4 The Comparison Tests We continue to study series in a “reverse-logical” order. That is, as we proceed, we will be encountering properties that are more and more basic. In this section we will meet another theorem that — like the Integral Test — applies only to series whose terms all have the same sign, the so-called “positive series”, or “series of positive terms”. Finally, in [1, §11.6] we will see a theorem [1, Theorem 3, p., 715] that will justify our continued investigation of series with positive terms. When we have covered all the sections on the syllabus, I hope to return to the various tests. The comparison tests are designed so that we can infer the convergence or divergence of a given positive series by comparison of its terms with those of another series that we know to be convergent or divergent. So, in order to use this test, we need to have available as large as possible a family of positive series whose convergence or divergence are known. Let’s remember for which series we know such facts; these include • positive geometric series of common ratio less than 1 (convergent) • positive geometric series of common ration 1 or more (divergent) • harmonic series (divergent) • p-series (convergent if and only if p > 1) • positive series whose terms do not approach 0 (divergent) Information for Students in Lecture Section 1 of MATH 141 2010 01 3226 • series obtained from a given series by deleting any finite number of terms at the beginning (have the same convergence properties as the original series) • series obtained from a given series by multiplying by a positive constant (have the same convergence properties as the original series) X Theorem C.93 (Comparison Test) Let an be a series with positive terms. X 1. If bn is a series of positive terms such that bn ≤ an for all n, then X X an converges ⇒ bn converges . 2. If X bn is a series of positive terms such that an ≤ bn for all n, then X X an diverges ⇒ bn diverges . 3. In either of the preceding results, the words “for all n” may be replaced by “for all n sufficiently large”, which means, in mathematical jargon, that all we require is that the inequality holds after some integer, which could be as large as you like; if the inequality fails a finite number of times, even an enormously large finite number of times, that is sufficient for the purposes of the hypotheses of the theorem. P Prior to applying this test we may first adjust the series an that we are using for comparison purposes by • deleting or otherwise changing a finite number of terms at the beginning; • multiplying by a positive constant The following related test is sometimes easier to use: X X Theorem C.94 (Limit Comparison Test) Let an be a series with positive terms. If bn is a series of positive terms such that an =c>0 lim n→∞ bn then X X an an diverges ⇔ converges ⇔ X X bn diverges . bn converges . (There is a sharper version of this theorem that permits the limit to be either 0 or ∞; but it is delicate, and difficult for students to remember, so we usually do not discuss it in this course, cf. [1, Exercises 40-42, p. 709].) Information for Students in Lecture Section 1 of MATH 141 2010 01 Estimating Sums cise 32, p. 709]. 3227 This question will be illustrated by the solution below of part of [1, Exer- Example C.95 ([7, Exercise 20, p. 735]) Determine whether the series ∞ X 1 + 2n 1 + 3n n=1 or diverges. converges 2n Solution: We could not compare the terms of this series naively with those of n , which is a 3 convergent geometric series because the terms here are slightly larger. However a number of other possibilities suggest themselves. For example, we have !n !n !n 1 + 2n 1 + 2n 1 2 2 ≤ = + <2 1 + 3n 3n 3 3 3 which is the general term of a geometric series with common ratio 23 < 1, which is convergent. Other comparisons are possible, some of them complicated to prove. For the purpose of proving only convergence (and not bounding the series) one tries to find the simplest possible comparison. This problem can also be solved by applying the Limit Comparison Test, again in comparP 2 n ison with : 3 1 + 2n n lim 1 + !3n n→∞ 2 3 3n 1 + 2n · = lim n→∞ 2n 1 + 3n ! = lim 2−n + 1 · 3−n + 1 n→∞ = 1 × 1 = 1. Since the limit is a positive real number, and since the geometric series given series must also converge. P 2 n 3 converges, the Example C.96 ([7, Exercise 34, p. 735]) Estimate the error when the sum of the first 10 terms ∞ X 1 + cos n . of the following sum is used to approximate the sum of the series: n5 n=1 Solution: The “tail” of the series will be estimated from above. Observe that 2 1 + cos n ≤ 5. 5 n n We know that 1 1 ≤ n5 x 5 for n − 1 ≤ x ≤ n. Information for Students in Lecture Section 1 of MATH 141 2010 01 3228 Comparing the sum of the terms on the left — interpreted as rectangles of width 1 lying under the graph of f (x) = x15 — we have Z ∞ ∞ X 1 dx 1 ≤ = 4. 5 5 n 10 10 x n=11 It follows that the “tail” of the series — the sum beginning with term n = 11 — is no greater than 2 × 10−4 . This gives an upper bound for the error if the series is truncated at term n = 10. This method will not give a useful lower bound for the error, however, since we know only that 1 + cos n 0 ≥ 5. 5 n n This is certainly not the best possible lower bound, but we can’t do better in this course. 11.4 Exercises In some of the discussions below I give only hints about the solution. [1, Exercise 12, p. 709] Determine whether the series converges or diverges: ∞ X 1 + sin n n=0 10n . Solution: The Limit Comparison Test cannot be used, since the terms behave erratically because of the sine function, and there is not likely to exist a limit, no matter what series you choose to compare with. But 1 + sin n 2 < n, n 10 10 so the terms of the series are smaller than the terms of a geometric series with common 1 ratio 10 , which is less than 1, and must converge. √ ∞ X n [1, Exercise 14, p. 709] Determine whether the series converges or diverges: . n − 1 n=0 Solution: The terms can be made smaller by making the denominator larger, into n. That yields a divergent p-series. [1, Exercise 18, p. 709] Determine whether the series converges or diverges: ∞ X n=0 1 . 2n + 3 Solution: The terms can be made smaller by making the denominator into 2n + 4. The 1 new series is a multiple 2 of a harmonic series. The harmonic series diverges, and that property doesn’t change if we discard 2 terms at the beginning, nor if we then multiply by 12 . Information for Students in Lecture Section 1 of MATH 141 2010 01 [1, Exercise 30, p. 709] Determine whether the series ∞ X n! n=1 nn 3229 converges or diverges. Solution: The general term is n! 1 = n n n 1 ≤ n 2 n 2 · n · 3 n−1 n · ... · · n n n 2 ·1= 2 n · for n ≥ 3, which is the general term in a convergent p-series. Hence the given series is convergent. It is impractical to use the limit comparison test here, because the limit would be difficult to compute, and because it would be zero or infinite, depending on which series is in the numerator; our version of this test does not permit the limit to be either 0 or ∞. ∞ X 1 sin converges or diverges. [1, Exercise 31, p. 709] Determine whether the series n n=1 sin 1n Solution: We know the value of lim 1 n n→∞ to be 1; so the Limit Comparison Test shows this series diverges because the harmonic series diverges. Could we have used the Comparison Test? Unfortunately, the geometric argument that we used to prove the result quoted earlier showed [1, p. 190-191] that 0 < sin 1 1 < . n n In its present form this inequality is not useful for proving the divergence of the given series, since the series known to be divergent occupies the outer position rather than lying between the terms of the given series and 0. [1, Exercise 32, p. 709] Determine whether the series ∞ X 1 converges or diverges. 1+ n1 n n=1 Solution: It appears that the series “resembles” the Harmonic Series, and we will try to compare it in the limit. The guess is a fortunate one, since the limit does exist. 1 1 1+ n lim n 1 n→∞ n = lim n→∞ = lim n→∞ = 1 lim en→∞ 1 1 nn 1 e ln n n ln n n by continuity of exponential Information for Students in Lecture Section 1 of MATH 141 2010 01 3230 1 = 1 by L’Hospital’s Rule e0 We apply the Limit Comparison Test: since the Harmonic Series diverges and the ratio of the terms of the given series to those of this divergent series approaches a non-zero positive real number, the given series also diverges. = The same argument could have been used to prove that the series ∞ X n=1 1 n 1+ ln1n is also divergent; the terms in this last series are smaller than those of the series we were given. C.32.2 Sketch of Solutions to Problems on the Final Examination in MATH 141 2006 01 (not for discussion at the lecture) The best way to prepare for the examination in MATH 141 2009 01 is to study the textbook, working problems, and verifying your solutions using the Student Solutions Manual: for several reasons I don’t recommend your studying exclusively from old examinations. However the old examinations do have some uses, and I am including a second one of them here, with solutions. I am hoping to discuss last April’s examination briefly in a final lecture; there will not be time for me to discuss the following examination in the lectures. 1. SHOW ALL YOUR WORK! For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X n=1 1 (tanh n)2 + 1 n −n −2n 1−e 1 = 1+e Solution: As n → ∞, tanh n = een −e −2n → 1 = 1 , so the terms summed in +e−n 1 1 this series have limit 12 +1 = 2 , 0. By the “Test for Divergence” this series cannot converge. Information for Students in Lecture Section 1 of MATH 141 2010 01 (b) [4 MARKS] ∞ X n2n e−n 3231 2 n=1 Solution: pn n2 n2n e−n2 = n2 e−n = n . e case) the limit of this nth root as n → ∞ is nth root of nth term = By l’Hospital’s Rule ( ∞ ∞ n2 2n 2 lim = lim n = lim n = 0 < 1 . n→∞ en n→∞ e n→∞ e By the Root Test, the series is convergent. ∞ X n2 − 85n + 12 (c) [4 MARKS] n(n + 6)2 n=1 Solution: The ratio of the nth term of this series to the nth term of the Harmonic ∞ P 1 series is n n=1 12 85 n2 − 85n + 12 1 − n + n2 = → 1 , 0. (n + 6)2 6 2 1+ n By the Limit Comparison Test the given series is divergent, since the Harmonic series is divergent. 2. SHOW ALL YOUR WORK! Z 2 (a) [4 MARKS] Evaluate |x| dx . −1 Solution: Z Z 2 |x| dx = −1 2 |x| dx + −1 Z 0 = Z 0 |x| dx 0 Z 2 (−x) dx + x dx −1 0 " 2 #0 " 2 #2 x x = − + 2 −1 2 0 5 1 +2= = 2 2 Ze3 (b) [3 MARKS] Evaluate 1 dt . √ t 1 + ln t Information for Students in Lecture Section 1 of MATH 141 2010 01 dt t Solution: Under the substitution u = ln t, du = Ze3 √ 1 Z dt 3 = t 1 + ln t 0 d (c) [3 MARKS] Evaluate dx 2 Solution: With u = x , d dx Z x2 0 3232 and i √ √ du 1 3 = 2(1 + u) 2 = 2( 4 − 1) = 2(2 − 1) = 2 √ 0 1+u Z x2 2 et dt . 0 d e dt = dx Z u t2 2 2 et dt = eu 2x = e( x ) 2x = 2xe x , 2 2 4 0 by the Fundamental Theorem of Calculus. (d) [4 MARKS] Evaluate !7 !7 !7 !7 1 2 n − 1 1 0 . + + + ... + lim n→∞ n n n n n R1 Solution: You are asked to evaluate the limit of a Riemann sum for 0 x7 dx = h 8 i1 x = 18 , (where the values of the function are taken at the left end-point of each 8 0 subinterval). Accordingly the value is 81 . SHOW ALL YOUR WORK! 3. For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X n=1 1 (tanh n)2 + 1 n −n −2n 1−e 1 Solution: As n → ∞, tanh n = een −e = 1+e −2n → 1 = 1 , so the terms summed in +e−n this series have limit 121+1 = 12 , 0. By the “Test for Divergence” this series cannot converge. Information for Students in Lecture Section 1 of MATH 141 2010 01 (b) [4 MARKS] ∞ X n2n e−n 3233 2 n=1 Solution: pn n2 n2n e−n2 = n2 e−n = n . e case) the limit of this nth root as n → ∞ is nth root of nth term = By l’Hospital’s Rule ( ∞ ∞ 2n 2 n2 = lim n = lim n = 0 < 1 . n→∞ e n→∞ e n→∞ en lim By the Root Test, the series is convergent. ∞ X n2 − 85n + 12 (c) [4 MARKS] n(n + 6)2 n=1 Solution: The ratio of the nth term of this series to the nth term of the Harmonic ∞ P 1 series is n n=1 85 12 n2 − 85n + 12 1 − n + n2 = → 1 , 0. (n + 6)2 6 2 1+ n By the Limit Comparison Test the given series is divergent, since the Harmonic series is divergent. 4. BRIEF SOLUTIONS Express the value of each of the following as a definite integral or a sum, product, or quotient of several definite integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems, and should be simplified as much as possible, except that you are not expected to evaluate the integrals. (a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the region bounded by the parabola y2 = 2x + 6 and the line y = x − 1. An answer involving integration along the y-axis will not be accepted. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Information for Students in Lecture Section 1 of MATH 141 2010 01 Solution: Z −1 3234 Z 5p p 2 2(x + 3) dx + 2(x + 3) − (x − 1) dx −1 −3 (b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 the region bounded by the curves y = x3 and y = x2 . For this question you are to use only the method of “washers”. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: Only the first integral was required. π Z 1 2 2 1 − x3 − 1 − x2 dx 0 Z 1 −2x3 + x6 + 2x2 − x4 dx 0 " #1 2x4 x7 2x3 x5 = π − + + − 4 7 3 5 0 ! 2 1 2 1 23π = π − + + − = . 4 7 3 5 210 = π (c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 the region bounded by the curves y = x3 and y = x2 . For this question you are to use only the method of “cylindrical shells”. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: Only the first integral was required. Z 1 2π 0 1 1 (1 − y) y 3 − y 2 dy Information for Students in Lecture Section 1 of MATH 141 2010 01 3235 Z 1 1 4 1 3 = 2π y 3 − y 3 − y 2 + y 2 dy 0 ! 3 3 2 2 23π = 2π − − + = 4 7 3 5 210 (d) [3 MARKS] The length of the curve whose equation is x2 y2 + = 1. 4 9 DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: This is a special case of [7, Exercise 20, p. 553]. x 2 y2 + = 1 ⇔ 9x2 + 4y2 = 36 4 9 dy 9x ⇒ =− dx 4y s p !2 81x2 + 16y2 dy 1+ ⇒ = dx 4|y| Z 2 r 1 16 + 5x2 ⇒ arc length = 2 dx 4 − x2 −2 2 where the factor 2 is needed because the integral is the length of either the top or the bottom arc of the ellipse; this integral is improper. A cleaner solution is found by parameterizing the curve as x = 2 cos θ, y = 3 sin θ (0 ≤ θ ≤ 2π). Then the arc length is Z 2π p 4 sin2 θ + 9 cos2 θ dθ . 0 5. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z 5 x +x dx . x4 − 16 Information for Students in Lecture Section 1 of MATH 141 2010 01 3236 Solution: The first step is to divide denominator into numerator: x5 + x = x(x4 − 16) + 17x ⇒ 17x x5 + x = x + . x4 − 16 x4 − 16 Next we could separate the resulting remainder fraction, whose numerator has degree less than that of its denominator, into partial fractions: 17x Ax + B Cx + D = + 2 − 16 x2 − 4 x +4 E F Cx + D = + + 2 x−2 x+2 x +4 2 ⇒ 17x = E(x + 2)(x + 4) + F(x − 2)(x2 + 4) +Cx(x − 2)(x + 2) + D(x − 2)(x + 2) x4 We can determine the constants by a combination of the standard methods: giving x convenient values, and equating coefficients of like powers of x: x = 2 ⇒ 34 = 32E ⇒ E = 17 16 17 16 x = 0 ⇒ 0 = 6E − 6F − 8D ⇒ D = 0 17 coeff x3 : ⇒ 0 = E + F + C ⇒ C = − 8 x = −2 ⇒ −34 = −32F ⇒ F = Hence 1 x5 + x 17 1 2x = x + + − x4 − 16 16 x − 2 x + 2 x2 + 4 Hence Z ! x5 + x x2 17 x2 − 4 dx = + ln +C x4 − 16 2 16 x2 + 4 A better tactic would have been to make a substitution u = x2 in the integral prior to separation into partial fractions: ! Z Z Z 17 du 17 1 1 17 u − 4 17x +C dx = = − du = ln x4 − 16 2 (u − 4)(u + 4) 16 u−4 u+4 16 u + 4 6. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Information for Students in Lecture Section 1 of MATH 141 2010 01 3237 Z (a) [4 MARKS] cos x · cosh x dx Solution: Using integration by parts, u = cos x ⇒ du = − sin x dx Z dv = cosh x dx ⇒ v = sinh x Z ⇒ cos x · cosh x dx = (cos x)(sinh x) + sinh x · sin x dx . A second integration by parts yields U = sin x ⇒ dU = cos x dx Z dV = sinh x dx ⇒ V = cosh x Z ⇒ sin x · sinh x dx = (sin x)(cosh x) − cosh x · cos x dx Z Z ⇒ cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) − cos x · cosh x dx. Now we can move the integral from the right side of the equation to join the same integral with opposite sign on the left side, yielding Z 2 cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) + C or Z cos x · cosh x dx = (b) [5 MARKS] Z1 √ (cos x)(sinh x) + (sin x)(cosh x) + C0 . 2 x2 + 2x + 5 dx −3 Solution: We begin by completing the square: Z1 √ −3 x2 + 2x + 5 dx = Z1 p (x + 1)2 + 4 dx . −3 √ Then the integrand can be changed to the form of X 2 + 1: s !2 Z1 p Z1 x+1 2 (x + 1) + 4 dx = 2 + 1 dx . 2 −3 −3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3238 equal to either tan θ or sinh θ; Here an appropriate substitution would be to set x+1 2 I have chosen to use the trigonometric substitution: x+1 x+1 = tan θ ⇒ θ = arctan 2 2 2 ⇒ dx = 2 sec θ dθ Z1 √ Z 2 ⇒ x + 2x + 5 dx = 2 −3 Z1 ⇒ √ π 4 − π4 Z x2 π 4 + 2x + 5 dx = 8 | sec θ| · 2 sec2 θ dθ sec3 θ dθ 0 −3 Finally we have to find an antiderivative of sec3 θ, which can be integrated by “standard” methods, e.g., by integration by parts with u = sec θ, dv = sec2 θ dθ: du = sec θ tan θ dθ v = tan θ Z Z sec3 θ dθ = sec θ tan θ − sec θ tan2 θ dθ Z sec θ(sec2 θ − 1) dθ Z Z 3 = sec θ tan θ − sec θ dθ + sec θ dθ Z = sec θ tan θ − sec3 θ dθ + ln | sec θ + tan θ| + C = sec θ tan θ − Z sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C ⇒2 Z ⇒ Here sec3 θ dθ = Z1 √ −3 (c) [4 MARKS] sec θ tan θ + ln | sec θ + tan θ| + C0 . 2 Z π 4 x2 + 2x + 5 dx = 8 √ √ sec3 θ dθ = 4 2 + 4 ln( 2 + 1) . 0 Z sin2 x · cos2 x dx Solution: One way to simplify this integral is to use the identity sin x cos x = sin 2x : 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z Z 1 sin x · cos x dx = 4 Z 1 = 8 x = − 8 2 2 3239 sin2 2x dx ! 1 sin 4x (1 − cos 4x) dx = x− +C 8 4 sin 4x +C. 32 This could also have been simplified using the algorithm proposed in the textbook: Z Z 1 2 2 sin x · cos x dx = (1 − cos 2x)(1 + cos 2x) dx 4Z 1 (1 − cos2 2x) dx = 4 ! Z 1 1 + cos 4x = 1− dx 4 2 Z 1 (1 − cos 4x) dx etc. = 8 7. SHOW ALL YOUR WORK! Consider the curve C defined by x = x(t) = 10 − 3t2 y = y(t) = t3 − 3t , where −∞ < t < +∞. d2 y (a) [8 MARKS] Determine the value of 2 at the points where the tangent is horizondx tal. Solution: dx = −6t dt dy = 3t2 − 3 dt dy ! 1 dy 1 t2 − 1 dt = =− t− = dx dx −2t 2 t dt ! 2 d dy d y = dx2 dx dx Information for Students in Lecture Section 1 of MATH 141 2010 01 3240 ! dy dt · dx dx ! dy dx dx dt !! d 1 1 − t− dt 2 t = dx dt ! 1 1d − t− 2 dt t = −6t 1 1 1 + t2 t2 + 1 = − = . 2 −6t 12t3 d = dt d dt = dy The tangent is horizontal when = 0, i.e., when t2 − 1 = 0 or t = ±1. At such dx parameter values d2 y 2 1 = = ± . dx2 ±12 6 (b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of the arc o n √ (x(t), y(t)) : − 3 ≤ t ≤ 0 . Solution: The area of the surface of revolution about the x-axis is s !2 !2 Z 0 Z 0 p dx dy 2π √ y(t) + dt = 2π √ (t3 − 3t) 36t2 + 9(t2 − 1)2 dt dt dt − 3 − 3 Z 0 = 2π √ (t3 − 3t)(3t2 + 3) dt − 3 Z 0 = 2π √ (t5 − 2t3 − 3t) dt " − 3 t6 2t4 3t2 − − = 6π 6 4 2 8. SHOW ALL YOUR WORK! #0 √ − 3 = 27π . Information for Students in Lecture Section 1 of MATH 141 2010 01 3241 (a) [5 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z 0 dx . 2 −1 x 3 Solution: This integral is improper because the integrand is undefined, hence discontinuous at the right endpoint. Z 0 Z b dx dx = lim− by definition 2 2 b→0 −1 x 3 −1 x 3 h 1 ib 1 = lim− 3x 3 = 3 lim− b 3 − (−3) = 0 + 3 = 3. −1 b→0 b→0 Since the limit exists, the improper integral is convergent. (b) [5 MARKS] Determine whether the following series is conditionally convergent, absolutely convergent, or divergent. ∞ X (−1)n n − ln n n=1 Solution: Let f (x) = 1 x − ln x . then 1 − 1x f (x) = − . (x − ln x)2 0 For x > 1, 1− 1x > 0, so f 0 (x) < 0. Thus the terms of the given series are monotonely decreasing in magnitude, and alternating in sign. We observe that 1 ln x lim = lim x = 0; x→∞ x x→∞ 1 it follows that the limit of the ratio of corresponding terms of the given series and P the series 1n is 1 n−ln n lim 1 n→∞ n 1 1 = ln n n→∞ 1 − 1 − lim n = lim ln n n→∞ n = 1. Since the Harmonic Series diverges, this implies (by the Limit Ratio Test) that the series of absolute values of the given series is divergent; and the given series is either conditionally convergent or divergent. The limit of the terms of the series is 1 1 1 = lim · lim = 0 · 1 = 0. n→∞ n n→∞ 1 − ln n n→∞ n − ln n n lim Information for Students in Lecture Section 1 of MATH 141 2010 01 3242 This, combined with the property that the terms of the given series are alternating in sign and monotonely decreasing in magnitude, permits us to conclude by the Leibniz Test that the series converges. Thus this alternating series is convergent, but not absolutely convergent: it is conditionally convergent. (c) [3 MARKS] Give an example of a sequence {an } with the property that lim an = 0 n→∞ ∞ X but an = +∞. You are expected to give a formula for the general term an of n=1 your sequence. Solution: The Harmonic Series an = desired property. 1 n is one example of a series which has the 9. SHOW ALL YOUR WORK! [12 MARKS] The arc r = 1 − cos θ (0 ≤ θ ≤ π) divides the area bounded by the curve r = 1 + sin θ (0 ≤ θ ≤ 2π) into two parts. Showing all your work, carefully find the area of the part that contains the point (r, θ) = 12 , π2 . Solution: Since the functions defining the curves are both periodic with period 2π, we can confine our attention to any parameter interval of length 2π. We need to determine where the curves intersect. Solving the two given equations algebraically gives tan θ = −1, implying that θ = 3π , 7π in the interval 0 ≤ θ ≤ 2π. The only point whose parameter 4 4 value is in the intervals given for the two arcs is θ = 3π . However, we always need to be 4 alert to the possibility that there are intersections of the curves in which the parameter values could be different on the two curves. One place where precisely that occurs for these two curves is the pole: on r = 1−cos θ the pole appears as (r, θ) = (0, 0), which is in the restricted interval 0 ≤ θ ≤ π; on the cardioid r = 1 + sin θ the pole appears as 0, 3π . 2 Could there be any other points of intersection? Any others could be found by solving the variants of the equations obtainable by applying the transformation (r, θ) = (−r, θ+π) repeatedly. We can show that there are no such points that appear in this way. It follows that the region whose area we seek is bounded by two arcs: • The arc of r = 1 − cos θ bounded by 0 ≤ θ ≤ 3π . 4 or 3π ≤ θ ≤ 3π . (In the first of these • One arc of r = 1 + sin θ: either − π2 ≤ θ ≤ 3π 4 4 2 I have, for convenience, used the negative value θ = − π2 . You might wish to object that the prescribed interval for θ was 0 ≤ θ ≤ 2π. I could, instead have described Information for Students in Lecture Section 1 of MATH 141 2010 01 3243 2 1.5 1 0.5 0 -2 -1.5 -1 -0.5 0 0.5 1 Figure 22: The curves with equations r = 1 − cos θ, (θ ≤ 0), and r = 1 + sin θ, and the point 1 π , 2 2 the first of these arcs as being made up of two pieces: one being 3π ≤ θ ≤ 2π, and 2 the second being 0 ≤ θ ≤ 3π .) 4 But the point 21 , π2 is only in the region whose boundary contains the arc 3π ≤ θ ≤ 3π of 4 2 the curve r = 1 − cos θ. To find the area, think of a line segment being drawn from the pole to 1 + √12 , 3π . The 4 area is then the sum of the area below the line segment and the area above it, i.e., Z 3π4 Z 3π2 1 1 2 (1 − cos θ) dθ + (1 + sin θ)2 dθ 2 0 2 3π4 ! ! Z 3π4 Z 3π2 1 1 + cos 2θ 1 1 − cos 2θ = 1 − 2 cos θ + dθ + 1 + 2 sin θ + dθ 2 0 2 2 3π4 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 " # 3π " # 3π 1 3θ sin 2θ 4 1 3θ sin 2θ 2 = − 2 sin θ + + − 2 cos θ − 2 2 4 0 2 2 4 3π 4 9π √ 1 = − 2− . 8 4 3244 Information for Students in Lecture Section 1 of MATH 141 2010 01 3245 C.33 Supplementary Notes for the Lecture of March 24th, 2010 Distribution Date: Wednesday, March 24th, 2010, subject to further revision C.33.1 §11.5 Alternating Series I continue to follow the syllabus in the order of topics in the textbook. This order is the reverse of that which I would prefer, in that the most basic results for this topic will appear in the next section, where the textbook considers theorems about positive series. As with the preceding section, we are investigating a specialized topic concerning series prior to meeting a more general theorem in [1, §11.6] at the next lecture. The present section will broader the class of series that we can work with, albeit to a very special class of series. An alternating series is one in which the signs alternate. Our main interest, as in the preceding sections will be in three questions, listed in the order of our priorities: 1. Does the series converge? 2. If the series converges, what is its sum? 3. If we truncate the series at a specific partial sum, can we obtain a “good” upper bound for the error — i.e., for the difference between the sum of the series and the given partial sum. Our interest in alternating series derives partly from some specific series that will arise later in [1, Chapter 11], in the portion of the chapter assigned to Calculus III. There we would meet several important power series which represent some familiar functions; and many of those series have alternating signs. By the “Test for Divergence”, an alternating series cannot converge unless the limit of the sequence of its terms is 0. For these series, and these series alone, we have a partial converse of that test: Theorem C.97 (Leibniz’s Alternating Series Test) An alternating series ∞ X (−1)n−1 bn for which: n=1 (i) 0 ≤ bn+1 ≤ bn for all n; and (ii) lim bn = 0 n→∞ converges. Information for Students in Lecture Section 1 of MATH 141 2010 01 3246 Estimating Sums While the estimation of errors can be complicated for other series, the following very simple bounds hold for alternating series: Corollary C.98 (to the Leibniz Alternating Series Test) Continuing with the same notation ∞ X as the theorem, the remainder upon truncation of an alternating series (−1)n−1 bn (bn ≥ 0; n = n=1 1, 2, . . .) at the N th term cannot exceed the very next term; more precisely, ∞ X (−1)n−1 bn < |bN+1 | . n=N+1 ∞ X 1 converges. 2n + 1 n=0 Solution: This is an alternating series. Since the denominators are increasing, while the numerators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the terms are approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course, but it can be shown that this series converges to arctan 1, i.e., to π4 , and thereby provides a way of approximating π to any desired accuracy. It is, however, not a good series for that purpose, since it converges rather slowly.) Example C.99 Determine whether the series (−1)n ∞ X 1 converges. n+1 n=0 Solution: This is an alternating series. Since the denominators are increasing, while the numerators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the terms are approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course, but this series can be proved to converge to ln(1 + 1), i.e., to ln 2.) Example C.100 Determine whether the series (−1)n Example C.101 ([7, Exercise 26, p. 740]) How many terms of the series need to add in order to find the sum to within an error of 0.002? Solution: As n → ∞ lim 4nn = lim 4n 1ln n = 0. The terms are decreasing, since ∞ X (−1)n n n=1 4n do we n n+1 1 > n+1 ⇔ 4n > n + 1 ⇔ n > . n 4 4 3 Thus the Leibniz Theorem is applicable, and we apply the error estimate associated with that Theorem. If we truncate the series at the Nth term, then we know that the error will be less N+1 than N+1 . If we can find N satisfying the condition that 4 N+1 < 0.002 , 4N+1 (96) Information for Students in Lecture Section 1 of MATH 141 2010 01 3247 we will have the desired accuracy. Since d x 1 · 4 x − x · 4 x · ln 4 1 − x ln 4 = = , dx 4 x 4x (4 x )2 which is negative provided x > ln14 , which is certainly true if x ≥ 1. Thus we need only find the first value of N for which inequality (96) holds. By repeated calculations we find the first value is N = 5. That is, the series may be truncated by summing so that the last term we include is (−1)5 5 and the error will be within the prescribed tolerance.59 45 The student should understand that the Leibniz estimate for error is not necessarily as sharp as it might be. It is often possible to obtain much better approximations than the Leibniz estimate might guarantee; such investigations are beyond this course. The Leibniz theorem states that an alternating series whose terms are non-decreasing in magnitude and are approaching 0 is convergent. However, an alternating series whose terms are not decreasing, but whose terms do approach 0, may still converge! For example, suppose that we modify the “alternat∞ P ing harmonic series” (−1)n+1 1n in the following way: term number 2n is the negative fraction n=1 n (−1)2 +1 21n . Let’s divide just these terms by 2. The resulting alternating series can be shown to still be convergent; but it will converge to a sum which is less than the sum of the original series; in fact, the reduction converges to a sum of 1 1 · 2 2 1− 1 2 = 1 . 2 But the condition of the Leibniz Theorem that requires that the terms be monotonely decreasing in magnitude is not satisfied, so the convergence cannot be inferred from the Leibniz Theorem. Nonetheless, the series does converge (since the partial sums are equal to the difference of the partial sums of two series known to converge). 11.5 Exercises [1, Exercise 14, p. 713] Test for convergence or divergence the series ∞ X n=1 (−1)n−1 ln n . n Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’s Rule to be lim 1n = 0. To determine whether the terms are decreasing, we can consider n→∞ 59 ! 1 − ln x d ln x <0 = dx x x2 We can actually sum this infinite series, using methods analogous to the method used to sum a geometric 4 . The sum of the first 5 terms is approximately -0.161133, so the progression. The sum can be shown to be − 25 error is approximately 0.001133, which is within the desired tolerance. Information for Students in Lecture Section 1 of MATH 141 2010 01 3248 for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s not important where the terms start to decrease — only that they ultimately decrease.60 √ ∞ X n n [1, Exercise 10, p. 713] Test for convergence or divergence the series (−1) √ . 1+2 n n=1 Solution: The general term of the given series is, √ n n (−1) √ = (−1)n 1+2 n √1 n 1 +2 1 whose magnitude → as n → ∞. Thus the sequence of terms of this series does not 2 have a limit as n → ∞, and the series cannot converge, by the “Test for Divergence”. In fact, the other condition of the Leibniz Theorem, that the terms decrease in magnitude, also fails to hold! But here we can say more: not only can we state that the Leibniz Test yields no information, but we can state that the Test for Divergence shows that the series actually diverges. [1, Exercise 14, p. 713] Test for convergence or divergence the series ∞ X n=1 (−1)n−1 ln n . n Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’s Rule to be lim 1n = 0. We have determined this limit for 2 reasons: n→∞ • as an application of “The” Test for Divergence — our result is inconclusive, and gives no information about divergence; • as a condition of the Leibniz Alternating Series Test — here our result is positive, and, when combined with the yet unproved fact that the terms are decreasing in magnitude, will show that the series does, indeed, converge. To determine whether the terms are decreasing, we can consider ! 1 − ln x d ln x = <0 dx x x2 for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s not important where the terms start to decrease — only that they ultimately decrease.61 60 61 However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease. However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease. Information for Students in Lecture Section 1 of MATH 141 2010 01 [1, Exercise 17, p. 713] Test for convergence or divergence the series 3249 ∞ X π (−1)n sin . n n=1 Solution: This problem should be compared with [1, Exercise 31, p. 709] (cf. these notes, p. 3229). In the earlier problem we proved the divergence of a series whose terms resemble those of the present series, except that it was not alternating. But the present series satisfies the conditions of the alternating series test, as the terms alternate in sign, are monotonely decreasing (because the sine function is an increasing function to the right of 0), and the limit of the terms is 0. Thus the series converges. When we have covered [1, §11.6] we will be able to describe the present series as being Conditionally Convergent — it is convergent, but loses that property if all terms are replaced by their absolute value. ∞ X π [1, Exercise 18, p. 713] Test for convergence or divergence the series (−1)n cos . n n=1 Solution: Since the magnitudes of the terms of this series approach 1 as n → ∞, the terms of the series have no limit! By “The” Test for Divergence, this series diverges! [1, Exercise 19, p. 713] Test for convergence or divergence the series ercise 30, p. 709][1, Exercise 18, p. 720]) ∞ X (−1)n n=1 nn (cf. [1, Exn! Solution: The terms may be factorized as (−1)n nn n n n n = (−1)n · · . . . · . n! 1 2 n−1 n While this is an alternating series, the limit of the terms cannot be 0, since the fractions into which I have factored the terms are, in magnitude, all greater than or equal to 1; thus the series must diverge, by “The” Test for Divergence; this also shows that not all conditions of the Leibniz Alternating Series Test are satisfied, but from that we can only conclude that that test does not apply, not that the series diverges. [1, Exercise 20, p. 713] Test for convergence or divergence the series ∞ X (−1)n n=1 n n 5 . Solution: This series also diverges, because the terms to not have limit 0. C.33.2 Solutions to Problems on the Final Examination in MATH 141 2007 01 Among the instructions for this examination was the following: There are two kinds of problems on this examination, each clearly marked as to its type. Information for Students in Lecture Section 1 of MATH 141 2010 01 3250 • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page. Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 1. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. Z 2 (a) [2 MARKS] Evaluate |x|2 dx. −1 Solution: Z 2 Z 2 " x3 |x| dx = x dx = 3 −1 −1 2 Z0 (b) [2 MARKS] Evaluate 1 #2 2 = −1 8 − (−1) = 3. 3 t4 dt . √ t5 + 1 5 Solution: Either √ by observation, or by applying a substitution like u = t , or u = t5 + 1, or u = t5 + 1, we can show that Z0 1 i0 2 √ 2 √5 2 √ t4 dt = t + 1 = (1 − 2) = − ( 2 − 1) . √ 1 5 5 t5 + 1 5 (c) [3 MARKS] Determine the value of !3 !3 !3 !3 1 0 1 2 n − 1 . + + + ... + n n n n n Information for Students in Lecture Section 1 of MATH 141 2010 01 3251 Solution: Note that the question did not ask for the value of the limit as n → ∞ — it asked simply for the value of the sum (as a function of n). !3 !3 !3 !3 1 0 1 2 n − 1 + + + ... + n n n n n n−1 X i3 = = i=0 n4 !2 (n − 1)n 2 (n − 1)2 = . n4 4n2 (Had the instructions been to evaluate the limit as n → ∞, then the given product could have been interpreted as a left Riemann sum with n intervals of equal length " 4 #1 Z1 x 1 3 1 x dx = ∆n = n . The value would have been = .) 4 0 4 0 (d) [3 MARKS] Suppose it is known that f 0 (x) = 4 cosh x for all x. Showing all your work, determine the value of f (1)− f (−1), expressed in terms of the values of either exponentials or hyperbolic functions. Solution: Integrating the given equation yields f (x) = 4 sinh x + C, where C is some (fixed) real constant C. Since f 0 exists, f is continuous everywhere, so it will be the same constant C that is involved throughout the interval [−1, +1]; hence f (1) − f (−1) = 4 sinh 1 + C − (sinh(−1) + C) = 4(sinh 1 + sinh 1) = 8 sinh 1 . Alternatively one could work with exponentials, beginning with f 0 (x) = 2 (e x + e−x ), ! 1 to show that f (1) − f (−1) = 4 e − . e Z x2 d t et dt when x = 1 . (e) [4 MARKS] Evaluate dx 12 Solution: Z Z x2 d d u tt du tt where u = x2 e dt = e dt · dx 12 du 12 dx 2 du u 2 (x ) = eu · = e( x ) · 2x dx Information for Students in Lecture Section 1 of MATH 141 2010 01 which simplifies to 2x · e x (2x2 ) 3252 ; when x = 1, the value is 2e. 2. SHOW ALL YOUR WORK! For each of the following series you are expected to apply one or more tests for convergence or divergence to determine whether the series is absolutely convergent, conditionally convergent, or divergent. All tests used must be named, and all statements must be carefully justified. (a) [4 MARKS] ∞ X (−n − 2)n (n − 2)n n=1 Solution: (2n2 + 1)n ∞ X (−n − 2)n (n − 2)n n=1 (2n2 + 1)n = ∞ X n=1 (−1)n (n2 − 4)n . (2n2 + 1)n This is an alternating series. Applying the Root Test to the series of absolute values of the terms of the given series, we have s 2 4 2 n n − 4 1 − n2 n (n − 4) = = (2n2 + 1)n 2n2 + 1 2 + 12 n for n ≥ 2. As n → ∞, this ratio → 12 < 1 . Since the limit of the ratio is less than 1, the original alternating series is absolutely convergent. ∞ X n! (−1)n+1 2 n (b) [4 MARKS] n2 n=1 Solution: If we define an = (−1)n+1 n! , we find, using the Ratio Test, that n2 2n 2 an+1 = (n + 1)n an (n + 1)2 1 n2 → ∞ > 1. = 2(n + 1) Hence the original series diverges. ∞ X 1 (c) [4 MARKS] (−1)n sin n n=1 Solution: The series of absolute values may be compared, using the Limit Comparison Test, with the Harmonic series, a positive series known to be divergent. Since Information for Students in Lecture Section 1 of MATH 141 2010 01 lim n→∞ sin 1n 1 n 3253 = 1 , 0, the series of absolute values is also divergent, and the original series is either (1) conditionally convergent, or (2) divergent. But ! ! ! d 1 1 1 sin = cos · − 2 < 0 dx x x x as x > 1. Hence the given series of alternating terms are monotonely decreasing in magnitude. As n → ∞, sin 1n → sin 0 = 0, by the continuity of the sine function. Thus the conditions of the Alternating Series Test are satisfied, and the series converges. We conclude that the given series is conditionally convergent. 3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum, product, or quotient of several definite integrals, simplified as much as possible; you are not expected to evaluate the integrals. R is defined to be the region enclosed by the curves x + y = 6 and y = x2 ; C is the arc y = 3 x (−1 ≤ x ≤ 2). (a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: By solving the equations of the line x + y = 6 and the parabola y = x2 , we find the points of intersection to be (−3, 9) and (2, 4). Either of the following integrals or sums of integrals was acceptable. (Students were not expected to evaluate the integrals; I have done so here in order to verify my own work, and to show you that the integrals weren’t really hard to evaluate.) i. Using the method of “Washers”, we find the volume to be Z 2 Z 2 2 2 2 π (6 − x) − (x ) dx = π −x4 + x2 − 12x + 36 dx −3 −3 500π . 3 ii. Using the method of cylindrical shells, we find the volume to be Z 4 Z 9 √ √ (2πy) · (2 y) dy + (2πy) · (6 − y + y) dy which can be shown to be equal to 0 4 Information for Students in Lecture Section 1 of MATH 141 2010 01 3254 500π . 3 (b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sum of integrals whose value is the volume of the resulting solid. whose value can again be shown to equal DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: Here again, students were expected only to state the integrals, but not to evaluate them. i. Using the method of cylindrical shells: We will be rotating about x = 5 vertical elements of area; the width can be taken to be ∆x, as integration will be with respect to x. The element extends from a point (x, x2 ) to (x, 6 − x), so its height is 6 − x − x2 . The volume is " 4 #2 Z 2 x 4x3 11x2 1375π 2 2π(5 − x) · (6 − x − x ) dx = 2π − − + 30x = . 4 3 2 6 −3 −3 ii. Using the method of washers: Z 9 Z 4 √ 2 √ 2 √ (5 + y) − (−5 + r) dy + π ((5 + y)2 − (5 − (6 − y))2 ) dy π 4 0 Z 4 Z 9 1375π √ √ = 20π y dy + π (24 + 10 y + 3y − y2 ) dy = . 6 0 4 (c) [3 MARKS] Express in terms of integrals — which you need not evaluate — the average length that R cuts off from the vertical lines which it meets. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: The length of the segment cut off from the line x = a has been shown above to be 6 − a − a2 . The average of this function over the interval −3 ≤ a ≤ 2 is the ratio R2 (6 − a − a2 ) da 25 −3 = . 2 − (−3) 6 Information for Students in Lecture Section 1 of MATH 141 2010 01 3255 Students weren’t asked to evaluate the integral. I have done so here to verify my work; in this case I can’t verify by comparing answers obtained in two different ways, but I can, at least, examine the magnitude of my answer, and decide whether it is reasonable for the given data. For example, if my answer had been 25, I would know something was wrong, by considering the sizes of the numbers being averaged, and the maximum value obtained by the function on the given interval. (d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: When y = 3 x , y0 = 3 x ln 3. The arc length is Z 2p 1 + (3 x ln 3)2 dx . −1 (e) [3 MARKS] Given an integral whose value is the area of the surface generated by rotating C about the line y = −1; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Solution: Z 2 2π p (3 x + 1) 1 + 32x (ln 3)2 dx −1 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x x2 − 4 (x − 2) + 4 dx . x2 − 4 (x − 2) Solution: The numerator of the integrand has degree not less than that of the denominator. Accordingly, the first step is to divide denominator into numerator and to integrate Information for Students in Lecture Section 1 of MATH 141 2010 01 3256 the quotient separately, and to factorize the denominator and group like factors together: ! Z x x2 − 4 (x − 2) + 4 Z 4 dx = x+ dx x2 − 4 (x − 2) (x + 2)(x − 2)2 Z x2 4 = + dx . 2 (x + 2)(x − 2)2 Next we apply to the new integrand the method of partial fractions: 4 A B C = + + (x + 2)(x − 2)2 x + 2 x − 2 (x − 2)2 ⇒ 4 = A(x − 2)2 + B(x + 2)(x − 2) + C(x + 2) ⇒ 4 = 4C when x = 2; and 4 = 16A when x = 2 implying that A = 41 and C = 1. Comparing coefficients of x2 on both sides of the identity yields 0 = A + B, implying that B = − 41 . Now we can continue integration of the original function: Z x x2 − 4 (x − 2) + 4 Z x2 4 dx = + dx x2 − 4 (x − 2) 2 (x + 2)(x − 2)2 ! Z Z 1 x2 1 1 dx = + − dx + 2 4 x+2 x−2 (x − 2)2 x2 1 x + 2 1 − = + ln +C. 2 4 x−2 x−2 5. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Z (a) [4 MARKS] e−x · cos x dx Solution: I will apply Integration by Parts twice. First, with u = e−x , dv = cos x dx (so du = −e−x dx, v = sin x) yields Z Z −x −x e cos x dx = e sin x + e−x sin x dx , and the second (applied to the integral resulting from the first application), with U = e−x , dV = sin x dx (so dU = −e−x dx, V = − cos x) yields Z Z −x −x e cos x dx = e (sin x − cos x) − e−x cos x dx . Information for Students in Lecture Section 1 of MATH 141 2010 01 3257 By collecting both integral terms on one side, and dividing by 2, we obtain Z 1 e−x cos x dx = e−x (sin x − cos x) + C . 2 Another way to solve this problem is to assume — from experience — that the indefinite integral is equal to a sum of the form Z e−x · cos x dx = e−x (A sin x + B cos x) + C , where A and B are “undetermined” constants, to be determined. One could then differentiate this equation, top obtain e−x cos x = −e−x (A sin x + B cos x)e−x (A cos x − B sin x) . This is an identity – true for all values of x. If we assign “convenient” values, for π example x = 0 and x = , we obtain equations 2 A − B = 1, A + B = 0, which we can solved, to obtain A = 12 , B = − 12 , yielding the same solution as found earlier. 5 Z2 x (b) [5 MARKS] dx √ 2 8 + 2x − x 1 −2 Solution: 5 Z2 − 12 5 x dx = √ 8 + 2x − x2 Z2 p − 12 x 9 − (x − 1)2 dx 5 = 1 3 Z2 q − 12 I will use the substitution u = 1 3 q − 12 x 1 − ( x−1 )2 3 1 − ( x−1 )2 3 dx x−1 , so du = . 3 3 5 Z2 x 1 dx = 3 Z 1 2 − 12 3u + 1 .3 du √ 1 − u2 dx . Information for Students in Lecture Section 1 of MATH 141 2010 01 i 21 √ −3 1 − u2 + arcsin u 1 2 r r 1 1 3 3 = −3 + arcsin − −3 − arcsin 4 2 4 2 1 π π = 2 arcsin = 2 · = . 2 6 3 ! Z 1 (c) [4 MARKS] cos2 x + · tan2 x dx cos2 x Solution: ! Z Z 1 2 2 · tan x dx = sin2 x + tan2 x · sec2 x dx cos x + 2 cos x ! Z 1 − cos 2x 2 2 = + tan x · sec x dx 2 x sin 2x tan3 x − + +C. = 2 4 3 = h 6. SHOW ALL YOUR WORK! Consider the arc C defined by x = x(t) = cos t + t sin t y = y(t) = sin t − t cos t , where 0 ≤ t ≤ π2 . (a) [6 MARKS] Determine as a function of t the value of Solution: d2 y . dx2 dx = − sin t + sin t + t cos t = t cos t dt dy = cos t − cos t + t sin t = t sin t dt dy dy t sin t dt = dx = = tan t when t , 0, π2 , dx t cos t dt ! ! 2 d dy d y d dy dt = = · dx2 dx dx dt dx dx d dy d tan t sec2 t 1 dt dx dt = = = = , dx t cos t t cos t t cos3 t dt 3258 Information for Students in Lecture Section 1 of MATH 141 2010 01 3259 when t , 0, π2 . (b) [6 MARKS] Determine the area of the surface generated by revolving C about the y-axis. Solution: The area of the surface generated is Z π 2 p x(t) (t cos t)2 + (t sin t)2 dt 0 Z π2 Z π2 = 2π (cos t + t sin t)|t| dt = 2π (t cos t + t2 sin t) dt . 2π 0 0 At this point I interrupt the calculations to integrate by parts. First, taking u = t, dv = cos t dt (so du = dt, v = sin t), Z Z t cos t dt = t sin t − sin t dt = t sin t + cos t + C1 . Then, taking u = t2 , dv = sin t dt (so du = 2t dt and v = − cos t), Z Z 2 2 t sin t dt = −t cos t + 2 t cos t dt = −t2 cos t + 2(t sin t + cos t) + C2 . Incorporating these results into the earlier integral yields Z π2 2π (t cos t + t2 sin t) dt h 0 = 2π (t sin t + cos t) + (−t2 cos t + 2t sin t + 2 cos t) π = 6π − 1 . 2 i π2 0 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z π sec x dx . π 2 Solution: The integrand has an infinite discontinuity at x = Z Z π π 2 sec x dx = lim + a→( π2 ) π sec x dx a π . 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 = = 3260 lim + [ln | sec x + tan x|]πa a→( π2 ) lim + (ln | − 1 + 0| − ln | sec a + tan a|) a→( π2 ) = 0 − lim + ln | sec a + tan a| . a→( π2 ) Here both sec a and tan a are negative, as the argument is in the second quadrant, so they both approach −∞. This implies that their sum also approaches −∞, and that ln | sec a + tan a| approaches +∞; hence the improper integral approaches −∞, and is not convergent. ∞ X 4 (b) [5 MARKS] Showing all your work, carefully determine whether the series n ln n n=3 is convergent. Solution: Since both factors in the denominator are increasing, the terms of this positive series are decreasing monotonely. We can apply the Integral Test with the function f (x) = x ln4 x . But Z ∞ dx = lim [ln ln x]b3 = +∞ x ln x b→∞ 3 is a divergent, improper integral, so the series is also divergent. (c) [3 MARKS] Showing all your work, determine whether the following sequence converges; if it converges, find its limit: a1 a2 a3 a4 a5 a6 = = = = = = 1. 1.23 1.2345 1.234545 1.23454545 1.2345454545 etc., where each term after a2 is obtained from its predecessor by the addition on the right of the decimal digits 45. Solution: ! 1 45 1 + + ... 1.23 + 45(0.0001 + 0.000001 + . . .) = 1.23 + 1+ 10000 100 1002 45 1 = 1.23 + 4 · 1 10 1 − 100 Information for Students in Lecture Section 1 of MATH 141 2010 01 3261 45 5 = 1.23 + 9900 1100 1358 679 = = . 1100 550 = 1.23 + 8. SHOW ALL YOUR WORK! [10 MARKS] The polar curves r = 2 + 2 sin θ and r = 6 − 6 sin θ (0 ≤ θ ≤ 2π) (0 ≤ θ ≤ 2π) divide the plane into several regions. Showing all your work, carefully find the area of the region bounded by these curves which contains the point (r, θ) = (1, 0). Solution: (see Figure 23 on page 3262)) Both of these curves are cardioids, but their sizes and orientations are different. The first step is to determine the points of intersection of the curves. Solving the equations, by eliminating r “between” them, we obtain sin θ = 12 , so θ = π6 + 2nπ, 5π +2nπ, etc. 6 For θ in the given interval, then, we have obtained the π 5π points of intersection 3, 6 , 3, 6 . (Are these the only points of intersection? We didn’tfind the pole to be a point of intersection — but it is! It lies on the first cardioid as the point 0, 3π 2 , and on the second cardioid as the π point 0, 2 — the same point, but appearing on the two curves with different coordinates! Could there be yet other points of intersection? These could be only at points which, like the pole, appeared with different sets of coordinates. The only other way in which different coordinates can occur is because of the convention that (r, θ) and (−r, θ + π) are the same point. If we apply the transformation (r, θ) → (−r, θ + π) to the cardioid r = 2 + 2 sin θ (0 ≤ θ ≤ 2π), we obtain the equation −r = 2 − 2 sin θ (0 ≤ θ ≤ 2π); and, when we solve this equation with the original equation of the second cardioid, i.e., r = 6 − 6 sin θ (0 ≤ θ ≤ 2π), we obtain θ = π2 and r = 0, i.e., the pole. Similarly, if we apply the transformation (r, θ) → (−r, θ + π) to the cardioid r = 6 − 6 sin θ (0 ≤ θ ≤ 2π), we obtain the equation −r = 6 + 6 sin θ (0 ≤ θ ≤ 2π); and, when we solve this equation with the original equation of the first cardioid, i.e., r = 2+2 sin θ (0 ≤ θ ≤ 2π), we obtain θ = 3π 2 and r = 0, i.e., again the pole. Finally, if we apply the transformation to both of the original equations, we obtain sin θ = − 12 , r = −3, which yields the same 2 points we found in the original solving of equations. A second application of the transformation (r, θ) → (−r, θ + π) takes either equation back to its original form. To summarize, we now know all the possible points of intersection: the pole, and the two points 3, π6 , 3, 5π 6 .) The region that interests us is bounded by the two arcs: r = 2 + 2 sin θ 13π 3π ≤θ≤ 2 6 ! (97) Information for Students in Lecture Section 1 of MATH 141 2010 01 3262 4 -8 -4 0 4 8 0 -4 -8 -12 Figure 23: The cardioids with equations r = 2 + 2 sin θ, r = 6 − 6 sin θ and r = 6 − 6 sin θ π π ≤θ≤ 6 2 (98) (99) Technically the first of these curves is described incorrectly, since the values shown for the angle are not in the given interval; so we should write, instead, ! 3π ≤ θ ≤ 2π , (100) r = 2 + 2 sin θ 2 π r = 2 + 2 sin θ 0≤θ≤ , (101) 6 and (102) Information for Students in Lecture Section 1 of MATH 141 2010 01 π π ≤θ≤ . 6 2 r = 6 − 6 sin θ 3263 (103) There are various ways in which the area can be calculated (see 24 on page Figure π 3263)). One method is to first draw a line joining (0, 0) and 3, 6 . The area of the 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 -0.5 Figure 24: The region bounded by cardioids r = 2 + 2 sin θ, r = 6 − 6 sin θ and containing the point (r, θ) = (1, 0) π subregion above the line is 1 2 π 6 1 2 Z2 (6 − 6 sin θ)2 dt; the area of the region below the line is π 6 Z (2 + 2 sin θ)2 dθ. To evaluate these definite integrals, observe that − π2 Z Z 2 (1 − sin θ) dθ = (1 − 2 sin θ + sin2 θ) dθ Information for Students in Lecture Section 1 of MATH 141 2010 01 3264 ! 1 − cos 2θ = 1 − 2 sin θ + dθ 2 3θ sin 2θ = + 2 cos θ − +C. 2 4 Z In a similar fashion we can prove that Z 3θ sin 2θ (1 + sin θ)2 dθ = − 2 cos θ − +C. 2 4 Hence π 1 2 " Z2 2 (6 − 6 sin θ) dt = π 6 sin 2θ 27θ + 36 cos θ − 9 2 # π2 π 6 √ √ 3 9 3 27π = − 9π + 36 · − · 2 2 2 2 √ 63 3 = 9π − 4 π #π " Z6 1 sin 2θ 6 2 (2 + 2 sin θ) dt = 3θ + 4 cos θ − 2 2 −π 2 − π2 √ √ ! π 3 3 −3π − − = + 4 · − 2 2 2 4 √ 7 3 = 2π + , 4 √ so the total area is 11π − 18 3. This area could have been computed in other ways. For example, half the area of the smaller cardiod can be seen to be 3π, and from this one could subtract the integral π 1 2 Z2 (2 + 2 sin θ)2 − (6 − 6 sin θ)2 2 dθ , π 6 √ which can be shown to equal 18 3 − 8π. The integration of this difference of squares depends on the fact that the two curves are “traced out” in the same direction; this should be verified before using this “combined” method. Information for Students in Lecture Section 1 of MATH 141 2010 01 3265 C.34 Supplementary Notes for the Lecture of March 26th, 2010 Distribution Date: Friday, March 26th, 2010, subject to further revision C.34.1 §11.6 Absolute Convergence and the Ratio and Root Tests Finally, after meeting, in [1, §§11.2–11.4] several tests for the convergence of positive series, we see a reason for our interest in this type of series. To state the theorem we require two definitions: P P Definition C.7 1. A series an is absolutely convergent if the series |an | is convergent. P P 2. A series an is conditionally convergent if an is convergent but not absolutely conP P vergent, i.e., if an converges but |an | diverges. Note that, thus far, you must not interpret the word absolutely as an adverb modifying the participle convergent; until the following theorem is available, you should treat absolutely convergent as a two-word name for a property, nothing more. Now we state the theorem that will permit a broader interpretation of the name. Theorem C.102 A series which is absolutely convergent is convergent. (I shall not prove the theorem in the lectures, but a short proof can be found in your textbook.) Now that we have this theorem, we can interpret the word absolutely as a modifier — if we drop the word from a statement, the resulting statement is still true; we didn’t need such a reservation with the term conditionally convergent, since the definition stated that a conditionally convergent series is convergent and . . . . While investigation of the convergence of general series, i.e., series whose terms have both plus and minus signs, can be difficult, we can begin by considering the series of absolute values: if the resulting series converges, we can infer that the original series is convergent. But, if the series of absolute values diverges, or if we are unable to say anything about it, then we can make no inference at all. The Ratio Test. The Ratio Test is a test for positive series which considers the limit of the ratio of a term to its predecessor. The textbook states the test in terms of the absolute values of terms of a general series, so we will present it in that variant: Theorem C.103 (Ratio Test) convergent. P |an+1 | = L < 1, then the series an is absolutely n→∞ |an | 1. If lim P |an+1 | = L > 1, then the series an is divergent. n→∞ |an | 2. If lim Information for Students in Lecture Section 1 of MATH 141 2010 01 3266 |an+1 | = 1, or if the limit does not exist, then this test provides no information n→∞ |an | P concerning the possible convergence of the series an . 3. If lim The Root Test. The Root Test is another test for positive series; here also the textbook presents it in a form that applies to all series. pn P Theorem C.104 (Root Test) 1. If lim |an | = L < 1, then the series an is absolutely n→∞ convergent. pn P 2. If lim |an | = L > 1, then the series an is divergent. n→∞ pn 3. If lim |an | = 1, or if the limit does not exist, then this test provides no information n→∞ P concerning the possible convergence of the series an . You will need to practice on many problems in order to become comfortable with all the tests you have met, and to know the limitations of each of them. You should not be surprised if some problems are amenable to the use of more than one test. Rearrangements The textbook reports on the result of Bernhardt Riemann that the terms of a conditionally convergent series may be written in another order to create series that will converge to any given real number, and even to diverge. Example C.105 ([7, Exercise 20, p. 746]) Determine whether the series ∞ X (−1)n (ln n)n n=2 is absolutely convergent, conditionally convergent, or divergent. Solution: Consider the series of absolute values. The ratio of the (n+1)st term to its predecessor is complicated, so we will look at the nth root of the nth term, obtaining 1 →0 ln n as n → ∞. The given series is absolutely convergent. If we were interested only in checking the convergence of the given series, then the Leibniz Alternating Series Test would have been sufficient, and easy to use; but it would have given no information about absolute convergence. Example C.106 ([7, Exercise 24, p. 746]) Determine whether the series ∞ X n=1 (−1)n (arctan n)n Information for Students in Lecture Section 1 of MATH 141 2010 01 3267 is absolutely convergent, conditionally convergent, or divergent. Solution: I will apply the Root Test. The nth root of the absolute value of the nth term is 1 2 , which approaches < 1. Hence the series is absolutely convergent. arctan n π 11.6 Exercises [1, Exercise 3, p. 719] I modify the problem: Determine whether the series ∞ X (−1) n n 1, 000, 000 n! n=0 is absolutely convergent, conditionally convergent, or divergent. Solution: Apply the Ratio Test to the series of absolute values. The ratio of the (n + 1)th 100000 100000 term to the nth is , and → 0 as n → ∞. Thus, by the Ratio Test, the n+1 n+1 series is absolutely convergent. This conclusion could also be proved in other ways. [1, Exercise 12, p. 719] Determine whether the series ∞ X sin 4n n=0 4n is absolutely convergent, conditionally convergent, or divergent. Solution: Neither the Root Test nor the Ratio Test is useful here, because the limits do P not exist. Compare the series of absolute values with the geometric series 41n . Since the latter converges, so does the former. Thus the given series is absolutely convergent. [1, Exercise 16, p. 720] Determine whether the series ∞ X 3 − cos n 2 n=1 n3 − 2 is absolutely convergent, conditionally convergent, or divergent. Solution: For n > 1000 — (we don’t need the best bound here) — this is a positive series; the numerator behaves erratically, and we will not be able to calculate a limit using the Ratio or Root Tests. But since, for large n, the denominator is larger than 2 n 3 , we would expect this series to diverge by comparison with the appropriate p-series. More precisely, the divergence of the p-series with p = 32 implies the divergence of the given series, since 3 − cos n 3−1 3−1 2 > 2 > = 2 . 2 2 n3 − 2 n3 − 2 n3 n3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3268 C.35 Supplementary Notes for the Lecture of March 29th, 2010 Distribution Date: Monday, March 29th, 2010, subject to further revision C.35.1 §11.6 Absolute Convergence and the Ratio and Root Tests (conclusion) Review of the last lecture After reviewing the Leibniz Alternating Series Test, I worked examples on these series, and mentioned the concept of conditional convergence, to be defined in the next section. In §11.6 we finally saw some justification for our development of tests for the convergence of positive series. • definitions of absolute and conditional convergence • Theorem: Absolute convergence implies convergence. • The Ratio Test and Root Test are essentially tests for the convergence of positive series, but they are formulated in your textbook for any series, since the limits involved are each applied to the absolute values of roots and/or ratios. • In each of these tests a limit is investigated, and that limit must be different from 1 for the test to yield any useful information. If the limit does not exist, or if the limit exists and is equal to 1, the test fails to give any information. • When the limit exceeds 1, that implies that the terms do not approach 0, and so divergence follows from “The” Test for Divergence. 11.6 Exercises (continued) [1, Exercise 4, p. 719] Determine whether the series conditionally convergent, or divergent. ∞ X n=1 (−1)n−1 2n is absolutely convergent, n4 Solution: This is an alternating series: the only test we have to apply directly to such series is the Leibniz Alternating Series Test (and the Test for Divergence, which forms a 2n part of the Leibniz Test). To determine lim 4 , where both numerator and denominator n→∞ n become infinite, we can use L’Hospital’s Rule. This will need to be applied 4 times, after 2n (ln 2)4 which we have lim = +∞ . The terms of the original series do not approach n→∞ 24 +∞, since they are alternating in sign; but the non-existence of a limit shows that, in particular, the limit of terms is not 0, so the Test for Divergence tells us that this series must diverge. We don’t even need to check the second condition of the Leibniz Test (which could be shown to fail). Thus, even though we thought we were applying the Information for Students in Lecture Section 1 of MATH 141 2010 01 3269 Leibniz Test, the failure of convergence derives from the Test for Divergence, which is one part of the Leibniz Test. The preceding was correct, but naive. It would have been better to apply the Ratio Test, since 2n+1 n 4 n 4 (n + 1)4 = 2 lim = 2, lim = 2 lim n→∞ n + 1 n→∞ n→∞ n + 1 2n n4 implying that the given series diverges. [1, Exercise 14, p. 719] Determine whether the series gent, conditionally convergent, or divergent. ∞ X n=1 (−1)n−1 n2 2n is absolutely convern! Solution: 1. This is an Alternating Series, and we could have begun by applying the Leibniz Test. We would find that the terms are decreasing only if (n + 1)2 2n+1 n! · 2 n <1 (n + 1)! n2 √ which is equivalent, after reduction, to n − 1 > 3; thus the condition of being monotonely decreasing is ultimately satisfied (here, for n ≥ 4). But how do n2 we determine lim (−1)n+1 n ? Simply observe that the (n + 1)st term is obtained n→∞ 2 2n + 2 from the nth by multiplying by . This factor will be less than, say, 12 , when 2 n (n − 2)2 > 8, i.e., when n ≥ 5. So, for such n, each term is less than 21 of its predecessor; this shows that the nth term is bounded by a constant multiple of 21n , which approaches 0 as n → ∞. Thus the terms approach 0, so the Test for Divergence reveals nothing in itself! However, as we have now proved that both conditions of the Leibniz Test apply to this Alternating Series, we can infer that the series is convergent. 2. But, to answer the given question, we need to know whether the series is absolutely convergent or conditionally convergent, so more information is required: we have proved only that it is at least conditionally convergent. 3. To check for absolute convergence we need to apply one of our tests to the positive series obtained by changing all the signs to +. Of the tests we have, the Integral Test is not useful, since we don’t have any “nice” function available to take on the values of n! at the integer points n; there does exist such a function, but its development Information for Students in Lecture Section 1 of MATH 141 2010 01 3270 is far beyond this course. The obvious test to try is the Ratio Test, since the terms are products, and the ratio of successive terms has many factors which cancel. ! We 1 1 2(n + 1) find that the ratio of the (n + 1)st term to the nth is = 2 + 2 which n2 n n approaches 0 as n → ∞. Since this limit exists, and is less than 1, the Ratio Test tells us that the series of absolute values is convergent — i.e., that the original series is absolutely convergent. (cf. [1, Exercise 17, p. 720]) Determine whether the series ∞ X (−1)n n=0 n ln n is absolutely convergent, conditionally convergent, or divergent. Solution: Here the ratio of the absolute value of the (n + 1)th term to that of the nth is ! 1 ln(n + 1) 1+ · n ln n in which the first factor approaches 1, and the second (by L’Hospital’s Rule) also approaches 1. Thus the Ratio Test fails to give useful information about this series. We are not helpless, however. Let’s first consider the series of absolute values. The terms are decreasing and approaching 0, so we may use the Integral Test to show that it is divergent. On the other hand, the original series is alternating, and the terms are decreasing and approaching 0; hence, by the Leibniz Test, that series converges; consequently it is conditionally convergent. ∞ X (−1)n . The preceding discussion [1, Exercise 17, p. 720] is concerned with the series n ln n n=0 shows that a series with terms which are smaller in magnitude is not absolutely convergent; hence, by the Comparison Test, the series of this exercise is surely not absolutely convergent; we could also probe the absence of absolute convergence by comparing with the harmonic series. But it is conditionally convergent, since the conditions of the Leibniz test are easily shown to be satisfied. (We could also probe the absence of absolute convergence by comparing with the harmonic series.) [1, Exercise 26, p. 720] Determine whether the series 2 2 · 6 2 · 6 · 10 2 · 6 · 10 · 14 + + + + ... 5 5 · 8 5 · 8 · 11 5 · 8 · 11 · 14 is absolutely convergent, conditionally convergent, or divergent. Information for Students in Lecture Section 1 of MATH 141 2010 01 3271 4n + 2 Solution: The ratio of the (n + 1)st term to the nth is , which approaches 43 > 1. 3n + 5 By the Ratio test, this series diverges. P [1, Exercise 30, p. 720] The terms of a series an are defined “recursively” by a1 = 1, an+1 = P 2 + cos n · an (n ≥ 1). Determine whether an converges or diverges. √ n √ Solution: The ratio of the (n + 1)st to the nth term is 2 + cos n n (n ≥ 1). As n → ∞ the numerator is bounded between 1 and 3, while the denominator becomes infinitely large; by the Squeeze Theorem, applied to 0≤ an+1 3 ≤ √ , an n an+1 = 0. By the Ratio Test, this shows that the series converges abson→∞ an we see that lim lutely. [1, Exercise 32, p. 720] For which positive integers k is the series ∞ X (n!)2 convergent? (kn)! n=1 Solution: The ratio of the (n + 1)st term to the nth is ((n + 1)!)2 (n + 1)2 (k(n + 1))! (n + 1)!(n + 1)!(kn)! = = n!n!(kn + k)! (kn + k)(kn + k − 1) · . . . · (kn + 1) (n!)2 (kn)! in which the factors in the denominator are k in number, and all of them lie between n + 1 and kn + k. As long as k > 2, this fraction has more 1st degree factors in the denominator than in the numerator, so it will approach 0 as n → ∞, and we note that (n + 1)2 0 < 1; when k = 2 the fraction is , which approaches 14 < 1 as n → ∞; (2n + 2)(2n + 1) when k = 1 the fraction is equal to n + 1, and approaches +∞. By the Ratio Test the series will converge for k ≥ 2, and will diverge when k = 1. Information for Students in Lecture Section 1 of MATH 141 2010 01 3272 C.36 Supplementary Notes for the Lecture of Wednesday, March 31st, 2010 Distribution Date: Wednesday, March 31st, 2010 subject to correction C.36.1 §11.7 Strategy for Testing Series This section is concerned with improving your methods for attacking series problems. Planning strategies really needs some experience: after you have worked a substantial number of questions from earlier sections in the chapter, you might wish to read the 8 points that the textbook gives on [1, p. 721]. These points will be of much more use to you after you have accumulated some experience in applying the individual tests. Exercise C.2 ([7, Problems 2 and 4, p. 748]) You are asked to test for convergence or diver∞ ∞ X X n−1 n−1 (−1)n−1 2 gence the series and . 2 n + n n + n n=1 n=1 Solution: 1. Consider the given positive series. (a) It is useful, where possible, to formulate a good conjecture (guess) about the outcome of the investigation, in order to know whether to concentrate efforts on attempting to prove convergence or divergence. I suggest that a valuable first strategic stepX is to replace the general term by a “simpler” term: in this case I would n look at . Since this gives the harmonic series, a first guess in this case would n2 n be that the series diverges. I would thus try to use some test that relates the given series to the harmonic series. (b) The simplest method would be to use the Limit Comparison Test: n−1 2 n−1 lim n + n = lim = 1 , 0. n→∞ n→∞ n + 1 1 n Hence the given series and the Harmonic Series converge or diverge together. We know that the Harmonic Series diverges; hence the given series also diverges. (c) We could also have attacked this problem naively by using the Integral Test, with x−1 . Since f (x) = 2 x +x −x2 + 3x x(x − 3) f 0 (x) = =− , 2 x2 + x x2 + x 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3273 the function is decreasing for n > 3; and 1 − 1n n−1 = lim = 0. n→∞ n2 + n n→∞ 1 + n lim Then the series will converge or diverge according as the following improper integral converges or diverges: ! Z ∞ Z ∞ x−1 1 1 dx = − + dx x2 + x x x+1 3 3 expanding by partial fractions ! Z b 2 1 = lim − + dx b→∞ 3 x x+1 = lim [− ln x + 2 ln(x + 1)]b3 b→∞ " #b (x + 1)2 = lim ln b→∞ x 3 ! (b + 1)2 16 = lim ln − ln b→∞ b 3 ! ! 1 16 = lim ln b + 2 + − ln b→∞ b 3 = ∞. Thus this integral diverges, and the given series must also diverge. ∞ X n−1 . We know from the preceding discussion 2+n n n=1 that the given alternating series does not converge absolutely. However, the Alternating Series Test may be applied to show that the alternating series does converge. Thus the series converges conditionally. So the answer to the question as stated is that the given series does converge. 2. Now consider the series (−1)n−1 11.7 Exercises [1, Exercise 10, p. 722] Test for convergence or divergence the series ∞ P 3 n2 e−n . n=1 a function with an obvious antiderivative. More precisely, Solution: The terms suggest R 3 2 −x3 if f (x) = x e , then f (x) dx = − 13 e−x + C. Thus it would appear that we could use the Integral Test here. But let’s check that the conditions for that test are satisfied: • Is the given series positive: YES Information for Students in Lecture Section 1 of MATH 141 2010 01 3274 • Does the function f match the series values at integer terms: YES • Is f (continuous and) positive (not just at the integer points): YES • Is f decreasing? NOT CHECKED YET! 3 f 0 (x) = x 2 − 3x3 e−x , a product of 3 factors. The first factor is positive for x > 0; the last factor is an exponential, so it is always positive; and the middle factor is negative 1 for x > 32 3 , so it is certainly negative for x ≥ 1. Thus f 0 < 0, and f is decreasing. The Integral RTest tells us that the series will converge or diverge according as the improper ∞ integral 1 f (x) dx converges or diverges. But Z a 3 x2 e−x dx = − 0 1 −a 1 e − e−1 → 3 3e as a → ∞. From the convergence of the integral we infer the convergence of the given series. Another approach would have been to use the Ratio Test: 3 (n + 1)2 e−(n+1) 1 lim = 1+ 3 2 −n n→∞ n ne !2 · 1 e3n2 +3n+1 → 12 · 0 = 0 < 1 as n → ∞. The Root Test could also be used, although the calculation is more difficult: pn 2 ln n 2 as n → ∞ , n2 · e−n3 = e n · e−n → e0 · 0 = 1 · 0 = 0 < 1 by l’Hospital’s Rule. In either of these cases the limit being less than 1 implies that the series is absolutely convergent. [1, Exercise 12, p. 722] Test for convergence or divergence the series ∞ P sin n. n=1 Solution: Here is a case where the Test for Divergence is needed. None of the other tests you know will be helpful. As n → ∞, sin n does not approach a limit. (However, students in this course could not be expected to prove that fact rigorously without some strong hint.) Thus the series cannot converge. [1, Exercise 20, p. 722] Test for convergence or divergence the series ∞ X k+5 k=1 5k . Solution: The series looks at first like a geometric series; but the numerators are increasing, so we can’t compare with a geometric series; the Limit Comparison Test will also X 1 not be helpful, at least not in comparing with the Geometric Series . But several 5k other possibilities suggest themselves. Information for Students in Lecture Section 1 of MATH 141 2010 01 3275 Compare with a geometric series which converges more slowly: e.g., with P 2 k 5 . Here k+5 k k+5 1 lim 5 !k = lim k = lim k = 0. k→∞ 2 k→∞ 2 k→∞ 2 ln 2 5 Unfortunately, this limit being 0, we cannot use the form of the Limit Comparison Test given in your book. We could use the version given in [1, Exercise 40(a), p. 709], but that was not studied in our course. Ratio Test: k+6 1 ak+1 1 = lim = <1 k→∞ ak 5 k→∞ k + 5 5 hence the given positive series converges. lim Root Test: lim k→∞ √k √k 1 1 ak = lim k + 5 = lim e 5 k→∞ 5 k→∞ ln(k + 5) ln(k + 5) lim 1 k→∞ k k . = e 5 It can be shown by l’Hospital’s Rule that the exponent approaches 0, so the kth root approaches 15 , which is less than 1. In fact one can determine the actual sum here, but it involves techniques slightly beyond the course. ∞ P [1, Exercise 23, p. 722] Test for convergence or divergence the series tan n1 . n=1 1 n 1 n Solution: As n → ∞, → 0, and tan → 0 (since it is the ratio of sin 1n to cos n1 , and cos n1 → cos 0 = 1, while sin 1n → 0). Thus the Test for Divergence does not eliminate the possibility that the series may converge, just as the same test did not eliminate the possibility that the Harmonic Series might converge. If you remember that lim n→∞ tan 1n 1 n = lim sin 1n n→∞ 1 n = lim n→∞ · cos 1n sin 1n 1 n · lim sec n→∞ 1 n = 1·1=1 you can apply the Limit Comparison Test, comparing the given series with the Harmonic Series. Since the limit is a non-zero real number, and since the Harmonic Series is known to diverge, the given series must also diverge. Information for Students in Lecture Section 1 of MATH 141 2010 01 [1, Exercise 36, p. 722] Test for convergence or divergence the series 3276 ∞ X n=2 1 . (ln n)ln n Solution: Observe that 1 1 1 1 1 = (ln ln n)(ln n) = = ln ln n . = ln n ln ln n ln n (ln n) e n eln ln n eln n We can arrange for the exponent, i.e., ln ln n, to be greater than, for example, 2, by taking P 2 n > ee = 1618.17799.... Thus we may compare the given series with the p-series n12 , which is known to converge, and infer that the given series also is convergent. [1, Exercise 38, p. 722] Test for convergence or divergence the series ∞ X √n 2−1 . n=1 Solution: I start with the observation that the following factorization holds: an − bn = (a − b) an−1 + an−2 b + . . . + abn−2 + an−1 , √n 2 and b = 1, we have √n n 2 −1 √n 2 − 1 = √ n−1 √ n−2 . √n n n 2 + 2 + ... + 2 + 1 for any real numbers a and b, Taking a = The is equal to 1. All of the n terms in the denominator are smaller than √n numerator n 2 ≤ 2, so the denominator is smaller than 2n. Thus the terms are greater than terms in a multiple of the harmonic series, so the series must diverge, by the Comparison Test. We could have used the Limit Comparison Test also, since we know that √n ln 2 ln 2 lim 2 = lim e n = elimn→∞ n e0 = 1 . n→∞ n→∞ Now apply l’Hospital’s Rule: ln 2 √n √n − 2 2 2−1 lim = lim n n→∞ n→∞ 1 1 − 2 n n√ n = lim (ln 2) 2 = ln 2 , n→∞ since lim n→∞ √n 1 ln 2 lim ln 2 2 = lim eln 2 n = lim e n = en→∞ n = e0 = 1 . n→∞ n→∞ Information for Students in Lecture Section 1 of MATH 141 2010 01 3277 By the Limit Comparison Test, the divergence of the harmonic series implies the divergence of the given series. (Note that, if we alternate the signs in this problem, we obtain a series whose terms are decreasing and approaching 0, so the Alternating Series Test tells us that that series converges, i.e., that it is conditionally convergent.) C.37 Supplementary Notes for the Lecture of Wednesday, April 7th, 2010 Distribution Date: Wednesday, April 07th, 2010; corrected April 17, 2010; subject to further correction The main topic of today’s lecture will be a discussion of one version of the final examination from April, 2009. However, before beginning a report of that discussion in these notes, I am including a set of draft solutions for the final examination from April, 2008. As always, I urge you not to base your studying on final examinations from previous years. C.37.1 Final Examination in MATH 141 2008 01 (one version) This examination was written during a labour disruption, when the services of Teaching Assistants were not available for grading purposes. The following additional instructions were distributed with the examination. VERSION n McGILL UNIVERSITY FACULTY OF SCIENCE FINAL EXAMINATION IMPORTANT ADDITIONAL INSTRUCTIONS MATHEMATICS 141 2008 01 CALCULUS 2 EXAMINER: Professor W. G. Brown ASSOCIATE EXAMINER: Mr. S. Shahabi DATE: Monday, April 14th, 2008 TIME: 09:00 – 12:00 hours A. Part marks will not be awarded for any part of any question worth [4 MARKS] or less. B. To be awarded part marks on a part of a question whose maximum value is 5 marks or more, a student’s answer must be deemed to be more than 75% correct. C. While there are 100 marks available on this examination 80 MARKS CONSTITUTE A PERFECT PAPER. You may attempt as many problems as you wish. All other instructions remain valid. Where a problem requires that all work be shown, that remains the requirement; where a problem requires only that an answer be written in a box without work being graded, that also remains the requirement. Information for Students in Lecture Section 1 of MATH 141 2010 01 3279 Students are advised to spend time checking their work; for that purpose you could verify your answers by solving problems in more than one way. Remember that indefinite integrals can be checked by differentiation. W. G. Brown, Examiner. Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 4. This examination booklet consists of this cover, Pages 1 through 9 containing questions; and Pages 10, 11, and 12, which are blank. Your neighbour’s version of this test may be different from yours. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed; in some of these problems you are instructed to write the answer in a box, but a correct answer alone will not be sufficient unless it is substantiated by your work, clearly displayed outside the box. When space provided for that work is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. Information for Students in Lecture Section 1 of MATH 141 2010 01 Z (a) [2 MARKS] Evaluate 3280 4 − 6x dx . 1 + x2 Solution: Z Z Z 4 − 6x dx 2x dx = 4 −3 dx 2 2 1+x 1+x 1 + x2 = 4 arctan x − 3 ln(1 + x2 ) + C Z2 (b) [3 MARKS] Evaluate p y2 y3 + 1 dy . 0 Solution: I apply the change of variable u = y3 (or v = du = 3y2 dy: Z2 y 2 p Z y3 8 + 1 dy = √ 0 0 u+1· p y3 + 1), under which 1 du 3 i 1 2h 3 8 · (u + 1) 2 0 3 3 52 2 23 3 9 − 12 = = . 9 9 = Z (c) [3 MARKS] Evaluate sin(18 θ) · cos(30 θ) dθ . Solution: i. The easiest way to solve this problem is by using a trigonometric identity which relates products to sums: Z Z 1 sin(18 θ) cos(30 θ) dθ = (sin 48θ + sin(−12θ)) dθ 2 1 1 = − cos 48θ + cos 12θ + C . 96 24 ii. The problem could also be solved laboriously by two applications of integration by parts: Take u = sin 18θ, dv = cos 30θ dθ, implying that du = 1 sin 30θ: 18 cos 18θ, v = 30 Z Z 18 1 sin 18θ · sin 30θ − cos 18θ · sin 30θ dθ . sin(18 θ) · cos(30 θ) dθ = 30 30 Information for Students in Lecture Section 1 of MATH 141 2010 01 3281 A second application of integration by parts, with U = cos 18θ, dV = sin 30θ dθ, 1 cos 30θ, yields implying that dU = −18 sin 18θ dθ, V = − 30 Z sin(18 θ) · cos(30 θ) dθ ! Z 1 1 18 18 = − cos 18θ · cos 30θ − sin 18θ · sin 30θ − sin 18θ · cos 30θ dθ . 30 30 30 30 This equation may be solved for the desired indefinite integral, yielding Z sin(18 θ) cos(30 θ) dθ ! 1 1 18 = sin 18θ · sin 30θ + 2 cos 18θ · cos 30θ + C 2 30 30 18 1 − 30 1 = (5 sin 18θ · sin 30θ + 3 cos 18θ · cos 30θ) + C . 96 The validity of this solution can be demonstrated by differentiation. 2. SHOW ALL YOUR WORK! d (a) [3 MARKS] Simplifying your answer as much as possible, evaluate dx Solution: Z √3 Z −x d d arcsin z e dz = − earcsin z dz dx −x dx √3 d = −earcsin(−x) · (−x) dx arcsin(−x) = e = e− arcsin x Z √ 3 earcsin z dz . −x (The last simplification was not required.) (b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for the function f (x) = 3 − x, where the sample points are the left end-point of each of n subintervals of equal length. 5−2 3 3i Solution: The intervals have length = ; f (2) = 1, xi = 2 + , f (xi ) = n n n ! 3i 3i 3− 2+ = 1 − . The Riemann sum is n n ! ! n−1 n 3X 3i 3X 3(i − 1) 1− or 1− n i=0 n n i=1 n Information for Students in Lecture Section 1 of MATH 141 2010 01 3282 (c) [4 MARKS] Determine the value of the preceding Riemann sum as a function of n, simplifying your work as much as possible. (NOTE: You are being asked to determine the value of the sum as a function of n, not the limit as n → ∞.) Solution: n−1 ! n−1 3 X 3 X 3 n(n − 1) 3 i = n− · 1− n i=0 n i=0 n n 2 3 9 = − + . 2 2n 3. SHOW ALL YOUR WORK! For each of the following series determine whether the series diverges, converges conditionally, or converges absolutely. All of your work must be justified; prior to using any test you are expected to demonstrate that the test is applicable to the problem. ! ∞ X 1 (a) [4 MARKS] √ n=3 n ln n √ Solution: √ Since x, and ln x are both increasing functions of x, so is their compo√ sition ln x; since x is also an increasing function of x, so is the product x ln x. Thus the denominators of these fractions are increasing, and the fractions must be decreasing; (this is because √the function 1x is a decreasing function of x). We observe also that the limit of n n as n → ∞ is infinite; hence the limit of the terms of the series is 0. As the function x ln1 x is continuous, we have proved that the conditions of the integral test have been satisfied: the series will converge iff the integral Z∞ 1 dx converges. (The fact that the function is decreasing could also have x ln x 3 been shown by differentiation.) Z a √ ia √ √ dx = 2 ln x = 2 ln a − 2 ln 3 → ∞ as a → ∞ . √ 3 3 x ln x We conclude, by the Integral Test that the given series diverges. (The proof that the terms were the values of a function which satisfied the conditions of the theorem was an essential part of the solution.) √ ∞ X 4n + 5 n+1 (b) [4 MARKS] (−1) 3n + 10 n=1 Solution:√We may begin by considering the corresponding series of absolute val∞ X 4n + 5 ues, . If we attempt to apply the Ratio Test, we find that the limit of 3n + 10 n=1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3283 the ratios of successive terms is 1, so that test is useless here. But apply the Xwe can − 12 Limit Comparison Test, comparing with the divergent p-series n : √ q 4n + 5 4 + 5n 2 3n + 10 lim = > 0. = lim n→∞ n→∞ 3 + 10 1 3 n 1 n2 Since this limit is positive, and since the p-series to which we have compared is divergent, we may conclude that the series of absolute values is also divergent; and hence that the original series we were given is not absolutely convergent. But that still leaves open the question of whether that series is conditionally convergent or divergent. I will check whether the conditions of the Leibniz Alternating Series Test are satisfied: q √ 5 4 + n 1 4n + 5 lim = lim √ · =0 n→∞ 3n + 10 n→∞ n 3 √ 1 1 ·2· √ · (3x + 10)) − ( 4x + 5 · 3) 2 4x + 5 (3x + 10)2 −9x − 5 = < 0, √ (3x + 10)2 4x + 5 √ d 4x + 5 = dx 3x + 10 hence we may apply the Leibniz Test, and conclude that the given series is convergent. Since it has been shown to not be absolutely convergent, it is conditionally convergent. ! !! ∞ X 1 −1 −1 1 (c) [4 MARKS] cot − cot n+1 n n=1 Solution: This is a telescoping series: N X n=0 cot −1 ! !! 1 1 −1 1 . − cot = − cot−1 1 + cot−1 n+1 n N+2 As N → ∞ this partial sum approaches − cot−1 1 + cot−1 0 = − π4 + π 2 = π4 , so this is 1 , the value to which the series converges. Since the derivative of cot−1 x is − 1 + x2 the function is monotonely decreasing. Hence the terms of the given series are all positive: a positive series which is convergent is absolutely convergent. Information for Students in Lecture Section 1 of MATH 141 2010 01 3284 (The problem did not require the student to find the precise value of the sum. If one did not notice that the given series was telescopic, one could still have applied the integral test to prove that it is absolutely convergent. It can be seen that ! Z ∞ 1 −1 −1 1 cot − cot dx x+1 x 0 Z ∞ (arctan(x + 1) − arctan(x)) dx = 0 " ! !#a ln(1 + (x + 1)2 ) ln(1 + x2 ) = lim arctan(x + 1) − − arctan x − a→∞ 2 2 1 #a " ! 2 1 1 + x 1 2 = lim arctan(x + 1) − arctan x + ln = 0 − ln . a→∞ 2 1 + (x + 1)2 1 2 5 4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the curves 2y = x, y = 2x, and x2 + y2 = 5. (a) [4 MARKS] The region R is rotated about the line x = −1. Give an integral or sum of integrals whose value is the volume of the resulting solid. Solution: Using “Washers”: Z 1 Z 2 p y 2 ! 2 ! 2 y 2 2 π (2y + 1) − + 1 dy + π 5−y +1 − + 1 dy 2 2 0 1 Using Cylindrical Shells: Z 1 Z 2 √ x x 2 2π (x + 1) 2x − dx + 2π (x + 1) 5 − x − dx 2 2 0 1 (b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies inside R. Express in terms of integrals — which you need not evaluate — the average of the positive lengths L(a). Solution: ! Z 1 Z 2p 1 y y 2 dy + 5−y − dy 2y − 2 0 2 2 1 (c) [4 MARKS] Let C1 be the curve x(t) = t, y(t) = cosh t (0 ≤ t ≤ ln 2). Simplifying your answer as much as possible, find the length of C1 . UPDATED TO April 17, 2010 Information for Students in Lecture Section 1 of MATH 141 2010 01 3285 Solution: dx = 1 dt dy = sinh t dt s !2 !2 p dx dy + = 1 + sinh2 t = | cosh t| = cosh t dt dt Z ln 2 Length = cosh t dt 0 2 = [sinh t]ln 0 = sinh ln 2 − sinh 0 3ln 2 − e− ln 2 = −0 2 1 2− 2 = 3. = 2 4 5. SHOW ALL YOUR WORK! (a) [8 MARKS] Evaluate the indefinite integral Z 36 dx . (x + 4)(x − 2)2 Solution: We observe that the degree of the numerator of the integrand is less than the degree of the denominator; thus we do not need to divide denominator into numerator and obtain a quotient and remainder. Next we need to find a Partial Fraction decomposition of the integrand. Assuming a decomposition of the form 36 A B C = + + 2 (x + 4)(x − 2) x + 4 x − 2 (x − 2)2 we obtain, by taking the right side to a common denominator, 36 = A(x − 2)2 + B(x + 4)(x − 2) + C(x + 4) . Students should know two methods to find the values of these constants: either, by assigning “convenient” values to x and thereby obtaining enough equations that can be solved for the coefficients, or by comparing coefficients of powers of x on Information for Students in Lecture Section 1 of MATH 141 2010 01 3286 the two sides of the equation. These methods, or a combination of them, yield A = 1, B = −1, C = 6. Hence ! Z Z 1 6 36 1 − + dx = dx (x + 4)(x − 2)2 x + 4 x − 2 (x − 2)2 x + 4 6 − = ln +C. x−2 x−2 Z∞ (b) [4 MARKS] Determine whether find its value. Z a 3 36 dx converges. If it converges, (x + 4)(x − 2)2 3 1 + 4 36 6 a − − (ln 7 − 6) ln dx = 2 a − 2 (x + 4)(x − 2)2 1 − a → 0 − 0 − (ln 7 − 6) = 6 − ln 7 as a → ∞. Since the limit exists, the integral converges to the limiting value, 6 − ln 7. 6. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Z p √ (a) [4 MARKS] e x dx Solution: This integral can be solved by a series of substitutions; for example, the √ x substitution u = , which implies that x = 4u2 , dx = 8u du yields 2 Z p Z √ Z √ x t e dt = e 2 dx = 8 u · eu du , which can then be integrated by parts, as follows: with u = u, dv = eu du, du = du, v = eu , so Z p Z √ t e dt = 8 u · eu du ! Z u u = 8 u · e − e du = 8 ((u − 1)eu ) + C √ √x = 4 x−8 e 2 +C. Information for Students in Lecture Section 1 of MATH 141 2010 01 Z0 (b) [5 MARKS] √ − 21 x 3287 dx 3 − 4x − 4x2 Solution: √ 3 − 4x − 4x2 = = Z0 − 12 x dx = √ 3 − 4x − 4x2 = = = = Z π (c) [4 MARKS] p 4 − (1 + 2x)2 s !2 1 + 2x 2 1− . 2 Z 1 0 x q dx 2 − 12 1+2x 2 1− 2 Z 1 1 1 2 u− 2 du √ 2 0 1 − u2 " # 12 1 √ 1 − 1 − u2 − · arcsin u 2 2 0 √ 3 1 1 1 − − arcsin − (−1 − 0) 2 2 2 2 √ √ 1 3 1 π 3 π 1 − − · + 1 = − − . 2 2 2 6 2 24 4 sin2 t cos4 t dt . 0 Solution: To evaluate integrals of this type the method of the textbook is to replace the square powers of sines and cosines by functions of the cosine of twice the angle. The following is a variant of that method, using both the sine and the cosine of the double angle. Z π Z π (2 sin t · cos t)2 2 4 sin t cos t dt = · cos2 t dt 4 0 0 ! ! Z π sin2 2t 1 + cos 2t = · dt 4 2 0 ! Z π sin2 2t sin2 2t · cos 2t + = dt 8 8 0 ! Z π 1 − cos 4t 3 · sin2 2t · cos 2t · 2 + = dt 16 32 0 UPDATED TO April 17, 2010 Information for Students in Lecture Section 1 of MATH 141 2010 01 " t sin 4t sin3 2t = − + 16 64 48 π . = 16 3288 #π 0 7. SHOW ALL YOUR WORK! x = x(t) = 1 + e−t , Consider the curve C2 defined by y = y(t) = t + t2 . (a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis. Solution: y = 0 ⇒ t + t2 = 0 ⇒ t = 0, −1 t = 0 ⇒ (x, y) = (2, 0) t = −1 ⇒ (x, y) + (1 + e, 0) , so the points of intersection with the x axis are (2, 0) and (1 + e, 0). (b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent is horizontal. Solution: dx dt dy dt = −e−t = 1 + 2t dy dt dx dt dy 1 =0⇒t=− dt 2 ! √ 1 1 t=− ⇒ (x, y) = 1 + e, − . 2 4 dy =0 ⇒ dx =0⇒ (c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis. Z2 Solution: The area is y dx = 1+ 1e Z 0 Z 0 t + t2 −e−t dt −1 h i0 = e−t (t + 1) + e−t t2 + 2t + 2 −1 h i0 −t 2 = e · t + 3t + 3 −1 = 3 − e(1 − 3 + 3) = 3 − e . dx y(t) · (t) dt = dt −1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3289 The preceding evaluation required two indefinite integrals that could be obtained by integration by parts: • Taking u = t, v0 = e−t , which imply that u0 = 1, v = −e−t , Z Z −t −t t · e dt = −te + e−t dt = −e−t (t + 1) + C • Taking U = t2 , V 0 = e−t , which imply that U 0 = 2t, V = −e−t , Z Z 2 −t 2 −t t · e dt = −t e + 2t · e−t dt = −(t2 + 2t + 2)e−t + C . 8. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing all your work, determine whether the series ∞ X √ √ √ n n+2− n−2 n=2 is convergent or divergent. Solution: This is a telescoping series. N X √ √ √ n n+2− n−2 n=2 = N X √ n· √ N−2 X √ √ n+2− m· m+2 n=2 m=0 p p √ = − 3 + N(N + 2) + (N − 1)(N + 1) → ∞ as N → ∞ . As the partial sums are approaching infinity, they do not have a finite limit, and the series is, by definition, divergent. Alternatively, one could argue that √ √ √ √ n (n + 2 − n + 2) n n+2− n−2 = √ √ n+2+ n−2 √ 4 n = √ √ n+2+ n−2 4 = q → 2 , 0 as n → ∞. q 2 2 1+ n + 1− n By “The” Test for Divergence, the series diverges. Information for Students in Lecture Section 1 of MATH 141 2010 01 3290 (b) [5 MARKS] Showing all your work, determine whether the following sequence converges; if it converges, find its limit: a1 a2 a3 a4 a5 a6 = = = = = = 3. 3.14 3.1414 3.141414 3.14141414 3.1414141414 etc., where each term after a2 is obtained from its predecessor by the addition on the right of the decimal digits 14. ! 1 1 1 + + ... + an = 3 + 0.14 1 + 100 1002 100n−1 1 0.14 1 − 100 n 0.14 311 →3+ = as n → ∞ . = 3+ 1 0.99 99 1 − 100 9. SHOW ALL YOUR WORK! Curves C3 and C4 , respectively represented by polar equations r = 4 + 2 cos θ and r = 4 cos θ + 5 (0 ≤ θ ≤ 2π) (104) (0 ≤ θ ≤ 2π) , (105) divide the plane into several regions. (a) [8 MARKS] Showing all your work, carefully find the area of the one region which is bounded by C3 and C4 and contains the pole. Solution: (cf. Figure 25 on page 3291 of these notes) Solving the given!equations ! 2π 4π 2π 4π yields θ = , and r = 3. The points of intersection are 3, , 3, . 3 3 3 3 (Strictly speaking, the student should also solve using equations −r = 4 + cos(−θ) and −r = 4 cos(−θ)+5, taking all 2×2 combinations of the equations, in case a point of intersection appeared on the two given curves with different sets of coordinates; that would not have yielded any new points of intersection in this problem. It could also have been necessary to check for the curves’ possibly passing through the pole — but that does not happen in this problem, since, when we set r = 0, we obtain an equation for θ which cannot be solved.) There are several different ways of finding the area in question. Information for Students in Lecture Section 1 of MATH 141 2010 01 3291 6 4 2 -2 0 2 4 6 8 0 -2 -4 -6 Figure 25: The curves with equations r = 4 + 2 cos θ, r = 4 cos θ + 5 i. Find the area of the small oval (a limac¸on) and subtract that of the region on the left. The area of the limac¸on is Z Z 1 2π 1 2π 2 (2(1 + cos 2θ) + 16 cos θ + 16) dθ (2 cos θ + 4) dθ = 2 0 2 0 1 [18θ + sin 2θ + 16 sin θ]2π = 0 = 18π . 2 The area of the region to be subtracted is Z 1 π 2· (4 + 2 cos θ)2 − (4 cos θ + 5)2 dθ 2 2π3 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z = 3292 π 2π 3 ((16 + 16 cos θ + 2(1 + cos 2θ)) − (8(1 + cos 2θ) + 40 cos θ + 25)) dθ = [−15θ − 3 sin 2θ − 24 sin θ]π2π 3 √ √ 3 3 = (−15π − 0 − 0) − −10π + 3 · − 24 · 2 2 √ 21 3 = − 5π , 2 √ √ 21 3 21 3 so the area of the region in question is 18π − + 5π = 23π − . 2 2 ii. Find the area of the larger outer oval and subtract the area of the region to the right of the desired region. The area bounded by the larger oval is Z Z 1 2π 1 2π 2 (8(1 + cos 2θ) + 40 cos θ + 25) dθ (4 cos θ + 5) dθ = 2 0 2 0 1 = [33θ + 4 sin 2θ + 40 sin θ]2π 0 = 33π . 2 The area to be subtracted is Z 2π3 2π 1 2· (4 cos θ + 5)2 − (4 + 2 cos θ)2 dθ = [15θ + 3 sin 2θ + 24 sin θ]03 2 0 √ √ 3 3 24 3 − 0 = 10π − + 2 2 √ 21 3 . = 10π + 2 Thus the net area is equal to √ √ 21 3 21 3 = 23π − 33π − 10π + . 2 2 iii. Join the pole by line segments to the points of intersection of the two curves. Then compute the areas of the two subregions of the desired region bounded by these line segments. The area to the right of the line segments is Z 2π3 Z 2π3 1 2 (2(1 + cos 2θ) + 16 cos θ + 16) dθ (2 cos θ + 4) dθ = 2 2 0 0 2π = [18θ + sin 2θ + 16 sin θ]03 √ √ √ 3 16 3 15 3 − 0 = 12π + + . = 12π − 2 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3293 The area to the left of the line segments is Z 1 π 2 (4 cos θ + 5)2 dθ 2 2π3 = [33θ + 4 sin 2θ + 40 sin θ]π2π 3 √ = 11π − 18 3 . √ 21 3 . Summing these two areas yields 23π − 2 (b) [4 MARKS] Find another equation — call it (105*) — that also represents C4 , and has the property that there do not exist coordinates (r, θ) which satisfy equations (104) and (105*) simultaneously. You are expected to show that equations (104) and (105*) have no simultaneous solutions. Solution: In my discussion of the preceding part of the problem I have shown (by replacing (r, θ) by (−r, θ + π)) that an alternative equation for the second curve is −r = 4 cos(−θ)+5, equivalently r = −4 cos θ−5. When this equation is solved with 3 r = 4 + 2 cos θ, we obtain, as a consequence, that cos θ = − , which is impossible. 2 C.37.2 Draft Solutions to the Final Examination in MATH 141 2009 01 (Version 4) Instructions 1. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 2. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 3. . . . A TOTAL OF 75 MARKS ARE AVAILABLE ON THIS EXAMINATION. 4. You are expected to simplify all answers wherever possible. • Most questions on this paper require that you SHOW ALL YOUR WORK! . . . To be awarded partial marks on a part of a question a student’s answer for that part must be deemed to be more than 50% correct. • Some questions on this paper require only BRIEF SOLUTIONS ; . . . you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. 1. SHOW ALL YOUR WORK! Information for Students in Lecture Section 1 of MATH 141 2010 01 3294 Z (a) [4 MARKS] Evaluate t3 cos t2 dt . Solution: One substitution is u = t2 , which implies that du = 2t dt, Z Z 1 3 2 t cos t dt = u cos u du . 2 Now we can integrate by parts, taking U = u, dV = cos u du, so dU = du, V = sin u. Z Z 1 3 2 t cos t dt = u cos u du 2 ! Z 1 = u sin u − sin u du 2 1 (u sin u + cos u) + C = 2 12 t sin t2 + cos t2 + C . = 2 This problem could also have been attacked by an immediate application of integration by parts: u = t2 , dv = t cos t2 dt ⇒ du = 2t dt, v = 21 sin t2 : Z Z 2 3 2 2 1 sin t2 · 2t dt t cos t dt = t · sin t − 2 1 2 = t · sin t2 + cos t2 + C . 2 Of course, whichever method you used, you should check your answer by differentiation. (b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivative Zt2 d tanh x2 dx . dt 0 Solution: Let u = t2 . d dt Z t2 0 Z u d tanh x dx = tanh x2 dx dt 0 ! Z u d du 2 = tanh x dx · du 0 dt du = tanh t4 · 2t . = tanh u2 · dt 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3295 2. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. 1 Z2 (a) [4 MARKS] Evaluate √1 2 dx . √ 1 − x2 · arcsin x dx Solution: Let u = arcsin x, so du = √ . Then 1 − x2 1 Z2 √1 2 arcsin dx = √ 1 − x2 · arcsin x Z arcsin 1 2 du u √1 2 π Z6 = π 4 du u π = ln |u|] π6 = ln 4 π π 2 − ln = ln . 6 4 3 1 1 It should be no surprise that this answer is negative, since √ > , and the inte2 2 h i 1 √1 grand is positive in the interval 2 , 2 . Z 2y dy (b) [4 MARKS] Evaluate p y2 − y + 1 Solution: By completion of the square we find that !2 1 3 2 y −y+1 = y− + 2 4 !!2 3 1 2 1 + √ y − = 4 2 3 !2 3 2y − 1 1 + √ . = 4 3 √ 2y − 1 2 u 3+1 We may thus substitute u = √ , obtaining du = √ · dy, y = . 2 3 3 √ Z Z √ u 3 + 1 2y 3 du dy = · p √3 √ y2 − y + 1 1 + u2 2 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z √ Z du u = 3 du + √ √ 1 + u2 Z 1 + u2 √ √ du = 3 · 1 + u2 + √ 1 + u2 Z p du = 2 y2 − y + 1 + . √ 1 + u2 TO BE CONTINUED 3296 Information for Students in Lecture Section 1 of MATH 141 2010 01 3297 C.38 Supplementary Notes for the Lecture of Friday, April 09th, 2010 Distribution Date: Friday, April 09th, 2010, subject to correction C.38.1 Final Examination in MATH 141 2009 01 (Version 4, continued) Z 2y 2. (b) [4 MARKS] Evaluate dy p y2 − y + 1 Solution: By completion of the square we find that !2 3 1 + y −y+1 = y− 2 4 !!2 3 2 1 1 + √ y − = 4 2 3 !2 3 2y − 1 . 1 + √ = 4 3 2 √ 2y − 1 2 u 3+1 We may thus substitute u = √ , obtaining du = √ · dy, y = . 2 3 3 √ Z √ Z u 3 + 1 3 2y √ √ du dy = · p 3 2 2 y2 − y + 1 1 + u 2 Z √ Z u du 3 du + = √ √ 1 + u2 Z 1 + u2 √ √ du = 3 · 1 + u2 + √ 1 + u2 Z p du = 2 y2 − y + 1 + . √ 1 + u2 Now we can apply a trigonometric (or hyperbolic) substitution, like tan θ = u = 2y − 1 2y − 1 √ , i.e., θ = arctan √ , so du = sec2 θ dθ. Thus 3 3 Z Z p 2y 2 dy = 2 y − y + 1 + sec θ dθ p y2 − y + 1 p = 2 y2 − y + 1 + ln | sec θ + tan θ| + C s !2 p 2y − 1 2y − 1 = 2 y2 − y + 1 + ln 1 + √ + √ + C 3 3 Information for Students in Lecture Section 1 of MATH 141 2010 01 3298 p p 2 y2 − y + 1 + 2y − 1 + C = 2 y2 − y + 1 + ln √ 3 p p = 2 y2 − y + 1 + ln 2 y2 − y + 1 + 2y − 1 + C 0 . SHOW ALL YOUR WORK! 3. For each of the following series determine whether the series diverges, converges conditionally, or converges absolutely. All of your work must be justified; prior to using any test you are expected to demonstrate that the test is applicable to the problem. (a) [4 MARKS] ∞ X n+1 (−1) cos n n 2 n=1 Solution: We can’t use the nth root test here, as the sequence cos n has no 2 n=1,2,... limit as n → ∞. But, if we first consider the positive series of the magnitudes of the terms of the given series, we have cos n n 1 ≤ n . 0 ≤ 2 2 1 Since the series whose general term is n is a convergent (geometric) series, we 2 may apply the Comparison Test to conclude that the series of absolute values of the terms of the original series is convergent; hence the original series is absolutely convergent. ∞ X 1 (−1)n √ (b) [4 MARKS] . n ln n n=2 √ 1 Solution: As x → ∞, x, ln x, and x are all increasing. Hence f (x) = √ is x ln x decreasing. As it is a continuous function for sufficiently large x, we may apply the Integral Test. The divergence of the improper integral Z∞ 2 h √ ia dx = lim 2 ln x = +∞ , √ 2 x ln x a→∞ ∞ X 1 implies the divergence of the series of absolute values. Thus the series (−1)n √ n ln n n=2 √ is at most conditionally convergent. Since x, ln x, and x are increasing functions Information for Students in Lecture Section 1 of MATH 141 2010 01 3299 of x, x ln1 x is a decreasing function; and, as x → ∞, this last function approaches 0. Since the conditions of the Leibniz Alternating Series Theorem are satisfies, the original series is convergent; but, as it is not absolutely convergent, is is therefore conditionally convergent. ∞ X 1 (c) [4 MARKS] (−1)n ln (3n + 1) n n=4 Solution: I apply the “ Test for Divergence”: 3n · ln 3 n ln (3n + 1) lim = lim 3 + 1 n→∞ n→∞ n 1 1 n→∞ 1 + 1n 3 = ln 3 > ln e = 1 . = (ln 3) · lim Hence the limit lim (−1) n n ln (3 + 1) n n→∞ ! does not exist, and the given series diverges. 4. [9 MARKS] SHOW ALL YOUR WORK! n rπ 1 X cos2 (a) [3 MARKS] Evaluate lim · . (Hint: This could be a Riemann n→∞ n n r=1 sum.) Solution: First consider the area under the graph of f (x) = cos2 x between x = 0 π and x = π. Divide the interval into n parts of equal width ∆x = . If we hang n rectangular elements of area from the curve by their upper right-hand corners, then π iπ the area of the ith rectangle will be · cos2 . In order to be able to interpret the n n given sum as a Riemann sum, we will need to scale the function by a constant. So cos2 x let’s use the function . Thus the area will be π Zπ 0 cos2 x dx = π Zπ 0 " #π 1 1 1 1 + cos 2x dx = x + sin 2x = . 2π 2π 2 2 0 (b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an = arctan(−2n) as n → ∞. π Solution: As n → ∞, 2n → ∞, and arctan(−2n) → − . 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3300 ∞ n ∞ X X 1 (c) [3 MARKS] Showing all your work, prove divergence, or find the value of . 5i n=1 Solution: Since TO BE CONTINUED n 1 1 ∞ 1 X 1 − 1 1 5 5 5 = = , = lim 1 4 i n→∞ 5 4 1 − i=1 5 5 ∞ n ! ∞ ∞ n 1 X X 1 X 1 1 = = 4 1 = . i 5 4 1− 4 3 n=1 i=1 n=1 i=1 Information for Students in Lecture Section 1 of MATH 141 2010 01 3301 C.39 Supplementary Notes for the Lecture of Monday, April 12th, 2010 Distribution Date: Monday, April 12th, 2010, subject to correction C.39.1 Final Examination in MATH 141 2009 01 (Version 4, continued) 5. SHOW ALL YOUR WORK! (a) [8 MARKS] Evaluate the indefinite integral Z t+1 dt . 2 2t − t − 1 Solution: We observe that ! 1 2t − t − 1 = (2t + 1)(t − 1) = 2 t + (t − 1) , 2 2 a product of two distinct linear factors. There must therefore exist a partial fraction decomposition of the form t+1 A B A(t − 1) + B(2t + 1) = + = . − t − 1 2t + 1 t − 1 (2t + 1)(t − 1) 2t2 The constants can be determined by comparing coefficients of corresponding powers of t in the numerator polynomial, or as follows, by assigning to t “convenient” values of t in the identity t + 1 = A(t − 1) + B(2t + 1) (or by a combination of the two methods): 2 t = 1 ⇒ 2 = 0 + 3B ⇒ B = 3 ! 1 1 3 1 t=− ⇒ = A − +B·0 ⇒ A = − . 2 2 2 3 Hence Z ! 1 1 2 1 − · + · dt 3 2t + 1 3 t − 1 2 1 = − ln |2t + 1| + ln |t − 1| + C 6 3 4 1 (t − 1) = ln +C. 6 |2t + 1| t+1 dt = 2 2t − t − 1 Z Information for Students in Lecture Section 1 of MATH 141 2010 01 3302 (b) [2 MARKS] Determine whether the following improper integral converges or diverges; if it converges, find its value: Z∞ t+1 dt . −t−1 2t2 4 Solution: We have found an antiderivative in the preceding part. Thus Z∞ 4 " #a t+1 1 (t − 1)4 dt = lim ln a→∞ 6 2t2 − t − 1 |2t + 1| 4 = In the factorization 1 (a − 1)4 1 81 lim ln − ln . 6 a→∞ |2a + 1| 6 9 1 4 (a − 1)4 3 1 − a = a · 2a + 1 2 + 1a (a − 1)4 the last factor has limit 12 as a → ∞. Thus lim = +∞; this implies that its a→∞ 2a + 1 logarithm also approaches ∞ as a → ∞ (cf. [1, Result 4, p. A50]). Thus the given improper integral diverges. 6. SHOW ALL YOUR WORK! x = 3t2 , Consider the curve C1 defined by y = 2t3 , (t ≥ 0). (a) [7 MARKS] Showing all your work, determine the area of the surface generated when the arc 0 ≤ t ≤ 1 of C1 is rotated about the y-axis. Solution: dx = 6t dt dy = 6t2 dt s !2 !2 √ dy dx + = 6|t| 1 + t2 dt dt Z1 dx Surface Area = 2π x · dt dt 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z1 = 2π 3303 √ 3t2 · 6t 1 + t2 dt 0 Z1 = 36π √ t3 1 + t2 dt 0 √ Z 2 = 36π u2 − 1 · u2 du 0 using substitution u = " 5 # √2 3 u u = 36π − 5 3 1 √ 24π = 2+1 . 5 √ 1 + t2 , so u du = t dt (b) [2 MARKS] Showing all your work, determine all points — if any — where the normal to the curve is parallel to the line x + y = 8. Solution: The normal has slope dx 6t 1 − dt = − 2 = − . dy 6t t dt 1 Line x + y = 8 has slope −1, so we may impose the condition that − = −1, which t tells us that the normal to the curve has the desired property only at the point with parameter value t = 1. That is the point (x, y) = (3 · 12 , 2 · 13 ) = (3, 2). 7. SHOW ALL YOUR WORK! Curves C3 and C4 , are respectively represented by polar equations r = 3 + 3 cos θ (0 ≤ θ ≤ 2π) and r = 9 cos θ (0 ≤ θ ≤ π) . (106) (107) (a) [7 MARKS] Showing all your work, carefully find the area of the region lying inside both of the curves. Solution: (cf. Figure 26 on page 3304 of these notes) When we solve the two given Information for Students in Lecture Section 1 of MATH 141 2010 01 3304 Figure 26: The curves with equations r = 3 + 3 cos θ, (0 ≤ θ ≤ 2π), and r = 9 cos θ, (0 ≤ θ ≤ π) equations simultaneously (for variables r and cos θ), we find that cos θ = and r = 9 . 2 1 2 (108) The only solution θ to equation (108) which satisfy the condition π 0 ≤ θ ≤ π is θ = : thus one of the points of intersection of the two curves 3 ! 9 π is the point (r, θ) = , . Another point of intersection could be identified 2 3 by symmetry, or by solving with equation (107) the equation −r = 3 − 3 cos θ (0 ≤ θ ≤ 2π), which we can obtain from equation (106) by replacing (r, θ) respectively by (−r, θ + π); this is another equation for the same cardioid; (there is no distinct second equation for the circle, as the change described here leaves the 1 equation invariant). Simultaneous solution of these equations yields cos θ = − , 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 3305 ! 9 2π and yields the point (r, θ) = − , , which point can also be represented as 2 3 ! 9 −1π (r, θ) = , . The curves also intersect at the pole, which appears on curve C3 2 3 π with coordinates (r, θ) = (0, π), and on C4 with coordinates (r, θ) = 0, . 2 I will find the portion of the desired area above the line θ = 0, and double it. And I will divide this upper area into two parts by!the line segment joining the pole to 9 π the upper point of intersection, (r, θ) = , . The upper area to the right of this 2 3 line segment has magnitude π 1 2 Z3 π ! 1 + cos 2θ 1 + 2 cos θ + dθ 2 0 " #π 9 3θ sin 2θ 3 = + 2 sin θ + 2 2 4 0 √ 9 π 9 3 + . = 2 2 8 9 (3 + 3 cos θ) dθ = 2 2 0 Z3 The area to the left of the line segment has area π 1 2 π Z2 (9 cos θ)2 dθ = π 3 81 2 Z2 π 3 1 + cos 2θ dθ 2 " #π 81 θ sin 2θ 2 = + 2 2 4 π 3 √ 27π 81 3 − . = 4 16 Hence the area of the region interior to both closed curves is 2(9π) = 18π. Suppose that you wished to solve the problem without appealing to symmetry. Then the area could be decomposed into 3: i. the sector subtended by the pole on the cardioid; and ii. the two sectors subtended by the pole on the circle from θ = π 3 to θ = 2π . 3 The area of the sector subtended on the cardioid can be represented by an integral from θ = − π3 to θ = + π3 . To the student who might object that this appears to involve values of θ outside of the given interval of values, one could respond that the single Information for Students in Lecture Section 1 of MATH 141 2010 01 3306 integral could also be given by two separate integrals, with the same integrand, the to 2π: first from 0 to π3 , and the second from 5π 3 π 1 2 2π Z3 1 2 (3 + 3 cos θ)2 dθ + − π3 Z3 (9 cos θ)2 dθ π 3 i.e., π 2π Z2π Z3 Z3 1 1 1 2 2 (3 + 3 cos θ) dθ + (9 cos θ)2 dθ . (3 + 3 cos θ) dθ + 2 2 2 π 3 0 5π 3 (b) [3 MARKS] Determine the length of the curves which form the boundary of the region whose area you have found. Solution: The region has a boundary formed by four arcs. I will find the lengths of p the two arcs above θ = 0, and double the sum. For the arc of the circle C4 , r2 + (r0 )2 = 9, a constant; for the arc of the cardioid C3 , q q √ √ θ 2 2 2 0 2 r + (r ) = 9(1 + cos θ) + 9 sin θ = 3 2 · 1 + cos θ = 6 cos . 2 Hence the length of the arc of the cardioid is π Z3 6 θ 3 θ cos dθ = 12 sin = 6. 2 2 0 π 0 The length of the arc of the circle is π Z2 9 dθ = π 3 3π . 2 The boundary of the region is therefore of length ! 3π = 12 + 3π . 2 6+ 2 Suppose that you wished to solve the problem without appealing to symmetry. The circumference could be represented by 4 separate integrals: 1 for the circle, from Information for Students in Lecture Section 1 of MATH 141 2010 01 3307 π3 to 2π3, and 1 for the cardioid, from θ = − π3 to θ = + π3 . Again, to the student who might object that this appears to involve values of θ outside of the given interval of values, one could respond that the single integral could also be given by two separate integrals, with the same integrand, the first from 0 to π3 , and the second from 5π to 2π: 3 π Z2π Z3 Z 2π3 θ θ 6 cos dθ + 6 cos dθ + 9 dθ . π 2 2 3 5π 3 0 Information for Students in Lecture Section 1 of MATH 141 2010 01 3308 C.40 Supplementary Notes for the Lecture of Wednesday, April 14th, 2010 Distribution Date: Wednesday, April 14th, 2010, subject to correction C.40.1 Final Examination in MATH 141 2009 01 (Version 4, concclusion) 8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the 8 curves y = 2 , x = y, x = 1. x (a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensional solid, S 1 . Give an integral or sum of integrals whose value is the volume of S 1 ; you are not asked to evaluate the integral(s). Solution: Solving the given equations, we find that the vertices of the region R are (1, 8), (1,1), and (2,2). I will solve the problem in two ways. Students were not asked to evaluate the integrals, but I will do so: Using Cylindrical Shells: Z2 Volume = ! 8 (x + 1) 2 − x dx x 1 ! 8 8 −x − x + + 2 dx x x Z2 2 = 2π 1 " ! 1 = π 16 ln 2 + . 3 1 s 8 . There will Using “Washers”: We need to express the first equation as x = y be two types of “washers”, so the volume will have to be expressed as the sum of two integrals; the change in description of the right boundary of the “washer” occurs at height y = 2. 1 1 8 = 2π − · x3 − · x2 + 8 ln |x| − 3 2 x Volume #2 Z2 = π (y − (−1))2 − (1 − (−1))2 dy 1 2 Z8 s 8 2 − (−1) − (1 − (−1)) dy +π y 2 Information for Students in Lecture Section 1 of MATH 141 2010 01 Z2 Z8 2 = π y + 2y − 3 dy + π 1 3309 √ 8 4 2 + √ − 3 dy y y 2 " #2 i8 h p 1 3 2 = π · y + y − 3y + π 8 ln y + 8 2y − 3y 2 3 1 ! 7π 1 = + π(16 ln 2 − 2) = π 16 ln 2 + . 3 3 (b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensional solid, S 2 . Give an integral or sum of integrals whose value is the volume of S 2 obtained only by the Method of Cylindrical Shells; you are not asked to evaluate the integral(s). Solution: Students were asked to express this volume of revolution only by using the method of cylindrical shells, and not to evaluate the integral(s). I will solve the problem both ways. Using CylindricalsShells: I will again need to express the non-linear boundary in the form x = 8 . y Z8 s 8 = 2π y · (y − 1) dy + 2π y − 1 dy y 1 2 " 3 #2 " # √ 2 2 y2 8 y y2 3 = 2π − + 2π 8 · · y − 3 2 1 3 2 2 49π = . 3 Z2 Volume Using “Washers”: Volume ! Z2 8 2 2 π 2 − πx dx = x 1 ! 64 2 = π − x dx x4 1 #2 " x3 49π 64 . = = π − 3− 3x 3 1 3 Z2 Information for Students in Lecture Section 1 of MATH 141 2010 01 (c) [3 MARKS] Calculate the area of R. Solution: The area is Z2 1 ! " #2 8 8 x2 5 − x dx = − − = ; 2 x x 2 1 2 or, integrating along the y-axis: Z8 s 8 (y − 1) dy + − 1 dy y 1 2 #2 h " 2 i8 p y − y + 4 2y − y = 2 2 1 1 5 = + (8 − 6) = . 2 2 Z2 3310 Information for Students in MATH 141 2010 01 5001 D Problem Assignments from Previous Years D.1 1998/1999 The problem numbers listed below refer to the textbook in use at that time, [31], [33]. For many of the problems there are answers in the textbook or in the Student Solution Manual [34]. D.1.1 Assignment 1 §5.2: 5, 11, 15, 21, 29 §5.3: 3, 9, 15, 35, 47 §5.4: none §5.5: 17, 27, 33, 41 §5.6: 47, 55, 59, 65 §5.7: 21, 27, 33, 39, 45, 51, 57 §5.8: 33, 39, 45, 51, 57 D.1.2 Assignment 2 §6.1: none §6.2: 3, 9, 15, 21, 27, 31, 35, 41 §6.3: 3, 9, 15, 21, 27, 31, 39, 43 §6.4: 3, 9, 15, 21, 27, 31, 35, 41 §3.8: none Chapter 7: none Information for Students in MATH 141 2010 01 D.1.3 Assignment 3 §8.2: 5, 13, 21, 29, 39, 45, 53 §9.2: 5, 13, 21, 29, 39 §9.3: 5, 13, 21, 29, 39, 41 §9.4: 5, 13, 21, 29, 39 §9.5: 5, 9, 17, 21, 29, 33 §9.6: 5, 9, 17, 21, 29, 33 D.1.4 Assignment 4 §9.7: 13, 17, 21, 25, 29, 33 §9.8: 21, 23, 29, 33, 39 §10.2: 39, 41, 43, 45, 47, 49, 51, 53, 57 §10.3: 9, 13, 17, 21, 23, 29, 33, 35 §10.4: 3, 5, 9, 13 D.1.5 Assignment 5 §11.2: 9, 17, 23, 33, 39 §11.3: 3, 9, 15, 21, 29, 35, 47 §11.4: 3, 9, 15, 21, 29, 35, 45, 47 §11.5: 3, 9, 15, 21, 23 §11.6: 3, 9, 15, 21, 29 §11.7: 3, 9, 15, 21, 27, 33 5002 Information for Students in MATH 141 2010 01 5003 D.2 1999/2000 (Students had access to brief solutions that were mounted on the web.) D.2.1 Assignment 1 Before attempting problems on this assignment you are advised to try some “easy” problems in the textbook. In most of the following problems there is a reference to a “similar” problem in the textbook. You should always endeavour to show as much of your work as possible, and to reduce your solution to “simplest terms”. Remember that the main reason for submitting this assignment is to have an opportunity for your tutor to grade your work; the actual grade obtained should be of lesser significance. In Exercises 1-5 below, evaluate the indefinite integral, and verify by differentiation: ! Z 3 1 2 −3 1. (cf. [31, Exercise 5.2.5, p. 294]) − 5x 2 − x + 4x dx x4 ! Z 3 2 2. − dx x 1 + x2 Z 2 3. (cf. [31, Exercise 5.2.13, p. 294]) xe x − e4x dx Z 4. (cf. [31, Exercise 5.2.19, p. 294]) (1 − √ x)(2x + 3)2 dx Z 5. (cf. [31, Exercise 5.2.27, p. 294]) (4 cos 8x − 2 sin πx + cos 2πx − (sin 2π)x) dx 6. (cf. [31, Example 5.2.8, p. 289]) Determine the differentiable function y(x) such that 1 π dy 1 = √ and y 2− 2 = . dx 2 1 − x2 7. (This is [31, Exercise 5.2.51, p. 295] ! written in purely mathematical terminology.) Solve dy d dy = sin x, where y = 0 and = 0 when x = 0. [Hint: the initial value problem: dx dx dx dy First use one of the initial values to determine the general value of from the given dx “differential equation”; then use the second initial value to determine y(x) completely.] 8. ([31, Exercise 5.3.4, p. 306]) Write the following in “expanded notation”, i.e. without 6 P X using the symbol : (2 j − 1). j=1 Information for Students in MATH 141 2010 01 5004 9. (cf. [31, Exercise 5.3.18, p. 306]) Write the following sum in “summation notation”: x3 x5 x7 x999 + − + ... ± 3 5 7 999 where the signs are alternating +, −, +, −, ... The sign of the last term has not been given — you should determine it. x− 10. (cf. [31, Example 5.3.6, p. 302]) Given that n X i= i=1 n(n + 1) , 2 n X n(n + 1)(2n + 1) , 6 i2 = i=1 n X i=1 i3 = n2 (n + 1)2 , 4 (n + 1)3 + (n + 2)3 + ... + (2n)3 . n→∞ n4 determine lim D.2.2 Assignment 2 1. Evaluate the following integrals: Z 3 (a) (x − 1)4 dx 1 Z 1 (b) (2e x − 1)2 dx Z0 π (c) sin 4x dx. 0 2. Interpreting the following integral as the area of a region, evaluate it using known area formulas: Z 6√ 36 − x2 dx. 0 3. Use properties of integrals to establish the following inequality without evaluating the integral: Z 1 Z 1 1 1 dx. √ dx ≤ 3 x 0 1+ 0 1+ x 4. Deduce the Second Comparison Property of integrals from the First Comparison Property [31, p. 325, §5.5]. 5. Apply the Fundamental Theorem of Calculus [31, p. 331, §5.6] to find the derivative of the given function: Z x −1 (t2 + 2)15 dt. Information for Students in MATH 141 2010 01 5005 6. Differentiate the functions Z x3 cos t dt (a) Z 0 3x (b) sin t2 dt. 1 √ dy 7. Solve the initial value problem = 1 + x2 , y(1) = 5 . Express your answer in dx terms of a definite integral (which you need not attempt to evaluate). This problem can be solved using the methods of [31, Chapter 5]. 8. Evaluate the indefinite integrals: Z √ 2x 3 − 2x2 dx (a) Z (b) x2 sin(3x3 ) dx Z x+3 (c) dx 2 x + 6x + 3 9. Evaluate the definite integrals: Z 8 √ (a) t t + 2 dt Z 0 π/2 (b) Z (1 + 3 sin η)3/2 cos η dη 0 π (c) sin2 2t dt. 0 10. Sketch the region bounded by the given curves, then find its area: (a) x = 4y2 , (b) y = cos x, x + 12y + 5 = 0 y = sin x, 0≤x≤ π . 4 x 2 y2 + = 1 is A = πab. This problem can be 11. Prove that the area of the ellipse a2 b2 solved using the methods of [31, Chapter 5]. It is not necessary to use methods of [31, Chapter 9]. UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 5006 D.2.3 Assignment 3 In all of these problems you are expected to show all your work neatly. (This assignment is only a sampling. Your are advised to try other problems from your textbook; solutions to some can be found in the Student Solution Manual [32].) 1. [31, Exercise 6.1.6, p. 382] As n → ∞, the interval [2, 4] is to be subdivided into n subintervals of equal length ∆x by n − 1 equally spaced points x1 , x2 , ..., xn−1 (where n X 1 x0 = 2, xn = 4). Evaluate lim ∆x by computing the value of the appropriate n→∞ x i i=1 related integral. 2. (a) [31, Exercise 6.2.6, p. 391] Use the method of cross-sections to find the volume of the solid that is generated by rotating the plane region bounded by y = 9 − x2 and y = 0 about the x-axis. (b) (cf. Problem 2a) Use the method of cylindrical shells to find the volume of the solid that is generated by rotating the plane region bounded by y = 9− x2 and y = 0 about the x-axis. (c) Use the method of cross-sections to find the volume of the solid that is generated by rotating the plane region bounded by y = 9 − x2 and y = 0 about the y-axis. (d) (cf. Problem 2c) Use the method of cylindrical shells to find the volume of the solid that is generated by rotating the plane region bounded by y = 9− x2 and y = 0 about the y-axis. 3. (a) [31, Exercise 6.2.24, p. 392] Find the volume of the solid that is generated by rotating around the line y = −1 the region bounded by y = 2e−x , y = 2, and x = 1. (b) (cf. Problem 3a) Set up an integral that would be obtained if the method of cylindrical shells were used to represent the volume of the solid that is generated by rotating around the line y = −1 the region bounded by y = 2e−x , y = 2, and x = 1. YOU ARE NOT EXPECTED TO EVALUATE THE INTEGRAL. 4. (cf. [31, Exercise 6.2.40, p. 392]) The base of a certain solid is a circular disk with diameter AB of length 2a. Find the volume of the solid if each cross section perpendicular to AB is an equilateral triangle. 5. (a) [31, Exercise 6.3.26, p. 401] Use the method of cylindrical shells to find the volume of the solid generated by rotating around the y-axis the region bounded by the 1 , y = 0, x = 0, x = 2. curves y = 1 + x2 Information for Students in MATH 141 2010 01 5007 (b) (cf. Problem 5a) Use the method of cross sections to find the volume of the solid generated by rotating around the y-axis the region bounded by the curves y = 1 , y = 0, x = 0, x = 2. 1 + x2 e x + e−x 2 6. (cf. [31, Exercise 7.3.69, p. 450]) Find the length of the arc of the curve y = between the points (0, 1) and (ln 2, 2). 7. (a) [31, Exercise 6.4.30, p. 411] Find the area of the surface of revolution generated by revolving the arc of the curve y = x3 from x = 1 to x = 2 around the x-axis. (b) (cf. 7a) Set up an integral for, BUT DO NOT EVALUATE, the area of the surface of revolution generated by revolving the arc of the curve y = x3 from x = 1 to x = 2 around the y-axis. 8. [31, Exercise 7.2.44, p. 442] Evaluate the indefinite integral Z x+1 dx 2 x + 2x + 3 Z x 9. (cf. [31, Exercise 7.2.36, p. 442]) Determine the value of the function f (x) = for any point x < 2. −1 t2 dt 8 − t3 10. (cf. [31, Exercise 7.3.70, p. 450]) Find the area of the surface generated by revolving around the x-axis the curve of Problem 6. D.2.4 Assignment 4 1. Differentiate the functions: (a) sin−1 (x50 ) (b) arcsin(tan x) (c) cot−1 e x + tan−1 e−x 2. Showing all your work, evaluate the integrals: Z dx (a) √ 1 − 4x2 Z dx (b) √ 2 x(1 + x) Information for Students in MATH 141 2010 01 5008 Z ex dx 1 + e2x √ √ Z cot y csc y (d) dy √ y Z (ln t)8 (e) dt t Z (f) tan4 2x sec2 2x dx (c) (g) THIS PROBLEM BE OMITTED. IT MAY BE INCLUDED IN ASZ SHOULD x2 dx SIGNMENT 5. √ 16x2 + 9 3. Use integration by parts to compute the following integrals. Show all your work. Z t cos t dt (a) Z √ (b) y ln y dy (c) THIS PROBLEM Z SHOULD BE OMITTED. IT MAY BE INCLUDED IN ASSIGNMENT 5. x2 arctan x dx Z (d) csc3 x dx Z (e) ln(1 + x2 ) dx 4. Showing all your work, evaluate the following integrals: Z (a) cos2 7x dx Z (b) cos2 x sin3 x dx Z (c) Z (d) sin3 2x dx cos2 2x sec6 2t dt Information for Students in MATH 141 2010 01 5009 D.2.5 Assignment 5 Z x3 dx. (Your solution should be valid for x x2 + x − 6 in any one of the intervals x < 3, −3 < x < 2, x > 2.) Z 1 2. [31, Exercise 9.5.8, p. 540] Find dx. (x + 1)(x2 + 1) Z x2 3. (a) [31, Exercise 9.5.23] Find dx. (x + 2)3 (b) Find the volume of the solid of revolution generated by the region bounded by x y= , y = 0, x = 1, and x = 2 about the x-axis. 3 (x + 2) 2 (c) Find the volume of the solid of revolution generated by the region bounded by x , y = 0, x = 1, and x = 2 about the y-axis. y= 3 (x + 2) 2 1. [31, Exercise 9.5.6, p. 540] Find 4. [31, Exercise 9.5.38, p. 540] Make a preliminary substitution before using the method of partial fractions: Z cos θ dθ 2 sin θ(sin θ − 6) 5. [31, 9.6.6, p. 547] Use trigonometric substitutions to evaluate the integral Z Exercise x2 dx. √ 9 − 4x2 6. [31, Z Exercise 9.6.26, p. 547] Use trigonometric substitutions to evaluate the integral 1 dx. 9 + 4x2 7. [31,√Exercise 9.6.35, p. 547] Use trigonometric substitutions to evaluate the integral Z x2 − 5 dx. x2 Z √ 8. [31, Exercise 9.7.14, p. 553] Evaluate the integral x 8 + 2x − x2 dx. 9. [31, Exercise 9.8.17, p. 561] Determine the following improper integral conZ whether ∞ x verges; if it does converge, evaluate it: dx. 2 −∞ x + 4 10. [31, Exercise 9.8.27, p. 561] Determine the following improper integral conZ whether ∞ verges; if it does converge, evaluate it: cos x dx. 0 Information for Students in MATH 141 2010 01 5010 11. (cf. [31, Exercise 9.8.14, p. 561]) Determine Z +8whether the following improper integral 1 dx. converges; if it does converge, evaluate it: 2 −8 (x + 4) 3 12. [31, Exercise 10.2.2, p. 580] Find two polar coordinate representations, one with r ≥ 0, and the other with r ≤ 0 for the points with the following rectangular coordinates: (a) (−1, −1), √ (b) ( 3, −1), (c) (2, 2), √ (d) (−1, 3), √ √ (e) ( 2, − 2), √ (f) (−3, 3). 13. For each of the following curves, determine — showing all your work — equations in both rectangular and polar coordinates: (a) [31, Exercise 10.2.20, p. 580] The horizontal line through (1, 3). (b) [31, Exercise 10.2.26, p. 580] The circle with centre (3, 4) and radius 5. 14. (a) [31, Exercise 10.2.56, p. 581] Showing all your work, find all points of intersection of the curves with polar equations r = 1 + cos θ and r = 10 sin θ. (b) Showing all your work, find all points of intersection of the curves with polar equations r2 = 4 sin θ and r2 = −4 sin θ. [Note: The procedure sketched in the solution of [31, Example 10.2.8, p. 579] for finding points of intersection is incomplete. Your instructor will discuss a systematic procedure in the lectures.] D.2.6 Assignment 6 1. Find the area bounded by each of the following curves. (a) r = 2 cos θ, (b) r = 1 + cos θ. 2. Find the area bounded by one loop of the given curve. (a) r = 2 cos 2θ, (b) r2 = 4 sin θ. Information for Students in MATH 141 2010 01 5011 3. Find the area of the region described. (a) Inside both r = cos θ (b) Inside both r = 2 cos θ and and r= √ 3 sin θ. r = 2 sin θ . 4. Eliminate the parameter and then sketch the curve. (a) x = t + 1, y = 2t2 − t − 1. (b) x = et , y = 4e2t . (c) x = sin 2πt, y = cos 2πt; 0 ≤ t ≤ 1. Describe the motion of the point (x(t), y(t)) as t varies in the given interval. 5. Find the area of the region that lies between the parametric curve x = cos t, y = sin2 t, 0 ≤ t ≤ π, and the x-axis. 6. Find the arc length of the curve x = sin t − cos t, y = sin t + cos t; π/4 ≤ t ≤ π/2. 7. Determine whether the sequence an converges, and find its limit if it does converge. n2 − n + 7 , 2n3 + n2 √ 1 + (−1)n n (b) an = , (3/2)n (c) an = n sin πn, !n n−1 (d) an = . n+1 (a) an = 8. Determine, for each of the following infinite series, whether it converges or diverges. If it converges, find its sum. (a) 1 + 3 + 5 + 7 + . . . + (2n − 1) + . . . , (b) 4 + 43 + . . . + 34n + . . . , ∞ X (c) (5−n − 7−n ), (d) n=1 ∞ X n=1 e n . π 9. Find the set of all those values of x for which the series ∞ n X x 3 ric series, then express the sum of the series as a function of x. n=1 is a convergent geomet- Information for Students in MATH 141 2010 01 5012 10. Find the Taylor polynomial in powers of x − a with remainder by using the given values of a and n. (a) f (x) = sin x; a = π/6, n = 3. 1 (b) f (x) = ; a = 5, n = 5 . (x − 4)2 11. Find the Maclaurin series of the function e−3x by substitution in the series for e x . 12. Find the Taylor series for f (x) = ln x at the point a = 1. 13. Use comparison tests to determine whether each of the following infinite series converge or diverge. ∞ X 1 , 1 + 3n n=1 √ ∞ X n (b) , 2 n +n n=1 (a) (c) ∞ X sin2 (1/n) n=1 n2 . D.3 2000/2001 (In the winter of the year 2001 Assignments based on WeBWorK were used, although the experiment had to be terminated in mid-term because of technical problems.) D.4 2001/2002 This was the first time WeBWorK assignments were used exclusively in this course. D.5 MATH 141 2003 01 WeBWorK assignments were used exclusively for assignments. The questions are not available for publication. D.6 MATH 141 2004 01 WeBWorK assignments were used exclusively for assignments. The questions are not available for publication. Information for Students in MATH 141 2010 01 5013 D.7 MATH 141 2005 01 WeBWorK assignments were used for online assignments; the questions are not available for publication. In addition, these written assignments were intended to provide students with individualized opportunities to work problems for which the textbook often provided examples; at the time, appropriate materials available from WeBWorK for this purpose were limited. The individualization was often based on the student number. D.7.1 Written Assignment W1 Your written assignments will usually be mounted on the WeBWorK site, and will usually be individualized, that is, your problems will not be exactly the same as those of other students. This first written assignment is based on your WeBWorK assignment R1 . Subsequent written assignments may be designed in other ways. Because of its general form, it was possible to release this assignment in the document Information for Students in MATH 141 2005 01; some of the other assignments may appear only on your WeBWorK site. Your completed assignment must be submitted together with your solutions to quiz Q1 , inside your answer sheet for that quiz. No other method of submission is acceptable. Purpose of the written assignments These assignments are designed to help you learn how to write full solutions to problems. While they carry a very small weight in the computation of your final grade, conscientious completion of the assignments should help you substantially in learning the calculus, and help prepare you for your final examination. Certificate Your assignment will not be graded unless you attach or include the following completed certificate of originality: I have read the information on the web page http://www.mcgill.ca/integrity/studentguide/, and assert that my work submitted for W1 and R1 does not violate McGill’s regulations concerning plagiarism. Signature(required) The assignment questions sion of R1 : Date(required) Your assignment consists of the following problems on your ver- Information for Students in MATH 141 2010 01 5014 ##3, 4, 5, 6, 7, 8 (Teaching Assistants are not primarily checking for plagiarism; but, if they detect it, they may be obliged to report any apparent violations to the Associate Deans.) Complete solutions are required It is not enough to give the correct answer; in fact, the numerical answer alone may be worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should look if you have doubts about the amount of detail required in a solution. Use of calculators You are expected to complete the entire assignment without the use of a calculator. In particular, you are expected to be familiar with the values of trigonometric functions at “simple” multiples and submultiples of π. Not all problems may be graded On all of the written assignments it is possible that the Teaching Assistant will grade only a small number of the solutions you submit. The numbers of the questions that will be graded will not be announced in advance, even to the tutor. For that reason you are advised to devote equal attention to all of the problems. D.7.2 Written Assignment W2 Written assignment W2 is based on your WeBWorK assignment R3 , but some problems are being modified. Your completed assignment must be submitted together with your solutions to quiz Q2 , inside your answer sheet for that quiz. No other method of submission is acceptable. Purpose of the written assignments These assignments are designed to help you learn how to write full solutions to problems. While they carry a very small weight in the computation of your final grade, conscientious completion of the assignments should help you substantially in learning the calculus, and help prepare you for your final examination. Certificate Your assignment will not be graded unless you attach or include the following completed certificate of originality, signed in ink: Information for Students in MATH 141 2010 01 5015 I have read the information on the web page http://www.mcgill.ca/integrity/studentguide/, and assert that my work submitted for W2 and R3 does not violate McGill’s regulations concerning plagiarism. Signature(required) The assignment questions sion of R1 : Date(required) Your assignment consists of the following problems on your ver- 1. Problem 1 of R3 , solved by integration with respect to x. Include in your solution a rough sketch of the region, showing a typical element of area. 2. Problem 1 of R3 solved by integration with respect to y. This will require rewriting the equations of the curves appropriately. You may assume without proof that Z ln x dx = x(ln x − 1) + C , a fact which you will see derived later in the course. Include in your solution a rough sketch of the region, showing a typical element of area. 3. Problem 6 of R3 , evaluated using the Method of Washers. Include in your solution a rough sketch of the plane region which generates the solid, showing a typical element of area which will generate a typical washer. 4. Problem 6 of R3 , evaluated using the Method of Cylindrical Shells. Include in your solution a rough sketch of the plane region which generates the solid, showing a typical element of area which will generate a typical cylindrical shell. Complete solutions are required It is not enough to give the correct answer; in fact, the numerical answer alone may be worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should look if you have doubts about the amount of detail required in a solution. Use of calculators calculator. You are expected to complete the entire assignment without the use of a Information for Students in MATH 141 2010 01 5016 Not all problems may be graded On all of the written assignments it is possible that the Teaching Assistant will grade only a small number of the solutions you submit. The numbers of the questions that will be graded will not be announced in advance, even to the tutor. For that reason you are advised to devote equal attention to all of the problems. D.7.3 Written Assignment W3 Unlike the preceding written assignments Written Assignment W3 is not directly based on your WeBWorK assignments, although some problems will be similar to WeBWorK assignment problems. Your completed assignment must be submitted together with your solutions to quiz Q3 , inside your answer sheet for that quiz. No other method of submission is acceptable. Certificate Your assignment will not be graded unless you attach or include the following completed statement of originality, signed in ink: I have read the information on the web page http://www.mcgill.ca/integrity/studentguide/, and assert that my work submitted for W3 does not violate McGill’s regulations concerning plagiarism. Signature(required) Date(required) The assignment questions The parameters in these problems are based on the digits of your 9-digit McGill student number, according to the following table: Parameter name: A B C D E F G H J Your student number: Before starting to solve the problems below, determine the values of each of these integer constants; then substitute them into the descriptions of the problems before you begin your solution. Z 1. Showing all your work, systematically determine Ax2 + Bx + C e−x dx by repeated integration by parts: no other method of solution will be accepted. Verify by differentiation that your answer is correct. Information for Students in MATH 141 2010 01 5017 2. Showing all your work, use trigonometric or hyperbolic substitutions to evaluate Z Z Z du du du each of , , . p p p u2 − (F + 1)2 u2 + (F + 1)2 −u2 + (F + 1)2 Verify that your answers are correct by differentiation. j k h i 3. Let K = J+4 = J+4 .62 Showing all your work, develop for this integer K a reduction 2 2 formulaZ of the following type that can be used to evaluate In (x) = xn (sin(K x)) dx in terms of In−2 (x): Z Z n n x (sin K x) dx = L · x cos K x + M · x n−1 sin K x + N xn−2 (sin K x) dx where n ≥ 2 and L, M, N are constants that you are expected to determine only by integration by parts. Again showing all your work, use the reduction formula you have R just determined to evaluate x2 sin K x dx, and test by differentiation the answer that it gives — you should recover x2 sin K x. 4. Showing all your work, evaluate both of the integrals Z Z F+1 2 sin x · cos x dx and sinF+2 x · cos2 x dx Complete solutions are required It is not enough to give the correct answer; in fact, the final answer alone may be worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should look if you have doubts about the amount of detail required in a solution. Not all problems may be graded On all of the written assignments it is possible that the Teaching Assistant will grade only a small number of the solutions you submit. The numbers of the questions that will be graded will not be announced in advance, even to the tutor. If for no other reason, you are advised to devote equal attention to all of the problems. D.7.4 Written Assignment W4 Your completed assignment must be submitted together with your solutions to quiz Q4 , inside your answer sheet for that quiz. No other method of submission is acceptable. 62 Determine Khfrom i J using the greatest integer function, defined j k in your textbook, page 110. Your textbook J+4 uses the notation 2 , but some authors write the function as J+4 2 . Information for Students in MATH 141 2010 01 5018 Certificate Your assignment will not be graded unless you attach or include the following completed statement of originality, signed in ink: I have read the information on the web page http://www.mcgill.ca/integrity/studentguide/, and assert that my work submitted for W4 does not violate McGill’s regulations concerning plagiarism. Signature(required) Date(required) The assignment questions Some of the parameters in these problems are based on the digits of your 9-digit McGill student number, according to the following table: Parameter name: A B M D E F G H J Your student number: Before starting to solve the problems below, determine the values of each of these integer constants; then substitute them into the descriptions of the problems before you begin your solution. It is not enough to give the correct answer; in fact, the final answer alone could be worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should look if you have doubts about the amount of detail required in a solution. Not all problems may be graded. 1. Showing all your work, systematically determine the value of the following integral: Z x dx . Verify your work by differentiation of your answer: you 2 (x + J) (x − H − 1)1 should recover the integrand. (Systematically means that you to use the methods you learned in this course for the treatment of problems of this type, even if you happen to see some other method that could be used in this particular example.) 2. Showing all your work, use a substitution to transform the integrand into a rational function, then integrate the particular integral that is assigned for your particular value of G: Z 1 If G = 1,4,or 7: dx √ x− x+2 Z cos 2x dx If G = 0, 2, 5, or 8: sin2 2x + sin 2x Information for Students in MATH 141 2010 01 Z If G = 3, 6, or 9: 5019 1 dx √ x−2−2 x+1 3. Problem 55, pages 504-505 of your textbook, describes a substitution discovered by Karl Weierstrass (1815-1897) for the evaluation of rational functions of sin x and cos x into an ordinary rational function. It states that if, for x such that −π < x < π, we define 1 − t2 2t and sin x = and that, at a consequence, dx = t = tan 2x , then cos x = 1 + t2 1 + t2 2 dt . You are not asked to verify these facts. You are asked to use the substitution 1 + t2 to transform the following integral and then to evaluate it: Z π 2 π 3 1 dx 1 + sin x − cos x 4. Use the trigonometric identities given in your textbook on page 487 to evaluate the folR lowing integral: cos Mx · cos Dx · sin x dx where M and D are the digits of your student number, defined above. 5. Showing all your work, determine whether each of the following integrals is convergent or divergent. If it is convergent, determine its value (again showing all your work): Z 4 1 (a) dx 0 2+E ·F Z ∞ 2 x −E−F (b) dx x2 + E + F 0 π Z B+4 1 (c) dx x sin((G + 2)x) 0 D.7.5 Written Assignment W5 Your completed assignment must be submitted together with your solutions to quiz Q5 , inside your answer sheet for that quiz. No other method of submission is acceptable. Certificate Your assignment will not be graded unless you attach or include the following completed statement of originality, signed in ink: Information for Students in MATH 141 2010 01 5020 I have read the information on the web page http://www.mcgill.ca/integrity/studentguide/, and assert that my work submitted for W5 does not violate McGill’s regulations concerning plagiarism. Signature(required) Date(required) The assignment questions Some of the parameters in these problems are based on the digits of your 9-digit McGill student number, according to the following table: Parameter name: A B D E F G H J K Your student number: Before starting to solve the problems below, determine the values of each of these integer constants; then substitute them into the descriptions of the problems before you begin your solution. It is not enough to give the correct answer; in fact, the final answer alone could be worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should look if you have doubts about the amount of detail required in a solution. Not all problems may be graded. π 1. For the point with polar coordinates H, − 2 , K +5 (a) find four other pairs of polar coordinates, two with r ≤ 0 and two with r > 0: (b) Find the cartesian coordinates, assuming that the positive x-axis is along the polar axis, the origin is at the pole, and the positive y-axis is obtained by turning the polar axis through a positive angle of π2 . 2. For the point whose cartesian coordinates are (F 2 +1, F 2 −2), determine polar coordinates (r, θ) with the following properties: (a) r > 0, 0 ≤ θ < 2π (b) r < 0, 0 ≤ θ < 2π (c) r > 0, 5π 2 ≤θ< 9π 2 3. For the following curve given in polar coordinates, determine, showing all your compuπ tations, the slope of the tangent at the point with θ = : 4 r = A + cos(B + 2)θ . Information for Students in MATH 141 2010 01 5021 4. Showing all your work, find the area contained between the outer loop and the inner loop of the curve r = 1 + 2 sin θ . Explain carefully how you have established the limits for your definite integral. 5. The curve C is given by the parametric equations x = 1 + (C + 2)t2 y = t − (E 2 − 2)t3 Showing all your work, determine the value of d2 y (t). dx2 6. Giving an explanation, determine whether or not the following sequences converge. Where a sequence converges, find its limit. (a) an = ln n − ln(n + A + 1) ln n (b) an = ln((A + 2)n) √ √ (c) an = n2 + A2 − n2 + A2 + 2 D.8 MATH 141 2006 01 D.8.1 Solution to Written Assignment W1 Release Date: Friday, 27 January, 2006 The Assignment was posted to the class via WeBCT on January 2nd, 2006 Instructions to Students 1. Your solution to this assignment should be brought to your regular tutorial, during the week 16-20 January, and folded inside your solution sheet to the quiz which will be written at that time. 2. Your TA may make special arrangements for submission at other times, but, no solutions are ever accepted after the end of the week. 3. Your solution must use the data on your own WeBWorK assignment, as described below. Information for Students in MATH 141 2010 01 5022 The Assignment Question Problem 7 of WeBWorK assignment R1 requires that you evaluate a definite integral of the form Z C (Ex2 − Ax + B) dx, D where A, B, C, D, E are various combinations of constants, individualized for each student. Your written assignment is to evaluate a simplified version of this integral as the limit of a sequence of Riemann sums, using left endpoints63 . You may simplify your problem only in the following way: you may take the lower limit of the integral (here called D) to be 0; this is purely to make the algebra a little easier. You should model your proof on that given in [7, Example 2(b), pp. 383-385]. You are expected to provide a full solution for your version of the problem — it is not enough to supply the correct answer, and you must solve the version of the problem on your own WeBWorK assignment R1 . In your solution you will need to use the well known formulæ for the sums of the 1st and 2nd powers of the natural numbers from 1 to n; these formulæ are in your textbook [7, Formulæ ##4, 5, p. 383], and you will not be expected to prove them here.64 (This is a time-consuming exercise; the purpose of the assignment is to ensure that you will have correctly solved a problem of this type during the term.) The Solution (This solution has been composed with variables, so that it will produce valid numerical solutions for all students. The particular solution for your version of the problem should look much simpler, because, in place of all variables except n, there will be specific integers.) The length of each of n subintervals is ∆x = Thus xi = D + i∆x = D + i formula, p. 381] Z C D = = 63 lim n→∞ lim n→∞ C − D n C−D . n . Since we are using left endpoints we apply [7, top Ex2 − Ax + B dx n X 2 Exi−1 − Axi−1 + B · ∆x i=1 n X i=1 2D(C − D) (C − D)2 E D + · (i − 1) + · (i − 1)2 n n2 ! 2 Note that the solution given in the textbook uses right endpoints. Of course, these formulæ may be proved by induction, and students who took MATH 140 2005 09 should know how to write up such proofs if they had to. 64 Information for Students in MATH 141 2010 01 5023 C−D −A D + · (i − 1) + B · ∆x n n n n X (2DE − A)(C − D) X E(C − D)2 X 2 2 = lim (ED − AD + B) 1+ (i − 1) + (i − 1) · ∆x 2 n→∞ n n i=1 i=1 i=1 Now n X 1 = sum of n 1’s i=1 n X = n n−1 X (i − 1) = j i=1 j=1 (n − 1)n 2 n−1 X j2 = = n X (i − 1)2 by [7, Formula 4, p. 383] j=1 i=1 (n − 1)n(2n − 1) = 6 Substituting the values of these sums yields Z C by [7, Formula 5, p. 383]. Ex2 − Ax + B dx D = = = = = C−D (2DE − A)(C − D) (n − 1)n · (ED2 − AD + B)n + · n→∞ n n 2 ! 2 E(C − D) (n − 1)n(2n − 1) + · 6 n2 ! 1 (2DE − A)(C − D)2 2 1− lim (ED − AD + B)(C − D) + n→∞ 2 n ! !! 3 E(C − D) 1 1 + · 1− 2− 6 n n (2DE − A)(C − D)2 E(C − D)3 (ED2 − AD + B)(C − D) + ·1+ ·1·2 2 6 (2DE − A)(C − D)2 E(C − D)3 (ED2 − AD + B)(C − D) + + 2 3 3 3 2 2 E C −D A C −D − + B(C − D) 3 2 lim It was suggested that you simplify your problem by taking D = 0; in that case the value would have worked out to be EC 3 AC 2 − + BC . 3 2 Information for Students in MATH 141 2010 01 The Grading Scheme 5024 The assignment was to be graded OUT OF A TOTAL OF 10 MARKS. (In the version of this solution circulated to TA’s, there followed some technical details about the grading.) D.8.2 Solution to Written Assignment W2 Release Date: Mounted on the Web on February 8th, 2006; corrected on March 16th, 2006 Solutions were to be submitted inside answer sheets to Quiz Q2 , at tutorials during the period January 30th – February 2nd, 2006. The Problem. In [7, Example 6, p. 441] of your textbook, the author solves the following problem: “Find the area enclosed by the line y = x − 1 and the parabola y2 = 2x + 6,” by integrating with respect to y. The textbook then states the following: “We could have found the area in Example 6 by integrating with respect to x instead of y, but the calculation is much more involved. It would have meant splitting the region in two, and computing the areas labelled A1 and A2 in Figure 14. The method we used in Example 6 is much easier.” Your assignment is to solve the problem by integrating with respect to x, and you may find the figure in the√textbook helpful. The computation √ is not really very hard, but it does involve working with 2. You should not approximate 2; just write it that way and work with it carefully, and the square roots will cancel by the time you finish your solution. You know the numerical answer to this question; the purpose of the assignment is to nudge you into solving problems in two ways, so that you can verify your work; and to show you that there really is nothing to fear, even if you do happen to choose the “wrong” way to approach a problem. Your TA will be alert to the possibility that students might be copying their solutions from others. Please write up your own solution, so that your TA will not have to waste everyone’s time by sending exact copies to the disciplinary officer of the Faculty. You need to know how to solve problems of this type, since you could be expected to demonstrate that ability at a future quiz or on the examination. The Solution. The integration must be carried out separately for two subregions. This is because the area to the left of the line x = −1 is bounded above and below by the parabola; while the area to the right of x = −1 is bounded above by the parabola, and below by the line y = x − 1. The method you know for finding the area between two curves requires that the equations of the curves be written as the graphs of functions of x. But the parabola y2 = 2x + 6 is not the graph of a function, since it crosses some vertical lines twice. However, we can factorize the equation in the form p p y − 2(x + 3) y + 2(x + 3) = 0 : Information for Students in MATH 141 2010 01 5025 the parabola is the graph of two functions — the upper arm √ of the parabola is the graph of √ = − 2(x + 3). To the left of x = −1 y = 2(x + 3), while the lower arm is the graph of y √ the region A is bounded above by the graph of y = 2(x + 3), and below by the graph of 1 √ y = − 2(x + 3); thus the vertical element √ the height of √ √ of area is the difference between these two functions, i.e., 2(x + 3) − (− 2(x + 3)) = 2 2(x + 3), and this will be the integrand for √ the integral; the area to the right of x = −1 is bounded above by the parabola √ y = 2(x + 3), and below by the line y = x−1, so the element of area for the integral will be 2(x + 3)−(x−1). The total area is thus Z −1 p Z 5p 2 2(x + 3) dx + 2(x + 3) − (x − 1) dx −3 −1 " #−1 " !#5 2 √ 2 √ 2 x 3 3 = 2 2 · · (x + 3) 2 + 2 · · (x + 3) 2 − −x 3 3 2 −3 −1 √ √ ! √ ! i 2 2 3 2 4 2 h 32 25 1 2 3 2 − 0 + · 82 − − 5 − · 22 − + 1 = 3 3 2 3 2 " # 56 16 38 16 + −6 = + = 18. = 3 3 3 3 The Grading Scheme. The assignment was to be graded out of a maximum of 15 marks. (In the version of this document prepared for TA’s there were additional details on the marking scheme.) D.8.3 Solutions to Written Assignment W3 Release Date: March 25th, 2006 Solutions were to be submitted inside answer sheets to Quiz Q3 , at tutorials during the period 13-16 February, 2006 The Problems Z 1. Use two integrations by parts to evaluate the integral (sin x) · (sinh x) dx. You will need, at the last step, to solve an equation. The solution should resemble the solution in R x your textbook to [7, Example 4, p. 478], where the author evaluates e sin x dx. 2. Write your student number (9 digits) in the following chart. A1 A2 A3 A4 A5 A6 A7 A8 A9 Student # : UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 5026 Some of the digits are to be used in solving the following problem. (a) You are to evaluate the following integral by integration by parts: Z A7 x2 + A8 x1 + A9 eA1 x dx Your answer should be simplified as much as possible. (b) Then you are to differentiate the function you have obtained, to verify that it is indeed an antiderivative of the given integrand. Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in the Handbook on Student Rights and Responsibilities (PDF English version - French version). Article 15(a) of the Code, which is devoted to plagiarism, reads as follows: No student shall, with intent to deceive, represent the work of another person as his or her own in any academic writing, essay, thesis, research report, project or assignment submitted in a course or program of study or represent as his or her own an entire essay or work of another, whether the material so represented constitutes a part or the entirety of the work submitted. Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you really need to learn how to solve these kinds of problems yourself. The Solutions 1. Let u = sin x, v0 = sin x. Then u0 = cos x, v = cosh x (or cosh x plus any real constant — I have taken the constant to be 0, as all choices of constant here will lead to the same solution.) Z Z (sin x) · (sinh x) dx = (cos x) · sinh x − (cosh x) · (cos x) dx . Z (cosh x) · (cos x) dx I will set U = cos x, V 0 = cosh x, so U 0 = − sin x, To evaluate V = sinh x. Z Z (cosh x) · (cos x) dx = (cos x) · sinh x + (sinh x) · sin x dx . Combining these results yields Z Z (sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) − (sin x) · (sinh x) dx Information for Students in MATH 141 2010 01 5027 which we may solve by moving the two integral terms to the same side of the equation: Z 2 (sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) + C or Z (sin x) · (sinh x) dx = 2. (sin x) · (cosh x) − (cos x) · (sinh x) + C0 2 (a) In the first integration by parts I set ⇒ Hence Z = = u = A7 x2 + A8 x1 + A9 dv = eA1 x dx du = (2A7 x + A8 )dx, 1 A1 x v = e . A1 A7 x2 + A8 x1 + A9 eA1 x dx Z 1 A1 x 1 (2A7 x + A8 ) · eA1 x dx A7 x + A8 x + A9 · e − A1 A1 ! ! Z A7 2 A8 1 A9 2A7 A8 A1 x x + x + ·e − ·x+ · eA1 x dx A1 A1 A1 A1 A1 2 To evaluate the subtracted integral we must apply Integration by Parts a second time: U = dV = ⇒ dU = V = Z ! 2A7 A8 ·x+ · eA1 x dx = A1 A1 = = 2A7 A8 ·x+ A1 A1 A1 x e dx 2A7 dx A1 1 A1 x e , so A1 ! Z 2A7 A8 1 A1 x 2A7 A1 x e dx ·x+ · e − A1 A1 A1 A21 ! 2A7 A8 2A7 · x + 2 · e A1 x − 3 e A1 x 2 A1 A1 A1 ! 2A7 A8 2A7 · x + 2 − 3 · eA1 x 2 A1 A1 A1 Information for Students in MATH 141 2010 01 5028 Combining our results yields Z A7 x2 + A8 x1 + A9 eA1 x dx !! ! A7 2 A8 2A7 1 A9 A8 2A7 = x + − 2 x + − + 3 · eA1 x + C A1 A1 A1 A21 A1 A1 (b) Remember that you were to differentiate the preceding product of polynomial and exponential by the Product Rule to show that, indeed, its derivative is the given integrand. D.8.4 Solutions to Written Assignment W4 Release Date: March 25th, 2006 Solutions were to be submitted at inside answer sheets to Quiz Q4 , at tutorials during the week 06 – 09 March, 2006 The Problems This assignment is based on your WeBWorK assignment R5 . You are asked to write out complete solutions to the following modifications of your versions of problems on that assignment. Note that, in some cases, the question asks for more than was asked on the assignment. You are not permitted to use a Table of Integrals. 1. Problem 10: (a) Evaluate the following integral with the specific values of the constants given in your own WeBWorK assignment: Z Ex2 + Ex + F dx . x3 + Gx2 + Hx + J HINT: −C is a root of the denominator. (b) No marks will be given unless you verify your integration by differentiating your answer. 2. Problem 12: (a) Evaluate the following integral with the specific values of the constants given in your own WeBWorK assignment: Z Bx + C dx . (x2 + A2 )2 Information for Students in MATH 141 2010 01 5029 (b) No marks will be given unless you verify your integration by differentiating your answer. 3. Problem 16: Determine whether the following integral (with the constants given in your own WeBWorK assignment) is divergent or convergent. If it is convergent, evaluate it. If not, prove that fact. Z A 1 dx . 2 −∞ x + 1 In each case your solution should begin by your writing out the full problem with your data, so your TA does not have to look up your data on the WeBWorK system. While these problems were generated by WeBWorK, they now constitute a conventional mathematics assignment, and it does not suffice to make unsubstantiated statements. You are expected to prove everything you state. Thus, for example, in Problem 10, you have to show how you use the fact that a certain number is stated to be a root of the denominator. Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in the Handbook on Student Rights and Responsibilities (PDF English version - French version). Article 15(a) of the Code, which is devoted to plagiarism, reads as follows: No student shall, with intent to deceive, represent the work of another person as his or her own in any academic writing, essay, thesis, research report, project or assignment submitted in a course or program of study or represent as his or her own an entire essay or work of another, whether the material so represented constitutes a part or the entirety of the work submitted. Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you really need to learn how to solve these kinds of problems yourself. The Solutions x 1. Under a change of variable of the form u = , the integral reduces to one of the form A Z Ku + L Ku du simplifies under a substitution v = u2 to a multiple 2 du. The term 2 2 Zu + 1 Zu + 1 L dv . The term of du simplifies under a substitution θ = arctan u, 2 (v + 1) u2 + 1 2 tan θ eventually proving to be a multiple of θ + , etc. 2(tan2 θ + 1) 2. An antiderivative is arctan x. This is evaluated between an upper limit of some constant A, and a lower limit we may call B. We need to observe that lim arctan B = − π2 . B→−∞ Information for Students in MATH 141 2010 01 5030 D.8.5 Solutions to Written Assignment W5 Release Date: Solutions were to be submitted inside answer sheets to Quiz Q5 , at tutorials during the week 20 – 23 March, 2006 The Problems This assignment will be graded out of a maximum of 20 MARKS. Write your student number (9 digits) in the following chart. A1 A2 A3 A4 A5 A6 A7 A8 A9 Student # : Some of the digits are to be used in solving the following problems. 1. Consider the curve in the plane defined parametrically by x(t) = t2 + 1 y(t) = A7 t2 + A8 t + A9 (a) [3 MARKS] Showing all your work, determine the slope of the tangent to the curve at the point with parameter value t = 1. d2 y (b) [6 MARKS] Showing all your work, determine the value of 2 at the point with dx general parameter value t (t , 0). For this problem you must not substitute in any formula from your class notes or any textbook; you are expected to determine the 2nd derivative by differentiation, for example in the manner similar to that done in your textbook, Example 1, page 661. (c) [1 MARK] Determine the range of values of t — if there are any such values — where the curve is concave upward. 2. This problem is based on Problem 12 on your WeBWorK assignment R8 . For the given arc of the given curve, (a) [8 MARKS] determine the area of the surface of revolution of that arc about the x-axis; (b) [2 MARKS] set up an integral for the area of the surface of revolution of that arc about the y-axis, but, in this case only, do not evaluate the integral. In each case you are expected to show all your work. Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in the Handbook on Student Rights and Responsibilities (PDF English version - French version). Article 15(a) of the Code, which is devoted to plagiarism, reads as follows: Information for Students in MATH 141 2010 01 5031 No student shall, with intent to deceive, represent the work of another person as his or her own in any academic writing, essay, thesis, research report, project or assignment submitted in a course or program of study or represent as his or her own an entire essay or work of another, whether the material so represented constitutes a part or the entirety of the work submitted. Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you really need to learn how to solve these kinds of problems yourself. The Solutions 1. (a) x(t) y(t) dx dt dy dt dy dx dy dx t=1 (b) = t2 + 1 = A7 t2 + A8 t + A9 = 2t = 2A7 t + A8 dy dt dx dt 2A7 t + A8 2t A8 = A7 + 2 = = ! d 2A7 t + A8 d2 y = dx2 dx 2t ! d 2A7 t + A8 dt = · dt 2t dx ! d 2A7 t + A8 1 = · dx dt 2t ! dt d 2A7 t + A8 1 · = dt 2t 2t (2A7 )(2t) − (2A7 t + A8 )(2) = (2t)3 A8 = − 3 4t (c) If the student’s A8 is 0, the curve is flat. Otherwise it is concave upwards precisely when t < 0. Information for Students in MATH 141 2010 01 2. 5032 CORRECTED ON 20 MARCH 2006. The following solution is valid only when K, the upper limit of integration, is sufficiently small. The correct solution is much more complicated, as we need to ensure that the respective factors cos θ + θ sin θ and sin θ − θ cos θ remain positive. I have corrected my original version by inserting absolute signs. However, the evaluation of these integrals is complicated when we attempt to break the integral up into parts based on the signs of these factors. This is certainly more difficult than was intended from students in this course. Please exercise good judgment in grading the two parts of this problem, and do not expect those students for whom the upper limit K is large to obtain the correct area. ⇒ p x(θ) y(θ) ⇒ x0 y0 = = = = a(cos θ + θ sin θ) a(sin θ − θ cos θ) aθ cos θ aθ sin θ (x0 )2 + (y0 )2 = |aθ| (a) about the x-axis, PROVIDED K IS SUFFICIENTLY SMALL, Z K Z K Z Area = 2πa| sin θ − θ cos θ| · |θ| dθ = 2πa θ sin θ dθ − 0 0 K ! 2 θ cos θ dθ 0 for positive K sufficiently small. We must integrate by parts. Z Z θ sin θ dθ = −θ cos θ + cos θ dθ = −θ cos θ + sin Z θ+C Z 2 θ cos θ dθ = θ2 sin θ − 2 θ sin θ dθ = θ2 sin θ + 2θ cos θ − 2 sin θ + C 0 Z (θ sin θ − θ2 cos θ) dθ = −3θ cos θ + 3 sin θ − θ2 sin θ + C 00 Hence the area of the surface of revolution is −3K cos K + 3 sin K − K 2 sin K. (b) about the y-axis, PROVIDED K IS SUFFICIENTLY SMALL, ! Z K Z K Z K 2 Area = 2πa| cos θ + θ sin θ| · |θ| dθ = 2πa θ cos θ dθ + θ sin θ dθ . 0 D.9 MATH 141 2007 01 The use of written assignments was discontinued in 2007. 0 0 Information for Students in MATH 141 2010 01 5033 E Quizzes from Previous Years E.1 MATH 141 2007 01 E.1.1 Draft Solutions to Quiz Q1 Distribution Date: Mounted on the Web on 4 February, 2007 corrected 19 January, 2010 — subject to further corrections There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version 1. [5 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z a g(x) = b tan(t) dt , x (where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluate g(x), by first verifying carefully that ln | sec x| is an antiderivative of tan x. Solution: (a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as a function of its upper index of integration. Here the variable is the lower index of integration. Z a d b tan(t) dt dx x ! Z x d = − b tan(t) dt dx a Information for Students in MATH 141 2010 01 5034 Z x d = − b tan(t) dt dx a = −b tan x . Some students may quote a variant of Part 1 which gives the derivative of a definite integral with respect to the lower index, and this should be accepted if work has been shown. (b) Students were expected to first find the derivative of ln | sec x|. Since this is a composition of 2 functions, the Chain Rule will be needed. Let u = sec x. Then d d d ln | sec x| = ln |u| · secx dx du dx 1 = · sec x tan x u 1 = · sec x tan x sec x = tan x . Hence g(x) = b ln | sec t|ax sec a cos x = b ln . = b ln sec x cos a 2. [5 MARKS] Find an antiderivative of the integrand of the integral Zb k + `y + my2 dy, a (where a, b, k, `, m are constants), and then use the Fundamental Theorem of Calculus to evaluate the integral. You are not expected to simplify your numerical answer. Solution: (a) One antiderivative of ky0 + `y1 + my2 is k· y1 y2 y3 +`· +m· . 1 2 3 (b) Hence Zb a " = = k + `y + my2 dy #b y2 y3 y1 k· +`· +m· 1 2 3 a ! ! 1 2 b b b3 a2 a3 a1 k· +`· +m· +`· +m· − k· . 1 2 3 1 2 3 Information for Students in MATH 141 2010 01 5035 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z √x cos t dt , g(x) = b t a where a, b are constants. Solution: (a) Denote the upper index of the integral by u(x) = √ x. (b) Then Z √x d d cos t b g(x) = dt dx dx a t Z u(x) d cos t b dt = dx a t Z u(x) d cos t du(x) = b dt · du a t dx cos u du(x) = b · u dx cos u 1 = b · √ u 2 x √ cos x 1 = b √ · √ x 2 x √ cos x = b 2x 4. [10 MARKS] Showing all your work, determine all values of x where the curve y = Zx 1 dt is concave upward, where a, b are constants. (Each version of this 1 + at + bt2 0 quiz contained specific values for the constants a, b.) Solution: (a) By Part 1 of the Fundamental Theorem, y0 (x) = 1 . 1 + ax + bx2 (b) Differentiating a second time yields d 1 y (x) = dx 1 + ax + bx2 00 ! Information for Students in MATH 141 2010 01 5036 d 2 · 1 + ax + bx 1 + ax + bx2 2 dx a + 2bx = − . 1 + ax + bx2 2 = − 1 (c) The curve is concave upward where y00 > 0: − a + 2bx 1 + ax + bx2 2 > 0 ⇔ −(a + 2bx) > 0 since the denominator is a square, hence positive ⇔ 2bx < −a a when b>0 x < − 2b a x > − 2b when b<0 ⇔ never concave upward when b = 0, a > 0 always concave upward when b = 0, a < 0 Tuesday version Zc 1. [5 MARKS] If Zc f (x) dx = k and a Zb f (x) dx = `, find f (x) dx. Show your work. a b Solution: (a) Zc Zb f (x) dx = (b) Hence Zc f (x) dx + f (x) dx . a a b Zc Zc Zb f (x) dx = b f (x) dx − a f (x) dx . a (c) = k −`. Z a 2. [5 MARKS] Find an antiderivative of the integrand of the integral 0 √ x dx, and then use the Fundamental Theorem of Calculus to evaluate the integral. You are not expected to simplify your numerical answer, but no marks will be given unless all your work is clearly shown. UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 5037 1 (a) One antiderivative of x 2 is 1 2 (b) Z a 0 √ 1 2 3 1 · x 2 +1 = x 2 . 3 +1 " 2 32 x dx = x 3 #a = 0 2 3 2 23 a − 0 = a2 . 3 3 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative Zx2 √ of the function b 1 + tc dt. a Solution: (a) Let the upper index of the integral be denoted by u = x2 . Then (b) d dx Zx2 √ b 1 + tc dt = a d dx Zu √ b 1 + tc dt a Zu √ d du = b 1 + tc dt · du dx a √ du = b 1 + uc · dx √ c = b 1 + u · 2x √ = b 1 + x2c · 2x Z 4. [10 MARKS] If F(x) = 00 find F (2). Z x t2 f (t) dt, where f (t) = 1 1 a + ub du and a, b are constants, u Solution: (a) Applying Part 1 of the Fundamental Theorem yields Z x2 0 F (x) = f (x) = 1 a + ub du . u Information for Students in MATH 141 2010 01 5038 (b) A second application of Part 1 of the Fundamental Theorem yields d F (x) = f (x) = dx 00 Z x2 0 1 a + ub du . u (c) Denote the upper index of the last integral by v = x2 . (d) d dx Z x2 1 Z v a + ub a + ub d du = du u dx 1 u Z v a + ub d dv = du · dv 1 u dx b a+v dv = · v dx a + vb = · 2x v a + x2b = · 2x 2 x 2 a + x2b = . x Wednesday version 1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral Zbπ cos θ dθ, where a, b are given integers. No marks will be given unless all your work aπ is clearly shown. Your answer should be simplified as much as possible. Solution: (a) One antiderivative of cos θ is sin θ. (b) Zbπ cos θ dθ = [sin θ]bπ aπ = sin(bπ) − sin(aπ) . aπ (c) Students were expected to observe that the value of the sine at the given multiples of π is 0, so the value of the definite integral is 0. Information for Students in MATH 141 2010 01 2. [5 MARKS] Express lim n→∞ n X 5039 axi sin xi ∆x as a definite integral on the interval [b, c], i=1 which has been subdivided into n equal subintervals. Solution: Z c ax sin x dx . b 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z bx 2 t +c g(x) = dt , 2 ax t − c where a, b, c are positive integers. Solution: (a) The Fundamental Theorem gives the derivative of a definite integral with respect to the upper limit of integration, when the lower limit is constant. The given integral must be expressed in terms of such specialized definite integrals. Z bx 2 Z 0 2 Z bx 2 t +c t +c t +c g(x) = dt = dt + dt 2 2 t2 − c ax t − c ax t − c 0 Z ax 2 Z bx 2 t +c t +c =− dt + dt 2 t −c t2 − c 0 0 Zbx (b) For the summand t2 + c dt, let u = bx. Then t2 − c 0 d dx Zbx t2 + c d dt = 2 t −c dx 0 Zu t2 + c dt t2 − c 0 = d du Zu du t2 + c dt · 2 t −c dx 0 u2 + c du = 2 · u − c dx u2 + c = 2 ·b u −c (bx)2 + c = ·b (bx)2 − c Information for Students in MATH 141 2010 01 Z ax 2 (c) For the summand step, 5040 0 t +c dt, let u = ax. Then, analogously to the preceding t2 − c Z ax 2 d t +c (ax)2 + c dt = · a. dx 0 t2 − c (ax)2 − c (d) g0 (x) = (bx)2 + c (ax)2 + c · b − · a. (bx)2 − c (ax)2 − c 4. [10 MARKS] Find a function f (x) such that Z x √ f (t) dt = ` x k+ 2 t a (109) for x > 0 and for some real number a; k and ` are constants given in the question, different constants to different students. (HINT: Differentiate the given equation.) Solution: A more proper wording of the problem would have been “Find a function f (x) and a real number a such that...”. (a) Assume that equation (109) holds. Then differentiation of both sides with respect to x yields f (x) 1 1 0+ 2 = ·`· √ . x 2 x 3 (b) We solve the preceding equation to obtain that f (x) = 2` x 2 . (c) Substitution into equation (109) yields Z √ ` x − 21 k+ t dt = ` x . 2 a We know how to integrate powers of t: √ ` k+ 2· t cot #x √ = ` x. a √ (d) The preceding reduces to k = ` a, which may be solved to obtain k a= ` !2 . Information for Students in MATH 141 2010 01 5041 Thursday version π Zb 1. [5 MARKS] Evaluate the integral sin t dt. π a Solution: (a) An antiderivative of sin t is − cos t. (b) π Zb π sin t dt = [− cos t] πb a π a (c) [1 MARK] Your answer should be simplified as much as possible. 2. [5 MARKS] Evaluate the Riemann sum for f (z) = a−x2 , (0 ≤ x ≤ 2) with 4 subintervals, taking the sample points to be the right endpoints. It is not necessary to simplify the final numerical answer. Solution: (a) The interval 0 ≤ x ≤ 2 is divided by 3 points into 4 subintervals of length ∆x = 2 = 12 . 4 ∗ (b) The point xi selected in the ith interval will always be the right end-point, i.e., xi = i 12 (i = 1, 2, 3, 4). (c) The Riemann sum is 4 X f (xi∗ )∆x i=1 ! 4 1X i2 = a− . 2 i=1 4 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative Zbx cos (tc ) dt, where a, b, c are real numbers. of the function cos x Solution: (a) The Fundamental Theorem gives the derivative of a definite integral with respect to the upper limit of integration, when the lower limit is constant. The given integral must be expressed in terms of such specialized definite integrals. Zbx Z0 cos (tc ) dt = cos x Zbx cos (tc ) dt + cos x cos (tc ) dt 0 Information for Students in MATH 141 2010 01 5042 cos x Z Zbx c = − cos (t ) dt + cos (tc ) dt 0 0 Zbx cos (tc ) dt, let u = bx. Then (b) For the summand 0 d dx Zbx d cos (t ) dt = dx Zu c 0 cos (tc ) dt 0 = d du Zu cos (tc ) dt · du dx 0 = cos (uc ) · b = cos ((bx)c ) · b cos x Z (c) For the summand cos (tc ) dt, let v = cos x. 0 d dx cos x Z cos (tc ) dt = d dx 0 Zv cos (tc ) dt 0 = d dv Zv cos (tc ) dt · dv dx 0 = cos (vc ) · (− sin x) = cos (cosc x) · (− sin x) (d) Zbx cos (tc ) dt = − cos (cosc x) · (− sin x) + cos ((bx)c ) · b . cos x 0 if x < 0 Z x if 0 ≤ x ≤ a x 4. [10 MARKS] Let f (x) = and g(x) = f (t) dt, where a is 2a − x if a < x < 2a 0 0 if x > 2a a positive constant. Showing all your work, find a formula for the value of g(x) when a < x < 2a. Information for Students in MATH 141 2010 01 5043 Solution: (a) The interval where we seek a formula is the third interval into which the domain has been broken. For x in this interval the integral can be decomposed into Z x Z a Z x f (t) dt = f (t) dt + f (t) dt . 0 0 a The portion of the definition of f for x < 0 is of no interest in this problem, since we are not finding area under that portion of the curve; the same applies to the portion of the definition for x > 2a. (b) Z Z a a f (t) dt = 0 t dt " = 0 # 2 t=a t 2 = t=0 a2 . 2 (c) Z Z x x f (t) dt = a (d) (2a − t) dt " #t=x t2 = 2at − 2 t=a ! ! x2 a2 2 = 2ax − − 2a − . 2 2 a ! ! a2 x2 a2 x2 2 g(x) = + 2ax − − 2a − = 2ax − − a2 . 2 2 2 2 E.1.2 Draft Solutions to Quiz Q2 Distribution Date: Posted on the Web on 28 February, 2007 Caveat lector! There could be misprints or errors in these draft solutions. There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes Information for Students in MATH 141 2010 01 5044 • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version Z 1. [5 MARKS] Evaluate the integral positive integers). ! 1 x +a+ 2 dx, (where a and b are given x +1 b Solution: Z xb+1 (a) xb dx = + C1 , b+1 Z (b) a dx = ax + C2 Z 1 (c) dx = arctan x + C3 2 x +1 ! Z 1 xb+1 b x +a+ 2 (d) dx = + ax + arctan x + C. x +1 b+1 Z 2. [5 MARKS] Use a substitution to evaluate the indefinite integral t2 cos a − t3 dt, (where a is a given real number). Solution: (a) Try the substitution u = t3 . (b) du = 3t2 dt ⇒ t2 dt = 13 du. (c) Z Z 1 3 cos(a − u) du t cos a − t dt = 3 1 = − sin(a − u) + C 3 1 = − sin(a − t3 ) + C. 3 (Some students may wish to employ a second substitution v = a − u. Alternatively, a better substitution for the problem would have been to take u = a − t3 .) 2 Information for Students in MATH 141 2010 01 5045 3. [10 MARKS] Find the area of the region bounded by the parabola y = x2 , the tangent line to this parabola at (a, a2 ), and the x-axis, (where a is a given real number). Solution: This area can be computed by integrating either with respect to y or with respect to x. Integrating with respect to y: (a) Since y0 = 2x, the tangent line through (a, a2 ) has equation y − a2 = 2a(x − a) ⇔ y = 2ax − a2 . (b) To integrate with respect to y we need to express the equations of the parabola and the line in the form x = function of y . √ The branch of the parabola to the right of the y-axis is x = y. The line has y a equation x = + . 2a 2 ! y + a2 √ (c) The area of the horizontal element of area at height y is − y ∆y. 2a (d) The area is the value of the integral ! Z a2 y + a2 √ − y dy . 2a 0 (e) Integration yields " 2 #a2 ! y 1 1 2 3 ay 2 32 1 3 + − y = + − a = a . 4a 2 3 0 4 2 3 12 Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2 . Its a intercept with the x-axis is at x = . 2 a (b) The area of the vertical element of area at horizontal position x ≤ is x2 − 0 ∆x. 2 (c) The area of the vertical element of area at horizontal position x ≥ 2a is x2 − (2ax − a2 ) dx = (x − a)2 ∆x. (d) The area of the region is the sum Z a2 Z a 2 (x − a)2 dx . x dx + a 2 0 (e) Integration yields " x3 3 # a2 0 " (x − a)3 + − 3 #a = a 2 a3 . 12 Information for Students in MATH 141 2010 01 5046 4. [10 MARKS] Find the volume of the solid√obtained by rotating the region bounded by the given curves about the line y = 1: y = n x, y = x, where n is a given positive integer. Solution: A favoured method of solution was not prescribed. Using the method of “washers”: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a horizontal line, the element of area being rotated is vertical. For√arbitrary x the lower point on the element is (x, x); the upper point is (x, n x). √n The distances of these points from the axis are, respectively 1 − x and 1 − x. (c) The volume of the “washer” is, therefore, √ π −(1 − x)2 + (1 − n x)2 ∆x . (d) Correctly evaluate the integral: Z 1 √ −(1 − x)2 + (1 − n x)2 dx π 0 Z 1 1 2 −2x + x2 + 2x n − x n dx = π 0 " #1 x2 2n n+1 n n+2 2 = π −x + + xn − xn 3 n+1 n+2 0 ! 1 2n n (n − 1)(n + 4)π = π −1 + + − = 3 n+1 n+2 3(n + 1)(n + 2) Using the method of cylindrical shells: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a horizontal line, the element of area being rotated is also horizontal. For arbitrary y the left endpoint on the element is (yn , y); the right endpoint is (y, y). The length of the element is, therefore, y−yn ; the distances of the element from the axis of symmetry is 1 − y. (c) The volume of the cylindrical shell element of volume is, therefore, 2π(1 − y) · (y − yn ) · ∆y . (d) Correctly evaluate the integral: Z 1 2π (1 − y)(y − yn ) dy 0 Information for Students in MATH 141 2010 01 = 2π Z 1 " 5047 −yn + yn+1 + y − y2 dy 0 #1 1 n+1 1 n+2 1 2 1 3 = 2π − y + y + y − y n+1 n+2 2 3 0 ! 1 1 1 1 = 2π − + + − −0 n+1 n+2 2 3 ! 1 (n − 1)(n + 4)π 1 = = 2π − 6 (n + 1)(n + 2) 3(n + 1)(n + 2) Tuesday version Z 1. [5 MARKS] Evaluate the integral (a − t)(b + t2 ) dt. Solution: (a) Expand the product in the integrand: Z Z 2 (a − t)(b + t ) dt = ab − bt + at2 − t3 dt . (b) Integrate term by term: Z a 1 b ab − bt + at2 − t3 dt = ab · t − · t2 + · t3 − · t4 + C . 2 3 4 Z 2. [5 MARKS] Using a substitution, evaluate the indefinite integral cosn x sin x dx, where n is a fixed, positive integer. Solution: (a) Use new variable u, where du = − sin x dx; one solution is u = cos x. (b) Z Z n cos x sin x dx = − un du un+1 +C n+1 1 = − cosn+1 x + C n+1 = − Information for Students in MATH 141 2010 01 5048 3. [10 MARKS] Find the area of the region bounded by the parabola x = y2 , the tangent line to this parabola at (a2 , a), and the y-axis, where a is a fixed, positive real number. Solution: The solution is analogous (under the exchange x ↔ y) to that given for Problem 3 of the Monday quiz. 4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded by y = xn and x = yn about the line x = −1, where n is a given positive integer. Solution: Case I: n is even Using the method of “washers”: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a vertical line, the element of area being rotated is horizontal. For 1 arbitrary y the farther endpoint on the element is (y n , y); the nearer endpoint is (yn , y). The distances of these points from the axis are, respectively √ 1 + n y and 1 + yn . (c) The volume of the “washer” is, therefore, √ π −(1 + yn )2 + (1 + n y)2 ∆y . (d) Correctly evaluate the integral: Z 1 √ −(1 + y)2 + (1 + n y)2 dy π 0 Z 1 1 2 = π 2y n + y n − 2yn − y2n dy 0 " #1 2n n+1 n 2 1 n+2 n+1 2n+1 = π yn + yn − y − y n+1 n+2 n+1 2n + 1 0 2(n − 1)(3n2 + 7n + 3)π = (n + 1)(n + 2)(2n + 1) Using the method of cylindrical shells: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (b) Since the axis of revolution is a vertical line, the element of area being rotated is also vertical. For arbitrary x the top endpoint on the element is 1 (x, x n ); the lower endpoint is (x, xn ). The length of the element is, there1 fore, x n − xn ; the distance of the element from the axis of symmetry is 1 + x. Information for Students in MATH 141 2010 01 5049 (c) The volume of the cylindrical shell element of volume is, therefore, 1 2π(1 + x) · x n − xn . (d) Correctly evaluate the integral: Z 1 1 2π (1 + x)(x n − xn ) dx 0 Z 1 1 n+1 = 2π x n − xn + x n − xn+1 dx 0 n 1 n 1 = 2π − + − n + 1 n + 1 2n + 1 n + 2 2(n − 1)(3n2 + 7n + 3)π = (n + 1)(n + 2)(2n + 1) ! Case II: n is odd Using the method of “washers”: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1, ±1). Here there is an issue of interpretation. The textbook usually permits the word region to apply to one that may have more than one component; some authors would not wish to apply the term in such a situation. I will follow the textbook, and permit a region here to have two components. (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a vertical line, the element of area being rotated is horizontal. But there are two kinds of elements, depending on whether y is positive or negative. For arbitrary, positive y the farther endpoint on the element is 1 (y n , y); the nearer endpoint is (yn , y). The distances of these points from √ the axis are, respectively 1 + n y and 1 + yn . For arbitrary, negative y the 1 nearer endpoint on the element is (y n , y); the farther endpoint is (yn , y). √ The distances of these points from the axis are, respectively 1 + n y and 1 + yn (both of which are less than 1). (c) The volume of the “washer” is, therefore, √ π −(1 + yn )2 + (1 + n y)2 ∆y . (d) Correctly evaluate the integral: Z 1 −(1 + y)2 + (1 + √n y)2 dy π −1 Z 1 1 2 = π 2y n + y n − 2yn − y2n dy 0 Information for Students in MATH 141 2010 01 5050 Z 0 1 2 +π −2y n − y n + 2yn + y2n dy −1 " #1 2n n+1 n 2 1 n+2 n+1 2n+1 = π yn + yn − y − y n+1 n+2 n+1 2n + 1 0 " #0 2n n+1 n 2 1 n+2 n+1 2n+1 +π − yn − yn + y + y n+1 n+2 n+1 2n + 1 −1 2(n − 1)(3n2 + 7n + 3)π 2(n − 1)(n2 + 3n + 1)π = + (n + 1)(n + 2)(2n + 1) (n + 1)(n + 2)(2n + 1) 4(n − 1)π = . n+1 Using the method of cylindrical shells: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1, ±1). (b) Since the axis of revolution is a vertical line, the element of area being rotated is also vertical. For arbitrary, positive x the top endpoint on the el1 ement is (x, x n ); the lower endpoint is (x, xn ); for arbitrary, negative x the 1 bottom endpoint on the element is (x, x n ); the upper endpoint is (x, xn ). 1 The length of the element is, therefore, x n − xn ; the distance of the element from the axis of rotation is 1 + x. (c) The volume of the cylindrical shell element of volume is, therefore, 1 2π(1 + x) · x n − xn . (d) Correctly evaluate the integral: Z 1 1 n n 2π (1 + x) x − x dx −1 Z 1 1 n+1 n n+1 = 2π x n − x + x n − x dx −1 ! 1 n 1 n − + − = 2π n + 1 n + 1 2n + 1 n + 2 ! n 1 n 1 +2π − − + n + 1 n + 1 2n + 1 n + 2 4(n − 1)π = n+1 Wednesday version Information for Students in MATH 141 2010 01 5051 Z 1. [5 MARKS] Showing all your work, evaluate the indefinite integral sin 2t dt. cos t Solution: (a) Apply a “double angle” formula: Z Z Z sin 2t 2 sin t cos t dt = dt = 2 sin t dt . cos t cos t (b) Complete the integration: Z 2 sin t dt = −2 cos t + C . Z ex dx, where a is a non-zero ex + a 2. Using a substitution, evaluate the indefinite integral real number. Solution: (a) [2 MARKS] Try the substitution u = e x + a, so du = e x dx. (b) Z Z ex du dx = = ln |u| + C = ln |e x + a| + C . x e +a u (If the constant a is positive, then the absolute signs are not required.) 3. [10 MARKS] Find the number b such that the line y = b divides the region bounded by the curves y = ax2 and y = k into two regions with equal area, where a, k are given positive constants. Solution: (a) Determine the range of values for integration by finding the of the r intersections k bounding curves: solving the equations yields the points ∓ , k . a (b) Determine the portion of the full area which is below the line y = b. We begin by repeating the calculation of the preceding part: the corner points have coordinates r b , b . The area is ∓ a Z √ ba " ax √ b (b − ax ) dx = 2 bx − 3 − a 2 3 # √ ba 0 4 = b 3 r b . a Information for Students in MATH 141 2010 01 5052 (c) As a special case of the foregoing, or r by a separate calculation, we can conclude 4 k that the area of the entire region is k . 3 a (d) The condition of the problem is that r r 4 b 1 4 k b = · k 3 a 2 3 a 3 3 which is equivalent to 4b = k , and implies that the line should be placed where 2 b = 2− 3 k. 4. [10 MARKS] Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. √ y= x−1 , y=0 , x=a √ about y = b, where a, b are fixed positive constants, and b ≥ a − 1 . Solution: (a) Solve equations to determine the limits of integration. √ Solving x = a with y = √ x − 1 yields the single point of intersection a, a − 1 . (b) The horizontal element of area at height y which generates the cylindrical shell has left endpoint (1 + y2 , y) and right endpoint (a, y), so its length is a − (1 + y2 ). (c) The distance of the horizontal element of area which generates the shell from the axis of rotation is b − y. (d) Set up the integral for the volume by cylindrical shells: Z √a−1 (b − y) a − (1 + y2 ) dy . 2π 0 (e) Evaluate the integral Z 2π Z √ a−1 0 = 2π 0 √ a−1 (b − y) a − (1 + y2 ) dy b(a − 1) − (a − 1)y − by2 + y3 dy # √a−1 a−1 2 b 3 1 4 y − y + y = 2π b(a − 1)y − 2 3 4 0 ! b a−1 1 3 3 2 2 2 = 2π b(a − 1) − (a − 1) − (a − 1) + (a − 1) 2 3 4 ! √ 1 2 3 = 2π(a − 1) 2 ·b− a−1 . 3 4 " UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 5053 Thursday version 1. [5 MARKS] Showing all your work, evaluate the integral by making a substitution: Z b dx , (1 + ax)3 where a, b are non-zero constants. Solution: (a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, so dx = 1a du. (b) Z b b dx = 3 (1 + ax) a Z du b b = − u−2 + C = − +C. 3 u 2a 2a(1 + ax)2 Z 2. [5 MARKS] Evaluate the indefinite integral seca x tan x dx, where a is a constant, positive integer. Solution: (a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x. (b) Z Z a sec x tan x dx = ua−1 du = ua seca x +C = +C. a a Some students may have integrated by sight. 3. [10 MARKS] Find the number b such that the line divides the region bounded by the curves x = ay2 and x = k into two regions with equal area. Solution: The solution is analogous (under the exchange x ↔ y) to that given for Problem 3 of the Wednesday quiz. 4. [10 MARKS] The region bounded by the given curves is rotated about the axis x = −1. Find the volume of the resulting solid by any method: y = 5, y = x2 − ax + b Solution: Because there are constraints on the constants, I will work just one variant, with a = 3, b = 7. Using the method of cylindrical shells: (a) To find the extremes of integration, we solve the equations y = 5 and y = x2 − 3x + 7, obtaining (x, y) = (1, 5), (2, 5). Information for Students in MATH 141 2010 01 5054 (b) The height of a vertical element of area which generates a cylindrical shell is, at horizontal position x, 5 − (x2 − 3x + 7) = −x2 + 3x − 2. (c) The distance of that vertical element of area from the axis of revolution is 1+ x. (d) The volume is given by the integral Z 2 2π (1 + x)(−x2 + 3x − 2) dx 1 (e) Evaluating the integral: Z 2 2π (1 + x)(−x2 + 3x − 2) dx 1 Z 2 = 2π −x3 + 2x2 + x − 2 dx 1 #2 " 1 4 2 3 1 2 = 2π − x + x + x − 2x 4 3 2 1 ! 16 1 2 1 = 2π −4 + +2−4+ − − +2 3 4 3 2 5π = 6 Using the method of “washers” (a) To find the lowest point on the parabola, we solve x2 − 3x + 7 ≥ 0. This can be done by completing the square, ! or by using the 3 19 calculus to find the local minimum. We find it to be , . 2 4 (b) The horizontal element generating the “washer” at height y extends between the solutions in x to the equation y = x2 − 3x + 7; these are p 3 ± 4y − 19 x= . 2 (c) The volume of the “washer” at height y is, therefore, 2 2 p p 4y − 19 4y − 19 3 + 3 − ∆y π 1 + − 1 + 2 2 p = 5π 4y − 19 ∆y Z 5p (d) The volume is given by the integral 5π 4y − 19 dy. 19 4 (e) Evaluation of the integral: " #5 Z 5p 2 1 5π 3 4y − 19 dy = 5π · · (4y − 19) 2 = . 5π 19 3 4 6 19 4 4 Information for Students in MATH 141 2010 01 5055 E.1.3 Draft Solutions to Quiz Q3 Distribution Date: Posted on the Web on 21 March, 2007 Caveat lector! There could be misprints or errors in these draft solutions. There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version Z 1. [6 MARKS] Evaluate the integral t3 eat dt, where a is a non-zero constant. Solution: This problem requires several consecutive applications of integration by parts to reduce the exponent of the power of t to 0: 1 u = t3 , dv = eat dt ⇒ du = 3t2 dt, v = eat Z Z a 1 3 ⇒ t3 eat dt = · t3 eat − t2 eat dt a a 1 U = t2 , dV = eat dt ⇒ dU = 2t dt, V = eat ! a Z Z 1 3 6 3 at 3 2 at ⇒ t e dt = · t − 2t e + 2 teat dt a a a 1 u˜ = t, d˜v = eat dt ⇒ du˜ = dt, v˜ = eat a ! Z Z 1 3 6 6 3 at 3 2 at ⇒ t e dt = · t − 2 · t + 3t e − 3 eat dt a a a a ! 1 3 3 2 6 6 at = · t − 2 · t + 3t − 4 e + C a a a a Information for Students in MATH 141 2010 01 5056 the correctness of which integration may be verified by differentiation of the product on the right. 2. [8 Z MARKS] Showing all your work, find a reduction formula for the indefinite integral cosn ax dx, where a is a non-zero constant, and n is an integer not less than 2. Solution: (a) Introduce a symbol for the general indefinite integral sought: Z In = cosn ax dx (b) Integration by parts: u = cosn−1 ax ⇒ du = −a(n − 1) cosn−2 ax · sin ax dx 1 dv = cos ax dx ⇒ v = · sin ax a Z 1 n−1 · cos ax · sin ax + (n − 1) cosn−2 ax · sin2 ax dx In = a Z 1 n−1 = · cos ax · sin ax + (n − 1) cosn−2 ax · 1 − cos2 ax dx a 1 = · cosn−1 ax · sin ax + (n − 1)In−2 − (n − 1)In . a (c) Solve the last equation for In n−1 1 · sin ax · cosn−1 ax + In−2 + C . an n Z 1 3. [4 MARKS] Showing all your work, evaluate the integral dx , where a is √ x 2 x 2 − a2 a non-zero constant. In = Solution: The following solution uses a trigonometric substitution; it is also possible to solve this problem using a hyperbolic substitution. As is customary, I will proceed mechanically, taking square roots where necessary without much attention to the sign choices; and then verify at the end by differentiation that this process has produced a valid indefinite integral. This procedure can be made rigorous by defining the new variable in terms of an inverse trigonometric function of x. This is a useful exercise, but becomes extremely complicated in this case, because we would have to work with either the inverse cosine or the inverse secant, and the textbook we Information for Students in MATH 141 2010 01 5057 are using chooses different domains for these two functions. So I will avoid the niceties and proceed as described. I propose to use a substitution which provides that x = a sec θ. Then dx = a sec θ · tan θ · dθ , and Z 1 dx = √ x2 x2 − a2 = = = = Z 1 cos θ dθ a2 1 sin θ + C a2 1 tan θ · cos θ + C a2 sec2 θ − 1 +C 2 √a sec θ x2 − a2 +C, a2 x the correctness of which integration may be verified by differentiation of the quotient on the right. Z ` x(x − a)(x − b) + c 4. [8 MARKS] Showing all your work, evaluate the integral dx, (x − a)(x − b) k where a, b are distinct constants, c is a non-zero constant, and the limits of integration k, ` are also prescribed. (The integrand was not presented to students in factored form.) Solution: (a) Since the degree of the numerator is not less than the degree of the denominator, begin by dividing the denominator into the numerator, obtaining a quotient and a remainder: Z Z x(x − a)(x − b) + c c dx = x+ dx . (x − a)(x − b) (x − a)(x − b) (b) Expand the fraction into partial fractions. Assume that c A B = + , (x − a)(x − b) x − a x − b take to a common denominator, and equate the resulting polynomials: c = A · (x − b) + B · (x − a) . Information for Students in MATH 141 2010 01 5058 (c) Now either equate coefficients of like powers of x, or, equivalently, give x successive values x = a and x = b: c c = A(a − b) ⇒A= = −B . c = B(b − a) a−b (d) The integration reduces to Z Z x(x − a)(x − b) + c dx = (x − a)(x − b) ! c 1 c 1 x+ · − · dx . a−b x−a a−b x−b (e) Complete the integration: Z x(x − a)(x − b) + c x2 c dx = + (ln |x − a| − ln |x − b|) + C (x − a)(x − b) 2 a−b c x − a x2 +C, + ln = 2 a − b x − b the correctness of which integration may be verified by differentiation of the function on the right. (f) Provided it is convergent the given definite integral can now be evaluated: Z ` k #` x2 c x − a + ln 2 a − b x − b k ! c ` − a k − a `2 − k2 − ln + ln = 2 a−b ` − b k − b ! `2 − k2 c (` − a)(k − b) = + ln 2 a−b (` − b)(k − a) x(x − a)(x − b) + c dx = (x − a)(x − b) " (g) All of the preceding is based on the integral being convergent. In some of the versions the integral was divergent. This was because at least one of the roots of the polynomial which is the denominator of the integrand was contained in the interval of integration. In such a case the integral can be seen to diverge. √1 Z3 5. [4 MARKS] Evaluate the integral −1 earctan y dy . I have stated the problem with just one 1 + y2 pair of possible limits for the integral; the variations of the problem included several possible limits in each case, for each of which students should have been familiar with the arctangent. Information for Students in MATH 141 2010 01 5059 Solution: For simplicity, I work a specific instance of this problem. Use the substitution dy u = arctan y, so du = . Then an antiderivative can be obtained as follows 1 + y2 Z arctan y Z e dy = eu du = eu + C = earctan y + C 1 + y2 so the definite integral is equal to h earctan y i √1 3 −1 π π = e 6 − e− 4 . Alternatively, the substitution may be executed in the definite integral, replacing the lower limit of −1 by arctan(−1) = − π4 , and the upper limit of √13 by arctan √13 = π6 . Tuesday version Z 1. [4 MARKS] Showing all your work, evaluate the integral x2 cos ax dx, where a is a non-zero constant. Solution: Two applications of integration by parts will be used to reduce the exponent of the power of x to 0. (a) u = x2 ⇒ du = 2x dx 1 dv = cos ax dx ⇒ v = sin ax a Z Z 2 x 2 2 x cos ax dx = · sin ax − x · sin ax dx a a (b) U = x ⇒ dU = dx 1 dV = sin ax dx ⇒ V = − · cos ax a ! Z Z 2 x 2 x 1 2 x cos ax dx = · sin ax − − · cos ax + cos ax dx a a a a Z x2 2 2 = · sin ax + 2 x · cos ax − 2 cos ax dx a a a x2 2 2 = · sin ax + 2 x · cos ax − 3 sin ax + C a a a Information for Students in MATH 141 2010 01 5060 The integration can be checked by differentiation of the alleged antiderivative. Z 2. [9 MARKS] Showing all your work, find a reduction formula for the integral xn eax dx, where a is a non-zero constant. Solution: (a) Introduce a symbol for the general indefinite integral sought: Z In = xn eax dx . (b) Integration by parts: u = xn ⇒ du = nxn−1 dx 1 dv = eax dx ⇒ v = · eax a Z 1 ax n In = ·e − xn−1 eax dx a a 1 ax n = · e − In−1 . a a which is the desired reduction formula. 3. [4 MARKS] Showing your work, evaluate the integral Z sin3 ax · cos2 ax dx, where a is a non-zero constant. Solution: This integral is easily evaluated by a substitution giving du = constant × sin ax dx. So a convenient substitution is u = cos ax, which yields du = −a sin ax dx. Z Z −du 3 2 sin ax · cos ax dx = sin2 ax · u2 · a Z −du = 1 − cos2 ax · u2 · a Z −du = 1 − u2 · u2 · a Z 1 = u4 − u2 du a ! 1 u5 u3 = − +C a 5 3 ! 1 cos5 ax cos3 ax − = +C, a 5 3 Information for Students in MATH 141 2010 01 5061 which integration may be checked by differentiation. (Of course, there are other, equivalent ways of expressing this indefinite integral.) Z x3 4. [4 MARKS] Showing all your work, evaluate the integral dx, where a is a √ x 2 + a2 given non-zero constant. Solution: To simplify the surd in the denominator one may use either a trigonometric or a hyperbolic substitution. For students in this course a trigonometric substitution is usually a better choice. To arrange that x = a tan u, we use a substitution x u = arctan , (110) a and dx = a sec2 u du. We may assume that a > 0. The interval of validity for substitution (110) is − π2 < x < π2 , in which the secant is positive. Z Z 3 3 x3 a tan u · a sec2 u du dx = √ 2 2 |a sec u| x +a Z 3 = a tan2 u · sec u tan u du Z d 3 = a sec2 u − 1 · sec u du du ! 3 3 sec u = a − sec u + C 3 effectively by substitution U = sec u p 1 (a tan u)2 − 2a2 (a tan u)2 + a2 + C = 3 √ 1 2 = x − 2a2 x2 + a2 + C 3 which integration may be verified by differentiation. 5. [9 MARKS] (To simplify the exposition of theZ solution, I work a specific example here.) x + 21 dx. Showing all your work, evaluate the integral (x + 9)(x − 5) Solution: Since the degree of the numerator is less than that of the denominator, we can dispense with the first step of dividing denominator into numerator. (a) We need to expand the integrand into a sum of partial fractions; fortunately the factorization of the denominator has been given. Assuming there are constants A, B such that A B x + 21 = + (x + 9)(x − 5) x + 9 x − 5 Information for Students in MATH 141 2010 01 5062 and transforming all fractions to have a common denominator, we find that x + 21 = A(x − 5) + B(x + 9) . (b) The values of A, B may be obtained either by comparing coefficients of like powers of x, or by assigning to x successively the “convenient” values 5, -9: we obtain that 13 7 6 12 = −14A ⇒ A = − . 7 26 = 14B ⇒ B = (c) We may now complete the integration: ! Z Z x + 21 1 6 13 − dx dx = + (x + 9)(x − 5) 7 x+9 x−5 1 = − (−6 ln(x + 9) + 13 ln(x − 5)) + C 7 which can also be expressed in other, equivalent ways. This integration may be verified by differentiation. Wednesday version Za (x2 + 1)e−x dx , where a is a given constant. 1. [4 MARKS] Showing your work, evaluate 0 Solution: I will integrate by parts twice, in order to reduce the degree of the polynomial factor of the integrand. (a) First integration by parts: u = x2 + 1 ⇒ du = 2x dx dv = e−x dx ⇒ v = −e−x Za h i Z a 2 −x 2 −x a (x + 1)e dx = (x + 1) −e + 2x · e−x dx . 0 0 0 (b) Second integration by parts: U = 2x ⇒ dU = 2 dx dV = e−x dx ⇒ V = −e−x Za h i Z a 2 −x 2 −x a (x + 1)e dx = (x + 1 + 2x) −e + 2 · e−x dx . 0 0 0 Information for Students in MATH 141 2010 01 5063 (c) Completion of the integration: Za (x2 + 1)e−x dx = h (x2 + 2 + 2x) −e−x ia 0 0 = −(a2 + 2a + 3)e−a + 3. π Z2 sinn ax dx, 2. [9 MARKS] Showing all your work, find a reduction formula for the integral 0 where a is a given integer. Solution: Assume n is an integer greater than 1. (a) Introduce a symbol for the definite integral sought: Z π2 sinn ax dx . In = 0 (b) Integration by parts: cos ax dx a u = sinn−1 ax ⇒ du = a(n − 1) sinn−2 ax · cos ax dx # π2 " Z π2 1 n−1 sinn−2 ax · cos2 ax dx In = − · sin ax · cos ax + (n − 1) a 0 0 dv = sin ax dx ⇒ v = − (c) Decomposition of integral: Z π2 Z π2 n−2 2 sin ax · cos ax dx = sinn−2 ax · 1 − sin2 ax dx 0 0 Z π2 Z π2 n−2 = sin ax dx − sinn ax dx 0 0 = In−2 − In (d) Solution of equation to obtain reduction formula # π2 " 1 n−1 In = − · sin ax · cos ax + (n − 1) (In−2 − In ) a 0 # π2 " 1 n−1 ⇒ nIn = − · sin ax · cos ax + (n − 1)In−2 a 0 1 n−1 ⇒ In = (0 − 0) + In−2 n n Information for Students in MATH 141 2010 01 5064 Because of the choice of limits and the fact that a is an integer, the “net change” is 0. Thus we obtain a very simple relationship, which can be solved. Students were not asked to complete this part of the solution. For example, it is possible to prove by induction that, if n = 2m, an even, positive integer, then 2m − 1 2m − 3 3 · · . . . · I0 2m 2m − 2 2 (2m)! π = m · . 4 m!m! 2 I2m = (This is Exercise 44, page 481 in the textbook.) Z x2 3. [4 MARKS] Showing all your work, evaluate the integral a2 − x 2 non-zero constant. 23 dx, where a is a Solution: Without limiting generality we take a > 0. A trigonometric substitution can simplify this integral. One such substitution would have x = a sin u; more precisely, u = arcsin ax , defined for − π2 ≤ x ≤ π2 , in which interval the cosine and secant are positive. Then dx = √a du 2 . 1−u Z x2 a2 − x 2 Z 32 a2 sin2 u · a cos u du a3 cos3 u Z Z 2 = tan u du = sec2 u − 1 du = = tan u − u + C sin u −u+C = cos u sin u = p −u+C 2 1 − sin u x x = q a − arcsin + C 2 a 1 − ax2 x x = √ − arcsin + C a a2 − x 2 which may be verified by differentiation. Z ` x(x − a)(x − b) + c dx, (x − a)(x − b) k where a, b, k, ` are distinct constants such that the integrand is defined throughout the 4. [9 MARKS] Showing all your work, evaluate the integral Information for Students in MATH 141 2010 01 5065 given interval, and c is a non-zero constant. (The integrand was not presented to students in factored form.) Solution: (a) Since the degree of the numerator is not less than the degree of the denominator, begin by dividing the denominator into the numerator, obtaining a quotient and a remainder: Z Z x(x − a)(x − b) + c c dx = x+ dx . (x − a)(x − b) (x − a)(x − b) (b) Expand the fraction into partial fractions. Assume that c A B = + , (x − a)(x − b) x − a x − b take to a common denominator, and equate the resulting polynomials: c = A · (x − b) + B · (x − a) . (c) Now either equate coefficients of like powers of x, or, equivalently, give x successive values x = a and x = b: c c = A(a − b) ⇒A= = −B . c = B(b − a) a−b (d) The integration reduces to Z Z x(x − a)(x − b) + c dx = (x − a)(x − b) ! c 1 c c x+ · − · dx . a−b x−a a−b x−b (e) Complete the integration: Z x2 c x(x − a)(x − b) + c dx = + (ln(x − a) − ln(x − b)) + C (x − a)(x − b) 2 a−b c x−a x2 + ln +C, = 2 a−b x−b the correctness of which integration may be verified by differentiation of the function on the right. Information for Students in MATH 141 2010 01 5066 (f) Evaluate the definite integral " 2 #` Z ` x c x(x − a)(x − b) + c dx = + (ln |x − a| − ln |x − b|) (x − a)(x − b) 2 a−b k k " 2 #` x c x−a = + ln 2 a−b x−b k 2 ` − k2 c (` − a)(k − b) = + ln 2 a − b (` − b)(k − a) Z 5. [4 MARKS] Showing all work, evaluate the integral sin3 ax dx, where a is a positive integer. Solution: Z Z 3 sin ax dx = 1 − cos2 ax · sin ax dx under substitution u = cos ax, where du = −a sin ax dx 1 = (1 − u2 )(− ) du a ! 3 u 1 +C = − u− a 3 cos ax cos3 ax = − + +C. a 3a Z Thursday version Z b √ 1. [4 MARKS] Showing all your work, evaluate the integral e t dt. a √ √ 2 Solution: Begin with a substitution t = u, so t = u , dt = 2u du. When t = a, u = a, etc.: Z √ Z t e dt = 2ueu du . Now apply integration by parts: U = u ⇒ dU = du dVZ = 2eu du ⇒ V = 2eu Z 2ueu du = u · 2eu − 2eu du = u · 2eu − 2eu + C = 2(u − 1)eu + C √ √ = 2( t − 1)e t + C Information for Students in MATH 141 2010 01 5067 h √ √ ib The definite integral given is then equal to 2( t − 1)e t . a Z 2. [9 MARKS] Showing all your work, find a reduction formula for the integral (ln(ax + 1))n dx, where a is a given, positive constant. Solution: This problem is a slight generalization of Exercise 45, p. 481 in the textbook, an odd-numbered problem for which there is a solution in the Student Solutions Manual, and also hints on one of the CD-Roms supplied with the textbook. (a) Introduce a symbol for the definite integral sought: In = (ln(ax + 1))n dx . (b) Change the variable (a step which is helpful, but not necessary) u = ax + 1 ⇒ duZ= a dx 1 (ln u)n du In = a (c) Integration by parts: U = (ln u)n ⇒ dU = n(ln u)n−1 · dV = 1 du ⇒ V = u In 1 u Z n = u(ln u) − n (ln u)n−1 du Z n = (ax + 1)(ln(ax + 1)) − na (ln(ax + 1))n−1 dx = (ax + 1)(ln(ax + 1))n − na · In−1 Z 3. [4 MARKS] Showing all your work, evaluate the integral cos4 at dt, where a is a given non-zero constant. Solution: We have to apply the following double angle identity twice: cos 2θ = 2 cos2 θ − 1 . Z !2 1 + cos 2at cos at dt = dt 2 ! Z Z Z 1 2 = cos 2at dt + 2 cos 2at dt + 1 dt 4 Z 4 Information for Students in MATH 141 2010 01 5068 ! Z Z Z 1 1 + cos 4at = dt + 2 cos 2at dt + 1 dt 4 2 1 1 1 1 3 = · sin 4at + · sin 2at + t + C 8 4a 2 2a 8 1 1 3 = sin 4at + sin 2at + t + C 32a 4a 8 which may be verified by differentiation. (Of course, the integral may be expressed in other ways under transformation by trigonometric identities.) Z dx 4. [4 MARKS] Showing your work, evaluate the integral , where a is a given √ x x2 + a positive integer, not a perfect square. Solution: The surd in the denominator may be simplified by either a trigonometric or a hyperbolic substitution. For students in this course the trigonometric substitutions are usually easier. √ √ x = a · tan θ ⇒ dx = a sec2 θ dθ Z Z dx sec2 θ dθ = . √ √ tan θ · a · | sec θ| x x2 + a The actual substitution is given by θ = arctan √xa , valid for − π2 < x < π2 . In that interval the secant function is positive, so the absolute signs may be dropped. The integral is equal to Z 1 1 csc θ dθ = √ ln | csc θ − cot θ| + C √ a a 1 sec θ − 1 +C = √ ln tan θ a √ tan2 θ − 1 1 + C = √ ln tan θ a √ x2 + a − √a 1 + C = √ ln x a which can be verified by differentiation. Z` 5. [9 MARKS] Showing all your work, evaluate the integral k x(x − a)(x − b) + c dx, (x − a)(x − b) where a, b, k, ` are distinct constants such that a, b are not contained in the interval whose end-points are k, `, and c is a non-zero constant. Solution: I will first determine an indefinite integral. Information for Students in MATH 141 2010 01 5069 (a) Since the degree of the numerator is not less than the degree of the denominator, begin by dividing the denominator into the numerator, obtaining a quotient and a remainder: Z Z c x(x − a)(x − b) + c dx = x+ dx . (x − a)(x − b) (x − a)(x − b) (b) Expand the fraction into partial fractions. Assume that c A B = + , (x − a)(x − b) x − a x − b take to a common denominator, and equate the resulting polynomials: c = A · (x − b) + B · (x − a) . (c) Now either equate coefficients of like powers of x, or, equivalently, give x successive values x = a and x = b: c c = A(a − b) ⇒A= = −B . c = B(b − a) a−b (d) The integration reduces to Z Z x(x − a)(x − b) + c dx = (x − a)(x − b) ! c 1 c c x+ · − · dx . a−b x−a a−b x−b (e) Complete the integration: Z x(x − a)(x − b) + c x2 c dx = + (ln |x − a| − ln |x − b|) + C (x − a)(x − b) 2 a−b x2 c x − a +C, = + ln 2 a − b x − b the correctness of which integration may be verified by differentiation of the function on the right. (f) Z ` k " 2 #` x c x − a x(x − a)(x − b) + c dx = + ln (x − a)(x − b) 2 a − b x − b k Information for Students in MATH 141 2010 01 5070 E.1.4 Draft Solutions to Quiz Q4 Distribution Date: Posted on the Web on 06 April, 2007; corrected on 09 April, 2007. Caveat lector! There could be misprints or errors in these draft solutions. There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. Each version of the quiz was graded out of a maximum of 30 marks, but 2 of the versions had 5 problems and 2 had 4 problems. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version 1. [6 MARKS] Showing all of your work, find the length of the following curve for the interval 0 < a ≤ u ≤ b : ! eu + 1 y = ln u . e −1 Solution: (a) dy eu − 1 eu (eu − 1) − (eu + 1)eu = u dx e +1 (eu − 1)2 2eu = − 2u e −1 !2 !2 dy e2u + 1 = coth2 u 1+ = 2u dx e −1 (b) s Z b arc length = a dy 1+ dx !2 Z ! b du = | coth u| du a Information for Students in MATH 141 2010 01 5071 (c) Successful completion of the integration: Z b Z b | coth u| du = coth u du a a since a < b [ln sinh u]ba sinh b eb − e−b = ln = ln a sinh a e − e−a ! 3.0 2.5 2.0 1.5 1.0 0.5 0.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Figure 27: The limac¸on r = 1 + 2 sin θ 2. [10 MARKS] (see Figure 27 on page 5071) The graph of the following curve is given. Information for Students in MATH 141 2010 01 5072 Showing detailed work, find the area that is enclosed between the inner and the outer loops: r = a(1 + 2 sin θ), where a is a positive constant. Solution: (a) Determination of the limits of integration: we need first to locate where the curve crosses itself. Since its formula is in terms of sin θ, the curve is periodic with period (at most) 2π. As θ ranges over the values from 0 to 2π, the values r(θ) range over uniquely determined values. How then can the curve cross itself? This can happen either i. at points (r(θ1 ), θ1 ) and (r(θ1 + π), θ1 + π) where r(θ1 + π) = −r(θ1 ); or ii. at the pole, where r = 0 for two distinct values of θ. The first possibility would, for the present curve, require that 1 + 2 sin(θ1 + π) = − (1 + 2 sin θ1 ) which is equivalent to 2 + 2 sin θ1 + 2 sin(θ1 + π) = 0 which is equivalent to 2 = 0, a contradiction. Thus the present curve can cross itself only at the pole. That occurs where 1 + 2 sin θ = 0, i.e., where sin θ = − 12 . The values of θ satisfying this equation are 2nπ− π6 and (2n+1)π+ π6 , where n is any integer. The outer loop of this lima¸con is traced out, for example, for − π6 ≤ θ ≤ 7π . 6 7π 11π The inner loop is traced for 6 ≤ θ ≤ 6 . (b) The area of the region bounded by the larger, outer loop is Z 7π 6 2 a − π6 Z 7π 6 2 = a − π6 Z 7π 6 2 = a a2 = 2 − π6 Z 7π 6 − π6 1 · (1 + 2 sin θ)2 dθ 2 1 · 1 + 4 sin θ + 4 sin2 θ dθ 2 1 · (1 + 4 sin θ + 2 − 2 cos 2θ) dθ 2 (3 + 4 sin θ − 2 cos 2θ) dθ 7π a2 [3θ − 4 cos θ − sin 2θ]−6 π 6 2 ! 2 7π 7π −π −π a 7π a2 π − 4 cos − sin − − 4 cos − sin = − 2 2 6 3 2 2 6 3 = Information for Students in MATH 141 2010 01 5073 √ 3 3 . = a2 2π + 2 (c) The area of the inner loop is Z 11π 6 1 2 a · (1 + 2 sin θ)2 dθ 7π 2 6 11π a2 [3θ − 4 cos θ − sin 2θ] 7π6 2 6 ! ! a2 11π 11π 11π a2 7π 7π 7π = − 4 cos − sin − − 4 cos − sin 2 2 6 3 2 2 6 3 √ 3 3 . = a2 π − 2 = (d) The area of the region between the loops is the excess of the area inside the outer loop over the area inside the inner loop, i.e., √ √ √ 3 3 3 3 2 − π − a2 2π + = a (π + 3 3) . 2 2 Note that a cleaner way of solving this problem would have been to first integrate from 0 to 2π, which would give the area between the loops plus twice the area inside the smaller loop; and then to subtract twice the area inside the smaller loop. This method is better because the first integral is very easy to evaluate, since the periodic terms contribute nothing. This curve is discussed in Exercise 10.4.21, on page 683 of the textbook, and is solved in the Student Solutions Manual and also on one of the CD-Roms which accompany the textbook. 3. [4 MARKS] Showing full details of your work, find the exact length of the curve x = et + e−t , y = a − 2t, 0 ≤ t ≤ b, where a, b are constants. Solution: dx dt !2 dy + dt !2 2 et − e−t + 4 = 2 et + e−t Z b arc length = et + e−t dt = = 0 t e − e−t b 0 = eb − e−b = 2 sinh b . Information for Students in MATH 141 2010 01 5074 4. [4 !) the value of the limit for the sequence. If it diverges, prove that fact: ( MARKS] Find 3n arctan . 3n + 1 3n 1 Solution: As n → ∞, 3n+1 = 1 − 3n+1 → 1. Since the arctangent function is continuous at the point 1, the limit of the sequence is the arctangent of 1, i.e., π4 . 5. [6 MARKS] The given curve is rotated about the y-axis. Find the area of the resulting surface: x = 2 √1 2 y2 − ln y , (1 ≤ y ≤ a), where a is a real constant greater than 1. Solution: (a) ! dx 1 1 = √ 2y − dy y 2 2 !2 ! dx 1 1 2 1+ = 1 + 4y + 2 − 4 dy 8 y ! 1 1 2 = 4y + 2 + 4 8 y !!2 1 1 = √ 2y + y 2 2 (b) surface area = = = = (c) One way to integrate R s !2 dx 2πx 1 + dy dy 1 ! Z π a 2 1 y − ln y 2y + dy 4 1 y ! Z π a ln y 3 2y + y − 2y ln y − dy 4 1 y #a " Z π y4 y2 1 π a 2 + − (ln y) − y ln y dy 4 2 2 2 2 1 1 Z a y ln y dy is by parts, with u = ln y, v0 = y: u0 = 1y , v = Z Z y2 y y ln y dy = · ln y − dy 2 2 y2 (2 ln y − 1) + C . = 4 y2 , 2 Information for Students in MATH 141 2010 01 5075 Another way is to use the substitution w = y2 , so dw = 2y dy, ln w = 2 ln y: Z Z Z ln w dw 1 1 y ln y dy = · = ln w dw = w (ln w − 1) + C etc. 2 2 4 4 Thus the surface area is #a " π 2 3 1 y2 2 y + ln y − (ln y) − (2 ln y − 1) 4 3 2 4 1 ! 4 2 2 π a a 1 a 2 = + − (ln a) − (2 ln a − 1) − 1 4 2 2 2 4 Tuesday version 3 1. [4 MARKS] Showing detailed work, find the arc length function for the curve y = ax 2 with starting point P0 (1, a), where a is a positive constant. That is, find a function f (x) whose value is the distance along the curve from the starting point to the point with abscissa x. Solution: (a) dy 3 √ = a· · x dx 2 !2 dy 9a2 x 1+ = 1+ dx 4 (b) Z x f (x) = r 1+ 1 9a2 t dt 4 ! 3 x 2 2 2 4 9a t = · 2 1 + 3 9a 4 1 3 3 2 2 4 + 9a x − 4 + 9a2 2 = . 27a2 2. [6 MARKS] Showing detailed work, find the area of the surface obtained by rotating the following curve about the x-axis: x2 ln x − (a ≤ x ≤ b) 4 2 where a, b are two positive real constants, a < b. y= Solution: Information for Students in MATH 141 2010 01 5076 (a) ! dy 1 1 = x− dx 2 x !2 ! dy 1 2 1 1+ = 1+ x + 2 −2 dx 4 x ! 1 2 1 = x + 2 +2 4 x !!2 1 1 = x+ 2 x (b) surface area = = = = (c) One way to integrate R s !2 dy 2πy 1 + dx dx a ! ! Z b 2 x ln x 1 − x+ dx π 4 2 x a ! Z b 3 x x x ln x ln x π + − − dx 4 4 2 2x a " #b Z b 1 4 x2 1 2 π x + − (ln x) − π x ln x dx 16 8 4 a a Z b x ln x dx is by parts, with u = ln x, v0 = x: u0 = 1x , v = Z Z x2 x x ln x dx = · ln x − dx 2 2 x2 (2 ln x − 1) + C . = 4 Another way is to use the substitution w = x2 , so dw = 2x dx, ln w = 2 ln x: Z Z Z ln w dw 1 1 x ln x dx = · = ln w dw = w (ln w − 1) + C etc. 2 2 4 4 Thus the surface area is " x2 x4 x3 1 + − (ln x)2 − (2 ln x − 1) π 16 8 4 4 #b a x2 , 2 Information for Students in MATH 141 2010 01 5077 3. [4 MARKS] Find the exact length of the curve given by x = et cos t y = et sin t (0 ≤ t ≤ a) where a is a positive constant. Solution: (a) dx = et (cos t − sin t) dt dy = et (sin t + cos t) dt !2 !2 dx dy + = e2t (2 cos2 t + 2 sin2 t) = 2e2t dt dt (b) The arc length is Z a √ t √ 2e dt = 2(ea − 1) . 0 4. [10 MARKS] Working only with polar coordinates, find the area of the region that lies inside the first curve and outside the second curve: r = b sin θ, r = a, where a and b are positive constants. Solution: (a) Both of these curves are circles; we need to determine the coordinates of the points of intersection. Solving the equations yields a . b One point of intersection will be (r, θ) = a, arcsin ba . Another point of intersection will be (r, θ) = a, π − arcsin ab — remember that the values of the arcsine function h i are in the interval − π2 , π2 . It appears from a drawing that we have all the points of intersection. If we solve the equation r = −a with r = b sin θ we obtain precisely the same points, albeit with different coordinates. If we attempt to replace the equation r = b sin θ with that obtained under the identification (r, θ) → (−r, θ + π) there is no change. This algebraic investigation discloses all possible points of intersection except the pole, which must be checked separately. But the pole cannot lie on r = a, since the pole has first coordinate 0 always. Thus we have, indeed, found all the points of intersection. r=a sin θ = Information for Students in MATH 141 2010 01 5078 (b) The area bounded by the arcs can be considered to consist of the disk r = b sin θ diminished by a sector of the circle r = a for arcsin ba ≤ θ ≤ π − arcsin ab , together with two small segments of the disk r = b sin θ bounded by the rays θ = arcsin ab and θ = π − arcsin ab . What I have provided is one prescription for computing the area. An easier way would be to take the integral a Z 1 π−arcsin b (b sin θ)2 − a2 dθ 2 arcsin ab Z π2 = (b sin θ)2 − a2 dθ arcsin a b by symmetry around the line θ = π2 ! ! Z π2 2 1 − cos 2θ 2 = b − a dθ 2 arcsin ba ! ! Z π2 b2 b2 2 = − a − cos 2θ dθ 2 2 arcsin ba " 2 ! # π2 b b2 2 = − a θ − sin 2θ 2 4 arcsin ab 2 2 b − 2a π a a√ 2 = − arcsin + b − a2 2 2 b 2 5. [6 MARKS] Determine whether the series is convergent or divergent. If it is convergent, ) ∞ ( X b find its sum. Otherwise prove that it is divergent: , where a, b are positive n(n + a) n=1 integers. Solution: (a) Expand the general term into partial fractions: there exist constants A, B such that b A B = + . n(n + a) n n + a To determine the coefficients A, B we can proceed in several ways. If we take the fractions to a common denominator n(n + a), we obtain the polynomial identity b = A · (N + A) + b · a . In this identity, if we set the variable n equal to −a, we obtain that b = B(−a), so B = − ab ; and, setting n = 0, we obtain A = ba , hence ! b 1 1 b = − . n(n + a) a n n + a Information for Students in MATH 141 2010 01 5079 (b) For sufficiently large N the Nth partial sum is equal to N X n=1 b = a b → a b n(n + a) ! 1 1 1 1 1 1 + + ... + − − − ... − 1 2 a−1 N+1 N+2 N+a ! 1 1 1 + + ... + − 0 − 0 − ... − 0 1 2 a−1 ! as N → ∞. Hence the series converges to the sum ! b 1 1 1 + + ... + . a 1 2 a−1 Convergence could be proved in other ways, thereby earning the student part marks. For example, using the Comparison or Limit Comparison Tests, or the Integral Test. Wednesday version 1. [10 MARKS] Showing detailed work, find all points different from the origin on the following curve where the tangent is horizontal; a is a positive constant: x = a(cos θ − cos2 θ), y = a(sin θ − sin θ cos θ) . Solution: (a) dx = dθ = dy = dθ = = a (− sin θ − 2 cos θ(− sin θ)) a(sin θ)(2 cos θ − 1) a − cos θ + 2 cos2 θ − 2 sin2 θ a − cos θ + 2 cos2 θ − 1 a(2 cos θ + 1)(cos θ − 1) Actually, you weren’t expected to find dx . dt (b) There will be a horizontal tangent at the point with parameter value t if i.e., if dy dθ dx dθ = 0, implying that dy dθ must be 0, provided dx dθ dy dx = 0, , 0 at the same value of t. Information for Students in MATH 141 2010 01 5080 (This last requirement is subtle, and you weren’t expected to actually check it. It is because of this restriction that I explicitly excluded the origin from consideration.) The equations we have to solve are cos θ = 1 and 1 cos θ = − . 2 They are not being solved simultaneously: we are looking for points t that satisfy at least one — meaning, here, either — of the equations. (c) The first of these equations is satisfied when θ is an even integer multiple of π. But this parameter value corresponds always to the origin, which was excluded from consideration. (d) The second is satisfied when θ is of the form 1 (2n + 1)π ± π . 3 (e) The functions defining this curve are all periodic with period 2π. Thus we can study the curve completely by examining its behavior for parameters θ chosen over an interval of length 2π, e.g. 0 ≤ θ < 2π. There are precisely 3 points here where dy = 0: dθ (x(0), y(0)) = (0, 0) √ ! !! a 3 3a 2π 2π x ,y = − , 3 3 4 4 √ ! !! a 3 3a 4π 4π . x ,y = − , − 3 3 4 4 Of these, you were specifically instructed to exclude the first, which is the origin. 2. [5 MARKS] Showing detailed work determine the total length of the portion of the following curve which is in the first quadrant: x = a cos3 θ, y = a sin3 θ, where a is a positive constant. Solution: dx = −3a cos2 θ · sin θ dθ dy = 3a sin2 θ · cos θ dθ Information for Students in MATH 141 2010 01 dx ⇒ dθ !2 dy + dθ !2 5081 = 9a2 sin2 θ · cos2 θ · cos2 θ + sin2 θ = 9a2 sin2 θ cos2 θ . The curve is in the first quadrant when both coordinates are positive; as each of these is a cube of a sine or cosine, this means that the portion of the curve in the first quadrant is that given by 0 ≤ θ ≤ π2 . The length of the arc is Z π 2 p Z 9a2 sin θ cos2 θ dθ = 0 " π 2 2 sin2 θ 3a sin θ · cos θ dθ = 3a 2 0 # π2 = 0 3a . 2 3. [10 MARKS] Find the area of the region that lies inside both of the following curves r = a + 2 sin θ, r = a − 1, where a is a suitable positive constant. Solution: (a) Determination of the limits of integration: we need first to locate where the curves cross. We begin by solving the two given equations, and find that 1 7π sin θ = − ⇒ θ = 2 6 or 11π 6 or any angle obtained from these by adding an integer multiple of 2π. This yields the points ! ! 7π 11π a − 1, , a − 1, . 6 6 Students weren’t expected to pursue this question further. Strictly speaking, they should then have solved r = a + 2 sin θ, r = −(a − 1), which would have yielded no points; then solved r = −a + 2 sin θ, r = a − 1, which would again yield no points; then r = −a + 2 sin θ, r = −(a − 1), which would have yielded the 2 points already found. (b) This problem could then be approached in several ways. To find the area “directly” would require finding the sum of the integrals 1 2 and 1 2 Z Z 7π 6 −pi 6 11π 6 7pi 6 (a − 1)2 dθ (a + 2 sin θ)2 dθ . Information for Students in MATH 141 2010 01 5082 The first of these is just 2/3 of the area of a disk of radius a − 1, i.e., second is Z 11π 6 1 (a + 2 sin θ)2 dθ 2 7pi6 Z 11π 6 1 = a2 + 4a sin θ + 2(1 − cos 2θ) dθ 2 7pi6 i 11π 1h 2 = a θ − 4a cos θ + 2θ − sin 2θ 7π6 2 6 √ 2π 1π 2 = · a − 2 3a + . 3 3 2π(a−1)2 . 3 The (Another way to solve this would be to find the area of the disk of radius a − 1 and subtract from it the portion that is cut off. These integrals could have been slightly more efficiently computed by taking only the area up to the y-axis and doubling it.) Hence the net area of the region inside both of the curves is ! √ 4π 5π 2 4π a − +2 3 + . 9 3 3 4. [5 MARKS] Showing detailed work, determine whether the following series is convergent or divergent. If it is convergent, find its sum. Otherwise explain why it diverges: ∞ X an + bn n=1 (ab)n , where a, b are integers greater than 1. Solution: The nth term is 1 an = b !n 1 + a !n . The nth partial sum is, therefore, the sum of the partial sums of two geometric series; both of the geometric series are convergent, since the common ratios are less than 1 in magnitude. Since the two separate partial sums approach a limit, the sum of these sequences approaches as its limit sum of the limits of the two sequences. In the case the 1 n of the series whose nth term is b , the limit of the partial sums is 1 b 1− 1 b = 1 ; b−1 Information for Students in MATH 141 2010 01 5083 1 . Hence the given series sums to the sum of these similarly, the second series sums to a−1 limits, i.e. 1 1 a+b−2 + = . b − 1 a − 1 (a − 1)(b − 1) Thursday version 1. [10 MARKS] The curve x = a(1 − 2 cos2 t), y = (tan t)(1 − 2 cos2 t), where a is a given positive integer, crosses itself at some point (x0 , y0 ). Showing all your work, find the point of crossing, and the equations of both tangents at the point. (In determining the point of crossing you are expected to investigate the parametric functions: it is not sufficient to simply plot a finite number of points on the curve.) Solution: (a) Since the functions are all periodic with period π, it suffices to take an interval of this length for t, and that will reveal all aspects of the behavior of this curve. (More precisely, the tangent function has period π, and, while the cosine function has period 2π, its square has period π.) So, without limiting generality, let’s consider − π2 ≤ t ≤ π2 : we have to exclude both end points of this interval, since the tangent function is not defined at either of them. Suppose that the curve crosses itself at the points with parameter values t = t1 and t = t2 ; without limiting generality, we can assume that these parameter values have been so labelled that t1 < t2 . Since the x-coordinates will need to be the same, a(1 − 2 cos2 t1 ) = a(1 − 2 cos2 t2 ) (111) cos t1 = ± cos t2 . (112) so Since the y-coordinates must also coincide, we also have (tan t1 )(1 − 2 cos2 t1 ) = (tan t2 )(1 − 2 cos2 t2 ) which implies that either cos2 t1 = cos2 t2 = 1 , 2 (113) (114) or tan t1 = tan t2 . (115) In the interval we have chosen for t, the cosines are always positive; the only solution to (114) is t1 = − π4 , t2 = + π4 . In that same interval for t there will be no solutions to (115), since the tangent function is increasing there. Thus the only possible Information for Students in MATH 141 2010 01 5084 crossing points are t = ± π4 , and the point of crossing is the origin, (x, y) = (0, 0). While it isn’t required in the solution, note that as t → ± π2 , x → 1: this curve is asymptotic to the vertical line x = 1. (b) Tangent at the point with parameter value t = − π4 : dx dt y dy dt dy dx = 4a(cos t)(sin t) = 2a sin 2t = −2a = (tan t)(1 − 2 cos2 t) = tan t − sin 2t = sec2 t − 2 cos 2t = 2 − 0 = dy dt dx dt =− 2 1 =− , 2a a and the tangent has equation y = − ax . (c) Tangent at the point with parameter value t = π4 : dx = 4a(cos t)(sin t) = 2a sin 2t = 2a dt dy = sec2 t − 2 cos 2t = 2 − 0 dt dy dy 2 1 dt = = dx = , dx 2a a dt and the tangent has equation y = ax . 2. [5 MARKS] Showing detailed work, find the surface area generated by rotating the following curve about the y-axis. x = at2 , y = bt3 , 0 ≤ t ≤ 5. Solution: dx = 2at dt dy = 3bt2 dt !2 !2 dx dy + = 4a2 t2 + 9b2 t4 dt dt Z 5 √ area about y-axis = 2π at2 4a2 t2 + 9b2 t4 dt 0 Z 5 √ = 2πa t3 4a2 + 9b2 t2 dt 0 Information for Students in MATH 141 2010 01 5085 Under the substitution u = 4a2 + 9b2 t2 , du = 18b2 t dt Z 4a2 +225b2 5 √ aπ 3 1 3 2 2 2 2πa t 4a + 9b t dt = (u 2 − 4a2 u 2 ) du 4 81b 4a2 0 " #4a2 +225b2 πa 2 52 8a2 32 = u − u 81b4 5 3 4a2 2πa 2 2 23 2 2 5 = (4a + 225b ) (2a + 675b ) − 16a (81)(15)b4 = ... Z 3. [10 MARKS] There is a region in the first quadrant that is bounded by arcs of both of the following curves. Showing your work in detail, find the area of the region: r2 = a sin 2θ r2 = a cos 2θ . Solution: (a) The given curves are expressed only in terms of sine and cosine of 2θ. The given functions are periodic with period π. When, for either of these curves, we permit θ to range over an interval of length π, we will trace out the entire curves. The intersections of the curves in the first quadrant will be where sin 2θ = cos 2θ is positive: thus the only point we have found by this algebraic solution of the two equations is at θ = π8 . However there are other ways in which curves can intersect, since points have infinitely many different sets of polar coordinates. If we transform either of the given equations under the substitution (r, θ) → (−r, θ + π), we find that there is no change in the equation. Thus we haven’t missed any points because of the convention that permits the first coordinate to be negative. But there is another situation that leads to multiple sets of coordinates; that is at the pole, where the second — angular — coordinate is totally arbitrary; the pole can lie on a curve simply because of the fact that its distance coordinate r = 0, with no reference to θ. To determine whether the pole lies on a curve we must investigate whether the equation is satisfied by r = 0 with any value of θ. We find the curve r2 = a sin 2θ does contain the pole: when r = 0 the equation is satisfied by any θ such that sin 2θ = 0; so two solutions are θ = 0 and θ = π2 . Similarly, the pole lies on the curve r2 = a cos 2θ with θ = π4 . I have given only the coordinates in the first quadrant. To summarize: there are 2 intersection points in the first quadrant: 1 1 π (r, θ) = a 2 2− 4 , 4 Information for Students in MATH 141 2010 01 5086 where the point lies on both of the curves with the same pair of coordinates; and the pole, which lies on the two curves with different sets of coordinates. √ (b) We can find the area by joining the point √4 a , π8 to the pole and calculating the 2 sum of two integrals: = π π πa πa [− cos 2θ]04 + [sin 2θ] π2 4 4 4 Z π Z π πa 2 πa 4 sin 2θ dθ + cos 2θ, dθ 2 0 2 π4 πa πa πa = + = . 4 4 2 4. [5 MARKS] Showing detailed work, express the number below as a ratio of integers. 0.ab = 0.abababab... where a, b are any two digits. You are expected to simplify your answer as much as possible. Solution: The repeating decimal is the sum of an infinite series ! ! ! ! 10a + b 10a + b 1 10a + b 1 10a + b 1 + + + + ... 100 100 100 100 1002 100 1003 ! N X 10a + b 1 = lim N→∞ 100 100n n=0 ! !n N X 1 10a + b lim = 100 N→∞ n=0 100 N+1 ! 1 1 − 100 10a + b = lim 1 100 N→∞ 1 − 100 ! 10a + b 1 = 1 100 1 − 100 ! 10a + b 100 10a + b = = 100 99 99 E.2 MATH 141 2008 01 E.2.1 Draft Solutions to Quiz Q1 Release Date: Mounted on the Web on Friday, 01 February, 2008 (but subject to correction) Information for Students in MATH 141 2010 01 5087 There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version Zc 1. [5 MARKS] If Zc f (x) dx = k and a Zb f (x) dx = `, find f (x) dx. Show your work. a b Solution: (a) Zc Zb f (x) dx = (b) Hence Zc f (x) dx . f (x) dx + a a b Zc Zc Zb f (x) dx = b f (x) dx − a f (x) dx . a (c) = k −`. Z a 2. [5 MARKS] Find an antiderivative of the integrand of the integral 0 √ x dx, and then use the Fundamental Theorem of Calculus to evaluate the integral. You are not expected to simplify your numerical answer, but no marks will be given unless all your work is clearly shown. Information for Students in MATH 141 2010 01 5088 1 (a) One antiderivative of x 2 is 1 2 (b) Z a √ 0 1 2 3 1 · x 2 +1 = x 2 . 3 +1 " 2 32 x dx = x 3 #a = 0 2 3 2 23 a − 0 = a2 . 3 3 Zx4 3. [10 MARKS] Showing all your work, differentiate the function g(x) = tan x 1 dt. √ 2 + t2 Solution: (a) First split the interval of integration into 2 parts at a convenient place: Z0 g(x) = tan x 1 dt + √ 2 + t2 Zx4 1 dt . √ 2 + t2 0 (b) Then reverse the limits in the first summand and change its sign, so that the variable limit is the upper one: tan x Z g(x) = − 0 1 dt + √ 2 + t2 Zx4 0 1 dt . √ 2 + t2 (c) Denote the upper limit of the first integral by u = tan x. Then d dx tan x Z 0 1 d dt = √ du 2 + t2 = √ Zu √ 0 1 u2 1 2 + t2 dt · du dx · sec2 x 2+ 2 sec x · tan x 2 sec x · tan x = √ = √ . 1 + sec2 x 2 + tan2 x (d) Denote the upper limit of the second integral by v = x4 . Then d dx Zx4 0 d dt = √ dv 2 + t2 1 Zv 0 dv 1 dt · √ dx 2 + t2 Information for Students in MATH 141 2010 01 5089 = = 1 · 4x3 √ 2 2+v 4x3 . √ 2 + x8 (e) Hence d 4x3 2 sec x · tan x . + √ g(x) = − √ dx 2 + x8 2 + tan2 x Z x Z t2 a + ub 4. [10 MARKS] If F(x) = f (t) dt, where f (t) = du and a, b are constants, u 1 1 find F 00 (2). Solution: (a) Applying Part 1 of the Fundamental Theorem yields Z x2 0 F (x) = f (x) = 1 a + ub du . u (b) A second application of Part 1 of the Fundamental Theorem yields d F (x) = f (x) = dx 00 Z x2 0 1 a + ub du . u (c) Denote the upper index of the last integral by v = x2 . (d) d dx Z x2 1 Z v a + ub a + ub d du = du u dx 1 u Z v d a + ub dv = du · dv 1 u dx b a+v dv = · v dx b a+v = · 2x v a + x2b · 2x = 2 x 2 a + x2b . = x Information for Students in MATH 141 2010 01 5090 Tuesday version 1. [5 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z a b tan(t) dt , g(x) = x (where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluate g(x), by first verifying carefully that ln | sec x| is an antiderivative of tan x. Solution: (a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as a function of its upper index of integration. Here the variable is the lower index of integration. Z a d b tan(t) dt dx x ! Z x d b tan(t) dt = − dx a Z x d = − b tan(t) dt dx a = −b tan x . Some students may quote a variant of Part 1 which gives the derivative of a definite integral with respect to the lower index, and this should be accepted if work has been shown. (b) Students were expected to first find the derivative of ln | sec x|. Since this is a composition of 2 functions, the Chain Rule will be needed. Let u = sec x. Then d d d ln | sec x| = ln |u| · secx dx du dx 1 · sec x tan x = u 1 = · sec x tan x sec x = tan x . Hence g(x) = b ln | sec t|ax sec a cos x = b ln . = b ln sec x cos a Information for Students in MATH 141 2010 01 5091 2. [5 MARKS] Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 1]: r r r r 1 7 14 21 7n lim + + + ... + . n→∞ n n n n n Solution: (a) We are told to take the interval of integration to be [0, 1]; when this is divided into 1 n equal parts, each has length ∆x = . Such a factor has been explicitly written in n the sum. r 7i (b) The typical summand is — aside from the common factor 1n — of the form . n Since the distance of the left end-point of the ith subinterval from 0 is i∆x = ni , we r 7i √ may interpret = 7x . n (c) Thus the limit must be equal to # Z 1√ √ 2 3 1 2√ 7x dx = 7 · · x 2 = 7. 3 3 0 0 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z √x cos t g(x) = b dt , t a where a, b are constants. Solution: (a) Denote the upper index of the integral by u(x) = √ x. (b) Then Z √x d d cos t g(x) = dt b dx dx a t Z u(x) d cos t = dt b dx a t Z u(x) d du(x) cos t = dt · b du a t dx cos u du(x) · = b u dx Information for Students in MATH 141 2010 01 5092 cos u 1 · √ u 2 x √ cos x 1 = b √ · √ x 2 x √ cos x = b 2x = b 4. [10 MARKS] Showing all your work, determine all values of x where the curve y = Zx 1 dt is concave downward, where a, b are constants. 1 + at + bt2 0 Solution: (a) By Part 1 of the Fundamental Theorem, y0 (x) = 1 . 1 + ax + bx2 (b) Differentiating a second time yields ! 1 d y (x) = dx 1 + ax + bx2 d 1 2 1 + ax + bx = − · 1 + ax + bx2 2 dx a + 2bx = − . 1 + ax + bx2 2 00 (c) The curve is concave downward where y00 < 0: − a + 2bx 1 + ax + bx2 2 > 0 ⇔ −(a + 2bx) > 0 since the denominator is a square, hence positive ⇔ 2bx < −a when b > 0 x < − 2ba a when b < 0 x>− ⇔ never concave2bupward when b = 0 Wednesday version Information for Students in MATH 141 2010 01 5093 π Zb 1. [5 MARKS] Evaluate the integral sin t dt. π a Solution: (a) An antiderivative of sin t is − cos t. (b) π Zb π sin t dt = [− cos t] πb a π a (c) Your answer should be simplified as much as possible. 2. [5 MARKS] Evaluate the following limit by first recognizing the sum as a Riemann n X i8 . A full solution is required — it is not sum for a function defined on [0, 1]: lim 9 n→∞ n i=1 sufficient to write only the value of the limit. Solution: (a) We are told to take the interval of integration to be [0, 1]; when this is divided into 1 n equal parts, each has length ∆x = . n i 8 1 (b) The typical summand is — aside from the common factor — of the form . n n i Since the distance of the left end-point of the ith subinterval from 0 is i∆x = , we n i 8 8 may interpret =x . n (c) Thus the limit must be equal to Z 1 0 1 x dx = x9 9 #1 8 = 0 1 . 9 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative Zbx of the function cos (tc ) dt, where a, b, c are real numbers. cos x Solution: Information for Students in MATH 141 2010 01 5094 (a) The Fundamental Theorem gives the derivative of a definite integral with respect to the upper limit of integration, when the lower limit is constant. The given integral must be expressed in terms of such specialized definite integrals. Zbx Z0 Zbx cos (tc ) dt = cos x cos (tc ) dt + cos x cos (tc ) dt 0 cos x Z Zbx cos (tc ) dt + = − 0 cos (tc ) dt 0 Zbx cos (tc ) dt, let u = bx. Then (b) For the summand 0 d dx Zbx d cos (t ) dt = dx Zu c 0 cos (tc ) dt 0 = d du Zu cos (tc ) dt · du dx 0 = cos (uc ) · b = cos ((bx)c ) · b cos x Z (c) For the summand cos (tc ) dt, let v = cos x. 0 d dx cos x Z cos (tc ) dt = d dx 0 Zv cos (tc ) dt 0 = d dv Zv cos (tc ) dt · dv dx 0 = cos (vc ) · (− sin x) = cos (cosc x) · (− sin x) (d) Zbx cos (tc ) dt = − cos (cosc x) · (− sin x) + cos ((bx)c ) · b . cos x Information for Students in MATH 141 2010 01 5095 0 if x < 0 Z x if 0 ≤ x ≤ a x and g(x) = f (t) dt, where a is 4. [10 MARKS] Let f (x) = 2a − x if a < x < 2a 0 0 if x > 2a a positive constant. Showing all your work, find a formula for the value of g(x) when a < x < 2a. Solution: (a) The interval where we seek a formula is the third interval into which the domain has been broken. For x in this interval the integral can be decomposed into Z x Z a Z x f (t) dt = f (t) dt + f (t) dt . 0 0 a The portion of the definition of f for x < 0 is of no interest in this problem, since we are not finding area under that portion of the curve; the same applies to the portion of the definition for x > 2a. (b) Z Z a a f (t) dt = 0 t dt " = 0 # 2 t=a t 2 = t=0 a2 . 2 (c) Z Z x x f (t) dt = a (d) Thursday version (2a − t) dt " #t=x t2 = 2at − 2 t=a ! ! x2 a2 2 = 2ax − − 2a − . 2 2 a ! ! a2 x2 a2 x2 2 g(x) = + 2ax − − a2 . − 2a − = 2ax − 2 2 2 2 Information for Students in MATH 141 2010 01 5096 1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral Zbπ cos θ dθ, where a, b are given integers. No marks will be given unless all your work aπ is clearly shown. Your answer should be simplified as much as possible. Solution: (a) One antiderivative of cos θ is sin θ. (b) Zbπ cos θ dθ = [sin θ]bπ aπ = sin(bπ) − sin(aπ) . aπ (c) Students were expected to observe that the value of the sine at the given multiples of π is 0, so the value of the definite integral is 0. 2. [5 MARKS] Express lim n→∞ n X axi sin xi ∆x as a definite integral on the interval [b, c], i=1 which has been subdivided into n equal subintervals. Solution: Z c ax sin x dx . b 3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function Z bx 2 t +c g(x) = dt , 2 ax t − c where a, b, c are positive integers. Solution: (a) The Fundamental Theorem gives the derivative of a definite integral with respect to the upper limit of integration, when the lower limit is constant. The given integral must be expressed in terms of such specialized definite integrals. Z 0 2 Z bx 2 Z bx 2 t +c t +c t +c dt = dt + dt g(x) = 2 2 t2 − c ax t − c 0 ax t − c Z ax 2 Z bx 2 t +c t +c =− dt + dt 2 t −c t2 − c 0 0 Information for Students in MATH 141 2010 01 Zbx (b) For the summand 5097 t2 + c dt, let u = bx. Then t2 − c 0 d dx Zbx t2 + c d dt = t2 − c dx 0 Zu t2 + c dt t2 − c 0 = d du Zu t2 + c du dt · 2 t −c dx 0 2 u + c du · u2 − c dx u2 + c = 2 ·b u −c (bx)2 + c ·b = (bx)2 − c = Z ax 2 (c) For the summand step, 0 t +c dt, let u = ax. Then, analogously to the preceding t2 − c Z ax 2 d t +c (ax)2 + c dt = · a. dx 0 t2 − c (ax)2 − c (d) g0 (x) = (bx)2 + c (ax)2 + c · b − · a. (bx)2 − c (ax)2 − c Zx3 4. [10 MARKS] Find the derivative of the function f (x) = √ √ t cos t dt. x Solution: (a) First split the interval of integration into 2 parts at a convenient place: Z0 f (x) = √ x √ t cos t dt + Zx3 0 √ t cos t dt Information for Students in MATH 141 2010 01 5098 (b) Then reverse the limits in the first summand and change its sign, so that the variable limit is the upper one: √ Zx f (x) = − √ t cos t dt + 0 Zx3 0 (c) Denote the upper limit of the first integral by u = √ d dx Zx √ √ t cos t dt . d t cos t dt = dx 0 Zu √ √ x. Then t cos t dt 0 = d du Zu √ du t cos t dt · dx 0 √ du = u cos u · dx q √ √ 1 = x cos x · √ 2 x √ cos x . = 1 2x 4 (d) Denote the upper limit of the second integral by v = x3 . Then d dx Zx3 √ t cos t dt = d dx 0 Zv √ t cos t dt 0 = d dv Zv √ dv t cos t dt · dx 0 √ dv v cos v · dx √ 3 = x cos x3 · 3x2 7 = 3x 2 cos x3 = (e) Hence √ cos x d 7 2 cos x3 . f (x) = − + 3x 1 dx 2x 4 Information for Students in MATH 141 2010 01 5099 E.2.2 Draft Solutions to Quiz Q2 Distribution Date: Mounted on the Web on Monday, March 03rd, 2008 Caveat lector! There could be misprints or errors in these draft solutions. There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version 1. [10 MARKS] Showing all your work, find the volume of the√solid obtained by rotating about the line y = 1 the region bounded by the curves y = n x and y = x, where n is a given positive integer. Solution: A favoured method of solution was not prescribed. Using the method of “washers”: (a) The solution I am giving is for the case where n is even. (b) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (c) It’s not clear from the wording of the problem whether it was intended, in the case of odd n, to permit the second intersection point (x, y) = (−1, −1); the decision was left to the individual TA’s. The remainder of this solution covers the case of even n; for odd n this solution does not consider the solid generated by rotating the region with vertices (x, y) = (−1, −1), (0, 0). (d) Find the inner and outer dimensions of the washer. Since the axis of revolution is a horizontal line, the element of area being rotated is vertical. For√arbitrary x the lower point on the element is (x, x); the upper point is (x, n x). √ The distances of these points from the axis are, respectively 1 − x and 1 − n x. Information for Students in MATH 141 2010 01 5100 (e) The volume of the “washer” is, therefore, √ π −(1 − x)2 + (1 − n x)2 ∆x . (f) Correctly evaluate the integral: Z 1 √ −(1 − x)2 + (1 − n x)2 dx π 0 Z 1 1 2 = π −2x + x2 + 2x n − x n dx 0 " #1 n 2n n+1 x2 n+2 2 = π −x + + xn − xn 3 n+1 n+2 0 ! 1 2n n (n − 1)(n + 4)π = π −1 + + − = 3 n+1 n+2 3(n + 1)(n + 2) Using the method of cylindrical shells: (a) The solution I am giving is for the case where n is even. (b) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (c) It’s not clear from the wording of the problem whether it was intended, in the case of odd n, to permit the second intersection point (x, y) = (−1, −1); the decision was left to the individual TA’s. The remainder of this solution covers the case of even n; for odd n this solution does not consider the solid generated by rotating the region with vertices (x, y) = (−1, −1), (0, 0). (d) Find the inner and outer dimensions of the washer. Since the axis of revolution is a horizontal line, the element of area being rotated is also horizontal. For arbitrary y the left endpoint on the element is (yn , y); the right endpoint is (y, y). The length of the element is, therefore, y−yn ; the distances of the element from the axis of symmetry is 1 − y. (e) The volume of the cylindrical shell element of volume is, therefore, 2π(1 − y) · (y − yn ) · ∆y . (f) Correctly evaluate the integral: Z 1 2π (1 − y)(y − yn ) dy 0 Z 1 = 2π −yn + yn+1 + y − y2 dy 0 Information for Students in MATH 141 2010 01 5101 #1 1 n+1 1 n+2 1 2 1 3 = 2π − y + y + y − y n+1 n+2 2 3 0 ! 1 1 1 1 = 2π − + + − −0 n+1 n+2 2 3 ! 1 (n − 1)(n + 4)π 1 = 2π − = 6 (n + 1)(n + 2) 3(n + 1)(n + 2) " Z 2. [5 MARKS] Showing all your work, evaluate the integral (a − t)(b + t2 ) dt. Solution: (a) Expand the product in the integrand: Z Z 2 (a − t)(b + t ) dt = ab − bt + at2 − t3 dt . (b) Integrate term by term: Z b a 1 ab − bt + at2 − t3 dt = ab · t − · t2 + · t3 − · t4 + C . 2 3 4 3. [10 MARKS] Showing all your work, determine a number b such that the line x = b divides into two regions of equal area the region bounded by the curves x = ay2 and x = k. Solution: The solution is analogous (under the exchange x ↔ y) to that given for Problem 1 of the Tuesday quiz. 4. [5 Showing all your work, use a substitution to evaluate the indefinite integral Z MARKS] ex dx, where a is a non-zero real number. ex + a Solution: (a) Try the substitution u = e x + a, so du = e x dx. (b) Z du ex dx = = ln |u| + C = ln |e x + a| + C . x e +a u (If the constant a is positive, then the absolute signs are not required.) Z Information for Students in MATH 141 2010 01 5102 Tuesday version 1. [10 MARKS] Showing all your work, find a number b such that the line y = b divides the region bounded by the curves y = ax2 and y = k into two regions with equal area, where a, k are given positive constants. Solution: (a) Determine the range of values for integration by finding the of the r intersections k bounding curves: solving the equations yields the points ∓ , k . a (b) Determine the portion of the full area which is below the line y = b. We begin by repeating r the calculation of the preceding part: the corner points have coordinates ∓ b , b . The area is a Z √ ba " ax3 2 (b − ax ) dx = 2 bx − √ 3 − ba # √ ba 0 4 = b 3 r b . a (c) As a special case of the foregoing, or r by a separate calculation, we can conclude 4 k that the area of the entire region is k . 3 a (d) The condition of the problem is that r r 4 b 1 4 k b = · k 3 a 2 3 a which is equivalent to 4b3 = k3 , and implies that the line should be placed where 2 b = 2− 3 k. 2. [10 MARKS] The region bounded by the curves y = 5 and y = x2 − ax + b is rotated about the axis x = −1. Showing all your work, find the volume of the resulting solid. Solution: Because there are constraints on the constants, I will work just one variant, with a = 3, b = 7. Using the method of cylindrical shells: (a) To find the extremes of integration, we solve the equations y = 5 and y = x2 − 3x + 7, obtaining (x, y) = (1, 5), (2, 5). (b) The height of a vertical element of area which generates a cylindrical shell is, at horizontal position x, 5 − (x2 − 3x + 7) = −x2 + 3x − 2. (c) The distance of that vertical element of area from the axis of revolution is 1+ x. Information for Students in MATH 141 2010 01 5103 (d) The volume is given by the integral Z 2 2π (1 + x)(−x2 + 3x − 2) dx 1 (e) Evaluating the integral: Z 2 2π (1 + x)(−x2 + 3x − 2) dx 1 = = = = Z 2 2π −x3 + 2x2 + x − 2 dx 1 " #2 1 4 2 3 1 2 2π − x + x + x − 2x 4 3 2 1 ! 16 1 2 1 2π −4 + +2−4+ − − +2 3 4 3 2 5π 6 Using the method of “washers” (a) To find the lowest point on the parabola, we solve x2 − 3x + 7 ≥ 0. This can be done by completing the square, ! or by using the 3 19 calculus to find the local minimum. We find it to be , . 2 4 (b) The horizontal element generating the “washer” at height y extends between the solutions in x to the equation y = x2 − 3x + 7; these are p 3 ± 4y − 19 x= . 2 (c) The volume of the “washer” at height y is, therefore, 2 2 p p 4y − 19 4y − 19 3 + 3 − − 1 + ∆y π 1 + 2 2 p = 5π 4y − 19 ∆y Z 5p (d) The volume is given by the integral 5π 4y − 19 dy. 19 4 (e) Evaluation of the integral: " #5 Z 5p 2 1 5π 3 5π 4y − 19 dy = 5π · · (4y − 19) 2 = . 19 3 4 6 19 4 4 Information for Students in MATH 141 2010 01 5104 3. [5 Z MARKS] Showing all your work, use a substitution to evaluate the indefinite integral cosn x sin x dx, where n is a fixed, positive integer. Solution: (a) Use new variable u, where du = − sin x dx; one solution is u = cos x. (b) Z Z n cos x sin x dx = − un du un+1 +C n+1 1 = − cosn+1 x + C n+1 Z 4. [5 MARKS] Showing all your work, evaluate the integral xb + a + = − a and b are given positive integers). ! 1 dx, (where x2 + 1 Solution: Z xb+1 (a) xb dx = + C1 , b+1 Z (b) a dx = ax + C2 Z 1 (c) dx = arctan x + C3 x2 + 1 ! Z 1 xb+1 b (d) x +a+ 2 dx = + ax + arctan x + C. x +1 b+1 Wednesday version 1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotating about the line x = −1 the region bounded by y = xn and x = yn , where n is a given positive integer. Solution: Case I: n is even Using the method of “washers”: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). Information for Students in MATH 141 2010 01 5105 (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a vertical line, the element of area being rotated is horizontal. For 1 arbitrary y the farther endpoint on the element is (y n , y); the nearer endpoint is (yn , y). The distances of these points from the axis are, respectively √ 1 + n y and 1 + yn . (c) The volume of the “washer” is, therefore, √ π −(1 + yn )2 + (1 + n y)2 ∆y . (d) Correctly evaluate the integral: Z 1 √ π −(1 + y)2 + (1 + n y)2 dy 0 Z 1 1 2 = π 2y n + y n − 2yn − y2n dy 0 " #1 n 2 2n n+1 1 n+2 n+1 2n+1 = π yn + yn − y − y n+1 n+2 n+1 2n + 1 0 2 2(n − 1)(3n + 7n + 3)π = (n + 1)(n + 2)(2n + 1) Using the method of cylindrical shells: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1). (b) Since the axis of revolution is a vertical line, the element of area being rotated is also vertical. For arbitrary x the top endpoint on the element is 1 (x, x n ); the lower endpoint is (x, xn ). The length of the element is, there1 fore, x n − xn ; the distance of the element from the axis of symmetry is 1 + x. (c) The volume of the cylindrical shell element of volume is, therefore, 1 2π(1 + x) · x n − xn . (d) Correctly evaluate the integral: Z 1 1 2π (1 + x)(x n − xn ) dx 0 Z 1 n+1 1 = 2π x n − xn + x n − xn+1 dx 0 n 1 n 1 = 2π − + − n + 1 n + 1 2n + 1 n + 2 2(n − 1)(3n2 + 7n + 3)π = (n + 1)(n + 2)(2n + 1) ! Information for Students in MATH 141 2010 01 5106 Case II: n is odd Using the method of “washers”: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1, ±1). Here there is an issue of interpretation. The textbook usually permits the word region to apply to one that may have more than one component; some authors would not wish to apply the term in such a situation. I will follow the textbook, and permit a region here to have two components. (b) Find the inner and outer dimensions of the washer. Since the axis of revolution is a vertical line, the element of area being rotated is horizontal. But there are two kinds of elements, depending on whether y is positive or negative. For arbitrary, positive y the farther endpoint on the element is 1 (y n , y); the nearer endpoint is (yn , y). The distances of these points from √ the axis are, respectively 1 + n y and 1 + yn . For arbitrary, negative y the 1 nearer endpoint on the element is (y n , y); the farther endpoint is (yn , y). √ The distances of these points from the axis are, respectively 1 + n y and 1 + yn (both of which are less than 1). (c) The volume of the “washer” is, therefore, √ π −(1 + yn )2 + (1 + n y)2 ∆y . (d) Correctly evaluate the integral: Z 1 −(1 + y)2 + (1 + √n y)2 dy π −1 Z 1 1 2 2y n + y n − 2yn − y2n dy = π 0 Z 0 1 2 −2y n − y n + 2yn + y2n dy +π −1 " #1 2n n+1 n n+2 2 n+1 1 2n+1 n n = π y + y − y − y n+1 n+2 n+1 2n + 1 0 " #0 2n n+1 n 2 1 n+2 n+1 2n+1 +π − yn − yn + y + y n+1 n+2 n+1 2n + 1 −1 2(n − 1)(3n2 + 7n + 3)π 2(n − 1)(n2 + 3n + 1)π + = (n + 1)(n + 2)(2n + 1) (n + 1)(n + 2)(2n + 1) 4(n − 1)π = . n+1 Using the method of cylindrical shells: (a) Find the intersections of the curves bounding the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1, ±1). Information for Students in MATH 141 2010 01 5107 (b) Since the axis of revolution is a vertical line, the element of area being rotated is also vertical. For arbitrary, positive x the top endpoint on the el1 ement is (x, x n ); the lower endpoint is (x, xn ); for arbitrary, negative x the 1 bottom endpoint on the element is (x, x n ); the upper endpoint is (x, xn ). 1 The length of the element is, therefore, x n − xn ; the distance of the element from the axis of rotation is 1 + x. (c) The volume of the cylindrical shell element of volume is, therefore, 1 2π(1 + x) · x n − xn . (d) Correctly evaluate the integral: Z 1 1 2π (1 + x) x n − xn dx −1 Z 1 1 n+1 n n+1 n n = 2π x − x + x − x dx −1 ! n 1 n 1 = 2π − + − n + 1 n + 1 2n + 1 n + 2 ! n 1 n 1 +2π − − + n + 1 n + 1 2n + 1 n + 2 4(n − 1)π = n+1 Z sin 2t 2. [5 MARKS] Showing all your work, evaluate the indefinite integral dt. cos t Solution: (a) Apply a “double angle” formula: Z Z Z sin 2t 2 sin t cos t dt = dt = 2 sin t dt . cos t cos t (b) Complete the integration: Z 2 sin t dt = −2 cos t + C . 3. [10 MARKS] Showing all your work, find the area of the region bounded by the parabola y = x2 , the tangent line to this parabola at (a, a2 ), and the x-axis, (where a is a given real number). Solution: This area can be computed by integrating either with respect to y or with respect to x. Information for Students in MATH 141 2010 01 5108 Integrating with respect to y: (a) Since y0 = 2x, the tangent line through (a, a2 ) has equation y − a2 = 2a(x − a) ⇔ y = 2ax − a2 . (b) To integrate with respect to y we need to express the equations of the parabola and the line in the form x = function of y . √ The branch of the parabola to the right of the y-axis is x = y. The line has y a equation x = + . 2a 2 ! y + a2 √ (c) The area of the horizontal element of area at height y is − y ∆y. 2a (d) The area is the value of the integral ! Z a2 y + a2 √ − y dy . 2a 0 (e) Integration yields " 2 #a2 ! y 1 1 2 3 ay 2 32 1 3 + − y = + − a = a . 4a 2 3 0 4 2 3 12 Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2 . Its a intercept with the x-axis is at x = . 2 a (b) The area of the vertical element of area at horizontal position x ≤ is x2 − 0 ∆x. 2 (c) The area of the vertical element of area at horizontal position x ≥ 2a is x2 − (2ax − a2 ) dx = (x − a)2 ∆x. (d) The area of the region is the sum Z a2 Z a 2 x dx + (x − a)2 dx . a 2 0 (e) Integration yields " x3 3 # a2 0 " (x − a)3 + − 3 #a = a 2 a3 . 12 4. [5 MARKS] Showing all your work, evaluate the indefinite integral where a is a constant, positive integer. Solution: Z seca x tan x dx, Information for Students in MATH 141 2010 01 5109 (a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x. (b) Z Z a sec x tan x dx = ua−1 du = ua seca x +C = +C. a a Some students may have integrated by sight. Thursday version 1. [10 MARKS] Showing all your work, use the method of cylindrical shells to find the volume by the curves √ generated by rotating about the axis x = b the region bounded √ y = x − 1, y = 0, x = a, where a, b are fixed real constants. b ≥ a − 1 . Solution: (a) Solve equations to determine the limits of integration. √ Solving x = a with y = √ x − 1 yields the single point of intersection a, a − 1 . (b) The horizontal element of area at height y which generates the cylindrical shell has left endpoint (1 + y2 , y) and right endpoint (a, y), so its length is a − (1 + y2 ). (c) The distance of the horizontal element of area which generates the shell from the axis of rotation is b − y. (d) Set up the integral for the volume by cylindrical shells: √ a−1 Z (b − y) a − (1 + y2 ) dy . 0 (e) Evaluate the integral √ a−1 Z Z 0 = 0 √ a−1 (b − y) a − (1 + y2 ) dy b(a − 1) − (a − 1)y − by2 + y3 dy # √a−1 a−1 2 b 3 1 4 y − y + y = b(a − 1)y − 2 3 4 0 b a − 1 1 3 3 (a − 1) − (a − 1) 2 + (a − 1)2 = b(a − 1) 2 − 2 4 !3 √ 1 2 3 = (a − 1) 2 ·b− a−1 . 3 4 " Information for Students in MATH 141 2010 01 2. [5 MARKS] Showing all your work, use a substitution to evaluate the integral where a, b are non-zero constants. 5110 R b (1+ax)3 dx , Solution: (a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, so dx = 1a du. (b) Z b b dx = 3 (1 + ax) a Z du b b = − u−2 + C = − +C. 3 u 2a 2a(1 + ax)2 3. [5 Showing all your work, use a substitution to evaluate the indefinite integral Z MARKS] t2 cos a − t3 dt, (where a is a given real number). Solution: (a) Try the substitution u = t3 . (b) du = 3t2 dt ⇒ t2 dt = 13 du. (c) Z Z 1 3 t cos a − t dt = cos(a − u) du 3 1 = − sin(a − u) + C 3 1 = − sin(a − t3 ) + C. 3 2 (Some students may wish to employ a second substitution v = a − u. Alternatively, a better substitution for the problem would have been to take u = a − t3 .) 4. [10 MARKS] Showing all your work, determine the area of the region bounded by the parabola x = y2 , the tangent line to this parabola at (a2 , a), and the y-axis, where a is a fixed, positive real number. Solution: The solution is analogous (under the exchange x ↔ y) to that given for Problem 3 of the Wednesday quiz. E.2.3 Draft Solutions to Quiz Q3 Release Date: Mounted on the Web on 05 April, 2008 These draft solutions could contain errors, and they must be subject to correction. Caveat lector! Information for Students in MATH 141 2010 01 5111 There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions • Time = 30 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version Z 1. [5 MARKS] Showing all your work, evaluate the integral x5 ln(20x) dx. Solution: (a) I will integrate by parts, setting u = ln(20x), dv = x5 dx. Then du = dx x6 ,v= . x 6 (b) ! Z x6 1 x ln(20x) dx = (ln(20x)) · − x5 dx 6 6 x6 ln(20x) x6 = − +C. 6 36 Z 1 − cot2 x 2. [5 MARKS] Showing all your work, evaluate the integral dx. csc2 x Solution: Z Z 1 − cot2 x sin2 x − cos2 x dx = dx 2 2x csc2 x sin x · csc Z = sin2 x − cos2 x dx Z = (− cos 2x) dx Z 5 1 = − sin 2x + C . 2 Information for Students in MATH 141 2010 01 5112 3. [10 MARKS] Use a substitution to transform this integral into the integral of a rational function; then integrate, and express your answer in terms of x: Z 1 dx e−3x − e−x Solution: (a) I would try the substitution u = e−x , so du = −e−x dx = −u dx. (b) Then Z 1 dx = −3x e − e−x = = = = Z du du − u2 ) u2 (1 Z 1 1 1 2 2 + + 2 du u 1+u 1−u 1 1 1 − + ln |1 + u| − ln |1 − u| + C u 2 2 1 1 1 − u +C − − ln u 2 1 + u r 1 − e−x x +C −e − ln 1 + e−x There are other, equivalent ways in which this last class of antiderivatives can be expressed. (For example, if we define K = eC , so C = ln K, we can bring the logarithm terms together.) Z x2 4. [10 MARKS] Showing all your work, evaluate the integral 3 dx. 4 − x2 2 Solution: (a) Use a trigonometric substitution, e.g., x = 2 cos θ, i.e., θ = arccos 2x . Then dx = −2 sin θ dθ. (b) Z x2 4 − x2 Z 32 dx = − = − = − Z cos2 θ sin θ dθ sin3 θ cot2 θ dθ Z csc2 θ − 1 dθ Information for Students in MATH 141 2010 01 5113 = cot θ + θ + C cos θ = +θ+C sin θ cos θ x = √ + arccos + C 2 1 − cos2 θ x x = q 2 + arccos + C 2 2 1 − x4 x x + arccos + C . = √ 2 4 − x2 Tuesday version Z 1. [10 MARKS] Showing all your work, evaluate the integral e5x cos(2x) dx . Solution: (a) Use integration by parts. In this case the factors e5x and cos 2x are rendered neither “more complicated” nor “simpler” under either integration or differentiation. Two applications of integration by parts, with appropriate choices of functions, will yield an equation that can be solved for the value of the indefinite integral. For the first application take, for example, u = e5x and dv = cos 2x. Then du = 5e5x dx, and v = 21 sin 2x. (b) Z Z 1 5 e cos(2x) dx = e · sin 2x − e5x sin 2x dx . (116) 2 2 R (c) Now we apply integration by parts to evaluate e5x sin 2x dx, taking U = e5x and dV = sin 2x. Then dU = 5e5x dx, and V = − 21 cos 2x. 5x (d) Z 5x ! Z 1 5 e sin(2x) dx = e · − cos 2x + e5x cos 2x dx . 2 2 5x 5x (117) (e) Combining equations (116), (117) yields ! Z Z 5 5 5x 5x 1 5x 1 5x e cos(2x) dx = e · sin 2x − −e · · cos 2x + e sin 2x dx 2 2 2 2 ! Z 1 5 25 5x = e · · sin 2x + cos 2x − e5x cos(2x) dx . 2 4 4 Information for Students in MATH 141 2010 01 5114 (f) Solving the last equation yields Z 1 e5x cos(2x) dx = e5x · (2 sin 2x + 5 cos 2x) + C . 29 2. [10 MARKS] Showing all your work, use a substitution to change this integral into the integral of a rational function; then integrate and express your solution in terms of x: Z 1 + 7e x dx . 1 − ex Solution: (a) Let u = e x , so du = e x dx. Z Z 1 + 7e x 1 + 7u du (b) Then dx = · and we proceed to expand the integrand into x 1−e 1−u u a sum of partial fractions. 1 + 7u A B (c) Assuming that = + , and multiplying both sides by u(1 − u), we u(1 − u) u 1−u obtain the identity in u 1 + 7u = A(1 − u) + Bu. Setting u = 0 and u = 1 yields the equations 1 = A and 8 = B. We can now continue integration. (d) Z ! Z 1 + 7u 1 8 du = + du 1−u u 1−u = ln |u| − 8 ln |1 − u| + C = ln e x − 8 ln |1 − e x | + C = x − 8 ln |1 − e x | + C which solution can be checked by differentiation. Z 3. [10 MARKS] Showing all your work, evaluate the integral dx dx. x2 + 8x + 17 2 Solution: (a) Completing the square of the polynomial in the denominator, we obtain !2 !2 8 8 2 2 + 17 − x + 8x + 17 = x + = (x + 4) + 1 . 2 2 Accordingly, we can simplify the integral by taking u = x + 4, du = dx. Information for Students in MATH 141 2010 01 5115 (b) The preceding substitution is not sufficient, however. All we obtain is Z Z dx du dx = . 2 x2 + 8x + 17 u2 + 1 2 We can simplify this further by taking u = tan θ, i.e., by taking θ = arctan u, so du dθ = . 1 + u2 (c) Z du 1 + u2 Z 2 = Z sec2 θ dθ sec4 θ cos2 θ dθ = Z 1 + cos 2θ θ sin 2θ dθ = + +C 2 2 4 u 1 arctan u + +C = 2 2(1 + u2 ) ! 1 x+4 = arctan(x + 4) + 2 +C 2 x + 8x + 17 = which can be verified by differentiation. You should always verify this type of integration by differentiation, in order to locate silly algebra mistakes (or worse). Wednesday version Z 1. [5 MARKS] Showing all your work, evaluate the indefinite integral x cos(18x) dx. Solution: (a) This can be solved using integration by parts. Define u = x, dv = cos(18x) dx, so sin(18x) du = dx and v = . 18 (b) Z Z sin(18x) sin(18x) x cos(18x) dx = x · − 18 18 x sin(18x) cos(18x) = + +C, 18 182 which can be verified by differentiation. Information for Students in MATH 141 2010 01 5116 Z 1 − sin x dx. cos x 2. [5 MARKS] Showing all your work, evaluate the integral Solution: Z Z 1 − sin x (sec x − tan x) dx dx = cos x = ln | sec x + tan x| + ln | cos x| + C = ln |(sec x + tan x) · (cos x)| + C = ln |1 + sin x| + C = ln(1 + sin x) + C . Note that the absolute signs are not needed, since 1 + sin x cannot be negative. π Z3 3. [10 MARKS] Showing all your work, evaluate the integral π 6 ln tan x dx. (sin x) · (cos x) Solution: (a) In view of the complicated nature of the integrand, one would be advised to seek a substitution that could render it more amenable. But the integrand involves both sines and cosines. However, note that tan x tan x = . sec2 x tan2 x + 1 √ π π Taking u = tan x, we have du = sec2 x dx. When x = , u = 3; when x = , 3 6 1 u= √ . 3 (sin x)(cos x) = (tan x)(cos2 x) = (b) π Z3 π 6 ln tan x dx (sin x) · (cos x) √ Z3 = √1 3 ln u u u2 +1 du u2 + 1 √ Z3 = √1 3 ln u du u Information for Students in MATH 141 2010 01 5117 (c) Now the integral looks as though it could be simplified by a substitution v = ln u, √ du . When u = 3, v = ln23 ; when u = √13 , v = − ln23 . so dv = u (d) √ Z3 ln u du = u √1 3 Z − ln23 " = ln 2 2 v2 2 v dv # ln23 = 0. − ln23 Z 4. [10 MARKS] Showing all your work, evaluate the integral √ et 49 − e2t dt. Solution: (a) Clearly a substitution of the form u = et is indicated, in order to simplify the integrand. We find that du = et dt. Z √ Z √ t (b) We obtain e 49 − e2t dt = 49 − u2 du. Now a trigonometric substitution is indicated. Take u = 7 sin θ — more precisely, θ = arcsin u7 (the inverse cosine could also have been used), so du = 7 cos θ dθ: Z √ Z 2 49 − u du = 49 cos2 θ dθ. (c) Z 49 2 cos θ dθ = = = = Z 49 (1 + cos 2θ) dθ 2 49 (θ + sin θ · cos θ) + C 2 u u√ 49 arcsin + 49 − u2 + C 2 7 2 √ 49 et et 49 − e2t arcsin + +C. 2 7 2 Thursday version Z 1. [5 MARKS] Showing all your work, evaluate the integral sin3 9x dx. Information for Students in MATH 141 2010 01 5118 Solution: The integrand is an odd power of the sine function. I will substitute u = cos 9x, so du = −9 sin 9x dx. Z Z 1! 3 2 sin 9x dx = du 1−u 9 u u3 = − + +C 9 27 cos x cos3 x = − + +C. 9 27 π Z2 cot2 x dx. Your answer 2. [5 MARKS] Showing all your work, evaluate the integral π 4 should be simplified as much as possible; the instructors are aware that you do not have the use of a calculator. Solution: (a) Recall that cot2 x = csc2 x − 1, and that d cot x = − csc2 x. dx (b) π π Z2 cot2 x dx = π 4 Z2 csc2 x − 1 dx π 4 π = [− cot x − x] π4 . 2 π π π (c) = − + 1 + = 1 − . 2 4 4 Z √ 9 − x2 3. [10 MARKS] Showing all your work, evaluate the integral dx. x Solution: (a) To simplify the square root, substitute x = 3 cos θ, i.e., θ = arccos 3x . Then dx = −3 sin θ dθ. (b) Z Z √ sin2 θ 9 − x2 dx = −3 dθ x cos θ Information for Students in MATH 141 2010 01 5119 Z 1 − cos2 θ dθ cos θ Z = −3 (sec θ − cos θ) dθ = −3 = −3 ln | sec θ + tan θ| + 3 sin θ + C (c) q r 1 + 1 − x2 9 + 3 1 − x2 + C −3 ln | sec θ + tan θ| + 3 sin θ + C = −3 ln x 9 3 √ 3 + 9 − x2 √ + 9 − x2 + C = −3 ln x 4. [10 MARKS] Showing all your work, determine whether the following integral is convergent or divergent. Evaluate it if it is convergent; in such a case you are expected to simplify your answer as much as is consistent with not having the use of a calculator: ! Z 1 ln 7x 6 √ dx x 0 Solution: Z ln 7x √ dx. The integrand x can be expressed as a product, in which one of the factors “simplifies” upon differentiation, while the other does not become significantly “more difficult” upon 1 integration. So we will integrate√by parts, taking u = ln 7x, and dv = x− 2 dx. Then du = 1x , and we may take v = 2 x. (a) Let’s look first at the associated indefinite integral, (b) Z Z √ ln 7x dx √ dx = (ln 7x)(2 x) − 2 √ x x √ √ = (ln 7x)(2 x) − 4 x + C . (c) The integrand is not defined at x = 0 in the given interval of integration. By the definition of an improper integral, we have Z 1 Z 1 ln 7x ln 7x √ dx = lim+ √ dx a→0 x x 0 a Information for Students in MATH 141 2010 01 5120 h √ √ i1 lim+ (ln 7x)(2 x) − 4 x a a→0 h i √ = lim+ (2 ln 7 − 4) − a · (2 ln(7a) − 4) a→0 √ = 2 ln 7 − 4 − lim+ (ln 7a)(2 a) = a→0 (d) The limit can be expressed as that of a ratio where numerator and denominator both become infinite. Thus l’Hospital’s Rule may be used: √ ln 7a lim+ (ln 7a)(2 a) = 2 lim+ 1 a→0 a→0 a− 2 = 2 lim+ 1 a 3 − 21 a− 2 √ = 2 lim+ (−2 a) = 0. a→0 a→0 Thus the original integral is convergent, and its value is 6(2 ln 7 − 4). Problems not used 1. [10 MARKS] Make a substitution to express the integrand as a rational function, and then evaluate the integral. Z 4 √ x dx . 1 x − 16 Solution: (a) Start by substituting u = x = 4, u = 2. √ x, so x = u2 , dx = 2u du. When x = 1, u = 1; when (b) Z 4 1 √ x dx = x − 16 Z 2 Z 1 Z 1 2 = 2 = = h 1 2u2 du u2 − 36 ! 72 du 2+ 2 u − 36 ! 6 6 − du 2+ u−6 u+6 i2 u2 + 6 ln |u − 6| − 6 ln |u + 6| 7 . = 3 + 6 ln 10 1 Information for Students in MATH 141 2010 01 5121 E.2.4 Draft Solutions to Quiz Q4 Release Date: Mounted on the Web on Wednesday, April 9th, 2008 These are draft solutions that were prepared when the quizzes were being designed. It was intended that Teaching Assistants would consult these draft solutions when they graded their students’ quizzes, and would report any errors or omissions. As the Teaching Assistants may believe that they are inhibited from communicating with the instructor who manages this course, it is not clear that the solutions have been thoroughly checked. The solutions are being released with the cautionary warning, Caveat lector! — Let the reader beware! Use them at your own risk. There were four different types of quizzes, for the days when the tutorials are scheduled. Each type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that included the instructions. • Time = 45 minutes • No calculators! • Show all your work: marks are not given for answers alone. • Enclose this question sheet in your folded answer sheet. In the following I will either provide a generic solution for all varieties, or a solution to one typical variety. Monday version x5 1 1. [10 MARKS] Showing all your work, find the length of the curve y = + 3 (1 ≤ 30 2x x ≤ 2). Simplify your answer as much as possible; the instructors are aware that you do not have the use of a calculator. Solution: dy x4 3 = − 4 dx 6 2x s r !2 dy x8 9 1 ⇒ 1+ 1+ + 8− = dx 36 4x 2 s r !2 8 9 1 3 x x4 + + = + = 36 4x8 2 6 2x4 4 x 3 = + 4 . 6 2x Information for Students in MATH 141 2010 01 5122 The absolute signs may be dropped, since the given square root is a sum of positive multiples of even powers, and must be non-negative. The length is Z 2 1 ! " 5 #2 x4 3 x 1 + − dx = 6 2x4 30 2x3 1 353 = . 240 2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating 3 1 2 x = y + 2 2 (7 ≤ y ≤ 10) about the x-axis. Simplify your answer as much as 2 possible; the instructors are aware that you do not have the use of a calculator. Solution: Thus 1 p dx 1 3 2 = · · y + 2 2 · 2y = y y2 + 2 . dy 3 2 s dx 1+ dy The area of the surface is Z 10 7 !2 = p 1 + y2 (2 + y2 ) = y2 + 2 . " y2 y4 1 + y · (2πy) dy = 2π + 2 4 7701π = . 2 2 #10 7 3. [10 MARKS] Showing all your work, find the area enclosed by the curve (in polar coordinates) r = 7 + 2 sin 6θ. Solution: The curve surrounds the pole, and is periodic with period 2π. The area may be expressed as an integral over an interval of length 2π; for example, as Z Z 1 2π 1 2π 2 (7 + 2 sin 6θ) dθ = (49 + 28 sin 6θ + 4 sin2 6θ) dθ 2 0 2 0 Z 1 2π = (49 + 28 sin 6θ + 2(1 − cos 12θ)) dθ 2 0 #2π " 2 1 28 cos 6θ + 2θ − sin 12θ = 49θ − 2 6 12 0 1 = (51)(2π) = 51π . 2 Information for Students in MATH 141 2010 01 5123 4. [10 MARKS] Showing all your work, find the exact length of the curve x = 6 + 3t2 , y = 2 + 2t3 (0 ≤ t ≤ 1). Solution: dx = 6t dt dy = 6t2 dt dy dy = dt = t dx dx dt s !2 √ dy = ⇒ 1+ 1 + t2 dx s !2 !2 Z 1 dx dy + Arc Length = dt dt 0 Z 1√ = 36t2 + 36t4 dt 0 Z 1 √ = 6 t 1 + t2 dt 0 3 1 3 1 1 2 2 2 2 2 1+t = 6· · =2 1+t 0 23 3 √ 0 = 2 22 − 1 = 4 2 − 2 . 5. [10 MARKS] Showing all your work, sum a series in order to express the following number as a ratio of integers: 0.35 = 0.35353535 . . .. Solution: 35 1 1 1 0.35 = 1+ + + + ... 100 100 1002 1003 = 35 100 1− 1 100 = ! 35 . 99 Tuesday version 1√ 1. [10 MARKS] Showing all your work, find the length of the curve x = y · (y − 3) 3 (49 ≤ y ≤ 64). Simplify your answer as much as possible; the instructors are aware that you do not have the use of a calculator. Information for Students in MATH 141 2010 01 5124 Solution: dx 1 3√ 3 1 1 1 √ 1 = · y − · y− 2 = y− √ dy 3 2 3 2 2 y s s !2 ! dx 1 1 ⇒ 1+ = 1+ y+ −2 dy 4 y s ! 1 1 = y+ +2 4 y s !2 1 √ 1 y+ √ = 4 y 1 √ 1 = y + √ . 2 y ! The absolute signs may be dropped, since the square root is non-negative. The length is ! " #64 " #64 Z 1 32 √ 1 64 √ 1 1 2 32 √ y + √ dy = ·y +2 y = ·y + y 2 49 y 2 3 3 49 49 172 = . 3 2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating the curve x = 6 + 2y2 (0 ≤ y ≤ 3) about the x-axis. Solution: dx = 4y dy s !2 p dx ⇒ 1+ = 1 + 16y2 dy Z 3p ⇒ Area = 1 + 16y2 · 2πy dy 0 i π h 3 3 = (1 + 16y2 ) 2 0 24 π 3 = (145) 2 − 1 . 24 3. [15 MARKS] Showing all your work, find the area of the region that lies inside the curve r = 15 cos θ, and outside the curve r = 5 + 5 cos θ. Information for Students in MATH 141 2010 01 5125 Solution: The first curve is a circle; the second is a cardioid whose axis of symmetry is the initial ray. If we solve the equations we find that the curves intersect at θ = ± arccos 12 = ± π3 . They also intersect at the pole, which appears on the circle when θ = π2 , etc., and on the cardioid when θ = π, etc. The region whose area we seek lies between the two curves when − π3 ≤ θ ≤ π3 and r is positive. Integration shows the area to be Z π 1 3 (15 cos θ)2 − (5 + 5 cos θ)2 dθ 2 − π3 Z π 25 3 = 8 cos2 θ − 2 cos θ − 1 dθ 2 − π3 Z π 25 3 (4(1 + cos 2θ) − 2 cos θ − 1) dθ = 2 − π3 Z π 25 3 (3 + 4 cos 2θ − 2 cos θ) dθ = 2 − π3 π = 25 [3θ + 2 sin 2θ − 2 sin θ]03 √ √ = 25 π + 3 − 3 = 25π . 4. [15 MARKS] Showing all your work, find the length of the loop of the curve x = 18t − 6t3 , y = 18t2 . Solution: dx = 18 − 18t2 dt dy = 36t dt s !2 !2 q dx dy + = 182 1 − t2 2 + 362 t2 = 18 1 + t2 . dt dt We must determine where the curve crosses itself. The student was expected to show that she knew how to find this crossing point systematically, not just by guessing or examining a rough graph. The crossing point can be found by solving for distinct parameter values t1 , t2 the equations x = 18t1 − 8t13 = 18t2 − 8t23 , y = 18t12 = 18t2 . Collecting terms and factorizing yields the system of equations (t1 − t2 ) 18 − 6 t12 + t1 t2 + t22 = 0, 18 (t1 − t2 ) (t1 + t2 ) = 0 . Information for Students in MATH 141 2010 01 5126 Since we are looking for a solution where t1 , t2 , we may divide by t1 − t2 , which cannot equal 0, and obtain the system 3 − t12 + t1 t2 + t22 = 0 , t1 + t2 = 0 . From the second equation we see that t2 = −t1 , and then the first equation yields 3 = t12 , √ so the solutions are t1 = −t2 = ± 3; we may take √ √ the loop as beginning with parameter value − 3 and ending with parameter value + 3. The length of the arc will be Z + √3 2 18 1 + t dt √ − 3 Z √3 = 2 18 1 + t2 dt 0 since the integrand is even and the interval is symmetric around 0 " # √3 √ 13 = 36 t + t = 72 3 . 3 0 Wednesday version 1. [10 MARKS] Showing all your work, find the length of the curve y = ln sec x. Simplify your answer as much as possible. Solution: dy 1 = · (sec x tan x) = tan x dx sec x s !2 p dy = 1 + tan2 x = | sec x| . ⇒ 1+ dx In the following integral I will drop the absolute signs because the secant is positive over the entire interval of integration; the length is Z π4 Z π4 | sec x| dx = sec x dx − π4 − π4 π = [ln | sec x + tan x|]−4 π √ √4 = ln( 2 + 1) − ln( 2 − 1) √ √ ( 2 + 1)2 2+1 = ln = ln √ 2−1 2−1 √ √ = ln ( 2 + 1)2 = 2 ln( 2 + 1) . Information for Students in MATH 141 2010 01 5127 It would be sufficient for a student to obtain the different of logarithms above. The subsequent steps simplify the argument, and would be useful if the user did not have the use of a calculator. √ 2. [10 MARKS] The curve y = 3 x (1 ≤ y ≤ 3) is rotated about the y-axis. Showing all your work, find the area of the resulting surface. Solution: The data are given partly in terms of x and partly in terms of y, so some care is needed. Since the limits are given in terms of y, I will integrate with respect to y; it will be convenient to rewrite the equation of the curve as x = y3 . dx = 3y2 dy s !2 p dx ⇒ 1+ = 1 + 9y4 dy Z 3p 1 + 9y4 · 2πy3 dy Area = 1 " 3 #3 1 1 2 4 2 = · · 2π · 1 + 9y 9 4 3 1 π 3 3 730 2 − 10 2 . = 27 3. [10 MARKS] Showing all your work, find the area of the region enclosed by the inner loop of the curve r = 9 + 18 sin θ. Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve is traced out as θ passes through an interval of that length. If, for example, we consider the interval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π and 6 11π at θ = 6 . Between these values the smaller loop is traced out; the larger loop is traced out, for example, for − π6 ≤ θ ≤ 7π . We can find the area of the small loop by integrating 6 between the appropriate limits; the area is Z 11π 6 1 (9 + 18 sin θ)2 dθ 2 7π6 Z 11π 6 81 = (1 + 4 sin θ + 4 sin2 θ) dθ 2 7π6 Z 11π 6 81 = (1 + 4 sin θ + 2(1 − cos 2θ)) dθ 2 7π6 Z 11π 6 81 = (3 + 4 sin θ − 2 cos 2θ) dθ 2 7π6 Information for Students in MATH 141 2010 01 5128 11π 81 [3θ − 4 cos θ − sin 2θ] 7π6 2 6 ! ! 81 11π 11π 11π 81 7π 7π 7π = − 4 cos − sin − − 4 cos − sin 2 2 6 3 2 2 6 3 √ √ √ √ 81 11π − 3 81 7π 3 − + 2 3 − = −2 3− 2 2 2 2 2 2 √ √ 243 3 81 (2π − 3 3) = 81π − . = 2 2 = 4. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating the curve x = 3t − t3 , y = 3t2 (0 ≤ t ≤ 4) about the x-axis. Solution: dx = 3 − 3t2 dt dy = 6t dt s !2 !2 q dx dy 32 1 − t2 2 + 62 t2 = 3 1 + t2 . + = dt dt The area of the surface of revolution about the x-axis will be Z 4 Z 4 2 2 t2 + t4 dt 3 1 + t · 2π 3t dt = 18π 0 0 " #4 13 15 6 × 64 × 53 20352 = 18π t + t = = . 3 5 0 5 5 5. [10 MARKS] Showing all your work, determine the value of c, if it is known that c)−n = 2. ∞ X (5+ n=2 Solution: We are told that the geometric series converges; this implies that its common ratio is less than 1 in magnitude, i.e., that |5 + c| < 1, which implies that −1 < 5 + c < 1, equivalently, that −6 < c < −4. The sum of the geometric series on the left is 1 1 . Equating this to 2, we obtain c2 + 9c + 18 = 0, implying that · 1−1 1 = (5+c)(4+c) (5+c)2 5+c (c + 3)(c + 6) = 0, so c = −3, −6. Of these two values, −6 lies outside of the admissible interval, and would yield a divergent series. Thus c can be equal only to −3. Information for Students in MATH 141 2010 01 5129 Thursday version 1. [10 MARKS] Showing all your work, find the exact length of the polar curve r = 7e4θ (0 ≤ θ ≤ 2π). Solution: dr = 28e4θ dθ s !2 q dr 2 r + = 72 + 282 e8θ dθ √ = 7 17e4θ . The length of the arc is then √ Z 7 17 2π 0 √ 2π 17 7 e4θ dθ = e4θ 4 0 √ 7 17 8π e −1 . = 4 2. [10 MARKS] The curve y = 4 − x2 (1 ≤ x ≤ 3) is rotated about the y-axis. Showing all your work, find the area of the resulting surface. Solution: s !2 √ dy dy = 1 + 4x2 = −2x ⇒ 1 + dx dx Z 3√ Area = 1 + 4x2 · 2πx dx 1 3 3 1 1 2 2 2 = 2π · · · · 1 + 4x 4 2 3 1 π 32 3 37 − 5 2 . = 6 3. [10 MARKS] Showing all your work, find the area of the region enclosed by the outer loop of the curve r = 9 + 18 sin θ: this region will include the entire inner loop. Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve is traced out as θ passes through an interval of that length. If, for example, we consider the and interval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π 6 11π at θ = 6 . Between these values the smaller loop is traced out; the larger loop is traced Information for Students in MATH 141 2010 01 5130 . We can find the area of the large loop by integrating out, for example, for − π6 ≤ θ ≤ 7π 6 between those limits, and the area will include the entire smaller loop. The area is 1 2 Z 81 = 2 81 = 2 = = = = = 11π 6 7π 6 Z Z (9 + 18 sin θ)2 dθ 7π 6 −π 6 7π 6 −π 6 7π 6 (1 + 4 sin θ + 4 sin2 θ) dθ (1 + 4 sin θ + 2(1 − cos 2θ)) dθ Z 81 (3 + 4 sin θ − 2 cos 2θ) dθ 2 −π6 7π 81 [3θ − 4 cos θ − sin 2θ]−6 π 6 2 ! 7π 7π 81 π π π 81 7π − 4 cos − sin − − − 4 cos + sin 2 2 6 3 2 2 6 3 √ √ √ √ 81 7π 3 81 π 3 + 2 3 − − − − 2 3 + 2 2 2 2 2 2 √ 243 3 . 162π + 2 4. [10 MARKS] Showing all your work, sum a series in order to express the following number as a ratio of integers: 4.645 = 4.645454545 . . .. Solution: ! 1 45 1 1 1 4.645 = 4.6 + 1+ + + + ... 10 100 100 1002 1003 = 4.6 + 45 1000 1 − 100 1 1 45 511 46 + · = . = 10 10 99 110 5. [10 MARKS] Showing all your work, find an equation for the tangent to the curve x = cos θ + sin 7θ, y = sin θ + cos 2θ (−∞ < θ < +∞) at the point corresponding to θ = 0. Solution: The slope of the tangent is dx = − sin θ + 7 cos 7θ dθ Information for Students in MATH 141 2010 01 5131 dy = cos θ − 2 sin θ dθ dy dy 1 cos θ − 2 sin θ = dθ = = dx dx − sin θ + 7 cos 7θ 7 dθ The line through (x(0), y(0)) = (1, 1) with slope x − 7y = −6. 1 7 when θ = 0 . has equation y − 1 = 1 (x − 1), i.e., 7 Problems prepared but not used p 1. The curve x = c2 − y2 (0 ≤ y ≤ 2c ) is rotated about the y-axis. (c is a fixed real number.) Showing all your work, find the area of the resulting surface. Solution: dx 1 1 y = · p · (−2y) = − p dy 2 c2 − y2 c2 − y2 s s !2 dx y2 |c| = 1+ 1+ 2 = p dy c − y2 c2 − y2 Z 2c p |c| Area = · 2π c2 − y2 dy p 0 c2 − y2 Z 2c dy = πc2 . = 2π|c| 0 2. Showing all your work, find the slope of the tangent line to the curve with equation in 12 polar coordinates r = , at the point corresponding to θ = π. θ Solution: 12 dr = − 2 dθ θ dr sin θ · dθ + r · cos θ dy = dr dx cos θ · dθ − r · sin θ − sin θ + θ cos θ . = − cos θ − θ sin θ At the point θ = π this ratio is equal to −π. Information for Students in MATH 141 2010 01 5132 3. Showing all your work, find the area of one of the regions bounded by the line θ = the closed curve r = 8 + 6 sin θ. π 2 and Solution: (The actual wording of the problem referred to a figure which it is not convenient to include in these notes.) The region can be interpreted as being swept out by a radius vector from the pole moving between − π2 and + π2 . The area is this Z π Z + π2 1 +2 2 (8 + 6 sin θ) dθ = 32 + 48 sin θ + 18 sin2 θ dθ 2 − π2 − π2 Z + π2 (32 + 48 sin θ + 9(1 − cos 2θ)) dθ = Z = " = − π2 + π2 − π2 (41 + 48 sin θ − 9 cos 2θ) dθ 9 41θ − 48 cos θ − sin 2θ 2 #+ π2 = 41π . − π2 4. Showing all your work, find the area enclosed by the curve (in polar coordinates) r = 9 + cos 2θ. Solution: The curve surrounds the pole, and is periodic with period 2π. The area may be expressed as an integral over an interval of length 2π; for example, as Z Z 1 2π 1 2π 2 (9 + cos 2θ) dθ = (81 + 18 cos 2θ + cos2 2θ) dθ 2 0 2 0 Z 1 2π 1 + cos 4θ = (81 + 18 cos 2θ + ) dθ 2 0 2 " #2π 1 1 1 = 81θ + 9 sin 2θ + θ + sin 4θ 2 2 8 0 163 163π = · 2π = . 4 2 5. Showing all your work, find the exact length of the polar curve r = 4θ2 (0 ≤ θ ≤ 2π). Solution: dr = 8θ dθ s !2 √ dr 2 r + 16θ4 + 64θ2 = dθ √ = 4|θ| 4 + θ2 . Information for Students in MATH 141 2010 01 5133 Over the interval in question θ is positive, and the absolute signs may be dropped. The length of the arc is Z 2π √ i 4h 3 2π 4θ 4 + θ s dθ = (4 + θ2 ) 2 0 3 0 32 3 = (1 + π2 ) 2 − 1 . 3 6. Showing all your work, find equations of the tangents to the curve x = 3t2 +4, y = 2t3 +3 that pass through the point (7, 5). Solution: We might, in error think that we need first to determine the parameter value associated with the given point. We would then solve the system of equations 3t2 + 4 = 7 2t3 + 3 = 5 to obtain t = +1. This would be an error. It happens that the given curve passes through the point (7, 5), but that is fortuitous: we want the tangents to pass through the point, not the curve! And we can’t find the points of contact of the tangents directly. So let’s first determine the general tangent to the curve, at the point with parameter value t. dx = 6t dt dy = 6t2 dt dy 6t2 dy = dt = = t, dx dx 6t dt 2 3 so the slope of the tangent at the point (3t +4, 2t + 3) on the curve is t; the equation of that tangent is y − 2t3 + 3 = t x − 3t2 + 4 , or y = tx − t3 − 4t + 3 . (118) We now impose the condition that this line pass through the point (x, y) = (7, 5), i.e., that its equation be satisfied by (x, y) = (7, 5), obtaining t3 − 3t + 2 = 0, whose left member factorizes to (t − 1)2 (t + 2) = 0, so the points of contact of the tangents are t = 1 and t = −2. The equations of the tangents through the given point are found by giving the parameter t these two values in equation (118): y= x−2 and y = −2x + 19 . Information for Students in MATH 141 2010 01 5134 7. Showing all your work, use methods of polar coordinates to find the length of the polar 4π curve r = 15 sin θ 0 ≤ θ ≤ 15 . Solution: dr = 15 cos θ dθ s !2 dr = 15 . r2 + dθ The length of the arc is, therefore Z 4π 15 15 dθ = 4π . 0 E.3 MATH 141 2009 01 E.3.1 Draft Solutions to Quiz Q1 Instructions to students 1. Show all your work. Marks may not be given for answers not supported by a full solution. For future reference, the form of your solutions should be similar to those shown in the textbook or Student Solutions Manual for similar problems. 2. In your folded answer sheet you must enclose this question sheet: it will be returned with your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.) All submissions should carry your name and student number. 3. Time = 20 minutes. 4. No calculators are permitted. Monday Versions 1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find the derivative of the function Z √x sin(t) f (x) = √ dt t5 3 Solution: Information for Students in MATH 141 2010 01 5135 • (This step may not be shown explicitly, but it underlies the successful implementation of the Chain Rule.) Introduction of an intermediate variable: If the new Zu √ d sin(t) du 0 variable/function is called u = u(x) = x, then f (x) = dt · 5 du √ t dx 3 • application of the Fundamental Theorem Zu d sin(t) sin(u) dt = . 5 du √ t u5 3 • completion √ √ sin( x) 1 − 12 sin x √ 5 · 2 x = 2x3 x 2. [10 MARKS] Compute (a) Z 0 8 sec2 (x) dx −π/3 Solution: [5 MARKS TOTAL] • state one antiderivative, e.g., 8 tan x • indicate that the value of the integral is the net change, 8 tan x]0− π 3 • compute the final answer. Students should know the trigonometric functions of simple submultiples of π. (b) Z 1 √ x(5x2 + 4x − 5) dx 0 Solution: [5 MARKS TOTAL] • state one antiderivative, here the obvious method is to express as a sum of fractional powers and to integrate each separately: Z 2 5 2 3 2 7 5 3 1 5x 2 + 4x 2 − 5x 2 dx = 5 · x 2 + 4 · x 2 − 5 · x 2 + C 7 5 3 • indicate that the value of the integral is the net change, " #1 2 52 2 32 2 72 5· x +4· x −5· x 7 5 3 0 • compute the final answer correctly = 10 7 + 85 − 10 3 Information for Students in MATH 141 2010 01 5136 Tuesday Versions 1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find the derivative of the function Z ex f (x) = 5 p 7 + ln6 t dt t Solution: • (This step may not be shown explicitly, but it underlies the successful implementation of the Chain Rule.) Introduction of an intermediate variable: If the new variable/function is called u = u(x) = e x , then Z u p d du 7 + ln6 t f 0 (x) = dt · du 5 t dx • application of the Fundamental Theorem p Z u p d 7 + ln6 t 7 + ln6 u dt = . du 5 t u • completion q 7 + (ln(e x ))6 ex √ 7 + x6 x √ ·e = · e = 7 + x6 . ex x While you may use some judgment about how much simplification you expect, I don’t believe it would not be appropriate to accept a composition like ln(e x ) not simplified. 2. [10 MARKS] Compute (a) Z 0 [8 sec(x) tan(x) + 7 cos(x)] dx −π/6 Solution: [5 MARKS TOTAL] • state one antiderivative, e.g., 8 sec x + 7 sin x • indicate that the value of the integral is the net change in the antiderivative, e.g., [8 sec x + 7 sin x]0− π 6 Information for Students in MATH 141 2010 01 5137 • compute the final answer correctly. ! 2 1 9 16 (8 + 0) − 8 · √ − 7 · = − √ 2 2 3 3 (b) Z 4 1 −3x−1 + 5x + 3 dx √ x Solution: [5 MARKS TOTAL] • state one antiderivative, here the obvious method is to express as a sum of fractional powers and to integrate each separately: ! ! Z −3x−1 + 5x + 3 2 − 12 2 3 2 1 dx = −3 · − x + 5 · x 2 + 3 · x 2 + C √ 1 3 1 x • indicate that the value of the integral is the net change, " ! #4 2 − 12 2 23 2 12 −3 · − x + 5 · x + 3 · x 1 3 1 1 • compute the final answer correctly ! ! 6 10 10 97 = + ·8+6·2 − 6+ +6 = 2 3 3 3 Wednesday Versions 1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find the derivative of the function Z ln x √ f (x) = et 1 + t2 dt 5 Solution: • (This step may not be shown explicitly, but it underlies the successful implementation of the Chain Rule.) Introduction of an intermediate variable: If the new variable/function is called u = u(x) = ln x, then Z u √ du d 0 et 1 + t2 dt · f (x) = du 5 dx Information for Students in MATH 141 2010 01 5138 • application of the Fundamental Theorem Z u √ √ d et 1 + t2 dt = eu 1 + u2 . du 5 • completion p 1 p eln x 1 + (ln x)2 · = 1 + (ln x)2 x ln x It is essential that e be simplified to x for full marks in this part. Z π4 2. [10 MARKS] Compute f (x) dx, where − π6 ( f (x) = 4 sin x if 5 sec x tan x if x≤0 0<x< π 2 Solution: • decompose the interval into subintervals matching the intervals where the 2 parts of the definition apply: Z π4 Z 0 Z π4 f (x) dx = f (x) dx + f (x) dx − π6 − π6 0 • matching the different functions to the appropriate subintervals: Z 0 Z π4 Z 0 Z π4 f (x) dx + f (x) dx = 4 sin x dx + 5 sec x tan x dx − π6 − π6 0 0 • shift the constants outside of the integration: Z 0 Z Z π4 4 sin x dx + 5 sec x tan x dx = 4 − π6 − π6 0 Z 0 sin x dx + 5 π 4 sec x tan x dx 0 • find antiderivatives for both of the 2 integrands, e.g., − cos x and sec x • indicate that the value of each integral is the net change, π 4[− cos x]0− π + 5[sec x]04 6 • correctly complete the computations √ √ √ √ π 3 π +5( 2−1) = 2 3+5 2−9 4 − cos 0 + cos +5 sec − sec 0 = 4 −1 + 6 4 2 Information for Students in MATH 141 2010 01 5139 Thursday Versions 1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find the derivative of the function Z 7x √ 8 + 9t2 f (x) = dt t 3x Solution: • The integral must be split into two, at a convenient place, each integral with one fixed and one variable limit; note that the point where the integral is split CANNOT BE 0, since the integrand is undefined there: Z 7x √ Z 1 √ Z 7x √ 8 + 9t2 8 + 9t2 8 + 9t2 f (x) = dt = dt + dt t t t 3x 3x 1 • one integral must be reversed so that the dependence on x is in the upper limit: Z 3x √ Z 7x √ 8 + 9t2 8 + 9t2 f (x) = − dt + dt t t 1 1 • differentiate each of the integrals separately, using the Fundamental Theorem, and multiply by the factor of the form du from the Chain Rule (see problems on earlier dx versions) p p 8 + 9(3x)2 d(3x) 8 + 9(7x)2 d(7x) d f (x) = − · + · dx 3x dx 7x dx • completion p p p √ 8 + 9(3x)2 8 + 9(7x)2 − 8 + 81x2 + 8 + 9(49)x2 − ·3+ ·7= 3x 7x x √ 3 Z 2. [10 MARKS] Compute f (x) dx, where 0 ( f (x) = Solution: 3x 6 1+x2 if if 0≤x≤1 x>1 Information for Students in MATH 141 2010 01 5140 • for decomposing the interval into subintervals matching the intervals where the 2 parts of the definition apply: Z √ Z 3 f (x) dx = √ 3 Z 1 f (x) dx + 0 f (x) dx 0 1 • for matching the different functions to the appropriate subintervals: √ 3 Z Z f (x) dx = 0 √ 3 Z 1 3x dx + 0 1 6 dx 1 + x2 • for shifting constants outside of the integration: Z Z 1 3x dx + 0 1 √ 3 6 3 dx = 2 1+x 2 Z Z 1 √ 3 2x dx + 6 0 1 1 dx 1 + x2 • for finding antiderivatives for both of the 2 integrands, e.g., x2 and arctan x • for indicating that the value of each integral is the net change, √ 3 21 [x ]0 + 6[arctan x]1 3 2 • for correctly completing the computations √ 3 π π 3 π √ 3 3 21 [x ]0 +6[arctan x]1 3 = (1−0)+6 arctan 3 − arctan 1 = +6 − = + 2 2 2 3 4 2 2 E.3.2 Draft Solutions to Quiz Q2 Instructions to students 1. Show all your work. Marks may not be given for answers not supported by a full solution. For future reference, the form of your solutions should be similar to those shown in the textbook or Student Solutions Manual for similar problems. 2. In your folded answer sheet you must enclose this question sheet: it will be returned with your graded paper. WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0. All submissions should carry your name and student number. 3. Time = 25 minutes. 4. No calculators are permitted. Information for Students in MATH 141 2010 01 5141 Monday Versions 1. [10 MARKS] Compute (a) Z sec2 x dx 1 + 6 tan x (b) Z 9/2 1/2 √ e 2x √ dx 2x Solution: (a) [4 MARKS] for this indefinite integral • • • • [1 MARK] for stating the substitution [1 MARK] for rewriting the indefinite integral in terms of the new variable [1 MARK] for finding an antiderivative in terms of the new variable [1 MARK] for restating the antiderivative in terms of the original variable (b) [6 MARKS] for this definite integral • [1 MARK] for stating the substitution • [3 MARKS] for transforming the definite integral, including the upper and lower limit • [1 MARK] for finding an antiderivative • [1 MARK] for the final answer (Some students may, instead, find an antiderivative [4 MARKS] and then find the net change [2 MARKS].) 2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtained π π by revolving the region bounded by the x-axis, the lines x = and x = , and the curve 6 3 √ y = 4 cos x. Solution: It was intended that students solve this problem using the “Method of Washers”. A solution using Cylindrical Shells would certainly be acceptable, but would be more difficult, as students do not yet know how to integrate arccos y, and may not have mastered integration by parts. If they complete part of such a solution, allocate the marks similarly to the scheme for Washers. • [4 MARKS] for the integrand Information for Students in MATH 141 2010 01 5142 • [2 MARKS] for the limits of integration • [2 MARKS] for finding an antiderivative • [2 MARKS] for completing the integration. 3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = π4 and x = π2 , and the curve y = 9 sin x. Compute the volume of the solid of revolution obtained by revolving the region R about the y-axis. Hint: Use the method of cylindrical shells. Solution: • [3 MARKS] for determining the integrand correctly • [2 MARKS] for determining the limits of integration correctly • [4 MARKS] for applying integration by parts and correctly determining the full antiderivative • [1 MARK] for apparently completing the integration correctly Tuesday Versions 1. [10 MARKS] Compute (a) Z sec x tan x dx −7 − 8 sec x (b) Z e4 e cos(9 ln x) dx x Solution: (a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions. (b) [6 MARKS] Same scheme as for Problem 1(b) on Monday Versions. 2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtained π π by revolving the region bounded by the lines y = 1, x = − , x = , and the curve 4 4 y = 3 sec x. Solution: Same scheme as for Problem 2 on Monday Versions. Information for Students in MATH 141 2010 01 5143 3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = 1 and x = 5, and the curve y = ln(5x). Compute the volume of the solid of revolution obtained by revolving the region R about the y-axis. Hint: Use the method of cylindrical shells. Solution: • [3 MARKS] for determining the integrand correctly • [2 MARKS] for determining the limits of integration correctly • [4 MARKS] for applying integration by parts and correctly determining the full antiderivative • [1 MARK] for apparently completing the integration correctly Wednesday Versions 1. [10 MARKS] Compute (a) Z √ 9x + 2 9x2 + 4x dx (b) Z 1 0 e3x dx e6x + 1 Solution: (a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions. (b) [6 MARKS] Same scheme as for Problem 1(b) on Monday Versions. 2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained by revolvingthe region bounded by the y-axis, the lines y = ln 3 and y = ln 4, and the x curve y = ln 2 . Solution: Note that this is a solid of revolution about the y-axis. To use the Method of Washers, which is intended, students will have to rewrite the equation of the curve in the form x = 2ey . A correct solution using the Method of Cylindrical Shells would certainly be acceptable. Follow the same grading scheme as shown above for Question 2 of Monday Versions. Information for Students in MATH 141 2010 01 5144 3. [10 MARKS] Find the average value of the function √ x 8+x on the interval [−8, −4]. Solution: • [2 MARKS] for setting up the integral correctly, with correct integrand and limits of integration • [7 MARKS] for the evaluation of this integral — more than one method is feasible: Integration by Parts: – [2 MARKS] for a correct selection of u and dv – [2 MARKS] for determining du and v – [2 MARKS] for applying integration by parts R – [1 MARK] for the integration of v du. Substitution: – [2 MARKS] for selection of an appropriate substitution u = u(x) – [2 MARKS] for transforming the integrand correctly into terms of u – [2 MARKS] for correctly changing the limits of integration into terms of the new variable – [1 MARKS] for correctly evaluating the new definite integral • [1 MARK] for dividing the weighted integral by the length of the interval, and obtaining the final answer. Thursday Versions 1. [10 MARKS] Compute (a) Z √ x3 x2 + 4 dx (b) Z 1 0 x+2 dx x2 + 1 Solution: (a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions. Information for Students in MATH 141 2010 01 5145 (b) [6 MARKS] Same scheme as for Problem 1(b) on Monday Versions. 2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained 3 by revolving the region bounded by the lines x = 1 and y = 2 and the curve y = . x Solution: Note that this is a solid of revolution about the y-axis; moreover, it has a hole in the middle. It was intended that students solve this problem using the “Method of Washers”. However, the Method of Cylindrical Shells would be acceptable, and is not difficult. • [4 MARKS] for the integrand • [2 MARKS] for the limits of integration; note that the data given are partly in terms of x-coordinates and partly in terms of y-. • [2 MARKS] for finding an antiderivative • [2 MARKS] for completing the integration. 3. [10 MARKS] Find the average value of the function √ x 3+x on the interval [−2, 1]. Solution: • [2 MARKS] for setting up the integral correctly, with correct integrand and limits of integration • [7 MARKS] for the evaluation of this integral — more than one method is feasible: Integration by Parts: – [2 MARKS] for a correct selection of u and dv – [2 MARKS] for determining du and v – [2 MARKS] for applying integration by parts R – [1 MARK] for the integration of v du. Substitution: – [2 MARKS] for selection of an appropriate substitution u = u(x) – [2 MARKS] for transforming the integrand correctly into terms of u – [2 MARKS] for correctly changing the limits of integration into terms of the new variable – [1 MARKS] for correctly evaluating the new definite integral • [1 MARK] for dividing the weighted integral by the length of the interval, and obtaining the final answer. Information for Students in MATH 141 2010 01 5146 E.3.3 Draft Solutions to Quiz Q3 Instructions to students 1. Show all your work. Marks may not be given for answers not supported by a full solution. For future reference, the form of your solutions should be similar to those shown in the textbook or Student Solutions Manual for similar problems. 2. In your folded answer sheet you must enclose this question sheet: it will be returned with your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.) All submissions should carry your name and student number. 3. Time = 25 minutes. 4. No calculators are permitted. Monday Versions 1. [10 MARKS] (a) Use integration by parts to compute the integral x3/5 ln x dx. (b) Make a substitution and then use integration by parts to compute the integral Z 2 x3 e x dx . Solution: (a) [4 MARKS] • [2 MARKS] for a correct choice of u and dv and correctly determining du and v • [2 MARKS] for correctly implementing the selection of u and v and completing the integration correctly (b) [6 MARKS] • [1 MARK] for correctly implementing an appropriate substitution • [2 MARKS] for a correct choice of u and dv and correctly determining du and v • [2 MARKS] for correctly implementing the selection of u and v and completing the integration in terms of the new variable • [1 MARK] for expressing the final, correct answer in terms of the original variable Information for Students in MATH 141 2010 01 5147 Z 2. [10 MARKS] Use a trigonometric substitution to compute your answer by differentiating it! 1 dx . Verify √ ( 25 − x2 )3 Solution: • [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do). (Strictly speaking, the substitution should be expressed first in terms of an inverse sine or inverse cosine, but it is common practice not to make that step explicit, so one can’t expect students to be better than the textbooks.) • [2 MARKS] for implementing the substitution correctly and writing the integral in terms of the the square of the secant or cosecant. • [2 MARKS] for correctly integrating in terms of the new variable • [2 MARKS] for transforming the integral into terms of the original variable x. • [2 MARKS] for correctly differentiating the antiderivative and thereby obtaining the original integrand item [10 MARKS] Find the arc length of the parameterized curve x(t) = e2t + e−2t , y(t) = 1 4t − 2 . for t between 0 and . 2 Solution: Grading instructions: • [1 MARKS] for an integral of the correct form • [4 MARKS] for correctly computing the derivatives of x and y • [3 MARKS] for correctly finding an antiderivative • [2 MARKS] for correctly completing the evaluation of the integral. Tuesday Versions 1. [10 MARKS] (a) Use integration by parts to compute the integral Z x sec2 (5x) dx (b) Make a substitution and then use integration by parts to compute the integral Z x3 cos(x2 ) dx Information for Students in MATH 141 2010 01 5148 Solution: (a) [4 MARKS] • [2 MARKS] for a correct choice of u and dv and correctly determining du and v • [2 MARKS] for correctly implementing the selection of u and v and completing the integration correctly (b) [6 MARKS] • [1 MARK] for correctly implementing an appropriate substitution • [2 MARKS] for a correct choice of u and dv and correctly determining du and v • [2 MARKS] for correctly implementing the selection of u and v and completing the integration in terms of the new variable • [1 MARK] for expressing the final, correct answer in terms of the original variable Z x2 2. [10 MARKS] Use a trigonometric substitution to compute dx . Verify your √ 4 − x2 answer by differentiating it! Solution: • [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do). (Strictly speaking, the substitution should be expressed first in terms of an inverse sine or inverse cosine, but it is common practice not to make that step explicit, so one can’t expect students to be better than the textbooks.) • [2 MARKS] for implementing the substitution correctly and writing the integral in terms of the the square of the secant or cosecant. • [2 MARKS] for correctly integrating in terms of the new variable • [2 MARKS] for transforming the integral into terms of the original variable x. This antiderivative does not include any term with plus/or/minus: there are no ambiguities of signs! If a student shows an ambiguity, this means he has not properly completed the differentiation of the next part, since only one of the signs will yield the correct derivative. • [2 MARKS] for correctly differentiating the antiderivative and thereby obtaining the original integrand 3. [10 MARKS] Find the arc length of the parameterized curve x(t) = −3 + e2t cos t, y(t) = 1 5 + e2t sin t for t between 0 and . 2 Information for Students in MATH 141 2010 01 5149 Solution: Grading instructions: • [1 MARKS] for an integral of the correct form • [4 MARKS] for correctly computing the derivatives of x and y • [3 MARKS] for correctly finding an antiderivative • [2 MARKS] for correctly completing the evaluation of the integral. Wednesday Versions Z 1. [10 MARKS] Compute the integral e9π cos(ln(x)) dx. Hint: A solution by integration 1 by parts could begin from the observation that cos(ln(x)) = 1 · cos(ln(x)). You could also apply integration by parts after making a substitution. Solution: There appear to be several ways of attacking this problem, but the attacks will require 2 applications of integration by parts, followed by the solving of an equation. Applying Integration by Parts immediately: • [1 MARK] for a correct selection of u and dv for the first integration by parts • [1 MARKS] for correctly determining du and v • [2 MARKS] for correctly applying integration by parts and expressing the given integral as the value of uv] minus a second integral, which will then require a second application of integration by parts • [1 MARK] for a correct selection of U and dV for the second integration by parts • [1 MARKS] for correctly determining dU and V • [2 MARKS] for correctly applying integration by parts and expressing the original integral as a sum of [uv + UV] minus the same integral • [2 MARKS] for solving the equation for the desired integral and completing all calculations apparently correctly Preceding Integration by Parts by a Substitution: • [0 MARKS] for selecting a correct substitution, and implementing that substitution correctly both in the integrand and the limits of integration, so that the integral is now written in a form where the use of integration by parts is well indicated. • [1 MARK] for a correct selection of u and dv for the first integration by parts • [1 MARKS] for correctly determining du and v • [2 MARKS] for correctly applying integration by parts and expressing the given integral as the value of uv] minus a second integral, which will then require a second application of integration by parts Information for Students in MATH 141 2010 01 5150 • [1 MARK] for a correct selection of U and dV for the second integration by parts • [1 MARKS] for correctly determining dU and V • [2 MARKS] for correctly applying integration by parts and expressing the original integral as a sum of [uv + UV] minus the same integral • [2 MARKS] for solving the equation for the desired integral and completing all calculations apparently correctly Z 13 2. [10 MARKS] Compute dx . (x + 3)(x2 + 4) Solution: • [2+2 MARKS] for correctly factorizing the denominator and expressing the need to expand the function into a sum of 2 partial fractions, one with a linear denominator, the other having a general numerator of degree 1 and denominator of degree 2. Reserve a full 2 MARKS for the numerator of the fraction with the quadratic denominator. • [3 MARKS] for determining correctly the 3 undetermined constants • [3 MARKS] for completing the integration correctly 3. [10 MARKS] Find the surface area of the solid of revolution obtained by revolving the graph of the parametric curve x(t) = 2t − 3, y(t) = t2 − 3t − 1, 0 ≤ t ≤ 3/2 about the y-axis. Solution: Grading instructions: • [2 MARKS] for an integral of the correct form • [3 MARKS] for correctly computing the derivatives of x and y • [3 MARKS] for correctly finding an antiderivative • [2 MARKS] for correctly completing the evaluation of the integral. Thursday Versions 1. [10 Use a substitution and then integration by parts to compute the integral Z 6 MARKS] 3/x e dx . x3 3 Solution: Information for Students in MATH 141 2010 01 5151 • [5 MARKS] for selecting an appropriate substitution to simplify the integral, and correctly changing the integrand and the limits of integration. Some students may arrive at the ultimate substitution through several composed substitutions. If they do not succeed in completing a substitution and thus cannot begin seriously integration by parts, you should grade their work out of a maximum of 5 MARKS. • [2 MARKS] for a correct choice of u and dv and correctly determining du and v • [2 MARKS] for correctly implementing the selection of u and v and completing the integration in terms of the new variable • [1 MARK] for expressing the final, correct answer in terms of the original variable Z 25 dx . 2. [10 MARKS] Compute x(x + 5)2 Solution: • [2+2 MARKS] for correctly factorizing the denominator and expressing the need to expand the function into a sum of either 2 or 3 partial fractions: – One expansion would have one partial fraction with a linear denominator, the other having a general numerator of degree 1 and denominator of degree 2 consisting of the 2nd power of a linear function. – Alternatively the function could be expressed as the sum of 3 partial fractions: the first being associated with the factor of multiplicity 1; the second having in its denominator the first power of the other linear factor; and the last having in its denominator the second power of the other linear factor In both cases there are 3 constants to be determined. Reserve a full 2 MARKS for the numerator of the fraction with the quadratic denominator, or, alternatively, the 2 terms associated with that linear factor of the denominator. • [3 MARKS] for determining correctly the 3 undetermined constants • [3 MARKS] for completing the integration correctly 3. [10 MARKS] Find the surface area of the solid of revolution obtained by revolving the √ 2 graph of the parametric curve x(t) = (t + 4)3/2 , y(t) = 2 t + 4, −4 ≤ t ≤ 0 about the 3 y-axis. Solution: Grading instructions: • [2 MARKS] for an integral of the correct form • [3 MARKS] for correctly computing the derivatives of x and y • [3 MARKS] for correctly finding an antiderivative • [2 MARKS] for correctly completing the evaluation of the integral. Information for Students in MATH 141 2010 01 5152 F Final Examinations from Previous Years F.1 Final Examination in Mathematics 189-121B (1996/1997) 1. [4 MARKS] Find the derivative of the function F defined by Z x4 F(x) = sin √ t dt . x2 Z 2. [4 MARKS] Evaluate π −π 2 f (x) dx , where ( f (x) = cos x, 3 x + 1, π −π ≤ x ≤ π3 2 π <x≤π 3 . Z x sin3 x2 cos x2 dx . 3. [7 MARKS] Evaluate Z (x5 + 4−x ) dx . 4. [7 MARKS] Evaluate 5. [10 MARKS] Calculate the area of the region bounded by the curves x − y = 2. x = y2 and 6. [10 MARKS] The region bounded by f (x) = 4x − x2 and the x-axis, between x = 1 and x = 4 , is rotated about the y-axis. Find the volume of the solid that is generated. Z 7. [6 MARKS] Evaluate x ln x dx . Z 8. [6 MARKS] Evaluate sin2 x cos5 x dx . 9. [6 MARKS] Determine the partial fraction decomposition of the following ratio of polynomials: x5 + 2 . x2 − 1 10. [4 MARKS] Determine whether or not the following sequence converges as If it does, find the limit: ) ( x 3n . 1+ n n→∞. Information for Students in MATH 141 2010 01 5153 11. [4 MARKS] Determine the following limit, if it exists: √ x lim+ √ √ . x→0 x + sin x ∞ X 12. [6 MARKS] Determine whether the series ke−k 2 converges or diverges. k=2 13. [6 MARKS] Test the following series for (a) absolute convergence, (b) conditional convergence. ∞ X √ k=10 (−1)k . k(k + 1) 14. [10 MARKS] Find the area of the region that consists of all points that lie within the circle r = 2 cos θ , but outside the circle r = 1 . 15. [10 MARKS] Determine the length of the curve r = 5(1 − cos θ) , (0 ≤ θ ≤ 2π) . F.2 Final Examination in Mathematics 189-141B (1997/1998) 1. [10 MARKS] (a) Sketch the region bounded by the curves y = x2 and y = 3 + 5x − x2 . (b) Determine the area of the region. 2. [10 MARKS] The triangular region bounded by the lines y = x, y= 3 x − , 2 2 and y=0 is revolved around the line y = 0. Determine the volume of the solid of revolution which is generated. Information for Students in MATH 141 2010 01 5154 3. [10 MARKS] Find the length of the curve x = 2. y = √ x2 − ln 4 x 2 from x = 1 to 4. [5 MARKS] Determine, at x = 12 , the value of the function sin−1 x and the slope of its graph. 5. [5 MARKS] Evaluate lim x→2 x3 − 8 . x4 − 16 6. [5 MARKS] Showing all your work, evaluate lim x x x→0+ . Z 2 x3 e−x dx . 7. [5 MARKS] Evaluate Z 8. [10 MARKS] Evaluate Z 9. [10 MARKS] Evaluate x3 − 1 dx . x3 + x x3 dx , where |x| < 1 . √ 1 − x2 10. [10 MARKS] Find the area of the region that lies within the limac¸on and outside the circle r = 2 . r = 1 + 2 cos θ 11. [5 MARKS] Z x Showing all your work, obtain a second-degree Taylor polynomial for f (x) = et(1−t) dt at x = 0 . 0 12. [5 MARKS] Showing all your work, determine whether the following infinite series converges or diverges. If it converges, find its sum. ∞ X 3n − 2n n=0 4n 13. [5 MARKS] Showing all your work, determine whether or not the following series converges: 1 ∞ X 2n n2 n=1 14. [5 MARKS] Showing all your work, determine whether the following series converges: ∞ X n=1 1 n · 2n Information for Students in MATH 141 2010 01 5155 F.3 Supplemental/Deferred Examination in Mathematics 189-141B (1997/1998) 1. [10 MARKS] (a) Sketch the region bounded by the curves y= 8 x+2 and x + y = 4. (b) Determine the area of the region. 2. [10 MARKS] The triangular region bounded by the lines y = x, y= 3 x − , 2 2 and y=0 is revolved around the line y = 0. Determine the volume of the solid of revolution which is generated. 3. [10 MARKS] Find the area of the surface of revolution generated by revolving the curve y= 1 x e + e−x 2 (0 ≤ x ≤ 1) about the x-axis. 4. [5 MARKS] Determine, at x = 12 , the value of the function cos−1 x and the slope of its graph. 5. [10 MARKS] Evaluate lim x→2 x − 2 cos πx . x2 − 4 ! x4 1 6. [5 MARKS] Evaluate lim cos 2 . x→∞ x Z e2x dx . 7. [5 MARKS] Evaluate 1 + e4x Z 8. [5 MARKS] Evaluate x2 cos x dx . Z 9. [10 MARKS] Evaluate x3 − 1 dx . x3 + x Z √ 10. [10 MARKS] Evaluate a2 − u2 du , where |u| < a. Information for Students in MATH 141 2010 01 5156 11. [10 MARKS] Find the area of the region that lies within the limac¸on and outside the circle r = 2 . r = 1 + 2 cos θ 12. [5 MARKS] Z x Showing all your work, obtain a second-degree Taylor polynomial for f (x) = e s(1−s) ds at x = 0 . 0 13. [5 MARKS] Showing all your work, determine whether the following infinite series converges or diverges. If it converges, find its sum. ∞ X 1 + 2n + 3n 5n n=0 14. [5 MARKS] Showing all your work, determine whether or not the following series converges. ∞ X ln n n n=1 15. [5 MARKS] Showing all your work, determine whether the following series convereges. ∞ X n=1 n2 + 1 en (n + 1)2 F.4 Final Examination in Mathematics 189-141B (1998/1999) 1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and (y−1)2 = 5− x. 2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about the line x = 1 the region bounded by the curve (x − 1)2 = 5 − 4y and the line y = 1 . 3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0 the region bounded by the curves and y y x x = = = = sin x −2 0 2π . 4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = x2 (0 ≤ √ x ≤ 2) about the y-axis. Information for Students in MATH 141 2010 01 5157 Zx 3 et dt . 5. Define the function F by F(x) = 0 (a) [4 MARKS] Showing all your work, explain clearly whether or not the following inequalities are true. 3 e < F(e) < ee +1 . (b) [4 MARKS] Determine following points: the value of d F(x3 ) dx at each of the i. at x = 0 . ii. at x = 2 . 6. [4 MARKS] Showing all your work, evaluate Z sin3 πx dx . 7. [4 MARKS] Showing all your work, evaluate Z x2 e−x dx . 8. [4 MARKS] Showing all your work, evaluate Z x−1 dx . 3 x − x2 − 2x 9. [4 MARKS] Showing all your work, evaluate Z 3 x + x2 + x − 1 dx . x2 + 2x + 2 10. [8 MARKS] Find the area of the region inside the curve r = 3 sin θ and outside the curve r = 2 − cos θ. 11. Showing all your work, determine whether each of the following integrals is convergent or divergent: Z∞ (a) [4 MARKS] sin x dx . 0 Information for Students in MATH 141 2010 01 Z2 (b) [4 MARKS] 5158 dx . 1 − x2 0 12. Showing all your work, determine whether each of the following sequences is convergent or divergent. π (a) [4 MARKS] n sin n (b) [4 MARKS] (2 n + 1) e−n 13. Showing all your work, determine whether each of the following infinite series is convergent or divergent: (a) [4 MARKS] ∞ X 1 . 3 4n n=1 ! ∞ X 1 1 (b) [4 MARKS] + . n n2 n=1 14. Showing all your work, determine whether each of the following series is convergent, divergent, conditionally convergent and/or absolutely convergent. (a) [4 MARKS] ∞ X (−1)n n+2 . n(n + 1) (−1)n cos n . n2 n=1 (b) [4 MARKS] ∞ X n=1 F.5 Supplemental/Deferred Examination in Mathematics 189-141B (1998/1999) 1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and y = 6 − x. 2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about the line x = 0 the region bounded by the curve y = 4 − x2 and the lines x = 0 and y=0 . 3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0 the region bounded by the curves and y y x x = = = = sin x 2 0 2π . Information for Students in MATH 141 2010 01 5159 4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = −x2 (0 ≤ √ x ≤ 2) about the y-axis. Zx sin10 t dt . 5. Define the function F by F(x) = 0 (a) [4 MARKS] Showing all your work, explain clearly whether or not the following inequalities are true. 0 < F(e) < e . (b) [4 MARKS] Determine following points: the value of d F(x) dx at each of the i. at x = 0 . π ii. at x = . 2 6. [8 MARKS] Showing all your work, evaluate Z 2 x5 e−x dx . 7. [4 MARKS] Showing all your work, evaluate Z 3 x − x2 + x + 1 dx . x2 − 2x + 2 8. [8 MARKS] Find the area of the region inside the curve r = 6 sin θ and outside the curve r = 4 − 2 sin θ. 9. Showing all your work, determine whether each of the following integrals is convergent or divergent: Z∞ (a) [4 MARKS] cos x dx . 0 Z4 (b) [4 MARKS] dx . 4 − x2 0 10. Showing all your work, determine whether each of the following sequences is convergent or divergent. Information for Students in MATH 141 2010 01 5160 π (a) [4 MARKS] n sin n (b) [4 MARKS] (2 n + 1) e−n 11. Showing all your work, determine whether each of the following infinite series is convergent or divergent: (a) [4 MARKS] ∞ X 1 . 5 4n n=1 ! ∞ X 1 1 (b) [4 MARKS] − 3 . n n n=1 F.6 Final Examination in Mathematics 189-141B (1999/2000) 1. [11 MARKS] Find the area of the region bounded by the curves −y2 + 12y − 16 . x = y2 and x= 2. [11 MARKS] Let C denote the arc of the curve y = cosh x for −1 ≤ x ≤ 1 . Find the volume of the solid of revolution generated by revolving about the line x = −2 e2 + 1 the region bounded by C and the line y = . 2e 3. (a) [5 MARKS] Showing all your work, evaluate Z 9 2 3 2 √ 6t − t2 dt . (b) [6 MARKS] Showing all your work, evaluate Z 3π 4 π 4 4. √ 1 − sin u du . (a) [7 MARKS] Showing all your work, determine a reduction Zformula which expresses, for any integer n not less than 2, the value of xn sin 2x dx in Z terms of xn−2 sin 2x dx. Z (b) [4 MARKS] Use your reduction formula to determine the indefinite integral x2 sin 2x dx. Information for Students in MATH 141 2010 01 5161 5. [11 MARKS] Showing all your work, evaluate Z 8x2 − 21x + 6 dx . (x − 2)2 (x + 2) 6. [11 MARKS] Find the area of the region r = 1 + cos θ and outside the curve r = 1 − cos θ . inside the curve 7. [11 MARKS] Determine whether the following integral is convergent or divergent. If it is convergent, find its value. Show all your work. Z 3 1 dx 4 0 5 (x − 1) 8. [11 MARKS] Showing all your work, determine whether the following infinite series is ∞ X 2 n! e−(n − 1) . convergent or divergent: n=1 9. Showing all your work, determine whether each of the following series is convergent, divergent, conditionally convergent and/or absolutely convergent. (a) [6 MARKS] ∞ X (−1)n √ n+2− √ n . n=1 (b) [6 MARKS] ∞ X n=1 (−1)n n . ln n2 F.7 Supplemental/Deferred Examination in Mathematics 189-141B (1999/2000) 1. [11 MARKS] Determine the area of the region bounded by the curves y = 2 − x2 . y = x4 and 2. [11 MARKS] Determine the volume of the solid generated by rotating the region bounded by the curves y = 2x2 and y2 = 4x around the x-axis. 3. Evaluate the integrals: Z x7 (a) [5 MARKS] dx . √ 1 − x4 Z x2 (b) [6 MARKS] dx . √ 4 − x2 Information for Students in MATH 141 2010 01 5162 Z π/2 4. [11 MARKS] Showing all your work, find e2x sin 3x dx . 0 Z 6x3 − 18x dx . (x2 − 1)(x2 − 4) 5. [11 MARKS] Determine 6. [11 MARKS] Find the area of the region inside the curve r = 2. r = 2 + 2 sin θ and outside Z 1 7. [11 MARKS] Determine whether the following improper integral converges: 0 ln x dx . x2 8. [11 MARKS] Showing all your work, determine whether the following infinite series ∞ X 1 converges: . √ 15n3 + 3 n=1 9. Showing all your work, determine, for each of the following series, whether it is convergent, divergent, conditionally convergent and/or absolutely convergent. (a) [6 MARKS] ∞ X (−1)n ln n n n=1 (b) [6 MARKS] ∞ X cos nπ n=1 n . . F.8 Final Examination in Mathematics 189-141B (2000/2001) 1. Showing all your work, determine, for each of the following infinite series, whether or not it converges. ∞ X n . n3 + 1 i=1 ∞ n X ln (b) [3 MARKS] . 3n + 1 n=1 (a) [3 MARKS] (c) [6 MARKS] ∞ X (−1)n (3n + 1)4 n=2 5n . 2. [12 MARKS] Determine the volume of the solid of revolution generated by revolving about the y-axis the region bounded by the curves 2 y = e−x , Information for Students in MATH 141 2010 01 5163 y = 0, x = 0, x = 1. 3. [12 MARKS] Determine the area of the surface of revolution generated by revolving about the x-axis the curve π y = cos x , 0≤x≤ . 6 [Hint: You may wish to make use of the fact that Z 2 sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C .] 4. [12 MARKS] Find the r = 3 cos θ and outside the curve 5. [14 MARKS] Evaluate the integral Z area that r = 2 − cos θ . the circle x = t3 + t2 + 1 , y = 1 − t2 , determine The equation of the tangent line at the point written in the form y = mx + b , where m and b are cond2 y dx2 (b) [6 MARKS] the value of 7. inside x dx . (x − 1)(x2 + 4) 6. For the curve given parametrically by (a) [6 MARKS] (x, y) = (1, 0) , stants; is at the point (x, y) = (1, 0) . (a) [10 MARKS] Use integration by parts to determine the value of Z e x cos x dx . Z 0 (b) [4 MARKS] Evaluate e x cos x dx . −∞ 8. [12 MARKS] Find the area of the region bounded by the curves y = −2x2 + 5x − 2 . y = x2 − 4 and Information for Students in MATH 141 2010 01 5164 F.9 Supplemental/Deferred Examination in Mathematics 189-141B (2000/2001) 1. (a) [6 MARKS] Showing all your work, find F 0 (1) when Z 2t x F(t) = dx . 3 x +x+7 1 (b) [6 MARKS] Showing all your work, evaluate R6 0 |x − 2| dx . 2. Showing all of your work, evaluate each of the following integrals: Z x+1 (a) [4 MARKS] dx; √ 9 − x2 Z 1 (b) [4 MARKS] dx; 3 2x + x Z (c) [4 MARKS] sin2 2x cos2 2x dx; Z (d) [4 MARKS] ln x dx 3. [15 MARKS] Showing all your work, find the area of the region bounded below by the 1 1 line y = , and above by the curve y = . 2 1 + x2 4. [15 MARKS] Showing all your work, find the volume generated by revolving about the y-axis the smaller region bounded by the circle x2 + y2 = 25 and the line x = 4 . 5. Showing all your work, (a) [2 MARKS] sketch the curve r = 1 − sin θ ; (b) [6 MARKS] find the length of the portion of the curve that lies in the region given π π by r ≥ 0 , − ≤ θ ≤ ; 2 2 (c) [5 MARKS] find the coordinates of the points on the curve where the tangent line is parallel to the line θ = 0 . 6. For each of the following integrals, determine whether it is convergent or divergent; if it is convergent, you are expected to determine its value. Show all your work. Z 2 1 dx ; (a) [7 MARKS] 3 −1 x Z ∞ 2 (b) [7 MARKS] xe−x dx . 1 Information for Students in MATH 141 2010 01 5165 7. Showing all your work, determine, for each of the following series, whether or not it converges: (a) [5 MARKS] ∞ X n=2 (b) [5 MARKS] ∞ X 1 ; n(ln n)2 ! n2 − 1 ; n2 + 1 n (−1) n=1 (c) [5 MARKS] ∞ X n+1 n=1 3n . F.10 Final Examination in Mathematics 189-141B (2001/2002) 1. Showing all your work, evaluate each of the following indefinite integrals: Z x3 (a) [3 MARKS] dx √ 4 − x2 Z 1 (b) [3 MARKS] dy p y ln y Z sec u (c) [3 MARKS] · tan u du 1 + sec u Z et (d) [3 MARKS] dt 1 + e2t 2. Let K denote the curve y = x2 , (0 ≤ x ≤ 1) . (a) [6 MARKS] Determine the area of the surface of revolution generated by revolving K about the y-axis. (b) [6 MARKS] Determine the volume of the solid of revolution formed by revolving about the line y = 0 the region bounded by K and the lines x = 1 and y = 0. 3. Consider the arc C given by r = θ2 (0 ≤ θ ≤ π). (a) [4 MARKS] Express the length of C as a definite integral. Then evaluate the integral. (b) [4 MARKS] Determine the area of the region subtended by C at the pole — i.e. of the region bounded by the arc C and the line θ = 0. Information for Students in MATH 141 2010 01 5166 (c) [4 MARKS] The given curve can be represented in cartesian coordinates parametrically as x = θ2 cos θ, y = θ2 sin θ. Determine the slope of the tangent to this curve 2 at the point (x, y) = 0, π2 . 4. [12 MARKS] Showing all your work, evaluate the integral Z 40 − 16x2 dx . 1 − 4x2 (1 + 2x) 5. [12 MARKS] Showing all your work, determine the area of the region bounded by the curves y = arctan x and 4y = π x in the first quadrant. 6. (a) [4 MARKS] Showing all your work, determine the value of Z sin3 x cos2 x dx . (b) [4 MARKS] Showing all your work, determine the value of Z tan4 x dx . π Z2 tan4 x dx . (c) [4 MARKS] Investigate the convergence of the integral 0 7. [12 MARKS] Showing all your work, determine the value of Z x Z e2t √ d2 dt u + 1 du dx2 0 1 when x = 0. 8. Showing all your work, determine, for each of the following infinite series, whether it is absolutely convergent, conditionally convergent, or divergent. (a) [4 MARKS] ∞ X n=5 (−1)n n2 − 1 . 6n2 + 4 ∞ X (−1)n (b) [4 MARKS] . n(ln n)2 n=2 (c) [4 MARKS] (d) [4 MARKS] n(n+1) ∞ X (−1) 2 n=2 ∞ X n=0 2n n+5 . 2n . Information for Students in MATH 141 2010 01 5167 F.11 Supplemental/Deferred Examination in Mathematics 189-141B (2001/2002) 1. Showing all your work, evaluate each of the following, always simplifying your answer as much as possible: Z (a) [3 MARKS] e x sin x dx Z 12 Z 12 sin−1 y arcsin y (b) [3 MARKS] dy equivalently, dy . p p 0 0 1 − y2 1 − y2 Z (c) [3 MARKS] (u2 + 2u)e−u du Z 1 + cos t (d) [3 MARKS] dt sin t 2. Let K denote the curve √ y = 2x − x2 , ! 1 0≤x≤ . 2 (a) [6 MARKS] Showing all your work, use an integral to determine the area of the surface of revolution generated by revolving K about the x-axis. (b) [6 MARKS] Determine the volume of the solid of revolution formed by revolving √ 3 about the line y = 2 the region bounded by K and the lines x = 0 and √ y = 23 . (You may assume that Z √ 2x − x2 dx = x−1√ 1 2x − x2 + arccos(1 − x) .) 2 2 3. A curve C in the plane is given by parametric equations x = t3 − 3t2 y = t3 − 3t . (a) [6 MARKS] Showing all your work, determine all points (x, y) on C where the tangent is horizontal. d2 y (b) [6 MARKS] By determining the value of as a function of t, determine all dx2 points (x, y) on C at which the ordinate (y-coordinate) is a (local) maximum, and all points at which the ordinate is a (local) minimum. Information for Students in MATH 141 2010 01 5168 4. [12 MARKS] Showing all your work, evaluate the indefinite integral Z 4x3 dx . x2 − 9 (3x + 9) 5. [12 MARKS] Showing all your work, determine the area of the region bounded by the curves x − 2y + 7 = 0 and y2 − 6y − x = 0 . 6. (a) [6 MARKS] Showing all your work, evaluate Z sin4 x cos2 x dx . (b) [4 MARKS] Showing all your work, evaluate Z tan5 x dx . π Z2 tan5 θ dθ . (c) [4 MARKS] Investigate the convergence of the integral π 4 7. [12 MARKS] Showing all your work, determine the value of ! Z x Z π3 p d2 4 + sin(−2u) du dt dx2 0 −2t when x = π4 . Your answer should be simplified, if possible. 8. Showing all your work, determine, for each of the following infinite series, whether it is absolutely convergent, conditionally convergent, or divergent. (a) [4 MARKS] ∞ X 1 (−1)n √ . n+1 n=5 ∞ X (−1)2n (b) [4 MARKS] . n(ln n)3 n=2 !3n ∞ X 2n (c) [4 MARKS] . 1 + 5n n=2 ∞ sin 1 X 1 n · . (d) [4 MARKS] 1 n n=1 cos n Information for Students in MATH 141 2010 01 5169 9. [10 MARKS] Prove or disprove the following statement: The point with polar coordinates √ r = 2( 2 − 1) √ θ = −π + arcsin(( 2 − 1)2 ) lies on the intersection of the curves with polar equations r2 = 4 sin θ, r = 1 + sin θ . You are expected to justify every statement you make, but you do not need to sketch the curves. F.12 Final Examination in MATH 141 2003 01 1. [10 MARKS] Find the area of the region bounded in the first quadrant by the curves y = ex , y = e−x , y = e2x−3 . Simplify your answer as much as possible. (Your instructors are aware that you do not have the use of a calculator.) 2. Showing all your work, evaluate each of the following indefinite integrals: Z 1 (a) [5 MARKS] dx 2 x + 2x + 17 Z ln(ln x) (b) [5 MARKS] dx x 3. [12 MARKS] For each of the following integrals, (a) [2 MARKS] Explain why the integral is improper. (b) [10 MARKS] Determine its value, or show that the integral does not converge. Show all your work. Z ∞ 2(x2 − x + 1) I1 = dx , 2 2 (x − 1)(x + 1) Z 1 I2 = 0 2(x2 − x + 1) dx (x − 1)(x2 + 1) 4. Let Ω denote the region in the first quadrant bounded by the curves x = x = 0, and y = 3. p 16 + y2 , y = 0, Information for Students in MATH 141 2010 01 5170 (a) [5 MARKS] Showing all your work, determine the volume of the solid of revolution obtained by rotating Ω about the y-axis. (b) [7 MARKS] Showing all your work, determine the area of the surface of revolution obtained by rotating the arc p x = 16 + y2 , (0 ≤ y ≤ 3) about the y-axis. You may assume that d (sec θ tan θ + ln |sec θ + tan θ|) = 2 sec3 θ . dθ . 5. Consider the arc C given parametrically by Z tp 4(1 − cos θ)θ2 dθ x = 0 y = cos t + t sin t (−π ≤ t ≤ 2π) . Showing all your work (a) [4 MARKS] Find the slope of the tangent to C at the point with parameter value 3π t= . 2 (b) [6 MARKS] Find the length of C. 6. Give, for each of the following statements, a specific example to show that the statement is not a theorem: (a) [3 MARKS] If {an }∞ n=0 is a sequence such that lim an = 0, then n→∞ (b) [3 MARKS] If the series n=0 divergent. (c) [3 MARKS] If a series ∞ P ∞ P n=0 an and ∞ P n=0 ∞ P n=0 an converges. bn are both divergent, then an converges, then ∞ P n=0 ∞ P (an + bn ) is n=0 a2n converges. 7. Showing all your work, determine, for each of the following infinite series, whether it is absolutely convergent, conditionally convergent, or divergent. ∞ X (−1)n (a) [4 MARKS] . 4n2 + 1 n=0 Information for Students in MATH 141 2010 01 5171 !n2 ∞ X n−1 (b) [4 MARKS] n n=2 (c) [4 MARKS] ∞ X n=2 1 . √ n(n + 1) 8. [10 MARKS] Showing all your work, determine the area of the part of one “leaf” of the “4-leafed rose” r = 2 cos(2θ) that is inside the circle r = 1. F.13 Supplemental/Deferred Examination in MATH 141 2003 01 1. [10 MARKS] Find the area of the region bounded by the curves y = e x − 1, y = x2 − x, x = 1. 2. Showing all your work, evaluate each of the following: Z 1 ex (a) [5 MARKS] dx x2 Z 5 (b) [5 MARKS] |x2 − 4x| dx 2 3. [12 MARKS] For each of the following integrals, (a) [2 MARKS] Explain precisely whether the integral is improper. (b) [10 MARKS] Determine its value, simplifying as much as possible; or show that the integral does not converge. (The examiners are aware that you do not have access to a calculator.) Show all your work. Z 1 2x I1 = dx , 2 0 (x − 1)(x + 1) Z 3 I2 = 2 2x dx (x − 1)(x2 + 1) 4. Let Ω denote the region bounded by the curves y = sin x, y = 0, x = π2 , x = π. (a) [6 MARKS] Showing all your work, determine the volume of the solid of revolution obtained by rotating Ω about the y-axis. (b) [6 MARKS] Showing all your work, determine the volume of the solid of revolution obtained by rotating Ω about the x-axis. Information for Students in MATH 141 2010 01 5172 5. Consider the arc C given parametrically by 2 x = 2t(t − 3) y = 6t(−t) (−1 ≤ t ≤ 1) . Showing all your work (a) [4 MARKS] Find the slope of the tangent to C at the point with parameter value t = − 12 . (b) [6 MARKS] Find the area of the surface obtained by rotating the curve about the x-axis. 6. Give, for each of the following statements, a specific example to show that the statement is not a theorem: (a) [3 MARKS] If {bn }∞ n=0 is a sequence such that lim bn = 1, then n→∞ (b) [3 MARKS] If a series ∞ P n=0 (c) [3 MARKS] If the series b2n converges, then ∞ P n=0 vergent. an and ∞ P n=0 ∞ P n=0 ∞ P n=0 bn converges. bn converges. bn are both divergent, then ∞ P (an bn ) is di- n=0 7. Showing all your work, determine, for each of the following infinite series, whether it is absolutely convergent, conditionally convergent, or divergent. (a) [4 MARKS] ∞ X (−1)n 1 n=3 ∞ X (b) [4 MARKS] n=0 (c) [4 MARKS] en . 1 (n + 5)(n + 6) ! ∞ X 2 (−1)n−1 √ . n − 1 n=2 8. [10 MARKS]Showing all your work, find the area of the region that lies inside the curve r = 2 − 2 sin θ and outside the curve r = 3 . F.14 Final Examination in MATH 141 2004 01 One of several versions Information for Students in MATH 141 2010 01 5173 1. BRIEF SOLUTIONS [3 MARKS EACH] Give the numeric value of each of the following limits, sums, integrals. If it does not exist write “DIVERGENT”. (a) ∞ X 1 + 2n−1 3n n=1 = ANSWER ONLY Z ∞ (b) 2 xe x dx = −∞ ANSWER ONLY (c) The limit of the Riemann sum lim n P 2 n→∞ i=1 n 3+ 2i 2 n 2i 5 −6 3+ n = ANSWER ONLY Z −∞ (d) 2 xe−x dx = ∞ ANSWER ONLY (e) ∞ X n=3 4 = (2n + 1)(2n + 3) ANSWER ONLY 2. BRIEF SOLUTIONS [3 MARKS EACH] Simplify your answers as much as possible. Information for Students in MATH 141 2010 01 5174 (a) For the point with polar coordinates 3, π7 give another set of polar coordinates (r, θ) in which r < 0 and θ > 2. ANSWER ONLY (b) Determine the length of the arc of the curve r = θ2 from (0, 0) to (1, 1). ANSWER ONLY Rt R4 2 2 (c) A curve is given parametrically by x(t) = 0 e−u du, y(t) = t eu du. Find the slope of the tangent to the curve at (x(1), y(1)). ANSWER ONLY (d) Give a definite integral whose value is the area of the surface obtained by rotating y3 1 1 ≤ y ≤ 1 about the y-axis. You need not evaluate the the curve x = + 2 6 2y integral. ANSWER ONLY (e) On the interval 0 ≤ x ≤ 4 the average value of the function ( √ 1 − x if 0 ≤ x ≤ 1 f (x) = is x−2 if 1 < x ≤ 4 Information for Students in MATH 141 2010 01 ANSWER ONLY 3. BRIEF SOLUTIONS [3 MARKS EACH] Give the value of each of the following indefinite integrals: Z x (a) dx = 2 3x + 1 ANSWER ONLY Z (b) √ e x 1 + e x dx = ANSWER ONLY Z (sin2 x − 3 cos2 x) dx = (c) ANSWER ONLY Z (d) tan2 3x dx = ANSWER ONLY 5175 Information for Students in MATH 141 2010 01 5176 Z (e) sec3 x tan3 x dx = ANSWER ONLY 4. SHOW ALL YOUR WORK! Let R be the finite region bounded by the curves x = y2 and x = 4 − 3y4 . (a) [5 MARKS] Find the area of R. (b) [5 MARKS] Find the volume of the solid generated by revolving R about the y-axis. 5. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the definite integral Z 12 4x 2x2 − x − 2 − 21 (x2 + 1)(x2 − 1) dx . 6. SHOW ALL YOUR WORK! (a) [4 MARKS] Show that, for any positive integer n, Z Z 2n 2n (ln x) dx = x(ln x) − 2n (ln x)2n−1 dx (b) [7 MARKS] Evaluate the integral Z 1 p 0 y 2y − y2 dy . 7. SHOW ALL YOUR WORK! [4 MARKS EACH] Determine for each of the following series whether it • converges absolutely; • converges conditionally; or • diverges. Information for Students in MATH 141 2010 01 5177 !2 1 −n (a) 1+ e n n=1 √ ∞ X (−1)n 2n (b) √ 1+2 n n=10 √ √ ∞ X n + 2 − n−1 (−1)n · (c) n n=2 ∞ X (d) ∞ X 2π + cos n 6n n=0 8. SHOW ALL YOUR WORK! [6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ. Another version 1. BRIEF SOLUTIONS [3 MARKS EACH] Give the numeric value of each of the following limits, sums, integrals. If it does not exist write “DIVERGENT”. (a) ∞ X 1 + 3n−1 n=1 4n = ANSWER ONLY Z ∞ (b) 2 yey dy = −∞ ANSWER ONLY (c) The limit of the Riemann sum lim n P 2 n→∞ i=1 n ANSWER ONLY 4+ 2i 2 n 5 − 7 4 + 2in = Information for Students in MATH 141 2010 01 Z −∞ (d) 5178 2 ye−y dy = ∞ ANSWER ONLY (e) ∞ X n=3 4 = (2n − 1)(2n + 1) ANSWER ONLY 2. BRIEF SOLUTIONS [3 MARKS EACH] Simplify your answers as much as possible. (a) On the interval 0 ≤ x ≤ 4 the average value of the function ( √ 1 − x if 0 ≤ x ≤ 1 f (x) = is x−2 if 1 < x ≤ 4 ANSWER ONLY (b) For the point with polar coordinates 3, π5 give another set of polar coordinates (r, θ) in which r < 0 and θ > 2. ANSWER ONLY (c) Determine the length of the arc of the curve r = θ2 from (0, 0) to (1, 1). ANSWER ONLY Information for Students in MATH 141 2010 01 5179 Rt R4 2 2 (d) A curve is given parametrically by x(t) = 0 e−v dv, y(t) = t ev dv. Find the slope of the tangent to the curve at (x(1), y(1)). ANSWER ONLY (e) Give a definite integral whose value is the area of the surface obtained by rotating y3 1 1 the curve x = ≤ y ≤ 1 about the y-axis. You need not evaluate the + 6 2y 2 integral. ANSWER ONLY 3. BRIEF SOLUTIONS [3 MARKS EACH] Give the value of each of the following indefinite integrals: Z (a) sec3 x tan3 x dx = ANSWER ONLY Z (b) x dx = 5x2 + 1 ANSWER ONLY Z (c) √ e x 1 + e x dx = Information for Students in MATH 141 2010 01 5180 ANSWER ONLY Z (3 sin2 x − cos2 x) dx = (d) ANSWER ONLY Z (e) tan2 4x dx = ANSWER ONLY 4. SHOW ALL YOUR WORK! Let S be the finite region bounded by the curves y = x2 and y = 4 − 3x4 . (a) [5 MARKS] Find the area of S . (b) [5 MARKS] Find the volume of the solid generated by revolving S about the xaxis. 5. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the definite integral Z 12 4x 4x2 − 2x − 4 − 12 (x2 + 1)(x2 − 1) dx . 6. SHOW ALL YOUR WORK! (a) [4 MARKS] Show that, for any positive integer m, Z Z 2m 2m (ln y) dy = y(ln y) − 2m (ln y)2m−1 dy Information for Students in MATH 141 2010 01 (b) [7 MARKS] Evaluate the integral Z 5181 1 √ 0 x 2x − x2 dx . 7. SHOW ALL YOUR WORK! [4 MARKS EACH] Determine for each of the following series whether it • converges absolutely; • converges conditionally; or • diverges. √ ∞ X (−1)n 2n (a) √ 1+2 n n=10 !2 ∞ X 1 −n (b) 1+ e n n=1 (c) ∞ X (cos n) − 2π 4n n=0 √ √ n+2− n−1 (−1) · (d) n n=2 ∞ X n 8. SHOW ALL YOUR WORK! [6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ. F.15 Supplemental/Deferred Examination in MATH 141 2004 01 1. BRIEF SOLUTIONS [3 MARKS EACH] Give the numeric value of each of the following limits, sums, integrals. If it does not exist write “DIVERGENT”. (a) ∞ X en − e−n n=1 3n = ANSWER ONLY Information for Students in MATH 141 2010 01 Z 0 5182 xe x dx = (b) −∞ ANSWER ONLY π X 2 iπ sin = (c) The limit of the Riemann sum lim n→∞ n n i=1 n ANSWER ONLY (d) ∞ X n=0 4 = (2n + 1)(2n + 3) ANSWER ONLY 2. BRIEF SOLUTIONS [3 MARKS EACH] Simplify your answers as much as possible. π (a) For the point with polar coordinates (r, θ) = − 10π , − give another set of polar 3 6 coordinates (r1 , θ1 ) in which r1 > 0 and θ1 > 2 . ANSWER ONLY (b) Find all points r = 1 − cos θ and — if r = − 32 there are intersect. any — where the curves Information for Students in MATH 141 2010 01 5183 ANSWER ONLY (c) Find the exact length of the curve r = e2θ , (0 ≤ θ ≤ π) . ANSWER ONLY (d) On the interval ln 12 ≤ x ≤ π the average value of the function ( sinh x if ln 12 ≤ x ≤ 0 f (x) = is sin x if 0<x≤π ANSWER ONLY 3. BRIEF SOLUTIONS [3 MARKS EACH] Give the value of each of the following indefinite integrals: Z 1 (a) dx = 4x2 + 1 ANSWER ONLY Information for Students in MATH 141 2010 01 Z (b) √ dx x2 − 25 5184 = ANSWER ONLY Z (cos x + 1)(cos x − 2) dx = (c) ANSWER ONLY Z (d) tan 3x dx = ANSWER ONLY 4. SHOW ALL YOUR WORK! 3 1 p 2 Let C be the arc x = y +2 , 3 √ (− 2 ≤ y ≤ 0) . (a) [6 MARKS] Find the area of the surface obtained by revolving C about the x-axis. (b) [6 MARKS] Find the volume of the solid generated by revolving√about the y-axis the region bounded by C, the coordinate axes, and the line y = − 2. 5. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x3 − 8x − 1 dx . (x2 − 1)(x + 1) Information for Students in MATH 141 2010 01 5185 6. SHOW ALL YOUR WORK! Simplify your answers as much as possible. (a) [6 MARKS] Evaluate the indefinite integral Z √ 1 − 9x2 dx . √ − 3 Z arctan x dx . (b) [6 MARKS] Evaluate the definite integral 0 7. SHOW ALL YOUR WORK! [4 MARKS EACH] Determine for each of the following series whether it • converges absolutely; • converges conditionally; or • diverges. √ n (−1) (a) ln(n2 ) n=2 ∞ X (b) n ln ∞ X 1 · 3 · 5 · . . . · (2n − 1) 3n n! n=1 (c) (d) ∞ X sin 2n 1 + 2n n=1 ∞ X n=1 (−1) n √ n 3n − 1 8. SHOW ALL YOUR WORK! [12 MARKS] Use polar coordinates — no other method will be accepted — to find the 1 area of the region bounded by the curve r = 2 and the line r = , and containing the cos θ pole. F.16 Final Examination in MATH 141 2005 01 1. SHOW ALL YOUR WORK! Z 3 (a) [4 MARKS] Evaluate |x − 1| dx . 0 Information for Students in MATH 141 2010 01 d (b) [3 MARKS] Evaluate dx Z Z 5 5186 √ 4 + t2 dt . x 2 x d (c) [3 MARKS] Evaluate sec t dt . dx 2π Z √3 (d) [3 MARKS] Evaluate x5 x3 + 1 dx . SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X 2 − cos n n=2 (b) [4 MARKS] ∞ X n(−3)n n=0 (c) [4 MARKS] n ∞ X n=2 4n 1 n ln n 3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral or a sum, product, or quotient of several definite integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems: (a) [6 MARKS] The area of the region bounded by the parabola y = x2 , the x-axis, and the tangent to the parabola at the point (1, 1). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Information for Students in MATH 141 2010 01 5187 (b) [3 MARKS] The volume of the solid obtained p by rotating about the line y = 4 the region bounded by x = 0 and the curve x = sin y (0 ≤ y ≤ π). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (c) [3 MARKS] The area of the surface obtained by revolving about the y-axis the curve y = e x , 1 ≤ y ≤ 2. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (d) [2 MARKS] The average value of the function 2. 2x over the interval 0 ≤ x ≤ (1 + x2 )2 DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z 2x3 + 3x2 + 3 dx . x2 + x − 12 5. SHOW ALL YOUR WORK! (a) [9 MARKS] Use m ≥ 2, Z integration by parts to prove that, for Z 1 m−1 m m−1 cos x dx = cos x · sin x + cosm−2 x dx m m integers Information for Students in MATH 141 2010 01 5188 (b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate Z π2 cos6 x dx. 0 6. SHOW ALL YOUR WORK! Consider the curve C defined by x = 2 cos t − cos 2t y = 2 sin t − sin 2t . (a) [8 MARKS] Determine the points where the arc of the curve given by π 7π ≤t≤ 4 4 has a vertical tangent. (b) [4 MARKS] Determine the length of the arc of the curve given by 0 ≤ t ≤ 2π . 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Determine whether the following integral is convergent; if it is convergent, determine its value: Z 1 dx √ −1 1 − x2 (b) [5 MARKS] Determine whether the following series is conditionally convergent, absolutely convergent, or divergent. ∞ X n=1 (−1)n n! nn (c) [3 MARKS] Determine whether the sequence an = ln(n + 1) − ln n is convergent; if it is convergent, carefully determine its limit. 8. SHOW ALL YOUR WORK! [12 MARKS] Find the area of the region bounded by the curves r = 4 + 4 sin θ r sin θ = 3 which does not contain the pole. Information for Students in MATH 141 2010 01 5189 F.17 Supplemental/Deferred Examination in MATH 141 2005 01 Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. Calculators are not permitted. 4. This examination booklet consists of this cover, Pages 1 through 7 containing questions; and Pages 8, 9, and 10, which are blank. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 85 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Z 0 (a) [4 MARKS] Evaluate |x2 − 1| dx . −2 d (b) [3 MARKS] Evaluate dx Z Z 4 2 et dt . x 2 x d dt . (c) [3 MARKS] Evaluate dx 1 1 + t5 Z (d) [3 MARKS] Evaluate x sin(x2 ) dx . SHOW ALL YOUR WORK! Information for Students in MATH 141 2010 01 5190 2. For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X n=2 2n 3n − 1 !n ∞ X (−1)n (b) [4 MARKS] √ n−1 n=2 ! ∞ X (−4)n (c) [4 MARKS] 3n + 2n n=2 3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral – possibly improper — or a sum, product, or quotient of several such integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems: (a) [6 MARKS] The area of the infinite region containing the point 0, 21 bounded by the curve y = e x , the x-axis, and the tangent to the curve at the point (1, e). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [3 MARKS] The volume generated by rotating the region bounded by the curves √ y = x − 1, y = 0, x = 5 about the line y = 3. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Information for Students in MATH 141 2010 01 5191 (c) [3 MARKS] The area of the surface obtained by revolving about the x-axis the curve x = ln y, 1 ≤ x ≤ 3. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z 3 x − x2 − 2x − 2 dx . x(x2 + x + 1) 5. SHOW ALL YOUR WORK! (a) [9 MARKS] Use integration by parts to prove that, for m , 1, Z Z 1 m−2 m m−2 sec x dx = sec x · tan x + secm−2 x dx m−1 m−1 integers (b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate Z π3 sec3 x dx. 0 6. SHOW ALL YOUR WORK! The curve C has equations x = t3 + 4t, y = 6t2 . (a) [8 MARKS] Determine the points on C where the tangent is parallel to the line with equations x = −7t, y = 12t − 5. (b) [4 MARKS] Determine a definite integral whose value is the length of the arc of C between the points with parameter values t = 1 and t = 2. YOU ARE NOT EXPECTED TO EVALUATE THE INTEGRAL! 7. SHOW ALL YOUR WORK! [12 MARKS] Find the area of the region between the inner loop and the outer loop of the curve r = 1 − 2 cos θ. Information for Students in MATH 141 2010 01 5192 F.18 Final Examination in MATH 141 2006 01 (One version) Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this test may be different from yours. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page. Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Z 2 (a) [4 MARKS] Evaluate |x| dx . −1 Ze3 (b) [3 MARKS] Evaluate 1 dt . √ t 1 + ln t Information for Students in MATH 141 2010 01 5193 Z x2 d 2 (c) [3 MARKS] Evaluate et dt . dx 0 (d) [4 MARKS] Evaluate !7 !7 !7 !7 1 0 1 2 n − 1 . lim + + + ... + n→∞ n n n n n SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • Name the test(s) that you are using. • Explain why the test(s) you have chosen is/are applicable to the given series. • Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X n=1 (b) [4 MARKS] ∞ X 1 (tanh n)2 + 1 n2n e−n 2 n=1 (c) [4 MARKS] ∞ X n2 − 85n + 12 n=1 n(n + 6)2 3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral or a sum, product, or quotient of several definite integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems, and should be simplified as much as possible, except that you are not expected to evaluate the integrals. (a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the region bounded by the parabola y2 = 2x + 6 and the line y = x − 1. An answer involving integration along the y-axis will not be accepted. (b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 the region bounded by the curves y = x3 and y = x2 . For this question you are to use only the method of “washers”. Information for Students in MATH 141 2010 01 5194 (c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 the region bounded by the curves y = x3 and y = x2 . For this question you are to use only the method of “cylindrical shells”. (d) [3 MARKS] The length of the curve whose equation is x2 y2 + = 1. 4 9 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z 5 x +x dx . x4 − 16 5. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Z (a) [4 MARKS] cos x · cosh x dx (b) [5 MARKS] Z1 √ x2 + 2x + 5 dx −3 Z (c) [4 MARKS] sin2 x · cos2 x dx 6. SHOW ALL YOUR WORK! Consider the curve C defined by x = x(t) = 10 − 3t2 y = y(t) = t3 − 3t , where −∞ < t < +∞. d2 y (a) [8 MARKS] Determine the value of 2 at the points where the tangent is horizondx tal. (b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of the arc o n √ (x(t), y(t)) : − 3 ≤ t ≤ 0 . Information for Students in MATH 141 2010 01 5195 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z 0 dx . 2 −1 x 3 (b) [5 MARKS] Determine whether the following series is conditionally convergent, absolutely convergent, or divergent. ∞ X (−1)n n − ln n n=1 (c) [3 MARKS] Give an example of a sequence {an } with the property that lim an = 0 n→∞ ∞ X an = +∞. You are expected to give a formula for the general term an of but n=1 your sequence. 8. SHOW ALL YOUR WORK! [12 MARKS] The arc r = 1 − cos θ (0 ≤ θ ≤ π) divides the area bounded by the curve r = 1 + sin θ (0 ≤ θ ≤ 2π) into two parts. Showing all your work, carefully find the area of the part that contains the point (r, θ) = 12 , π2 . F.19 Supplemental/Deferred Examination in MATH 141 2006 01 Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. Information for Students in MATH 141 2010 01 5196 4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10, and 11, which are blank. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Simplify your answers as much as possible. Z ee dt (a) [4 MARKS] Evaluate . e2 t ln t Z + π2 (b) [4 MARKS] Evaluate | cosh x| dx . − π2 (c) [4 MARKS] Evaluate !2 !2 !2 !2 1 1 2 n − 1 0 . lim 4 + + 4+ + 4+ + ... + 4 + n→∞ n n n n n d (d) [3 MARKS] Evaluate dx Z −3x √ 1 + t2 dt when x = 1. 0 SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests for convergence or divergence and determine whether the series is convergent. In each case you must answer 3 questions: • [1 MARK] Name the test(s) that you are using. Information for Students in MATH 141 2010 01 5197 • [1 MARK] Explain why the test(s) you have chosen is/are applicable to the given series. • [2 MARKS] Use the test(s) to conclude whether or not the series is convergent. (a) [4 MARKS] ∞ X n=1 (b) [4 MARKS] ∞ X n=1 (c) [4 MARKS] n n2 + 4 n (−1)n ln 3n + 2 ∞ X 3n + 6 n=1 5n 3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral or a sum, product, or quotient of several definite integrals, but do not evaluate the integral(s). It is not enough to quote a general formula: your integrals must have integrand and limits specific to the given problems, and should be simplified as much as possible, except that you are not expected to evaluate the integrals. (a) [4 MARKS] The length of the curve whose equation is x = 1 + et , y = t2 , (−3 ≤ t ≤ 3). (b) [4 MARKS] The volume of the solid obtained by rotating about the x-axis the region bounded by the curves y = x and y = x2 . For this question you are expected to use only the method of “cylindrical shells”. (c) [4 MARKS] The volume of the solid obtained by rotating about the line y = 2 the region bounded by the curves y = x and y = x2 . For this question you are expected to use only the method of “washers”. 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x3 dx . (x2 + 4)(x − 2) 5. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following. Simplify your answers as much as possible. Information for Students in MATH 141 2010 01 5198 Z (a) [4 MARKS] 8x cos 2x dx √ 2− Z 2 (b) [4 MARKS] 0 1 dx √ 4x − x2 Z (c) [4 MARKS] e x sin x dx 6. SHOW ALL YOUR WORK! Consider the arc C defined by x = x(t) = 3t − t3 y = y(t) = 3t2 , where 0 ≤ t ≤ 1. d2 y 1 at the point with parameter value t = . 2 dx 2 (b) [6 MARKS] Determine the area of the surface of revolution of C about the x-axis. (a) [6 MARKS] Determine the value of 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z 6 x xe 3 dx . −∞ (b) [4 MARKS] Give an example of a series which is convergent but not absolutely convergent. Justify all of your statements. (c) [4 MARKS] Give an example of 2 divergent sequences {an }, {bn } with the property that the sequence {an bn } is convergent. You are expected to give formulas for the general terms an , bn of both of your sequences. 8. SHOW ALL YOUR WORK! [12 MARKS] The curves r = 2 cos 2θ and r = 2 sin θ define a number of regions in the plane. Let R denote the region containing the point (r, θ) = (1, 0), bounded by arcs of both of the curves. Showing all your work, carefully find the area of R. Information for Students in MATH 141 2010 01 5199 F.20 Final Examination in MATH 141 2007 01 (One version) Instructions 1. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 2. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 3. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this test may be different from yours. 4. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 5. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. Z 2 (a) [2 MARKS] Evaluate |x|2 dx . −1 Z0 (b) [2 MARKS] Evaluate 1 t4 dt . √ t5 + 1 (c) [3 MARKS] Determine the value of !3 !3 !3 !3 1 0 1 2 n − 1 . + + + ... + n n n n n Information for Students in MATH 141 2010 01 5200 (d) [3 MARKS] Suppose it is known that f 0 (x) = 4 cosh x for all x. Showing all your work, determine the value of f (1)− f (−1), expressed in terms of the values of either exponentials or hyperbolic functions. Z x2 d t (e) [4 MARKS] Evaluate et dt when x = 1 . dx 12 SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests for convergence or divergence to determine whether the series is absolutely convergent, conditionally convergent, or divergent. All tests used must be named, and all statements must be carefully justified. (a) [4 MARKS] ∞ X (−n − 2)n (n − 2)n n=1 (b) [4 MARKS] ∞ X n=1 (c) [4 MARKS] ∞ X n=1 (2n2 + 1)n (−1)n+1 n! n2 2n (−1)n sin 1 n 3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum, product, or quotient of several definite integrals, simplified as much as possible; you are not expected to evaluate the integrals. R is defined to be the region enclosed by the curves x + y = 6 and y = x2 ; C is the arc y = 3 x (−1 ≤ x ≤ 2). (a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sum of integrals whose value is the volume of the resulting solid. Information for Students in MATH 141 2010 01 5201 DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (c) [3 MARKS] Express in terms of integrals — which you need not evaluate — the average length that R cuts off from the vertical lines which it meets. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (e) [3 MARKS] Given an integral whose value is the area of the surface generated by rotating C about the line y = −1; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x x2 − 4 (x − 2) + 4 dx . x2 − 4 (x − 2) 5. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Information for Students in MATH 141 2010 01 5202 Z e−x · cos x dx (a) [4 MARKS] 5 Z2 (b) [5 MARKS] √ − 21 Z (c) [4 MARKS] x 8 + 2x − x2 dx ! 1 cos x + · tan2 x dx cos2 x 2 6. SHOW ALL YOUR WORK! Consider the arc C defined by x = x(t) = cos t + t sin t y = y(t) = sin t − t cos t , where 0 ≤ t ≤ π2 . d2 y . dx2 (b) [6 MARKS] Determine the area of the surface generated by revolving C about the y-axis. (a) [6 MARKS] Determine as a function of t the value of 7. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z π sec x dx . π 2 (b) [5 MARKS] Showing all your work, carefully determine whether the series ∞ X n=3 4 n ln n is convergent. (c) [3 MARKS] Showing all your work, determine whether the following sequence converges; if it converges, find its limit: a1 a2 a3 a4 a5 a6 = = = = = = 1. 1.23 1.2345 1.234545 1.23454545 1.2345454545 Information for Students in MATH 141 2010 01 5203 etc., where each term after a2 is obtained from its predecessor by the addition on the right of the decimal digits 45. 8. SHOW ALL YOUR WORK! [10 MARKS] The polar curves r = 2 + 2 sin θ and r = 6 − 6 sin θ (0 ≤ θ ≤ 2π) (0 ≤ θ ≤ 2π) divide the plane into several regions. Showing all your work, carefully find the area of the region bounded by these curves which contains the point (r, θ) = (1, 0). F.21 Supplemental/Deferred Examination in MATH 141 2007 01 (One version) Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. The use of calculators is not permitted. This is a closed book examination. Use of regular and translation dictionaries is permitted. 4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may be different from yours. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. Information for Students in MATH 141 2010 01 5204 You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Z 0 1 dx . √ 2 − √1 1 − x 2 Z (b) [2 MARKS] Evaluate t3 cosh t4 dt . (a) [2 MARKS] Evaluate (c) [3 MARKS] Determine one antiderivative of x ln x. Z x 2 (d) [3 MARKS] Evaluate the integral tet dt. (e) [4 MARKS] Evaluate d dx Z −x 1 (ln | sec t + tan t|) dt when x = sin x π . 4 SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests to determine whether the series is convergent or divergent. All tests used must be named, and all statements must be carefully justified. ∞ X 1 (n + 2)(n − 2) n=3 ∞ ∞ X X −i 3 (b) [4 MARKS] (a) [4 MARKS] n=1 (c) [4 MARKS] ∞ X n=1 i=n n+1 n !n2 3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum, product, or quotient of several definite integrals, simplified as much as possible; you are not expected to evaluate the integrals. R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2 ; C is the arc x = t, y = e3t (−2 ≤ t ≤ 1). Information for Students in MATH 141 2010 01 5205 (a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (c) [3 MARKS] Let f (t) denote the vertical distance of the point t, e3t from the xaxis. Express in terms of integrals — which you need not evaluate — the average value of f (t) over the interval −2 ≤ t ≤ 1. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (e) [3 MARKS] Given an integral whose value is the area of the surface generated by rotating C about the line x = 1; you need not evaluate the integral. Information for Students in MATH 141 2010 01 5206 DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x3 x4 − 4x2 − 16 x4 − 4x2 dx . 5. SHOW ALL YOUR WORK! Z sin2 (3x) cos2 (3x) dx . (a) [4 MARKS] Showing all your work, evaluate Z (b) [4 MARKS] Showing all your work, evaluate (c) [5 MARKS] Assume that Z x f (x) = x2 1 dx . √ 9x2 − 16 sec100 t dt 0 Z x is known. Showing all your work, express the value of f (x). (You are not expected to determine f (x) explicitly.) sec102 t dt in terms of 0 6. SHOW ALL YOUR WORK! Consider the closed arc C defined by x = x(t) = 3t2 y = y(t) = t3 − 3t , √ √ where − 3 ≤ t ≤ 3. (a) [3 MARKS] Determine the area bounded by C. (b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter 1 value t = . 2 (c) [6 MARKS] Determine the area of the surface generated by revolving C about the y-axis. 7. SHOW ALL YOUR WORK! (a) [6 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z ∞ x dx . 2 −∞ x + 4 (b) [7 MARKS] Showing all your work, carefully determine whether the series √ ∞ X ln n n (−1) n n=3 is conditionally convergent, absolutely convergent, or divergent. 8. SHOW ALL YOUR WORK! [10 MARKS] Find the area inside the larger loop and outside the smaller loop of the limac¸on r = 2 sin θ − 1. F.22 Final Examination in MATH 141 2008 01 (one version) This examination was written during a labour disruption, when the services of Teaching Assistants were not available for grading purposes. The following additional instructions were distributed with the examination. VERSION n McGILL UNIVERSITY FACULTY OF SCIENCE FINAL EXAMINATION IMPORTANT ADDITIONAL INSTRUCTIONS MATHEMATICS 141 2008 01CALCULUS 2 EXAMINER: Professor W. G. Brown ASSOCIATE EXAMINER: Mr. S. Shahabi DATE: Monday, April 14th, 2008 TIME: 09:00 – 12:00 hours A. Part marks will not be awarded for any part of any question worth [4 MARKS] or less. Information for Students in MATH 141 2010 01 5208 B. To be awarded part marks on a part of a question whose maximum value is 5 marks or more, a student’s answer must be deemed to be more than 75% correct. C. While there are 100 marks available on this examination 80 MARKS CONSTITUTE A PERFECT PAPER. You may attempt as many problems as you wish. All other instructions remain valid. Where a problem requires that all work be shown, that remains the requirement; where a problem requires only that an answer be written in a box without work being graded, that also remains the requirement. Students are advised to spend time checking their work; for that purpose you could verify your answers by solving problems in more than one way. Remember that indefinite integrals can be checked by differentiation. W. G. Brown, Examiner. Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 4. This examination booklet consists of this cover, Pages 1 through 9 containing questions; and Pages 10, 11, and 12, which are blank. Your neighbour’s version of this test may be different from yours. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed; in some of these problems you are instructed to write the answer in a box, but a correct answer alone will not be sufficient unless it is substantiated by your work, clearly displayed outside the box. When space provided for that work is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. Information for Students in MATH 141 2010 01 5209 You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. Z 4 − 6x (a) [2 MARKS] Evaluate dx . 1 + x2 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) Z2 (b) [3 MARKS] Evaluate 0 p y2 y3 + 1 dy . Information for Students in MATH 141 2010 01 5210 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) Z (c) [3 MARKS] Evaluate sin(18 θ) cos(30 θ) dθ . ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 2. SHOW ALL YOUR WORK! d (a) [3 MARKS] Simplifying your answer as much as possible, evaluate dx Z √ 3 −x earcsin z dz . Information for Students in MATH 141 2010 01 5211 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) (b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for the function f (x) = 3 − x, where the sample points are the left end-point of each of n subintervals of equal length. ANSWER ONLY (c) [4 MARKS] Determine the value of the preceding Riemann sum as a function of n, simplifying your work as much as possible. (NOTE: You are being asked to determine the value of the sum as a function of n, not the limit as n → ∞.) Information for Students in MATH 141 2010 01 5212 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 3. SHOW ALL YOUR WORK! For each of the following series determine whether the series diverges, converges conditionally, or converges absolutely. All of your work must be justified; prior to using any test you are expected to demonstrate that the test is applicable to the problem. ! ∞ X 1 (a) [4 MARKS] √ n ln n n=3 √ ∞ X 4n + 5 (−1)n+1 (b) [4 MARKS] 3n + 10 n=1 ! !! ∞ X 1 −1 −1 1 (c) [4 MARKS] cot − cot n + 1 n n=1 4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the curves 2y = x, y = 2x, and x2 + y2 = 5. (a) [4 MARKS] The region R is rotated about the line x = −1. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies inside R. Express in terms of integrals — which you need not evaluate — the average of the positive lengths L(a). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) Information for Students in MATH 141 2010 01 5213 (c) [4 MARKS] Let C1 be the curve x(t) = t, y(t) = cosh t (0 ≤ t ≤ ln 2). Simplifying your answer as much as possible, find the length of C1 . ANSWER ONLY 5. SHOW ALL YOUR WORK! (a) [8 MARKS] Evaluate the indefinite integral Z 36 dx . (x + 4)(x − 2)2 Z∞ (b) [4 MARKS] Determine If it converges, find its value. whether 36 dx (x + 4)(x − 2)2 3 6. SHOW ALL YOUR WORK! Showing all your work, evaluate each of the following: Z p √ (a) [4 MARKS] e x dx ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) converges. Information for Students in MATH 141 2010 01 Z0 (b) [5 MARKS] √ − 21 x 3 − 4x − 4x2 dx ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) Z π (c) [4 MARKS] sin2 t cos4 t dt . 0 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 7. SHOW ALL YOUR WORK! 5214 Information for Students in MATH 141 2010 01 Consider the curve C2 defined by x = x(t) = 1 + e−t , 5215 y = y(t) = t + t2 . (a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis. ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) (b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent is horizontal. ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) Information for Students in MATH 141 2010 01 5216 (c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis. ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 8. SHOW ALL YOUR WORK! (a) [5 MARKS] Showing all your work, determine whether the series ∞ X √ √ √ n n+2− n−2 n=2 is convergent or divergent. (b) [5 MARKS] Showing all your work, determine whether the following sequence converges; if it converges, find its limit: a1 a2 a3 a4 a5 a6 = = = = = = 3. 3.14 3.1414 3.141414 3.14141414 3.1414141414 etc., where each term after a2 is obtained from its predecessor by the addition on the right of the decimal digits 14. Information for Students in MATH 141 2010 01 5217 9. SHOW ALL YOUR WORK! Curves C3 and C4 , respectively represented by polar equations r = 4 + 2 cos θ and r = 4 cos θ + 5 (0 ≤ θ ≤ 2π) (119) (0 ≤ θ ≤ 2π) , (120) divide the plane into several regions. (a) [8 MARKS] Showing all your work, carefully find the area of the one region which is bounded by C3 and C4 and contains the pole. (b) [4 MARKS] Find another equation — call it (120*) — that also represents C4 , and has the property that there do not exist coordinates (r, θ) which satisfy equations (119) and (120*) simultaneously. You are expected to show that equations (119) and (120*) have no simultaneous solutions. F.23 Supplemental/Deferred Examination in MATH 141 2008 01 (one version) Instructions 1. Fill in the above clearly. 2. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 3. The use of calculators is not permitted. This is a closed book examination. Use of regular and translation dictionaries is permitted. 4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may be different from yours. 5. There are two kinds of problems on this examination, each clearly marked as to its type. • Most of the questions on this paper require that you SHOW ALL YOUR WORK! Their solutions are to be written in the space provided on the page where the question is printed. When that space is exhausted, you may write on the facing page . Any solution may be continued on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! • Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you are expected to write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. Information for Students in MATH 141 2010 01 5218 You are expected to simplify your answers wherever possible. You are advised to spend the first few minutes scanning the problems. (Please inform the invigilator if you find that your booklet is defective.) 6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION. 1. SHOW ALL YOUR WORK! Z 0 1 dx . √ 2 − √1 1 − x 2 Z (b) [2 MARKS] Evaluate t3 cosh t4 dt . (a) [2 MARKS] Evaluate (c) [3 MARKS] Determine one antiderivative of x ln x. Z x 2 (d) [3 MARKS] Evaluate the integral tet dt. (e) [4 MARKS] Evaluate d dx Z −x 1 (ln | sec t + tan t|) dt when x = sin x π . 4 SHOW ALL YOUR WORK! 2. For each of the following series you are expected to apply one or more tests to determine whether the series is convergent or divergent. All tests used must be named, and all statements must be carefully justified. ∞ X 1 (n + 2)(n − 2) n=3 ∞ ∞ X X −i 3 (b) [4 MARKS] (a) [4 MARKS] n=1 (c) [4 MARKS] ∞ X n=1 i=n n+1 n !n2 3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum, product, or quotient of several definite integrals, simplified as much as possible; you are not expected to evaluate the integrals. R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2 ; C is the arc x = t, y = e3t (−2 ≤ t ≤ 1). Information for Students in MATH 141 2010 01 5219 (a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sum of integrals whose value is the volume of the resulting solid. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (c) [3 MARKS] Let f (t) denote the vertical distance of the point t, e3t from the xaxis. Express in terms of integrals — which you need not evaluate — the average value of f (t) over the interval −2 ≤ t ≤ 1. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluate the integral. DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (e) [3 MARKS] Given an integral whose value is the area of the surface generated by rotating C about the line x = 1; you need not evaluate the integral. Information for Students in MATH 141 2010 01 5220 DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) 4. SHOW ALL YOUR WORK! [12 MARKS] Evaluate the indefinite integral Z x3 x4 − 4x2 − 16 x4 − 4x2 dx . 5. SHOW ALL YOUR WORK! Z sin2 (3x) cos2 (3x) dx . (a) [4 MARKS] Showing all your work, evaluate Z (b) [4 MARKS] Showing all your work, evaluate (c) [5 MARKS] Assume that Z x f (x) = x2 1 dx . √ 9x2 − 16 sec100 t dt 0 Z x is known. Showing all your work, express the value of f (x). (You are not expected to determine f (x) explicitly.) sec102 t dt in terms of 0 6. SHOW ALL YOUR WORK! Consider the closed arc C defined by x = x(t) = 3t2 y = y(t) = t3 − 3t , √ √ where − 3 ≤ t ≤ 3. (a) [3 MARKS] Determine the area bounded by C. (b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter 1 value t = . 2 (c) [6 MARKS] Determine the area of the surface generated by revolving C about the y-axis. Information for Students in MATH 141 2010 01 5221 7. SHOW ALL YOUR WORK! (a) [6 MARKS] Showing detailed work, determine whether the following integral is convergent; if it is convergent, determine its value: Z ∞ x dx . 2 −∞ x + 4 (b) [7 MARKS] Showing all your work, carefully determine whether the series √ ∞ X ln n n (−1) n n=3 is conditionally convergent, absolutely convergent, or divergent. 8. SHOW ALL YOUR WORK! [10 MARKS] Find the area inside the larger loop and outside the smaller loop of the limac¸on r = 2 sin θ − 1. F.24 Final Examination in MATH 141 2009 01 (one version) Instructions 1. Do not tear pages from this book; all your writing — even rough work — must be handed in. You may do rough work for this paper anywhere in the booklet. 2. Calculators are not permitted. This is a closed book examination. Regular and translation dictionaries are permitted. 3. This examination booklet consists of this cover, Pages 1 through 8 containing questions; and Pages 9, 10 and 11, which are blank. A TOTAL OF 75 MARKS ARE AVAILABLE ON THIS EXAMINATION. 4. You are expected to simplify all answers wherever possible. • Most questions on this paper require that you SHOW ALL YOUR WORK! Solutions are to be begun on the page where the question is printed; a correct answer alone will not be sufficient unless substantiated by your work. You may continue your solution on the facing page , or on the last pages, or the back cover of the booklet, but you must indicate any continuation clearly on the page where the question is printed! To be awarded partial marks on a part of a question a student’s answer for that part must be deemed to be more than 50% correct. Most of these questions will require that the answer be written in a box provided on the page where the question is printed; even if you continue your work elsewhere, the answer should be in the box provided. Information for Students in MATH 141 2010 01 5222 • Some questions on this paper require only BRIEF SOLUTIONS ; for these you must write the correct answer in the box provided; you are not asked to show your work, and you should not expect partial marks for solutions that are not correct. Check your work! 1. SHOW ALL YOUR WORK! Z (a) [4 MARKS] Evaluate t3 cos t2 dt . ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) (b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivative Z t2 d tanh x2 dx . dt 0 Information for Students in MATH 141 2010 01 ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 2. SHOW ALL YOUR WORK! Your answers must be simplified as much as possible. 1 Z2 (a) [4 MARKS] Evaluate √1 2 dx . √ 1 − x2 · arcsin x ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) 5223 Information for Students in MATH 141 2010 01 Z (b) [4 MARKS] Evaluate p 2y y2 − y + 1 5224 dy ANSWER (SHOW YOUR WORK OUTSIDE THE BOX) SHOW ALL YOUR WORK! 3. For each of the following series determine whether the series diverges, converges conditionally, or converges absolutely. All of your work must be justified; prior to using any test you are expected to demonstrate that the test is applicable to the problem. (a) [4 MARKS] ∞ X n+1 (−1) n=1 ∞ X cos n n 2 1 (−1)n √ . n ln n n=2 ∞ X 1 (c) [4 MARKS] (−1)n ln (3n + 1) n n=4 (b) [4 MARKS] 4. [9 MARKS] SHOW ALL YOUR WORK! n rπ 1 X 2 (a) [3 MARKS] Evaluate lim · cos . (Hint: This could be a Riemann n→∞ n n r=1 sum.) (b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an = arctan(−2n) as n → ∞. UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 5225 ∞ n ∞ X X 1 (c) [3 MARKS] Showing all your work, prove divergence, or find the value of . 5i n=1 i=1 5. SHOW ALL YOUR WORK! (a) [8 MARKS] Evaluate the indefinite integral Z t+1 dt . 2 2t − t − 1 (b) [2 MARKS] Determine whether the following improper integral converges or diverges; if it converges, find its value: Z∞ t+1 dt . −t−1 2t2 4 6. SHOW ALL YOUR WORK! Consider the curve C1 defined by x = 3t2 , y = 2t3 , (t ≥ 0). (a) [7 MARKS] Showing all your work, determine the area of the surface generated when the arc 0 ≤ t ≤ 1 of C1 is rotated about the y-axis. (b) [2 MARKS] Showing all your work, determine all points — if any — where the normal to the curve is parallel to the line x + y = 8. 7. SHOW ALL YOUR WORK! Curves C3 and C4 , are respectively represented by polar equations r = 3 + 3 cos θ (0 ≤ θ ≤ 2π) and r = 9 cos θ (0 ≤ θ ≤ π) . (121) (122) (a) [7 MARKS] Showing all your work, carefully find the area of the region lying inside both of the curves. (b) [3 MARKS] Determine the length of the curves which form the boundary of the region whose area you have found. 8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the 8 curves y = 2 , x = y, x = 1. x Information for Students in MATH 141 2010 01 5226 (a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensional solid, S 1 . Give an integral or sum of integrals whose value is the volume of S 1 ; you are not asked to evaluate the integral(s). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensional solid, S 2 . Give an integral or sum of integrals whose value is the volume of S 2 obtained only by the Method of Cylindrical Shells; you are not asked to evaluate the integral(s). DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE) (c) [3 MARKS] Calculate the area of R. ANSWER ONLY Information for Students in MATH 141 2010 01 6001 G WeBWorK G.1 Frequently Asked Questions (FAQ) G.1.1 Where is WeBWorK? WeBWorK is located on Web servers of the Department of Mathematics and Statistics, and is accessible at the following URL: http://msr02.math.mcgill.ca/webwork2/MATH141 WINTER2010/ If you access WeBWorK through WebCT, the link on your page will have been programmed to take you to the correct WeBWorK server automatically. G.1.2 Do I need a password to use WeBWorK? You will need a user code and a password. Your user code. Your user code will be your 9-digit student number. Your password. The WeBWorK system is administered by the Mathematics and Statistics Department, and is not accessible directly through the myMcGill Portal; your initial password will be different from your MINERVA password, but you could change it to that if you wish. Your initial password will be your 9-digit student ID number. You will be able to change this password after you sign on to WeBWorK.65 Your e-mail address. The WeBWorK system requires each user to have an e-mail address. After signing on to WeBWorK, you should verify that the e-mail address shown is the one that you prefer. You should endeavour to keep your e-mail address up to date, since the instructors may send messages to the entire class through this route. We suggest that you use either your UEA66 or your po-box address. You may be able to forward your mail from these addresses to another convenient address, (cf. §4.) G.1.3 Do I have to pay an additional fee to use WeBWorK? WeBWorK is available to all students registered in the course at no additional charge. 65 If you forget your password you will have to send a message to Professor Brown so that the system administrator may be instructed to reset the password at its initial value. 66 Uniform E-mail Address UPDATED TO April 17, 2010 Information for Students in MATH 141 2010 01 6002 G.1.4 When will assignments be available on WeBWorK? Each assignment will have a begin date and a due date. The assignment is available to you after the begin date; solutions will be made available soon after the due date. G.1.5 Do WeBWorK assignments cover the full range of problems that I should be able to solve in this course? The questions on the WeBWorK assignments (A1 through A6 ) are a sampling of some types of problem you should be able to solve after successfully completing this course. Some types of calculus problems do not lend themselves to this kind of treatment, and may not appear on the WeBWorK assignments. Use of WeBWorK does not replace studying the textbook — including the worked examples, attending lectures and tutorials, and working exercises from the textbook — using the Student Solutions Manual [3] to check your work. Students are cautioned not to draw conclusions from the presence, absence, or relative frequencies of problems of particular types, or from particular sections of the textbook. Certain sections of the textbook remain examination material even though no problems are included in the WeBWorK assignments. G.1.6 May I assume that the distribution of topics on quizzes and final examinations will parallel the distribution of topics in the WeBWorK assignments? No! While the order of topics on WeBWorK assignments should conform to the order of the lectures, there are some topics on the syllabus that will not appear in WeBWorK questions. Use WeBWorK for the areas it covers, and supplement it by working problems from your textbook. Also, remember that WeBWorK — which checks answer only — cannot ascertain whether you are using a correct method for solving problems. But, if you write out a solution to an odd-numbered textbook problem, you can compare it with the solution in the Solutions Manual; and, if in doubt, you can show your work to a Teaching Assistant at one of the many office hours that they hold through the week. G.1.7 WeBWorK provides for different kinds of “Display Mode”. Which should I use? “Display mode” is the mode that you enter when you first view a problem; and, later, when you submit your answer. You may wish to experiment with the different formats. The default is jsMath mode, which should look similar to the version that you print out (cf. next question). Information for Students in MATH 141 2010 01 6003 G.1.8 WeBWorK provides for printing assignments in “Portable Document Format” (.pdf), “PostScript” (.ps) and “TEXSource” forms. Which should I use? Most newer home computers have already been loaded with the Acrobat Reader for .pdf files; if the Reader has not been installed on your computer67 , you will find instructions for downloading this (free) software in §1.6.5 of these notes. Most computers available to you on campus should be capable of printing in .pdf format. G.1.9 What is the relation between WeBWorK and WebCT? There is none. WebCT is the proprietary system of Web Course Tools that has been implemented by McGill University. You may access the web page for this course, and WeBWorK through your WebCT account68 , and WebCT will link you to the appropriate server for WeBWorK. If you follow this route to WeBWorK, you will still have to log in when you reach the WeBWorK site. At the present time we will be using WebCT primarily for the posting of grades, and as a convenient repository for links to notes and announcements in the course. We are not planning to use the potential WebCT sites that exist for the tutorial sections: use only the site for the lecture section in which you are registered. G.1.10 What do I have to do on WeBWorK? After you sign on to WeBWorK, and click on “Begin Problem Sets”, you will see a list of Assignments, each with a due date. Since there is no limit to the number of attempts at problems on P0 or the other “Practice” assignments, you may play with these assignments to learn how to use the WeBWorK software. You may print out a copy of your assignment by clicking on “Get hard copy”. This is your version of the assignment, and it will differ from the assignments of other students in the course. You should spend some time working on the assignment away from the computer. When you are ready to submit your solutions, sign on again, and select the same assignment. This time click on “Do problem set”. You can expect to become more comfortable with the system as you attempt several problems; but, in the beginning, there are likely to be situations where you cannot understand what the system finds wrong with some of your answers. It is useful to click on the Preview Answers button to see how the system interprets an answer that you have typed in. As the problems may become more difficult, you may have to refer to the “Help” page, and also to the “List of functions” which appears on the page listing the problems. Don’t submit an answer until you are happy with the interpretation that the Preview Answers button shows that the system will be taking of your answer. 67 At the time these notes were written, the latest version of the Reader was 9.0, but recent, earlier versions should also work properly. 68 http://mycourses.mcgill.ca Information for Students in MATH 141 2010 01 6004 G.1.11 How can I learn how to use WeBWorK? As soon as your instructor announces that the WeBWorK accounts are ready, sign on and try assignment P0 , which does not count. The system is self-instructive, so we will not burden you with a long list of instructions. You will need to learn how to enter algebraic expressions into WeBWorK as it is coded to read what you type in a way that may different from what you expect. For example, the symbol ˆ is used for writing exponents (powers). If you type 2ˆ3, WeBWorK will interpret this as 23 = 8. However, if you type 2ˆ3+x, WeBWorKwill interpret it as 23 + x, i.e. as 8 + x; if you wish to write 23+x , you have to type 2ˆ(3+x). You may obtain more information from the List of Available Functions, available online, or at http://webwork.maa.org/wiki/Mathematical notation recognized by WeBWorK G.1.12 Where should I go if I have difficulties with WeBWorK ? If you have difficulties signing on to WeBWorK, or with the viewing or printing functions on WeBWorK, or with the specific problems on your version of an assignment, you may send an e-mail distress message directly from WeBWorK by clicking on the Feedback button. You may also report the problem to your instructor and/or your tutor, but the fastest way of resolving your difficulty is usually the Feedback . Please give as much information as you can. (All of the instructors and tutors are able to view from within WeBWorK the answers that you have submitted to questions.) If your problem is mathematical, and you need help in solving a problem, you should consult one of the tutors at their office hours; you may go to any tutor’s office hours, not only to the hours of the tutor of the section in which you are registered. G.1.13 Can the WeBWorK system ever break down or degrade? Like all computer systems, WeBWorK can experience technical problems. The systems manager is continually monitoring its performance. If you experience a difficulty when online, please click on the Feedback button and report it. If that option is not available to you, please communicate with either instructor by e-mail. If you leave your WeBWorK assignment until the hours close to the due time on the due date, you should not be surprised if the system is slow to respond. This is not a malfunction, but is simply a reflection of the fact that other students have also been procrastinating! To benefit from the speed that the system can deliver under normal conditions, do not delay your WeBWorK until the last possible day! If a systems failure interferes with the due date of an assignment, arrangements could be made to change that date, and an e-mail message could Information for Students in MATH 141 2010 01 6005 be broadcast to all users (to the e-mail addresses on record), or a message could be posted on WeBCT or the WeBWorK sign-on screen.69 G.1.14 How many attempts may I make to solve a particular problem on WeBWorK? Practice Assignments P1 — P6 are intended to prepare you for Assignments A1 — A6 , and permit unlimited numbers of attempts; your grades on these “Practice” do not count in your term mark. For the problems on assignments A1 — A6 you will normally be permitted about 5 tries: read the instructions at the head of the assignment. G.1.15 Will all WeBWorK assignments have the same length? the same value? The numbers of problems on the various assignments may not be the same, and the individual problems may vary in difficulty. Assignments A1 — A6 will count equally in the computation of your grade. G.1.16 Is WeBWorK a good indicator of examination performance? A low grade on WeBWorK has often been followed by a low grade on the examination. A high grade on WeBWorK does not necessarily indicate a likely high grade on the examination. To summarize: WeBWorK alone is not enough to prepare this course; but students who don’t do WeBWorK appear to have a poor likelihood of success in MATH 141: that is one reason why we have made the WeBWorK assignments compulsory. 69 But slowness of the system just before the due time will not normally be considered a systems failure. Information for Students in MATH 141 2010 01 6101 H Contents of the DVD disks for Larson/Hostetler/Edwards These excellent disks were produced to accompany the textbook, Calculus of a Single Variable: Early Transcendental Functions, 3rd Edition[28] (called LHE in the charts below). The correspondence shown to sections of [7] are only approximate. (NOTE THAT THIS BOOK DOES NOT FOLLOW STEWART’S CONVENTIONS FOR INVERSE SECANT/COSECANT!) [All references in this table are to the 5th edition of Stewart, [7].] DVD LHE # Section 1 P 1 P.1 1 P.2 1 P.3 1 P.4 1 P.5 1 P.6 Subject Chapter P: Preparation for Calculus Graphs and Models Linear Models and Rates of Change Functions and Their Graphs Fitting Models to Data Inverse Functions Exponential and Logarithmic Functions Minutes Stewart Section 45 27 48 21 48 30 A10 1.1 1.2 1.6 1.5 DVD LHE # Section 1 1 1 1.1 1 1.2 1 1.3 1 1.4 1 1.5 Subject Chapter 1: Limits and Their Properties A Preview of Calculus Finding Limits Graphically and Numerically Evaluating Limits Analytically Continuity and One-Sided Limits Infinite Limits Minutes Stewart Section 11 25 28 22 18 2.1 2.2, 2.4 2.3 2.5 2.6 Subject Chapter 2: Differentiation The Derivative and the Tangent Line Problem Basic Differentiation Rules and Rates of Change The Product and Quotient Rules and Higher Order Derivatives Minutes Stewart Section 68 34 2.1 2.3 25 3.2, 3.7 DVD LHE # Section 1 2 1 2.1 1 2.2 1 2.3 Information for Students in MATH 141 2010 01 DVD # 2 2 2 2 2 2 LHE Section 2 2.4 2.5 2.6 2.7 2.8 DVD LHE # Section 2 3 2 3.1 2 3.2 2 3.3 2 2 2 2 2 3.4 3.5 3.6 3.7 3.8 DVD LHE # Section 3 4 3 4.1 DVD LHE # Section 4 7 4 7.7 Subject Chapter 2 (continued): Differentiation The Chain Rule Implicit Differentiation Derivatives of Inverse Functions Related Rates Newton’s Method Subject Chapter 3: Applications of Differentiation Extrema on an Interval Rolle’s Theorem and the Mean Value Theorem Increasing and Decreasing Functions and the First Derivative Test Concavity and the Second Derivative Test Limits at Infinity A Summary of Curve Sketching Optimization Problems Differentials Subject Chapter 4: Integration Antiderivatives and Indefinite Integration 6102 Minutes Stewart Section 44 50 17 34 26 3.5 3.6 3.5, 3.8, 3.9 3.10 4.9 Minutes Stewart Section 41 15 4.1 4.2 19 4.3 24 23 43 37 51 4.3 2.6 4.5 4.7 3.11 Minutes Stewart Section 40 4.10 Subject Minutes Chapter 7: Integration by Parts Trigonometric Substitution Partial Fractions L’Hˆopital’s Rule Indeterminate Forms and L’Hˆopital’s Rule 22 (The coverage extends to part of the material for Math 141 as well.) Stewart Section 4.4