math 141 long document

McGILL UNIVERSITYFACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 141 2010 01CALCULUS 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.

PRELIMINARY VERSION

W. G. Brown

January 6, 2010

Information for Students in MATH 141 2010 01

Contents1 General Information 1

1.1 Force Majeure . . . . . . . . . . 11.2 Instructors and Times . . . . . . 21.3 Calendar Description . . . . . . 2

1.3.1 Calendar Description . . 21.3.2 Late transfer from MATH

151/MATH 152 . . . . . 31.4 Tutorials . . . . . . . . . . . . . 3

1.4.1 Tutorial Times, Locations,and Personnel (subject tochange) . . . . . . . . . 3

1.4.2 Teaching Assistants (TA’s) 31.4.3 Friday, April 02nd, 2010

and Monday, April 05th,2010 . . . . . . . . . . 4

1.5 Evaluation of Your Progress . . 51.5.1 Your final grade (See Ta-

ble 3, p. 11) In the eventof extraordinary circum-stances beyond the Uni-versity’s control, the con-tent and/or evaluation schemein this course is subjectto change. . . . . . . . . 5

1.5.2 WeBWorK . . . . . . . 61.5.3 Written Submissions. . . 71.5.4 Quizzes at the Tutorials. 71.5.5 Final Examination . . . 81.5.6 Supplemental Assessments 81.5.7 Machine Scoring: “Will

the final examination bemachine scored?” . . . . 9

1.5.8 Plagiarism. . . . . . . . 91.5.9 Corrections to grades . . 10

1.6 Published Materials . . . . . . . 101.6.1 Required Text-Book . . 101.6.2 Optional Reference Books 101.6.3 Recommended Video Ma-

terials . . . . . . . . . . 121.6.4 Other Calculus Textbooks 13

1.6.5 Website . . . . . . . . . 131.7 Syllabus . . . . . . . . . . . . . 141.8 Preparation and Workload . . . 15

1.8.1 Prerequisites. . . . . . . 151.8.2 Calculators . . . . . . . 151.8.3 Self-Supervision . . . . 161.8.4 Escape Routes . . . . . 171.8.5 Terminology . . . . . . 18

1.9 Communication with Instructorsand TA’s . . . . . . . . . . . . . 18

1.10 Commercial tutorial and exampreparation services . . . . . . . 19

1.11 Special Office Hours and Tutorials 20

2 References 2012.1 Stewart Calculus Series . . . . . 2012.2 Other Calculus Textbooks . . . . 202

2.2.1 R. A. Adams . . . . . . 2022.2.2 Larson, Hostetler, et al. . 2032.2.3 Edwards and Penney . . 2032.2.4 Others, not “Early Tran-

scendentals” . . . . . . 2042.3 Other References . . . . . . . . 204

A Timetable for Lecture Section 001 ofMATH 141 2010 01 1001

B Timetable for Lecture Section 002 ofMATH 141 2009 01 2001

C Supplementary Notes for Students inSection 001 of MATH 141 2010 01 3001C.1 Lecture style in Lecture Section

001 . . . . . . . . . . . . . . . 3001C.2 Supplementary Notes for the Lec-

ture of January 04th, 2010 . . . 3002C.2.1 §5.1 Areas and Distances. 3002

C.3 Supplementary Notes for the Lec-ture of January 06th, 2010 . . . 3005C.3.1 §5.1 Areas and Distances

(conclusion). . . . . . . 3005C.3.2 §5.2 The Definite Integral 3007


D Problem Assignments from Previous Years 5001D.1 1998/1999 . . . . . . . . . . . . 5001

D.1.1 Assignment 1 . . . . . . 5001D.1.2 Assignment 2 . . . . . . 5001D.1.3 Assignment 3 . . . . . . 5002D.1.4 Assignment 4 . . . . . . 5002D.1.5 Assignment 5 . . . . . . 5002

D.2 1999/2000 . . . . . . . . . . . . 5003D.2.1 Assignment 1 . . . . . . 5003D.2.2 Assignment 2 . . . . . . 5004D.2.3 Assignment 3 . . . . . . 5006D.2.4 Assignment 4 . . . . . . 5007D.2.5 Assignment 5 . . . . . . 5009D.2.6 Assignment 6 . . . . . . 5010

D.3 2000/2001 . . . . . . . . . . . . 5012D.4 2001/2002 . . . . . . . . . . . . 5012D.5 MATH 141 2003 01 . . . . . . . 5012D.6 MATH 141 2004 01 . . . . . . . 5012D.7 MATH 141 2005 01 . . . . . . . 5013

D.7.1 Written Assignment W1 5013D.7.2 Written Assignment W2 5014D.7.3 Written Assignment W3 5016D.7.4 Written Assignment W4 5017D.7.5 Written Assignment W5 5019

D.8 MATH 141 2006 01 . . . . . . . 5021D.8.1 Solution to Written As-

signment W1 . . . . . . 5021D.8.2 Solution to Written As-

signment W2 . . . . . . 5024D.8.3 Solutions to Written As-



signment W5 . . . . . . 5030D.9 MATH 141 2007 01 . . . . . . . 5032

E Quizzes from Previous Years 5033E.1 MATH 141 2007 01 . . . . . . . 5033

E.1.1 Draft Solutions to Quiz Q1 5033E.1.2 Draft Solutions to Quiz Q2 5043E.1.3 Draft Solutions to Quiz Q3 5055

E.1.4 Draft Solutions to Quiz Q4 5070E.2 MATH 141 2008 01 . . . . . . . 5086

E.2.1 Draft Solutions to Quiz Q1 5086E.2.2 Draft Solutions to Quiz Q2 5099E.2.3 Draft Solutions to Quiz Q3 5110E.2.4 Draft Solutions to Quiz Q4 5121

E.3 MATH 141 2009 01 . . . . . . . 5134E.3.1 Draft Solutions to Quiz Q1 5134E.3.2 Draft Solutions to Quiz Q2 5140E.3.3 Draft Solutions to Quiz Q3 5146

F Final Examinations from Previous Years 5152F.1 Final Examination in Mathemat-

ics 189-121B (1996/1997) . . . 5152F.2 Final Examination in Mathemat-

ics 189-141B (1997/1998) . . . 5153F.3 Supplemental/Deferred Examina-

tion in Mathematics 189-141B(1997/1998) . . . . . . . . . . . 5155

F.4 Final Examination in Mathemat-ics 189-141B (1998/1999) . . . 5156

F.5 Supplemental/Deferred Examina-tion in Mathematics 189-141B(1998/1999) . . . . . . . . . . . 5158







F.12 Final Examination in MATH 1412003 01 . . . . . . . . . . . . . 5169


F.13 Supplemental/Deferred Examina-tion in MATH 141 2003 01 . . . 5171





F.18 Final Examination in MATH 1412006 01 (One version) . . . . . 5192


F.20 Final Examination in MATH 1412007 01 (One version) . . . . . 5199

F.21 Supplemental/Deferred Examina-tion in MATH 141 2007 01 (Oneversion) . . . . . . . . . . . . . 5203

F.22 Final Examination in MATH 1412008 01 (one version) . . . . . . 5207

F.23 Supplemental/Deferred Examina-tion in MATH 141 2008 01 (oneversion) . . . . . . . . . . . . . 5217

F.24 Final Examination in MATH 1412009 01 (one version) . . . . . . 5221

G WeBWorK 6001G.1 Frequently Asked Questions (FAQ) 6001

G.1.1 Where is WeBWorK? . 6001G.1.2 Do I need a password to

use WeBWorK? . . . . 6001G.1.3 Do I have to pay an ad-

ditional fee to use WeB-WorK? . . . . . . . . . 6001

G.1.4 When will assignmentsbe available on WeBWorK? 6002

G.1.5 Do WeBWorK assign-ments cover the full rangeof problems that I shouldbe able to solve in thiscourse? . . . . . . . . . 6002

G.1.6 May I assume that thedistribution of topics onquizzes and final exam-inations will parallel thedistribution of topics inthe WeBWorK assign-ments? . . . . . . . . . 6002

G.1.7 WeBWorK provides fordifferent kinds of “Dis-play Mode”. Which shouldI use? . . . . . . . . . . 6002

G.1.8 WeBWorK provides forprinting assignments in“Portable Document Format”(.pdf), “PostScript” (.ps)and “TEXSource” forms.Which should I use? . . 6003

G.1.9 What is the relation be-tween WeBWorK and We-bCT? . . . . . . . . . . 6003

G.1.10 What do I have to do onWeBWorK? . . . . . . 6003

G.1.11 How can I learn how touse WeBWorK? . . . . 6004

G.1.12 Where should I go if Ihave difficulties with WeB-WorK ? . . . . . . . . . 6004

G.1.13 Can the WeBWorK sys-tem ever break down ordegrade? . . . . . . . . 6004

G.1.14 How many attempts mayI make to solve a partic-ular problem on WeB-WorK? . . . . . . . . . 6005

G.1.15 Will all WeBWorK as-signments have the samelength? the same value? 6005

G.1.16 Is WeBWorK a good in-dicator of examination per-formance? . . . . . . . . 6005

H Contents of the DVD disks for


Larson/Hostetler/Edwards 6101

I Instructions and Suggestions for Teach-ing Assistants in MATH 141 2010 01 7001I.1 “Information for students in MATH

141 2010 01” . . . . . . . . . . 7001I.2 Workload . . . . . . . . . . . . 7001I.3 “Contingency/Other Duties” . . 7001I.4 Tutorial Sessions . . . . . . . . 7002I.5 WeBWorK . . . . . . . . . . . 7003I.6 Quizzes . . . . . . . . . . . . . 7004I.7 Your students . . . . . . . . . . 7005

I.7.1 Students from Other Tu-torial Sections . . . . . 7005

I.7.2 Your grading and yourcolleagues’ grading . . . 7005

I.8 Keeping Careful Records . . . . 7006I.8.1 Submission of quiz grades 7006I.8.2 Corrections to submitted

records . . . . . . . . . 7007I.8.3 Special Cases . . . . . . 7007I.8.4 Publication of Student Grades. 7007I.8.5 Confidentiality. . . . . . 7008

I.9 Office Hours . . . . . . . . . . . 7008I.9.1 Rescheduling Office Hours 7009

I.10 Grading the Final Examination . 7009I.11 Textbooks, Printed, and Web Ma-

terials . . . . . . . . . . . . . . 7010I.11.1 Textbooks . . . . . . . . 7010I.11.2 Printed Materials . . . . 7010I.11.3 Other duplication . . . . 7010I.11.4 Course Web Page . . . . 7010I.11.5 The “TA Page” on myCourses 7010I.11.6 Access to WeBWorK . 7011

I.12 Communications . . . . . . . . 7011I.12.1 Communications to TA’s

from the Instructors . . . 7011I.12.2 myCourses and E-mail . 7011

I.13 Professionalism . . . . . . . . . 7011I.14 Quality Control . . . . . . . . . 7013

I.14.1 Online evaluation of TA’s 7013I.14.2 Grading Standards . . . 7013

I.15 Midterm meeting with instructor(s) 7013I.16 Tutorial schedules . . . . . . . . 7014

List of Tables1 Schedule and Locations of Tu-

torials, as of January 6, 2010. . . 42 Tutors’ Coordinates, as of Jan-

uary 6, 2010 . . . . . . . . . . . 53 Summary of Course Requirements,

as of January 6, 2010; (all datesare subject to change) . . . . . . 11

6 Workload common to all teach-ing assistants in MATH 141 201001 (subject to correction or change) 7002

List of Figures1 The limacon r = 1 + 2 sin θ . . . 5071

Information for Students in MATH 141 2010 01 1

1 General InformationDistribution 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 defini-tive version will be the version accessible at

http://www.math.mcgill.ca/brown/math141b.html

or on myCourses, at

http://www.mcgill.ca/mycourses/

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 aboutthis course.All details of the course could be subject to discretionary change in case of forcemajeure.

1

1.1 Force MajeureIn the event of extraordinary circumstances beyond the University’s control, all details of thiscourse, including the content and/or evaluation scheme are subject to change.

1Please note that the statements about MATH 141 in an SUS publication called Absolute Zero were not givento instructors of this course to check, and some of them may not be currently correct.


1.2 Instructors and TimesINSTRUCTOR: Prof. W. G. Brown Dr. S. Shahabi Dr. A. Hundemer

(Course Coordinator)LECTURE SECTION: 1 2 3

CRN: 576 577 578OFFICE: BURN 1224 BURN 1243 BURN 1128

OFFICE HOURS: W 13:15→14:15 F 09:30→11:30 MW 15:30→16:25(subject to change) F 10:00→11:00 (tentative)

or by appointmentTELEPHONE: (514)-398-3836 (514)-398-3803 (514)-398-5318

E-MAIL:2 BROWN@ SHAHABI@ [email protected] MATH.MCGILL.CA MATH.MCGILL.CA

CLASSROOM: ADAMS AUD LEA 219 ADAMS AUDCLASS HOURS: MWF 11:35–12:25 h. MWF 11:35–12:25 h. MW 16:35–17:55 h.

1.3 Calendar Description1.3.1 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 takenMATH 121 or CEGEP objective 00UP or equivalent; not open to students who have taken orare taking MATH 122 or MATH 130 or MATH 131, except by permission of the Department ofMathematics 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 Officeas being equivalent to MATH 140) should not assume that, in possibly permitting MINERVAto accept their registration for MATH 141, the University was tacitly approving their regis-tration without the prerequisite. In particular, students who obtained a grade of F in MATH139/140/150 are expressly excluded from registration in MATH 141, even if they registered inthe course before the failed or missed examination.

2Please 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.

3E-mail messages generated by the Feedback command in WeBWorK should be used sparingly, and con-fined to specific inquiries about WeBWorK assignments.

4The previous designation for this course was 189-141, and the version given in the winter was labelled189-141B; an earlier number for a similar course was 189-121.


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 141after the end of the Change of Course Period. If your instructor in MATH 151 or 152 advisesyou that you are in this category, please send an e-mail message to Professor Brown as soonas your transfer has been approved.5

1.4 Tutorials1.4.1 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; thelast tutorial in Monday tutorial sections will be on Monday, April 12th, 2010; the date of thelast 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 butsometimes we are not informed in advance.6

You are expected to write quizzes only in the tutorial section in which you are reg-istered.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 perma-nently inconvenient; in the latter case the onus is on you to transfer formally to another tutorialsection, to change your other classes, or to drop MATH 141 2010 01. Please remember thattransfers 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. Likeyou, they are students, albeit at the graduate level; they have deadlines and commitments andpersonal lives, and the time they have available for MATH 141 is limited and controlled by acollective agreement (union contract). Please respect the important functions that our tutorsprovide, and do not ask them for services they are not expected to perform:

5This is to ensure that your WeBWorK account is opened, and that your date of entry to the course is recorded.6The current room for your tutorial should always be available by clicking on “Class Schedule” on MINERVA

FOR STUDENTS, http://www.mcgill.ca/minerva-students/.7In some time slots there may be several tutorial sections, meeting in different rooms.8Your instructors do not have the ability to change the maximum capacities of tutorials.


# CRN Day Begins Ends Room TutorT004 579 Fri 01:35 03:25 BURN 1B39 D. Attwell-DuvalT005 580 Tue 14:05 15:55 ARTS 260 H. BigdelyT006 581 Tue 16:05 17:55 BURN 1B23 L. CandeloriT007 582 Tue 16:05 17:55 BURN 1B24 F. CastellaT008 583 Thurs 14:05 15:55 BURN 1B39 Y. CanzaniT009 584 Thurs 16:05 17:55 BURN 1B39 J. FeysT010 585 Thurs 16:05 17:55 BURN 1B23 A.-P. GrecianuT011 586 Mon 13:35 15:25 ARTS W-20 J. MacdonaldT012 587 Mon 14:35 16:25 BURN 1B36 M. PrevostT013 588 Mon 14:35 16:25 BURN 1B24 P. RempelT014 589 Wed 13:35 15:25 ARTS W-20 J. RestrepoT015 590 Wed 14:35 16:25 LEA 14 B. TajiT016 591 Wed 14:35 16:25 BURN 1B36 A. TchengT017 2071 Mon 13:35 15:25 ENGMD 276 A. TombergT018 2072 Wed 13:35 15:25 BURN 1B24 J. Tousignant-BarnesT019 8194 Tue 08:05 09:55 BURN 1B23 A. FarooquiT020 8840 Tue 15:35 17:25 ENGMD 279 M. WongT021 8841 Fri 15:35 17:25 ENGMD 256 Y. Zhao

Some of these room assignments could change before or early in the beginning of the term, as wehave 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 January 6, 2010.

• Outside of the normal quiz times in their tutorials, tutors are neither expected nor au-thorized to administer a special quiz or a quiz that has already been administered toothers.

• Tutors in MATH 141 2010 01 are not permitted to offer paid, private tuition to studentsin 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 willresume on Wednesday, April 07th, 2010, and the total number of lecture hours is similar topast years, there will be some disruption to Monday and Friday tutorials: the Monday tutorialswill meet on Monday, April 12th, 2010, but there is no scheduled Friday available to completethe Friday tutorials; alternative arrangements for students in Friday tutorials will be announcedlater in the term.


Tutor E-mail address Office Office HoursBURN Day(s) Begins Ends Day Begins Ends

Attwell-Duval, D. [email protected] 1018 : : : :Bigdely, H. [email protected] 1030 : : : :

Candelori, L. [email protected] 1032 W 08:30 11:30Canzani, Y. [email protected] 1133 T 12:30 14:30 Th 13:00 14:00Castella, F. [email protected] 1008 : : : :Farooqui, A. [email protected] 1032 : :

Feys, J. [email protected] 1020 TTh 17:00 18:30 : :Grecianu, A.-P. [email protected] 1007 Th 14:30 16:00 F 13:00 14:30Macdonald, J. [email protected] 1030 : :

Prevost, M. [email protected] 1117 TTh 13:00 14:30Rempel, P. [email protected] 1140 Th 10:00 13:00 : :Restrepo, J. [email protected] 1117 T 10:30 12:00 T 14:30 16:00

Taji, B. [email protected] 1031 : : : :Tcheng, A. [email protected] 1029 TTh 14:30 16:00Tomberg, A. [email protected] 1031 W 12:30 15:30

Tousignant-Barnes, J. [email protected] 1032 TTh 09:00 10:30Wong, M. [email protected] 1031 : :Zhao, Y. [email protected] 1034 F 12:30 15:30

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 January 6, 2010

1.5 Evaluation of Your Progress1.5.1 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 thiscourse is subject to change.

Your grade in this course will be a letter grade, based on a percentage grade computed fromthe 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

9Numbers of assignments, quizzes, etc., are as planned as of the date of this version of these notes. Studentsmust be prepared for the possibility that it could be necessary to adjust these numbers during the term. If thereare any changes, these will be announced on myCourses, by broadcast e-mail messages, or by announcements atthe lectures.

10But be warned: students who fail to write quizzes are often at risk in this course. The quizzes are mainlya learning, rather than a testing experience. You need the information that comes from writing quizzes in a


3. The final examination — counting for 70%.

Where a student’s performance on the final examination is superior to her performance on thetutorial 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 notplanned 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 toexpose 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 WeBWorKare contained in Appendix G to these notes, page 6001.

Only answers submitted by the due date and time will count. The WeBWorK assign-ments which count in your term mark will be labelled A1, . . ., A6.

2. Due dates and times for WeBWorK assignments. Most due dates for WeBWorKassignments will be on specified Sundays, about 23:30h; last minute changes in the duedates 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 WeBWorKassignment until the hours close to the due time on the due date, you should not besurprised if the system is slow to respond. This is not a malfunction, but is simplya reflection of the fact that other students have also been procrastinating! To benefitfrom the speed that the system can deliver under normal conditions, do not delay yourWeBWorK until the last possible day! If a systems failure interferes with the due date ofan assignment, arrangements may be made to change that date, and an e-mail messagemay be broadcast to all users (to the e-mail addresses on record), or a note posted in thecourse announcements on myCourses; but slowness in the system just before the duetime will not normally be considered a systems failure.12

3. Numbers of permitted attempts at WeBWorK questions. While the number of timesyou may attempt each problem on WeBWorK An will be limited, there will be a com-panion “Practice” Assignment Pn (n = 1, 2, . . . , 6) with an unlimited number of attemptsat 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 thisexperience often undergo a rude awakening at the final examination.

11Be sure that your e-mail addresses are correctly recorded. See 4, p. 19 of these notes.12Should 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 systemsfailure.


the opportunity to prepare yourself on the Practice assignment before attempting the ac-tual assignment. The practice assignments DO NOT COUNT in your term mark, eventhough a grade is recorded.. Practice assignment Pn is normally due one week beforeassignment An. Another assignment which will not count will be Practice AssignmentP0, 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 theright to submit in English or in French any written work that is to be graded; course materialsare 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. Thesequizzes will be graded, and returned. The primary purpose of a quiz is to diagnosepossible gaps in your understanding. In the grading formula the quiz component ofthe final grade will be replaced by the final examination grade, if that is to a student’sadvantage.

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 (med-ical or otherwise), please communicate directly with Professor Brown, providing a copyof the medical or other supporting documents; do not contact your TA. Authorized med-ical absences can be accommodated only through averaging, as students in MATH 141are never permitted to write a quiz in any tutorial section other than the one in whichthey 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 ma-terial currently under discussion at the lectures, and you should have attempted any openWeBWorK assignments. But, unlike the WeBWorK assignments — where the empha-sis is on correct answers alone — students may be expected to provide full solutions to

13While it is not required for grading purposes, students are urged to keep a systematic record of writtensolutions to problems in the textbook. This could be in the form of a workbook, or a file folder, but should beorderly enough that you can look back at a later time to see your solutions. You are invited to bring such a file toTA’s or instructors at their office hours, to receive advice about the quality and correctness of your solutions.


some or all problems on quizzes.14

The quizzes may examine on only a sampling of topics. Students should not assume thattopics 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 andstudent numbers in unsupervised locations; you may be able to recover an unclaimedquiz from the tutor who graded it, during her/his regular office hours. Be sure to attendthe tutorial following a quiz15, as claims of incorrect recording of a quiz or assignmentgrade will need to be substantiated by a graded paper.

6. Your quiz grades on assignments and quizzes will be posted on myCourses within about2 weeks after they become available. Your WeBWorK grades may not be transferred tomyCourses 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 forthe winter term (April 15th, 2010 through April 30th, 2010). You are advised not to make anytravel arrangements that would prevent you from being present on campus at any time duringthis 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 theoption of doing additional work to upgrade their mark?” No. (“Additional Work” refersto an option available in certain Arts and Science courses, but not available in MATH141 2010 01.)

14In Math 141 the general rule for quizzes is that full solutions are expected to all problems, unless you receiveexplicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in the Student SolutionsManual [9] to the textbook can serve as a guide to what should be included in a “full” solution.

15The return of Quiz Q1 may be delayed to the 2nd week after the quiz was written.16Your instructors learn the date of your examination at the same time as you do — when the Provisional

examination timetable is published.


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 bemachine scored. Multiple choice problems, possibly machine scored, could also appear onsome quizzes. (Machine grading, if implemented in whole or in part, would be a change fromthe practice of past years in this course. Such a change was introduced into the most recentexamination in MATH 140, which included a substantial number of machine-graded multiplechoice 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 maynot 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 assign-ment problems with their colleagues, all work that you submit must be your own. The Senateof the University requires the following message in all course outlines:

“McGill University values academic integrity. Therefore all students must understand themeaning and consequences of cheating, plagiarism and other academic offences under theCode of Student Conduct and Disciplinary Procedures. (See http://www.mcgill.ca/integrityfor more information).

“L’universite McGill attache une haute importance a l’honnetete academique. Il incombepar consequent a tous les etudiants de comprendre ce que l’on entend par tricherie, plagiatet autres infractions academiques, ainsi que les consequences que peuvent avoir de tellesactions, selon le Code de conduite de l’etudiant et des procedures disciplinaires. (Pour deplus amples renseignements, veuillez consulter le site http://www.mcgill.ca/integrity).”

It is a violation of University regulations to permit others to solve your WeB-WorK problems, or to extend such assistance to others; you could be askedto sign a statement attesting to the originality of your work. The Handbookon Student Rights and Responsibilities17 states in ¶A.I.15(a) that

“No student shall, with intent to deceive, represent the work of another personas his or her own in any academic writing, essay, thesis, research report, projector assignment submitted in a course or program of study or represent as his orher own an entire essay or work of another, whether the material so representedconstitutes a part or the entirety of the work submitted.”

17http://upload.mcgill.ca/secretariat/greenbookenglish.pdf


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 tutorunder 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 recordedincorrectly, 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. Itis hoped that grades will be posted within 2 weeks of the due date. You will have to present thegraded quiz to support your claim, which must be submitted to the tutor that graded the quiz. Ifhe/she believes there has been an error, the tutor will advise Professor Brown. New correctionsto the myCourses posting will appear the next time grades are uploaded to myCourses.

1.6 Published Materials1.6.1 Required Text-Book

The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: Early Tran-scendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01169-X, [1]. This book is thefirst 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 soldin the McGill Bookstore bundled with its Student Solutions Manual (see below). The ISBNnumber 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 STEW-ART’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 January 6, 2010


Item # Due Date DetailsP0 DOES NOT COUNT: introduces WeBWorK

WeBWorK P1 17 Jan 10 DOES NOT COUNT; practice for A1Assignments A1 24 Jan 10(cf. §1.5.2) P2 31 Jan 10 DOES NOT COUNT; practice for A2

10% A2 07 Feb 10P3 14 Feb 10 DOES NOT COUNT; practice for A3A3 21 Feb 10P4 28 Feb 10 DOES NOT COUNT; practice for A4A4 07 Mar 10P5 14 Mar 10 DOES NOT COUNT; practice for A5A5 21 Mar 10P6 28 Mar 10 DOES NOT COUNT; practice for A6A6 05 Apr 10 A1–A6 count equally, but may have different num-

bers of problems.Quizzes Q1 18–22 Jan 10 Quizzes Q1 — Q4 count equally, but(cf. §1.5.4) Q2 08–12 Feb 10 the quizzes may be of different lengths.20% or 0% Q3 08–12 Mar 10

Q4 22–26 Mar 10Final Exam 15–30 Apr 10 Date of exam to be announced by Faculty70% or 90%

SupplementalExam(cf. §1.5.6.1)

18–19 Aug 10 Only for students who do not obtain standing at thefinal. Supplemental exams count in your averagelike taking the course again; exam counts for 100%.

Table 3: Summary of Course Requirements, as of January 6, 2010; (all dates are subject tochange)

• a “Study Guide”, designed to provide additional help for students who believe theyrequire it: R. St. Andre, STUDY GUIDE FOR STEWART’S SINGLE VARIABLECALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01239-4, [5]. (The “Study Guide” resembles the Student Solution Manual in ap-pearance: be sure you know what you are buying.)

• a “Companion” which integrates a review of pre-calculus concepts with the contentsof Math 140, including exercises with solutions: D. Ebersole, D. Schattschneider, A.Sevilla, K. Somers, A COMPANION TO CALCULUS. Brooks/Cole (1995), ISBN 0-534-26592-8 [39].


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 [videorecord-ing]. The publisher of Stewart’s Calculus has produced a series of videodisks, [?]. These willinitially be available for reserve loan at the Schulich Library. There may not be DVD viewingequipment freely available in the library; the intention is that interested students borrow disksfor 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 pro-duced a series of videotapes, [14] Video Outline for Stewart’s Calculus (Early Transcenden-tals), Fifth Edition. These will be available for reserve loan at the Schulich Library. There maynot be VCR viewing equipment in the library; the intention is that interested students borrowa tape for viewing on their own equipment at home.

Larson/Hostetler/Edwards DVD Disks A set of video DVD disks produced for anothercalculus book, [28] Calculus Instructional DVD Program, for use with (inter alia) Larson /

Hostetler / Edwards, Calculus of a Single Variable: Early Transcendental Functions, ThirdEdition [29] is produced by the Houghton Mifflin Company. A copy has been requested to beplaced on reserve in the Schulich Library. In Appendix H of these notes there are charts thatindicate the contents of these disks that pertain to MATH 141.

Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 6thEdition, (similar to [15]) 19This CD-ROM is included with certain new copies of the text-book. It contains, after an enlightening “pep-talk” by the author, a discussion of some of theworked examples in the text-book, followed by a quiz for each section in the book. Some stu-dents may find the animations of the examples helpful, although the examples are all workedin the book. You might wish to try some of the quiz questions using paper and pencil, andthen check your answers with those given on the CD. It is not recommended that you attemptto enter your answers digitally, as this is a time-consuming process, and uses a different inputmethod from your WeBWorK assignments, which serve the same purpose.

18No one will check whether you have used any of these aids; a student can obtain a perfect grade in the coursewithout ever consulting any of them. No audio-visual or calculator aid can replace the systematic use of paperand pencil as you work your way through problems. But the intelligent use of some of these aids can deepen yourunderstanding of the subject. However, the most important aid is the Student Solutions Manual to the textbook!

19The version of this CD-ROM for the 6th edition is being catalogued by the Library; it may not be availableat the beginning of the term.


1.6.4 Other Calculus Textbooks

While students may wish to consult other textbooks, instructors and teaching assistants inMath 141 will normally refer only to the prescribed edition of the prescribed textbook for thecourse. Other books can be very useful, but the onus is on you to ensure that your book coversthe syllabus to at least the required depth; where there are differences of terminology, you areexpected to be familiar with the terminology of the textbook.20

In your previous calculus course(s) you may have learned methods of solving problems thatappear to differ from those you find in the current textbook. Your instructors will be pleasedto discuss any such methods with you personally, to ascertain whether they are appropriate tothe present course. In particular, any methods that depend upon the use of a calculator, or theplotting of multiple points, or the tabulation of function values, or the inference of a trend froma graph should be regarded with scepticism.

1.6.5 Website

These notes, and other materials distributed to students in this course, will be accessiblethrough 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, whichmany users have on their computers. This free software may be downloaded from the followingURL:

http://www.adobe.com/prodindex/acrobat/readstep.html 21

The questions on some old examinations will also be available as an appendix to these noteson the Web.22

Where revisions are made to distributed printed materials — for example these informationsheets — 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

20There should be multiple copies of the textbook on reserve in the Schulich library.21At the time of this writing the current version appears to be 8.n.22There 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 writingat the end of the term.


1.7 SyllabusSection numbers in the following list refer to the text-book [1]. The syllabus will include allof 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 fromtime to time subsequently, is not examination material.

Chapter 6: Applications of Integration. §§6.1 – 6.3; §6.5. (§6.4 is not examination mate-rial, but Science students are urged to read it.)

Chapter 7: Techniques of Integration. §§7.1 – 7.3; §7.4, excluding the Weierstrass substitu-tion [1, Exercises 57-61]; §7.5 §7.8. (§7.6, intended for use in conjunction with integraltables and/or computer algebra systems, is not examination material, but students areadvised to try to solve the problems manually; §7.7 requires the use of a calculator or acomputer, and consequently is not examination material.)

Chapter 8: Further Applications of Integration. §8.1, §8.2 only. (§§8.3, 8.4 are not exam-ination material, but students are urged to read the applications relevant to their courseof study; §8.5 is not examination material.)

Chapter 9: Differential Equations. (No part of this chapter is examination material; how-ever, 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 examina-tion material; however, students are urged to peruse these sections.)

Appendices Appendix G contains material shifted from [22, §5.6]. Students are expected toknow 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

thenceforth be taken to be that given in the Appendix, as

x∫

0

dtt

.

23If a textbook section is listed below, you should assume that all material in that section is examinationmaterial even if the instructor has not discussed every topic in his lectures; however, the instructors may give youinformation 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 assign-ment or a quiz, or if it has not appeared on any of the old examinations in the course! Some topics to not lendthemselves 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.


Please do not ask the tutors to provide information as to which textbook sections should beemphasized. Unless you are informed otherwise by the instructors in the lecture sections orpublished notes — printed, or mounted on the Web — you should assume that all materialslisted are included in the syllabus. You are not expected to be able to reproduce proofs of thetheorems in the textbook. However, you could be expected to solve problems in which theremight be unspecific real variables, rather than specific numbers, and which problems mightlook like textbook theorems.24

1.8 Preparation and Workload1.8.1 Prerequisites.

It is your responsibility as a student to verify that you have the necessary prerequisite. It wouldbe 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 advisedto make a special effort to reinforce their foundations in differential calculus; if weaknessin MATH 139 or MATH 140 was a consequence of poor preparation for that course, it isnot too late to strengthen those foundations as well.26 The fact that MINERVA may permityou to register does not relieve you from the responsibility to observe university regulationsconcerning prerequisites, and exposes you to the risk of failure in a course for which youare nor properly prepared; students with an F in MATH 139 or MATH 140 could have theirregistration 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 advisedto 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 todo all arithmetic by hand. Students who use calculators when they answer their WeBWorKproblems are undermining the usefulness of the programme to themselves: learn to use thebuilt-in calculation capabilities that are present in WeBWorK.

24The intention is that you should be learning how to solve problems, but should not have to memorize wholeproofs from the textbook.

25and contrary to McGill regulations26The 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 adequateskills to succeed in MATH 141! The onus is on you to seek help and to take remedial actions where necessary.


1.8.3 Self-Supervision

This is not a high-school course, and McGill is not a high school. The monitoring of yourprogress before the final examination is largely your own responsibility. Students must notassume that they will be exposed in lectures and tutorials to detailed model solutions for everytype of Calculus 2 problem. It is essential that you supplement these classes with serious workon 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 untilyou 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: youmay attend the office hours of any tutor in the course, and are not restricted to those of the tutorof the tutorial in which you are registered.

Time Demands of your Other Courses. Be sure to budget enough time to attend lecturesand tutorials, for private study, and for the solution of many problems. Don’t be tempted todivert calculus study time to courses which offer instant gratification. While the significance ofthe tutorial quizzes in the computation of your grade is minimal, these are important learningexperiences, and can assist you in gauging your progress in the course. This is not a coursethat can be crammed for: you must work steadily through the term if you wish to develop thefacilities needed for a strong performance on the final examination.

Lecture Times, and Preparation for the Lectures The lecture sections in MATH 141 201001 meet at the times that have been made available to us: early in the morning, or late inthe afternoon. While these times may not appeal to you, you should not underestimate thedamage 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. Toextract maximum benefit from the lectures, you should peruse the scheduled material beforecoming to class, trying some of the textbook problems; your instructors invite you to drawtheir attention to specific difficulties that you encounter before class in the textbook — it maybe possible to respond to these difficulties during the lecture.

Working Problems on Your Own. An effective way to master the calculus is through work-ing large numbers of problems from the textbook. Your textbook was selected partly becauseof the availability of an excellent Student Solutions Manual [9]; this manual has brief but com-plete solutions to most of the odd-numbered exercises in the textbook. The skills you acquirein solving textbook problems could have much more influence on your final grade than eitherWeBWorK or the quizzes.


When to do the WeBWorK assignment. I recommend that you defer working WeBWorKproblems until you have tried some of the easier odd-numbered problems in the textbook. Forthese you (should) have the Student Solutions Manual to help you check your work. Onceyou know that you have the basic concepts mastered, then is a good time to start workingWeBWorK 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 signif-icance of WeBWorK assignments and the quizzes. While both contribute to your grade, theycan help you estimate the quality of your progress in the course. Quizzes are administered un-der examination conditions, so poor performance or non-performance on quizzes can providean indicator of your expectations at the final examination; take proper remedial action if youare obtaining low grades on quizzes27. Since WeBWorK is not completed under examinationconditions, the grades you obtain may not be a good indicator of your expectations on the ex-amination; 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 WeBWorKgrades themselves have little predictive use, unless they are unusually low. However, whileboth WeBWorK and the quizzes have a role to play in learning the calculus, neither is as im-portant as reading your textbook, working problems yourself, and attending and listening atlectures and tutorials.

What to strive for on WeBWorK assignments. Since the practice assignments give youample opportunity to experiment, your success rate on the assignments “that count” should beclose to 100%. If you are needing more than 2 attempts to solve a WeBWorK problem, thenyou are probably not ready to work the assignment. In order to be able to solve a WeBWorKproblem successfully on the first attempt you will need to check your work, and this is a skillthat you will need on the final examination, and in the advanced studies or the real world whereyou 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 experiencingmedical or personal difficulties should not hesitate to consult their advisors or the StudentAffairs office of their faculty. Don’t allow yourself to be overwhelmed by such problems; theUniversity has resource persons who may be able to help you.

27The worst action is to miss the quizzes, and thereby block out an unwelcome message.


1.8.5 Terminology

Do not be surprised if your instructors and tutors use different terminology from what youhave 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 asnegative 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 thereplacement of x by “its negative”; this is not entirely consistent with the usual practice, but isan “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 ofx — 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 ofthe functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolic functions, which wehave met on general functions, so a formula like f 2(x) does not have an obvious meaning, andwe 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 taughtto interpret log x as being the logarithm to base 10, you should now forget that — although itcould be the labelling convention of your calculator. Most often, if your instructor speaks of alogarithm, and writes log x, he will be referring to the base e, i.e. to loge; that is, he is referringto 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 atthe beginning of the discussion, as “all logarithms in this discussion are to the base 2”. Yourinstructors try to think like mathematicians even when lecturing to their classes, and so we usethe language and terminology we use when talking to each other.

1.9 Communication with Instructors and TA’s1. 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 wecan clearly identify you.

2. The only messages sent through WeBWorK should be those generated by the Feedbackfacility: this means a message that refers to a specific problem on a specific WeBWorK


assignment, generated by clicking on the Feedback button while you are working thatproblem, 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. Thisfacility is difficult for instructors to use, since it is not integrated with the other mailservices. We normally disable myCourses mail for that reason. If you had a need to senda message while you are connected to myCourses, just open another window and send amessage with your regular e-mail client.

4. Keep your e-mail address up to date Both myCourses and WeBWorK contain an e-mail address where we may assume you can be reached. If you prefer to use anothere-mail address, the most convenient way is to forward your mail from your studentmailbox, leaving the recorded addresses in these two systems unchanged.

1.10 Commercial tutorial and exam preparation servicesWe do not endorse any commercial tutorial service, nor any service that claims to prepare stu-dents 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!

28This facility should be used sparingly; you should not expect instant response, so questions sent close to thedue time on the due date will not likely receive a reply before the assignment becomes due.


1.11 Special Office Hours and TutorialsThe following chart will show any special activities that are scheduled during the term. Thistable was last updated on December 28th, 2009 TO BE UPDATED.

Review Tutorial TA/Instructor location Date Time



2 References

2.1 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 Sin-gle 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 Vari-able 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 Transcenden-tals), 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 Sin-gle 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 Vari-able 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.


[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 0-534-10307-3.

[13] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcenden-tals), 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, 5thEdition. 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). ISBN0-534-36298-2.

[19] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s SingleVariable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN0-534-36301-6.

[20] J. Stewart, L. Redlin, S. Watson, Precalculus: Mathematics for Calculus, EnhancedReview Edition. Thomson * Brooks/Cole. (2006). ISBN: 0-495-39276-6.

[21] J. Stewart, Trigonometry for Calculus. Thomson * Brooks/Cole. ISBN: 0-17-641227-1.

2.2 Other Calculus Textbooks

2.2.1 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.


[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, FifthEdition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79803-4.

[26] R. A. Adams, Calculus: A Complete Course, with Solution Manual, Fifth Edition. Ad-dison, 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.

2.2.2 Larson, Hostetler, et al.

[28] Calculus Instructional DVD Program, for use with (inter alia) Lar-son/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 Transcenden-tal Functions, Third Edition. Houghton Mifflin Company, Boston (2003). ISBN 0-618-22307-X.

2.2.3 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 Tran-scendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN0-13-793076-3.

[32] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus with Ana-lytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, EnglewoodCliffs, NJ (1997). ISBN 0-13-079875-4.

[33] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with Analytic Geome-try, 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 VariableCalculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Pren-tice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095247-1.


2.2.4 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 andCompany, 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.

2.3 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 athttp://coursecalendar.mcgill.ca/ug200910/wwhelp/wwhimpl/js/html/wwhelp.htm

Information for Students in Lecture Section 1 of MATH 141 2010 01 1001

A Timetable for Lecture Section 001 of MATH 141 2010 01Distribution Date: Monday, January 04th, 2010

(Updated on January 06th, 2010. Subject to further correction and change.)Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYJANUARY

04 §5.1, §5.2 06 §5.1, §5.2 08 §5.3, §5.4Tutorials begin week of January 11th, 2010

11 §5.4, §5.5 13 §5.5 15 §6.1Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010

18 §6.2 Q1 20 §6.2, §6.3 Q1 22 §6.3 Q1

WeBWorK Assignment A1 due Jan. 24, 2010Deadline for withdrawal with fee refund = Jan. 24, 2010

Verification Period: January 25 – 29, 201025 §6.5 A1 27 §6.5, §7.1 29 §7.1

FEBRUARY01 §7.2 03 §7.2, §7.3 05 §7.3

WeBWorK Assignment A2 due Feb. 07, 201008 §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, 201015 §7.8 17 §8.1, §8.2 19 §8.2

Study Break: February 21 – 27, 2010No lectures, no regular office hours, no regular tutorials!

WeBWorK Assignment A3 due Feb. 21, 201022 NO LECTURE, NO TU-

TORIALS A3

24 NO LECTURE, NO TU-TORIALS


Notation: An = Regular WeBWorK Assignment An due about 23:30 hourson the Sunday preceding this Monday

Qn = Quiz Qn will be administered at the tutorials this week.X = reserved for eXpansion or review


Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYMARCH

01 §10.1,§10.2 03 §10.3 05 §10.3, §10.4WeBWorK Assignment A4 due Mar. 07, 2010

08 §10.4 A4 10 §11.1, §11.2 12 §11.215 §11.3 Q3 17 §11.4 Q3 19 §11.4, §11.5 Q3

WeBWorK Assignment A5 due Monday, Mar. 21, 201022 §11.5 A5 24 §11.6 26 §11.629 §11.7 Q4 31 X Q4

APRIL02 NO LECTURE Q4

This week’s tutorials are the last.WeBWorK Assignment A6 due Apr. 04, 2010

05 NO LECTURE 08 X 10 X12 X 14 X




B Timetable for Lecture Section 002 of MATH 141 2009 01Distribution Date: Monday, January 04th, 2010

(Subject to further correction and change.)Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYJANUARY

04 §5.1, §5.2 06 §5.3 08 §5.4Tutorials begin week of January 11th, 2010

11 §5.4, §5.5 13 §5.5 15 §6.1Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010

18 §6.2 Q1 20 §6.2, §6.3 Q1 22 §6.3 Q1

WeBWorK Assignment A1 due Jan. 24, 2010Deadline for withdrawal with fee refund = Jan. 24, 2010

Verification Period: January 25 – 29, 201025 §6.5 A1 27 §6.5, §7.1 29 §7.1

FEBRUARY01 §7.2 03 §7.2, §7.3 05 §7.3

WeBWorK Assignment A2 due Feb. 07, 201008 §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, 201015 §7.8 17 §8.1, §8.2 19 §8.2

Study Break: February 21 – 27, 2010No lectures, no regular office hours, no regular tutorials!

WeBWorK Assignment A3 due Feb. 21, 201022 NO LECTURE, NO TU-

TORIALS A3






Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYMARCH

01 §10.1,§10.2 03 §10.3 05 §10.3, §10.4WeBWorK Assignment A4 due Mar. 07, 2010

08 §10.4 A4 10 §11.1, §11.2 12 §11.215 §11.3 Q3 17 §11.4 Q3 19 §11.4, §11.5 Q3

WeBWorK Assignment A5 due Monday, Mar. 21, 201022 §11.5 A5 24 §11.6 26 §11.629 §11.7 Q4 31 X Q4

APRIL02 NO LECTURE Q4

This week’s tutorials are the last.WeBWorK Assignment A6 due Apr. 04, 2010

05 NO LECTURE 08 X 10 X12 X 14 X




C Supplementary Notes for Students in Section 001 of MATH141 2010 01

C.1 Lecture style in Lecture Section 001Lecture content. The timetable on pages 1001, 1002 will show you approximately what Iplan to discuss at each lecture. I suggest that you look through the material in advance. Ifyou have time to try some of the exercises, and find some that cause you difficulty, you arewelcome to bring them to my attention; perhaps I may be able to work some of these examplesinto the lecture.

What goes on the chalkboard? — Should I take notes? I believe strongly that studentsshould 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 takenotes, 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 whenthe material on the board is equivalent to something in your textbook, the act of writing mayhelp you remember it. But much of the material will be restatements of your textbook, so youshould normally not panic if you miss something.

Graphs My emphasis is on qualitative properties of the graphs of functions, but not on theproduction of extremely precise graphs. You can expect to see me draw on the chalkboardsketches that are extremely crude approximations of functions, sometimes even caricaturesof the true graph. Mathematicians do not base proofs on sketches of graphs — the role of asketch is usually only to assist the reader to visualize the verbal or symbolic reasoning whichaccompanies it. Sometimes a graph is used help one discover a phenomenon, but the resultwould 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 inthe textbook. While some of the comments or explanations may be helpful in understanding thebook, the notes are not required reading for examination purposes. Sometimes it may happenthat the discussion of a topic or an exercise evolves during the lecture into one which requires

29This is why I usually avoid problems in the textbook that appear to be drawing inferences from graphs.


more detail than is practical to write on the chalkboard. In such cases you may be referredat the following lecture to supplementary material that will be contained in the notes placedon the Web. Such evolutions are spontaneous and not planned, and cannot be announced inadvance.

Timing and corrections The notes will usually not be posted until after the lecture. WhileI do try to check the notes before posting them, there will inevitably be errors: if you seesomething that doesn’t look right, please ask. The notes will be progressively corrected asmisprints and other errors are discovered.

C.2 Supplementary Notes for the Lecture of January 04th, 2010Release 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 appli-cations: “The Tangent Problem”, which was geometric; and “The Velocity Problem”, whichwas physical. Now, in motivating the integral calculus, the author presents two analogousproblems: “The Area Problem”, which is geometric; and “The Distance Problem”, which isphysical.

The Area Problem. The textbook discusses approximation of the area under the graph ofy = 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 boundedabove and below by easily computable sums, which together converge to the same value astheir number approaches ∞ and their width approaches 0. You should become comfortableusing the “sigma notation”, where, for integers n1 and n2 with n1 ≤ n2, we write

n2∑

i=n1

f (i)

to mean the sum f (n1) + f (n1 + 1) + f (n1 + 2) + . . . + f (n2) . The textbook then proposes thefollowing definition of the area between the graph of a function, two vertical lines, and thex-axis:


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 = aand x = b (where a ≤ b) is the limit of a sum of the areas of rectangles “hanging” from thegraph, 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 x∗i

(i = 1, . . . , n) such that xi ≤ x∗i+1 ≤ xi+1, and consider the sum Rn =

n∑

i=1

f(x∗i

)(xi − xi−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. Wealso need to clarify what restrictions hold for the points x1, x2, . . . xn−1, and how we select thepoints x∗1, x∗2, . . . x∗n.)

Example C.1 (cf. [1, Example 5.1.2, p. 357]) Let a and b be non-negative real numbers, andconsider the area under the parabola y = x2 between the vertical lines x = a, x = b, and abovethe 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.

2. Divide the interval [0, b] into n intervals of equal widthb − 0

n; thus the points x1, x2, . . .

are chosen to be xi =b − 0

n· i (i = 1, 2, . . . , n). For the sample points x∗i , let’s consider

the points at the right end of each of the subintervals: so x∗i = xi (i = 1, 2, . . . , n). Then

Rn =

n∑

i=1

(b − 0

n· i)2

·((

b − 0n· i)−

(b − 0

n· (i − 1)

))(1)

where the second factor is simply the common length of all the subintervals, i.e.,bn

.

The first factor is squared because we are evaluating the function f (x) = x2 at the pointb − 0

n· i. In order to evaluate this sum we need to know the sum of the squares of the

first n positive integers (=natural numbers). If you didn’t learn this in high school, hereit is:

n∑

i=1

i2 =n(n + 1)(2n + 1)

6. (2)

We won’t be able to prove this formula here: a proof requires use of a tool like Mathe-matical Induction, which we are omitting from the syllabus. Applying (2) to (1) yields

Rn = b3 ·n(n + 1)(2n + 1)

6n3 = b3 ·

(2 +

3n

+1n2

)

6.


As n is permitted to become arbitrarily large positively, Rn → b3

3.

3. If the region starts at the origin and extends to the line x = a, then the area will bea3

3.

So, if the region we wish to study starts at the line x = a, and ends at the line x = b,

where b ≥ a, we need only subtract one of these areas from the other, obtainingb3 − a3

3.


C.3 Supplementary Notes for the Lecture of January 06th, 2010Release 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 typeof problem, and shows that the solution is the same type of sum met in the Area Problemabove. In this case the problem is to determine the distance travelled by a particle moving sothat its velocity at time t is a given function f (t). It will be seen that the distance can again beinterpreted as a limit of a sum — the same sum that would be seen if we attempted to determinethe area in the xt-plane under the graph x = f (t).

If a particle is known to be moving along the x-axis at a velocity of v(t) = x′(t) =dxdt

(t)cm/s, how much distance is traversed between time t = a and time t = b? If, by distance, wemean displacement or “signed distance”, where movement to the right counts positively, andto the left negatively30, then the distance is x(b)− x(a); we shall see that this can be interpreted

as the area under the graph y =ddt

x(t) between x = a and x = b, which we will be denoting

by

b∫

a

ddt

x(t) dt . When it is intended to consider all motion as non-negative — the way the

odometer of an automobile measures distance, then we would want to find the area under thegraph not of the velocity, v(t) = x′(t), but of the speed, |v(t)| = |x′(t)|, and the distance travelledwould be ∫ b

a|v′(t)| dt =

∫ b

a

∣∣∣∣∣ddt

x(t)∣∣∣∣∣ dt .

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 dis-

tance travelled by the moving particle. But we expressed the velocity as the derivative of thedisplacement of the particle relative to some fixed origin; and the distance travelled can beexpressed as the difference between two values of the displacement. This is a special case ofthe Fundamental Theorem of the (Integral) Calculus, which will be introduced below.

5.1 Exercises30Note that, with this definition, a particle that moves around and then returns to the same point will have

travelled a distance of 0.


[1, Exercise 20, p. 365] Determine a region whose area is equal to the limit

limn→∞

n∑

i=1

2n

(5 +

2in

)10

.

Do not evaluate the limit at this time.

Solution: Take the widths of the approximating rectangles to be constant,

4x =2n

;

then n of these constant widths would span a distance of length 2. The limit can be

seen to be limn→∞4x ·

n∑i=1

(5 + i4x)10 . The rectangles could be interpreted, for example,

as “hanging by their upper right hand corners” from the curve y = (5 + x)10 abovethe interval 0 ≤ x ≤ 2. (Later in the course we shall see that the area is (5+2)11−511

11 =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

from the graph. In that case the sum could have been written as limn→∞

n∑

i=1

2n

(5 +

2(i − 1)n

)10

;

or, alternatively, as limn→∞

n−1∑

i=0

2n

(5 +

2in

)10

.

The original limit sum, or the limits of either of the latter sums, could also be interpretedas being the area under the graph of f (x) = x10 between x = 5 and x = 7. The transitionfrom this interpretation to the earlier one is represented by moving the location of they-axis — without changing the area of the region.

[1, Exercise 21, p. 365] Determine a region whose area is equal to the limit

limn→∞

n∑

i=1

π

4ntan

iπ4n

.

Do not evaluate the limit at this time.

Solution: You should read the discussion of this problem in your Student Solution Man-ual. What is different from the preceding example to this one, is that we don’t have anystraightforward algebraic way of evaluating the limit of the sum of rectangles obtainedto approximate this area. Thus there is going to be something new in our theory if wewill be able to determine this area exactly. And, in fact, we shall be able to evaluate this


area exactly! The upper boundary of the region in this case could be the graph of thefunction y = tan x, and the other boundaries of the region could be, in addition to thex-axis, the vertical lines x = 0 and x =

π

4.

C.3.2 §5.2 The Definite Integral

The formal definition of the integral involves a number of technical difficulties which I shallnot consider in detail in this course. You should read the definition the textbook gives of theintegral [1, p. 366], but you are not going to be asked to work with it in full generality; in factthe definition given in the textbook is simpler than the definition that is normally used for theRiemann Integral. We would need to appeal to this definition if we wished to formally proveall the properties that the author is going to ascribe to the integral; but we shall not attempt toprovide such proofs.

The usual definition of the integral would permit the widths of the subintervals, here de-noted by ∆x, to be different: ∆x1 for the first subinterval, ∆x2 for the second, etc., and wouldthen require that the largest of them should approach zero. This technicality is needed forgeneral 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 func-tion on the subintervals, we have what is called the upper Riemann sum; analogously, we mayspeak of the lower Riemann sum. To prove the existence of the definite integral we would wantto show that the difference between these sums approaches 0 in the limit. This can be shownto be the case in particular when the function is continuous everywhere, and it is even true incertain more general situations. In this course we will normally be taking the functions to beeither continuous or, more generally, “piecewise continuous”; that is, we will permit functionswhich can be obtained by “gluing” together functions which are continuous over adjacent in-tervals. As long as there are only a finite number of such components, it can be shown that the


integral exists; it doesn’t matter if the function is discontinuous at the finite number of loca-tions were the functions are “glued”. But some of the properties we will be using will applyonly to continuous functions, and we may have to break a problem up into parts in order tosolve it. More about this later.

Evaluating Integrals. Among the integrals discussed in this subsection are several that re-quire the following formulæ for sums of powers of the natural numbers:

n∑

i=1

i1 =n(n + 1)

2(3)

n∑

i=1

i2 =n(n + 1)(2n + 1)

6(4)

n∑

i=1

i3 =n2(n + 1)2

4(5)

to which we could add the following trivial result31:

n∑

i=1

i0 = n . (6)

Formulæ 3, 4, 5 are proved in [1, Appendix E], but you are not expected to have read thoseproofs. 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 tofollow any general pattern is not because there is no pattern, but simply that we are “asking the wrongquestion”. If, instead, we had asked for the sums of what are called falling factorials, i.e., productsof an integer with successive integers immediately less than it, we would obtain the following, muchprettier results. You do not need to remember these formulæ:

n∑

i=1

i0 =n1

(7)

n∑

i=1

i =(n + 1)n

2(8)

31A definition that the product of an empty set of numbers is equal to 1 is consistent with the definition ofmultiplication of real numbers.

32Mathematical Induction was not an examination topic in MATH 140 2009 09; in the present course you arenot expected to know how to apply Mathematical Induction, but interested students are urged to read about it inthe textbook [1, pp. 77].


n∑

i=1

i(i − 1) =(n + 1)n(n − 1)

3(9)

n∑

i=1

i(i − 1)(i − 2) =(n + 1)n(n − 1)(n − 2)

4(10)

A glance at these formulæ, which are certainly “prettier” than the formulæ for the sums of the powers,shows that the first one, (7) 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 ineach of the sums; only in the case of the 0th powers does this make any difference, since 00 is definedto be 1:

n∑

i=0

i0 =n + 1

1(11)

n∑

i=0

i =(n + 1)n

2(12)

n∑

i=0

i(i − 1) =(n + 1)n(n − 1)

3(13)

n∑

i=0

i(i − 1)(i − 2) =(n + 1)n(n − 1)(n − 2)

4. (14)

Now we can see much further; we can even conjecture that there is a general result that encompassesall of these particular cases:

n∑

i=0

i(i − 1)(i − 2) · . . . · (i − r + 1) =(n + 1)n(n − 1)(n − 2) · . . . · (n − r + 1)

r. (15)

And finally, of what use are these formulæ if we need the sums of the powers of the integers, notthe sums of “falling factorials”. Any power of n can be expressed in terms of “falling factorials”, forexample

n1 = n

n2 = n(n − 1) + n

n3 = n(n − 1)(n − 2) + 3n(n − 1) + n ,

so property (15) 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


cases of the following:k+∑

i=k

(rai + sbi) = rk+∑

i=k

ai + sk+∑

i=k

bi .

I may have more to say about the sigma notation after I discuss [1, §5.5], where we shallencounter properties of the integral that have analogues for sums. For the present let it be

noted that the symbol i ink+∑i=k

ai is not a “free” variable, in that you cannot assign any values to

it: it performs a function in the symbol, but that function would be performed equally well if

we replaced i by any other symbol that is not already in use, e.g.,k+∑u=k

au,k+∑λ=k

aλ,k+∑♥=k

a♥ .

The Midpoint Rule. The “Midpoint Rule” is an approximation formula for definite inte-grals. Use of an approximation formula entails a willingness to accept an error in the cal-culation. Mathematicians normally expect to see an estimate of how good or how bad anapproximation can be before recommending their use. A partial justification of the MidpointRule is contained in [1, §7.7], a section that is to be omitted from the syllabus. For that reasonyou are asked to omit this subsection: you will not be expected to know anything about theMidpoint Rule.

Properties of the Definite Integral – Linearity Properties. The textbook lists many prop-erties of the Definite Integral, proving some of them.

1.

b∫

a

c dx = c(b − a)

2.

b∫

a

[ f (x) + g(x)] dx =

b∫

a

f (x) dx +

b∫

a

g(x) dx

3.

b∫

a

c f (x) dx = c

b∫

a

f (x) dx

4.

b∫

a

[ f (x) − g(x)] dx =

b∫

a

f (x) dx −b∫

a

g(x) dx

for any real numbers a, b, c, and any continuous functions f , g.


Some of these properties can be derived from others, or can be combined into a more generalformula. So, for example we can prove that

∫ b

a(r · f (x) + s · g(x)) dx = r

∫ b

af (x) dx + s

∫ b

ag(x) dx (16)

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx (17)

for any real numbers r, s, a, b, c, and these two equations are equivalent to the properties thatthe textbook numbers ##2, 3, 4, etc. [1, p. 387]:

∫ a

af (x) dx = 0 (18)

∫ b

a( f (x) ± g(x)) dx =

∫ b

af (x) dx ±

∫ b

ag(x) dx (19)

∫ b

ac f (x) dx = c

∫ b

af (x) dx (20)

The first property in the textbook list,∫ b

ac dx = c(b − a) (21)

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 graphabove the x-axis, where the lower limit of the integral is not greater than the upper limit. Whenthe curve is below the x-axis, or the lower limit of the integral is not greater than the upperlimit, the area is negative. (This is the case for part of [1, Exercise 22, p. 377] which is solvedbelow: there the portion of the area that was below the x-axis cancelled part of the area abovethe x-axis; while the net result we obtained — 21 — was positive, it was not equal to thetotal of the magnitudes of the two areas above and below the x-axis, but was equal to theirdifference. The graph of the integrand crosses the x-axis at the points ±√6 − 1. The regionunder the interval [1,

√6−1] can be shown to have (negative) area −4

√6+ 28

3 ; while the regionover the interval [

√6 − 1, 4] can be shown to have (positive) area 35

3 + 4√

6.)

Properties of the Definite Integral – Additivity of the Interval. A second type or propertylisted states, in principle, that the area under a curve is the sum of the areas under any two partsinto which the curve can be decomposed:

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx (22)


for any real numbers a, b, c; here again, the constants are not necessarily positive, so one caninterpret the point c as lying outside of the interval [a, b] when a ≤ b. This property impliesanother property listed on [1, p. 373]:

∫ b

af (x) dx = −

∫ a

bf (x) dx (23)


D Problem Assignments from Previous Years

D.1 1998/1999The problem numbers listed below refer to the textbook in use at that time, [31], [33]. Formany 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


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


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” problemsin the textbook. In most of the following problems there is a reference to a “similar” problemin the textbook. You should always endeavour to show as much of your work as possible, andto reduce your solution to “simplest terms”. Remember that the main reason for submittingthis assignment is to have an opportunity for your tutor to grade your work; the actual gradeobtained should be of lesser significance.

In Exercises 1-5 below, evaluate the indefinite integral, and verify by differentiation:

1. (cf. [31, Exercise 5.2.5, p. 294])∫ (

3x4 − 5x

12 − x2 + 4x−3

)dx

2.∫ (

3x− 2

1 + x2

)dx

3. (cf. [31, Exercise 5.2.13, p. 294])∫ (

xex2 − e4x)

dx

4. (cf. [31, Exercise 5.2.19, p. 294])∫

(1 − √x)(2x + 3)2 dx

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 thatdydx

=1√

1 − x2and y

(2−

12

)=π

2.

7. (This is [31, Exercise 5.2.51, p. 295] written in purely mathematical terminology.) Solve

the initial value problem:ddx

(dydx

)= sin x, where y = 0 and

dydx

= 0 when x = 0. [Hint:

First use one of the initial values to determine the general value ofdydx

from the given“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

using the symbol∑

:6∑

j=1

(2 j − 1).


9. (cf. [31, Exercise 5.3.18, p. 306]) Write the following sum in “summation notation”:

x − x3

3+

x5

5− x7

7+ ... ± x999

999where the signs are alternating +, −, +, −, ... The sign of the last term has not been given— you should determine it.

10. (cf. [31, Example 5.3.6, p. 302]) Given thatn∑

i=1

i =n(n + 1)

2,

n∑

i=1

i2 =n(n + 1)(2n + 1)

6,

n∑

i=1

i3 =n2(n + 1)2

4,

determine limn→∞

(n + 1)3 + (n + 2)3 + ... + (2n)3

n4 .

D.2.2 Assignment 2

1. Evaluate the following integrals:

(a)∫ 3

1(x − 1)4 dx

(b)∫ 1

0(2ex − 1)2 dx

(c)∫ π

0sin 4x dx.

2. Interpreting the following integral as the area of a region, evaluate it using known areaformulas: ∫ 6

0

√36 − x2 dx.

3. Use properties of integrals to establish the following inequality without evaluating theintegral: ∫ 1

0

11 +√

xdx ≤

∫ 1

0

11 + x3 dx.

4. Deduce the Second Comparison Property of integrals from the First Comparison Prop-erty [31, p. 325, §5.5].

5. Apply the Fundamental Theorem of Calculus [31, p. 331, §5.6] to find the derivative ofthe given function: ∫ x

−1(t2 + 2)15 dt.


6. Differentiate the functions

(a)∫ x3

0cos t dt

(b)∫ 3x

1sin t2 dt.

7. Solve the initial value problemdydx

=√

1 + x2 , y(1) = 5 . Express your answer interms of a definite integral (which you need not attempt to evaluate). This problem canbe solved using the methods of [31, Chapter 5].

8. Evaluate the indefinite integrals:

(a)∫

2x√

3 − 2x2 dx

(b)∫

x2 sin(3x3) dx

(c)∫

x + 3x2 + 6x + 3

dx

9. Evaluate the definite integrals:

(a)∫ 8

0t√

t + 2 dt

(b)∫ π/2

0(1 + 3 sin η)3/2 cos η dη

(c)∫ π

0sin2 2t dt.

10. Sketch the region bounded by the given curves, then find its area:

(a) x = 4y2, x + 12y + 5 = 0

(b) y = cos x, y = sin x, 0 ≤ x ≤ π

4.

11. Prove that the area of the ellipsex2

a2 +y2

b2 = 1 is A = πab. This problem can besolved using the methods of [31, Chapter 5]. It is not necessary to use methods of [31,Chapter 9].



D.2.3 Assignment 3

In all of these problems you are expected to show all your work neatly. (This assignment isonly a sampling. Your are advised to try other problems from your textbook; solutions to somecan 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 nsubintervals of equal length ∆x by n − 1 equally spaced points x1, x2, ..., xn−1 (where

x0 = 2, xn = 4). Evaluate limn→∞

n∑

i=1

1xi

∆x by computing the value of the appropriate

related integral.

2. (a) [31, Exercise 6.2.6, p. 391] Use the method of cross-sections to find the volume ofthe solid that is generated by rotating the plane region bounded by y = 9 − x2 andy = 0 about the x-axis.

(b) (cf. Problem 2a) Use the method of cylindrical shells to find the volume of the solidthat is generated by rotating the plane region bounded by y = 9−x2 and y = 0 aboutthe x-axis.

(c) Use the method of cross-sections to find the volume of the solid that is generatedby 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 solidthat is generated by rotating the plane region bounded by y = 9−x2 and y = 0 aboutthe y-axis.

3. (a) [31, Exercise 6.2.24, p. 392] Find the volume of the solid that is generated byrotating 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 cylin-drical shells were used to represent the volume of the solid that is generated byrotating 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 di-ameter AB of length 2a. Find the volume of the solid if each cross section perpendicularto AB is an equilateral triangle.

5. (a) [31, Exercise 6.3.26, p. 401] Use the method of cylindrical shells to find the volumeof the solid generated by rotating around the y-axis the region bounded by the

curves y =1

1 + x2 , y = 0, x = 0, x = 2.


(b) (cf. Problem 5a) Use the method of cross sections to find the volume of the solidgenerated by rotating around the y-axis the region bounded by the curves y =

11 + x2 , y = 0, x = 0, x = 2.

6. (cf. [31, Exercise 7.3.69, p. 450]) Find the length of the arc of the curve y =ex + e−x

2between 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 generatedby 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 surfaceof revolution generated by revolving the arc of the curve y = x3 from x = 1 to x = 2around the y-axis.

8. [31, Exercise 7.2.44, p. 442] Evaluate the indefinite integral∫

x + 1x2 + 2x + 3

dx

9. (cf. [31, Exercise 7.2.36, p. 442]) Determine the value of the function f (x) =

∫ x

−1

t2

8 − t3 dt

for any point x < 2.

10. (cf. [31, Exercise 7.3.70, p. 450]) Find the area of the surface generated by revolvingaround 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 ex + tan−1 e−x

2. Showing all your work, evaluate the integrals:

(a)∫

dx√1 − 4x2

(b)∫

dx2√

x(1 + x)


(c)∫

ex

1 + e2x dx

(d)∫ cot

√y csc

√y√

ydy

(e)∫

(ln t)8

tdt

(f)∫

tan4 2x sec2 2x dx

(g) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5.∫

x2

√16x2 + 9

dx

3. Use integration by parts to compute the following integrals. Show all your work.

(a)∫

t cos t dt

(b)∫ √

y ln y dy

(c) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5.∫

x2 arctan x dx

(d)∫

csc3 x dx

(e)∫

ln(1 + x2) dx

4. Showing all your work, evaluate the following integrals:

(a)∫

cos2 7x dx

(b)∫

cos2 x sin3 x dx

(c)∫

sin3 2xcos2 2x

dx

(d)∫

sec6 2t dt


D.2.5 Assignment 5

1. [31, Exercise 9.5.6, p. 540] Find∫

x3

x2 + x − 6dx. (Your solution should be valid for x

in any one of the intervals x < 3, −3 < x < 2, x > 2.)

2. [31, Exercise 9.5.8, p. 540] Find∫

1(x + 1)(x2 + 1)

dx.

3. (a) [31, Exercise 9.5.23] Find∫

x2

(x + 2)3 dx.

(b) Find the volume of the solid of revolution generated by the region bounded byy =

x

(x + 2)32

, y = 0, x = 1, and x = 2 about the x-axis.

(c) Find the volume of the solid of revolution generated by the region bounded byy =

x

(x + 2)32

, y = 0, x = 1, and x = 2 about the y-axis.

4. [31, Exercise 9.5.38, p. 540] Make a preliminary substitution before using the methodof partial fractions: ∫

cos θsin2 θ(sin θ − 6)

dθ

5. [31, Exercise 9.6.6, p. 547] Use trigonometric substitutions to evaluate the integral∫x2

√9 − 4x2

dx.

6. [31, Exercise 9.6.26, p. 547] Use trigonometric substitutions to evaluate the integral∫1

9 + 4x2 dx.

7. [31, Exercise 9.6.35, p. 547] Use trigonometric substitutions to evaluate the integral∫ √x2 − 5x2 dx.

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 whether the following improper integral con-

verges; if it does converge, evaluate it:∫ ∞

−∞

xx2 + 4

dx.

10. [31, Exercise 9.8.27, p. 561] Determine whether the following improper integral con-

verges; if it does converge, evaluate it:∫ ∞

0cos x dx.


11. (cf. [31, Exercise 9.8.14, p. 561]) Determine whether the following improper integral

converges; if it does converge, evaluate it:∫ +8

−8

1

(x + 4)23

dx.

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 inboth 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 intersectionof 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 equa-tions 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 findingpoints of intersection is incomplete. Your instructor will discuss a systematic procedurein 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.


3. Find the area of the region described.

(a) Inside both r = cos θ and r =√

3 sin θ.

(b) Inside both r = 2 cos θ and 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)) ast 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.

(a) an =n2 − n + 72n3 + n2 ,

(b) an =1 + (−1)n√n

(3/2)n ,

(c) an = n sin πn,

(d) an =

(n − 1n + 1

)n

.

8. Determine, for each of the following infinite series, whether it converges or diverges. Ifit converges, find its sum.

(a) 1 + 3 + 5 + 7 + . . . + (2n − 1) + . . . ,

(b) 4 + 43 + . . . + 4

3n + . . . ,

(c)∞∑

n=1

(5−n − 7−n),

(d)∞∑

n=1

( eπ

)n.

9. Find the set of all those values of x for which the series∞∑

n=1

( x3

)nis a convergent geomet-

ric series, then express the sum of the series as a function of x.


10. Find the Taylor polynomial in powers of x − a with remainder by using the given valuesof a and n.

(a) f (x) = sin x; a = π/6, n = 3.

(b) f (x) =1

(x − 4)2 ; a = 5, n = 5 .

11. Find the Maclaurin series of the function e−3x by substitution in the series for ex.

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 convergeor diverge.

(a)∞∑

n=1

11 + 3n ,

(b)∞∑

n=1

√n

n2 + n,

(c)∞∑

n=1

sin2(1/n)n2 .

D.3 2000/2001(In the winter of the year 2001 Assignments based on WeBWorK were used, although theexperiment had to be terminated in mid-term because of technical problems.)

D.4 2001/2002This was the first time WeBWorK assignments were used exclusively in this course.

D.5 MATH 141 2003 01WeBWorK assignments were used exclusively for assignments. The questions are not avail-able for publication.

D.6 MATH 141 2004 01WeBWorK assignments were used exclusively for assignments. The questions are not avail-able for publication.


D.7 MATH 141 2005 01WeBWorK assignments were used for online assignments; the questions are not available forpublication. In addition, these written assignments were intended to provide students with in-dividualized opportunities to work problems for which the textbook often provided examples;at the time, appropriate materials available from WeBWorK for this purpose were limited. Theindividualization 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 beindividualized, 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 writtenassignments may be designed in other ways. Because of its general form, it was possible torelease this assignment in the document Information for Students in MATH 141 2005 01; someof 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 howto write full solutions to problems. While they carry a very small weight in the computation ofyour final grade, conscientious completion of the assignments should help you substantially inlearning the calculus, and help prepare you for your final examination.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted 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’sregulations concerning plagiarism.

Signature(required) Date(required)

The assignment questions Your assignment consists of the following problems on your ver-sion of R1:


##3, 4, 5, 6, 7, 8

(Teaching Assistants are not primarily checking for plagiarism; but, if they detect it, they maybe 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, thenumerical answer alone may be worth 0 marks. You should submit a full solution, similar tosolutions in Stewart’s textbook or the Student Solutions Manual, which are where you shouldlook 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 ofa calculator. In particular, you are expected to be familiar with the values of trigonometricfunctions at “simple” multiples and submultiples of π.

Not all problems may be graded On all of the written assignments it is possible that theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. Forthat reason you are advised to devote equal attention to all of the problems.


Written assignment W2 is based on your WeBWorK assignment R3, but some problems arebeing modified. Your completed assignment must be submitted together with your solutions toquiz 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 howto write full solutions to problems. While they carry a very small weight in the computation ofyour final grade, conscientious completion of the assignments should help you substantially inlearning the calculus, and help prepare you for your final examination.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted certificate of originality, signed in ink:




and assert that my work submitted for W2 and R3 does not violate McGill’sregulations concerning plagiarism.


The assignment questions Your assignment consists of the following problems on your ver-sion of R1:

1. Problem 1 of R3, solved by integration with respect to x. Include in your solution a roughsketch 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 theequations of the curves appropriately. You may assume without proof that

∫ln x dx = x(ln x − 1) + C ,

a fact which you will see derived later in the course. Include in your solution a roughsketch of the region, showing a typical element of area.

3. Problem 6 of R3, evaluated using the Method of Washers. Include in your solution arough sketch of the plane region which generates the solid, showing a typical element ofarea which will generate a typical washer.

4. Problem 6 of R3, evaluated using the Method of Cylindrical Shells. Include in yoursolution a rough sketch of the plane region which generates the solid, showing a typicalelement of area which will generate a typical cylindrical shell.

Complete solutions are required It is not enough to give the correct answer; in fact, thenumerical answer alone may be worth 0 marks. You should submit a full solution, similar tosolutions in Stewart’s textbook or the Student Solutions Manual, which are where you shouldlook 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 acalculator.


Not all problems may be graded On all of the written assignments it is possible that theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. Forthat reason you are advised to devote equal attention to all of the problems.


Unlike the preceding written assignments Written Assignment W3 is not directly based on yourWeBWorK assignments, although some problems will be similar to WeBWorK assignmentproblems. Your completed assignment must be submitted together with your solutions to quizQ3, 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 followingcompleted statement of originality, signed in ink:



and assert that my work submitted for W3 does not violate McGill’s regula-tions concerning plagiarism.


The assignment questions The parameters in these problems are based on the digits of your9-digit McGill student number, according to the following table:

Parameter name: A B C D E F G H JYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution.

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 differenti-ation that your answer is correct.


2. Showing all your work, use trigonometric or hyperbolic substitutions to evaluate

each of∫

du√u2 − (F + 1)2

,

∫du√

u2 + (F + 1)2,

∫du√

−u2 + (F + 1)2.

Verify that your answers are correct by differentiation.

3. Let K =⌊

J+42

⌋=

[[J+4

2

]].33 Showing all your work, develop for this integer K a reduction

formula of the following type that can be used to evaluate

In(x) =

∫xn (sin(Kx)) dx in terms of In−2(x):

∫xn (sin Kx) dx = L · xn cos Kx + M · xn−1 sin Kx + N

∫xn−2 (sin Kx) dx

where n ≥ 2 and L, M,N are constants that you are expected to determine only byintegration by parts. Again showing all your work, use the reduction formula you havejust determined to evaluate

∫x2 sin Kx dx, and test by differentiation the answer that it

gives — you should recover x2 sin Kx.

4. Showing all your work, evaluate both of the integrals∫

sinF+1 x · cos2 x dx and∫

sinF+2 x · cos2 x dx

Complete solutions are required It is not enough to give the correct answer; in fact, thefinal answer alone may be worth 0 marks. You should submit a full solution, similar to solu-tions in Stewart’s textbook or the Student Solutions Manual, which are where you should lookif 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 theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. If forno other reason, you are advised to devote equal attention to all of the problems.


Your completed assignment must be submitted together with your solutions to quiz Q4, insideyour answer sheet for that quiz. No other method of submission is acceptable.

33Determine K from J using the greatest integer function, defined in your textbook, page 110. Your textbookuses the notation

[[J+4

2

]], but some authors write the function as

⌊J+4

2

⌋.







The assignment questions Some of the parameters in these problems are based on the digitsof your 9-digit McGill student number, according to the following table:

Parameter name: A B M D E F G H JYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution. It is not enough to give the correct answer; in fact, the final answer alone could beworth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook orthe Student Solutions Manual, which are where you should look if you have doubts about theamount 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:∫x

(x + J)2(x − H − 1)1 dx . Verify your work by differentiation of your answer: you

should recover the integrand. (Systematically means that you to use the methods youlearned in this course for the treatment of problems of this type, even if you happen tosee 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 rationalfunction, then integrate the particular integral that is assigned for your particular valueof G:

If G = 1,4,or 7:∫

1

x − √x + 2dx

If G = 0, 2, 5, or 8:∫

cos 2xsin2 2x + sin 2x

dx


If G = 3, 6, or 9:∫

1

x − 2 − 2√

x + 1dx

3. Problem 55, pages 504-505 of your textbook, describes a substitution discovered by KarlWeierstrass (1815-1897) for the evaluation of rational functions of sin x and cos x intoan ordinary rational function. It states that if, for x such that −π < x < π, we define

t = tan x2 , then cos x =

1 − t2

1 + t2 and sin x =2t

1 + t2 and that, at a consequence, dx =

21 + t2 dt . You are not asked to verify these facts. You are asked to use the substitutionto transform the following integral and then to evaluate it:

∫ π2

π3

11 + sin x − cos x

dx

4. Use the trigonometric identities given in your textbook on page 487 to evaluate the fol-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 convergentor divergent. If it is convergent, determine its value (again showing all your work):

(a)∫ 4

0

12 + E · F dx

(b)∫ ∞

0

x2 − E − Fx2 + E + F

dx

(c)∫ π

B+4

0

1x sin((G + 2)x)

dx


Your completed assignment must be submitted together with your solutions to quiz Q5, insideyour answer sheet for that quiz. No other method of submission is acceptable.







The assignment questions Some of the parameters in these problems are based on the digitsof your 9-digit McGill student number, according to the following table:

Parameter name: A B D E F G H J KYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution. It is not enough to give the correct answer; in fact, the final answer alone could beworth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook orthe Student Solutions Manual, which are where you should look if you have doubts about theamount of detail required in a solution. Not all problems may be graded.

1. For the point with polar coordinates(H,− π

K2 + 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 polaraxis, the origin is at the pole, and the positive y-axis is obtained by turning the polaraxis through a positive angle of π

2 .

2. For the point whose cartesian coordinates are (F2+1, F2−2), determine polar coordinates(r, θ) with the following properties:

(a) r > 0, 0 ≤ θ < 2π

(b) r < 0, 0 ≤ θ < 2π

(c) r > 0, 5π2 ≤ θ < 9π

2

3. For the following curve given in polar coordinates, determine, showing all your compu-tations, the slope of the tangent at the point with θ =

π

4:

r = A + cos(B + 2)θ .


4. Showing all your work, find the area contained between the outer loop and the inner loopof 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 − (E2 − 2)t3

Showing all your work, determine the value ofd2ydx2 (t).

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)

(b) an =ln n

ln((A + 2)n)

(c) an =√

n2 + A2 −√

n2 + A2 + 2

D.8 MATH 141 2006 01D.8.1 Solution to Written Assignment W1

Release Date: Friday, 27 January, 2006The 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 theweek 16-20 January, and folded inside your solution sheet to the quiz which will bewritten at that time.

2. Your TA may make special arrangements for submission at other times, but, no solutionsare ever accepted after the end of the week.

3. Your solution must use the data on your own WeBWorK assignment, as described be-low.


The Assignment Question Problem 7 of WeBWorK assignment R1 requires that you eval-uate a definite integral of the form

∫ C

D(Ex2 − Ax + B) dx,

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 asequence of Riemann sums, using left endpoints34. You may simplify your problem only inthe following way: you may take the lower limit of the integral (here called D) to be 0; thisis 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 theproblem — it is not enough to supply the correct answer, and you must solve the version ofthe problem on your own WeBWorK assignment R1. In your solution you will need to use thewell 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 expectedto prove them here.35 (This is a time-consuming exercise; the purpose of the assignment is toensure 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 validnumerical solutions for all students. The particular solution for your version of the problemshould look much simpler, because, in place of all variables except n, there will be specificintegers.)

The length of each of n subintervals is

∆x =C − D

n.

Thus xi = D + i∆x = D + i(C − D

n

). Since we are using left endpoints we apply [7, top

formula, p. 381]∫ C

D

(Ex2 − Ax + B

)dx

= limn→∞

n∑

i=1

(Ex2

i−1 − Axi−1 + B)· ∆x

= limn→∞

n∑

i=1

(E

(D2 +

2D(C − D)n

· (i − 1) +(C − D)2

n2 · (i − 1)2)

34Note that the solution given in the textbook uses right endpoints.35Of 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.


−A(D +

C − Dn· (i − 1)

)+ B

)· ∆x

= limn→∞

(ED2 − AD + B)n∑

i=1

1 +(2DE − A)(C − D)

n

n∑

i=1

(i − 1) +E(C − D)2

n2

n∑

i=1

(i − 1)2

· ∆x

Nown∑

i=1

1 = sum of n 1’s

= nn∑

i=1

(i − 1) =

n−1∑

j=1

j

=(n − 1)n

2by [7, Formula 4, p. 383]

n∑

i=1

(i − 1)2 =

n−1∑

j=1

j2

=(n − 1)n(2n − 1)

6by [7, Formula 5, p. 383].

Substituting the values of these sums yields∫ C

D

(Ex2 − Ax + B

)dx

= limn→∞

C − Dn·((ED2 − AD + B)n +

(2DE − A)(C − D)n

· (n − 1)n2

+E(C − D)2

n2 · (n − 1)n(2n − 1)6

)

= limn→∞

((ED2 − AD + B)(C − D) +

(2DE − A)(C − D)2

2

(1 − 1

n

)

+E(C − D)3

6·(1 − 1

n

) (2 − 1

n

))

= (ED2 − AD + B)(C − D) +(2DE − A)(C − D)2

2· 1 +

E(C − D)3

6· 1 · 2

= (ED2 − AD + B)(C − D) +(2DE − A)(C − D)2

2+

E(C − D)3

3

=E

(C3 − D3

)

3−

A(C2 − D2

)

2+ B(C − D)

It was suggested that you simplify your problem by taking D = 0; in that case the value wouldhave worked out to be

EC3

3− AC2

2+ BC .


The Grading Scheme 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 grad-ing.)

D.8.2 Solution to Written Assignment W2

Release Date: Mounted on the Web on February 8th, 2006; corrected on March 16th, 2006Solutions 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 followingproblem: “Find the area enclosed by the line y = x − 1 and the parabola y2 = 2x + 6,” byintegrating with respect to y. The textbook then states the following: “We could have foundthe area in Example 6 by integrating with respect to x instead of y, but the calculation ismuch more involved. It would have meant splitting the region in two, and computing theareas labelled A1 and A2 in Figure 14. The method we used in Example 6 is much easier.” Yourassignment is to solve the problem by integrating with respect to x, and you may find the figurein the textbook helpful. The computation is not really very hard, but it does involve workingwith

√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 thatthere really is nothing to fear, even if you do happen to choose the “wrong” way to approach aproblem.

Your TA will be alert to the possibility that students might be copying their solutions fromothers. Please write up your own solution, so that your TA will not have to waste everyone’stime by sending exact copies to the disciplinary officer of the Faculty. You need to know howto solve problems of this type, since you could be expected to demonstrate that ability at afuture quiz or on the examination.

The Solution. The integration must be carried out separately for two subregions. This isbecause 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 liney = x − 1. The method you know for finding the area between two curves requires that theequations of the curves be written as the graphs of functions of x. But the parabola y2 = 2x + 6is not the graph of a function, since it crosses some vertical lines twice. However, we canfactorize the equation in the form

(y −

√2(x + 3)

) (y +

√2(x + 3)

)= 0 :


the parabola is the graph of two functions — the upper arm of the parabola is the graph ofy =

√2(x + 3), while the lower arm is the graph of y = −√2(x + 3). To the left of x = −1

the region A1 is bounded above by the graph of y =√

2(x + 3), and below by the graph ofy = −√2(x + 3); thus the height of the vertical element of area is the difference between thesetwo 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∫ −1

−32√

2(x + 3) dx +

∫ 5

−1

( √2(x + 3) − (x − 1)

)dx

=

[2√

2 · 23· (x + 3)

32

]−1

−3+

[√2 · 2

3· (x + 3)

32 −

(x2

2− x

)]5

−1

=4√

23

[2

32 − 0

]+

2√

23· 8 3

2 −(252− 5

) −2√

23· 2 3

2 −(12

+ 1)

=163

+

[563− 6

]=

163

+383

= 18.

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 markingscheme.)

D.8.3 Solutions to Written Assignment W3

Release Date: March 25th, 2006Solutions were to be submitted inside answer sheets to Quiz Q3, at tutorials during the period

13-16 February, 2006

The Problems

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 inyour textbook to [7, Example 4, p. 478], where the author evaluates

∫ex sin x dx.

2. 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 problem.

(a) You are to evaluate the following integral by integration by parts:∫ (

A7x2 + A8x1 + 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 isindeed an antiderivative of the given integrand.

Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears inthe Handbook on Student Rights and Responsibilities (PDF English version - French version). Article15(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 orher own in any academic writing, essay, thesis, research report, project or assignmentsubmitted in a course or program of study or represent as his or her own an entire essay orwork of another, whether the material so represented constitutes a part or the entirety ofthe work submitted.

Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you reallyneed to learn how to solve these kinds of problems yourself.

The Solutions

1. Let u = sin x, v′ = sin x. Then u′ = 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 samesolution.)

∫(sin x) · (sinh x) dx = (cos x) · sinh x −

∫(cosh x) · (cos x) dx .

To evaluate∫

(cosh x) · (cos x) dx I will set U = cos x, V ′ = cosh x, so U′ = − sin x,

V = sinh x.∫

(cosh x) · (cos x) dx = (cos x) · sinh x +

∫(sinh x) · sin x dx .

Combining these results yields∫

(sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) −∫

(sin x) · (sinh x) dx


which we may solve by moving the two integral terms to the same side of the equation:

2∫

(sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) + C

or ∫(sin x) · (sinh x) dx =

(sin x) · (cosh x) − (cos x) · (sinh x)2

+ C′

2. (a) In the first integration by parts I set

u = A7x2 + A8x1 + A9

dv = eA1 xdx⇒ du = (2A7x + A8)dx,

v =1A1

eA1 x .

Hence∫ (

A7x2 + A8x1 + A9

)eA1 x dx

=(A7x2 + A8x + A9

)· 1

A1eA1 x − 1

A1

∫(2A7x + A8) · eA1 x dx

=

(A7

A1x2 +

A8

A1x1 +

A9

A1

)· eA1 x −

∫ (2A7

A1· x +

A8

A1

)· eA1 x dx

To evaluate the subtracted integral we must apply Integration by Parts a secondtime:

U =2A7

A1· x +

A8

A1

dV = eA1 xdx

⇒ dU =2A7

A1dx

V =1A1

eA1 x , so∫ (

2A7

A1· x +

A8

A1

)· eA1 x dx =

(2A7

A1· x +

A8

A1

)· 1

A1eA1 x −

∫2A7

A21

eA1 x dx

=

(2A7

A21

· x +A8

A21

)· eA1 x − 2A7

A31

eA1 x

=

(2A7

A21

· x +A8

A21

− 2A7

A31

)· eA1 x


Combining our results yields∫ (

A7x2 + A8x1 + A9

)eA1 x dx

=

(A7

A1x2 +

(A8

A1− 2A7

A21

)x1 +

(A9

A1− A8

A21

+2A7

A31

))· eA1 x + C

(b) Remember that you were to differentiate the preceding product of polynomial andexponential by the Product Rule to show that, indeed, its derivative is the givenintegrand.


Release Date: March 25th, 2006Solutions 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 askedto write out complete solutions to the following modifications of your versions of problems onthat assignment. Note that, in some cases, the question asks for more than was asked on theassignment. 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 inyour own WeBWorK assignment:

∫Ex2 + Ex + F

x3 + Gx2 + Hx + Jdx .

HINT: −C is a root of the denominator.

(b) No marks will be given unless you verify your integration by differentiating youranswer.

2. Problem 12:

(a) Evaluate the following integral with the specific values of the constants given inyour own WeBWorK assignment:

∫Bx + C

(x2 + A2)2 dx .


(b) No marks will be given unless you verify your integration by differentiating youranswer.

3. Problem 16: Determine whether the following integral (with the constants given in yourown WeBWorK assignment) is divergent or convergent. If it is convergent, evaluate it.If not, prove that fact. ∫ A

−∞

1x2 + 1

dx .

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 conventionalmathematics assignment, and it does not suffice to make unsubstantiated statements. You areexpected to prove everything you state. Thus, for example, in Problem 10, you have to showhow you use the fact that a certain number is stated to be a root of the denominator.




The Solutions

1. Under a change of variable of the form u =xA

, the integral reduces to one of the form∫Ku + L(u2 + 1

)2 du. The termKu

u2 + 1du simplifies under a substitution v = u2 to a multiple

of∫

dv(v + 1)2 . The term

∫L

(u2 + 1

)2 du simplifies under a substitution θ = arctan u,

eventually proving to be a multiple of θ +tan θ

2(tan2 θ + 1), etc.

2. An antiderivative is arctan x. This is evaluated between an upper limit of some constantA, and a lower limit we may call B. We need to observe that lim

B→−∞arctan B = −π2 .



Release Date: Solutions were to be submitted inside answer sheets to Quiz Q5, at tutorialsduring 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 + 1y(t) = A7t2 + A8t + A9

(a) [3 MARKS] Showing all your work, determine the slope of the tangent to the curveat the point with parameter value t = 1.

(b) [6 MARKS] Showing all your work, determine the value ofd2ydx2 at the point with

general parameter value t (t , 0). For this problem you must not substitute in anyformula from your class notes or any textbook; you are expected to determine the2nd derivative by differentiation, for example in the manner similar to that done inyour 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 givenarc of the given curve,

(a) [8 MARKS] determine the area of the surface of revolution of that arc about thex-axis;

(b) [2 MARKS] set up an integral for the area of the surface of revolution of that arcabout the y-axis, but, in this case only, do not evaluate the integral.

In each case you are expected to show all your work.





The Solutions

1. (a)

x(t) = t2 + 1y(t) = A7t2 + A8t + A9

dxdt

= 2t

dydt

= 2A7t + A8

dydx

=

dydtdxdt

=2A7t + A8

2tdydx

∣∣∣∣∣t=1

= A7 +A8

2

(b)

d2ydx2 =

ddx

(2A7t + A8

2t

)

=ddt

(2A7t + A8

2t

)· dt

dx

=ddt

(2A7t + A8

2t

)· 1

dxdt

=ddt

(2A7t + A8

2t

)· 1

2t

=(2A7)(2t) − (2A7t + A8)(2)

(2t)3

= − A8

4t3

(c) If the student’s A8 is 0, the curve is flat. Otherwise it is concave upwards preciselywhen t < 0.


2. CORRECTED ON 20 MARCH 2006. The following solution is valid onlywhen K, the upper limit of integration, is sufficiently small. The correct so-lution is much more complicated, as we need to ensure that the respectivefactors cos θ + θ sin θ and sin θ − θ cos θ remain positive. I have corrected myoriginal version by inserting absolute signs. However, the evaluation of theseintegrals is complicated when we attempt to break the integral up into partsbased on the signs of these factors. This is certainly more difficult than wasintended from students in this course. Please exercise good judgment in grad-ing the two parts of this problem, and do not expect those students for whomthe upper limit K is large to obtain the correct area.

x(θ) = a(cos θ + θ sin θ)y(θ) = a(sin θ − θ cos θ)

⇒ x′ = aθ cos θy′ = aθ sin θ

⇒√

(x′)2 + (y′)2 = |aθ|(a) about the x-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

∫ K

02πa| sin θ − θ cos θ| · |θ| dθ = 2πa

(∫ K

0θ sin θ dθ −

∫ K

0θ2 cos θ dθ

)

for positive K sufficiently small. We must integrate by parts.∫θ sin θ dθ = −θ cos θ +

∫cos θ dθ

= −θ cos θ + sin θ + C∫θ2 cos θ dθ = θ2 sin θ − 2

∫θ sin θ dθ

= θ2 sin θ + 2θ cos θ − 2 sin θ + C′∫(θ sin θ − θ2 cos θ) dθ = −3θ cos θ + 3 sin θ − θ2 sin θ + C′′

Hence the area of the surface of revolution is −3K cos K + 3 sin K − K2 sin K.(b) about the y-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

∫ K

02πa| cos θ + θ sin θ| · |θ| dθ = 2πa

(∫ K

0θ cos θ dθ +

∫ K

0θ2 sin θ dθ

).

D.9 MATH 141 2007 01The use of written assignments was discontinued in 2007.


E Quizzes from Previous Years

E.1 MATH 141 2007 01E.1.1 Draft Solutions to Quiz Q1

Distribution Date: Mounted on the Web on 4 February, 2007corrected 12 February, 2009 — subject to further corrections

There were four different types of quizzes, for the days when the tutorials are scheduled. Eachtype of quiz was generated in multiple varieties for each of the tutorial sections. The order ofthe problems in the varieties was also randomly assigned. All of the quizzes had a heading thatincluded 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 onetypical variety.

Monday version

1. [5 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivativeof the function

g(x) =

∫ a

xb tan(t) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluateg(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 afunction of its upper index of integration. Here the variable is the lower index ofintegration.

ddx

∫ a

xb tan(t) dt

=ddx

(−

∫ x

ab tan(t) dt

)


= − ddx

∫ x

ab tan(t) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definiteintegral with respect to the lower index, and this should be accepted if work hasbeen shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

ddx

ln | sec x| =ddu

ln |u| · ddx

secx

=1u· sec x tan x

=1

sec x· sec x tan x

= tan x .

Henceg(x) = b ln | sec t|ax = b ln

∣∣∣∣∣sec asec x

∣∣∣∣∣ = b ln∣∣∣∣∣cos xcos a

∣∣∣∣∣ .

2. [5 MARKS] Find an antiderivative of the integrand of the integralb∫

a

(k + `y + my2

)dy,

(where a, b, k, `,m are constants), and then use the Fundamental Theorem of Calculus toevaluate the integral. You are not expected to simplify your numerical answer.

Solution:

(a) One antiderivative of ky0 + `y1 + my2 is

k · y1

1+ ` · y2

2+ m · y3

3.

(b) Henceb∫

a

(k + `y + my2

)dy

=

[k · y1

1+ ` · y2

2+ m · y3

3

]b

a

=

(k · b1

1+ ` · b2

2+ m · b3

3

)−

(k · a1

1+ ` · a2

2+ m · a3

3

).



g(x) =

∫ √x

ab

cos tt

dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u(x) =√

x.

(b) Then

ddx

g(x) =ddx

∫ √x

ab

cos tt

dt

=ddx

∫ u(x)

ab

cos tt

dt

=d

du

∫ u(x)

ab

cos tt

dt · du(x)dx

= bcos u

u· du(x)

dx

= bcos u

u· 1

2√

x

= bcos√

x√x· 1

2√

x

= bcos√

x2x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y =x∫

0

11 + at + bt2 dt is concave upward, where a, b are constants. (Each version of this

quiz contained specific values for the constants a, b.)

Solution:

(a) By Part 1 of the Fundamental Theorem,

y′(x) =1

1 + ax + bx2 .

(b) Differentiating a second time yields

y′′(x) =ddx

(1

1 + ax + bx2

)


= − 1(1 + ax + bx2)2

· ddx

(1 + ax + bx2

)

= − a + 2bx(1 + ax + bx2)2 .

(c) The curve is concave upward where y′′ > 0:

− a + 2bx(1 + ax + bx2)2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive⇔ 2bx < −a

⇔

x < − a2b when b > 0

x > − a2b when b < 0

never concave upward when b = 0, a > 0always concave upward when b = 0, a < 0

Tuesday version

1. [5 MARKS] If

c∫

a

f (x) dx = k and

c∫

b

f (x) dx = `, find

b∫

a

f (x) dx. Show your work.

Solution:

(a)c∫

a

f (x) dx =

b∫

a

f (x) dx +

c∫

b

f (x) dx .

(b) Hencec∫

b

f (x) dx =

c∫

a

f (x) dx −b∫

a

f (x) dx .

(c)= ` − k .

2. [5 MARKS] Find an antiderivative of the integrand of the integral∫ a

0

√x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expectedto simplify your numerical answer, but no marks will be given unless all your work isclearly shown.


(a) One antiderivative of x12 is

112 + 1

· x 12 +1 =

23

x32 .

(b) ∫ a

0

√x dx =

[23

x32

]a

0=

23

(a

32 − 0

)=

23

a32 .

3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative

of the function

x2∫

a

b√

1 + tc dt.

Solution:

(a) Let the upper index of the integral be denoted by u = x2. Then

(b)

ddx

x2∫

a

b√

1 + tc dt =ddx

u∫

a

b√

1 + tc dt

=ddu

u∫

a

b√

1 + tc dt · dudx

= b√

1 + uc · dudx

= b√

1 + uc · 2x= b

√1 + x2c · 2x

4. [10 MARKS] If F(x) =

∫ x

1f (t) dt, where f (t) =

∫ t2

1

a + ub

udu and a, b are constants,

find F′′(2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F′(x) = f (x) =

∫ x2

1

a + ub

udu .


(b) A second application of Part 1 of the Fundamental Theorem yields

F′′(x) = f ′(x) =ddx

∫ x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

ddx

∫ x2

1

a + ub

udu =

ddx

∫ v

1

a + ub

udu

=ddv

∫ v

1

a + ub

udu · dv

dx

=a + vb

v· dv

dx

=a + vb

v· 2x

=a + x2b

x2 · 2x

=2(a + x2b

)

x.

Wednesday version

1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ∫

aπ

cos θ dθ, where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ.

(b)bπ∫

aπ

cos θ dθ = [sin θ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiplesof π is 0, so the value of the definite integral is 0.


2. [5 MARKS] Express limn→∞

n∑

i=1

axi sin xi ∆x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: ∫ c

bax sin x dx .


g(x) =

∫ bx

ax

t2 + ct2 − c

dt ,

where a, b, c are positive integers.

Solution:

(a) The Fundamental Theorem gives the derivative of a definite integral with respect tothe upper limit of integration, when the lower limit is constant. The given integralmust be expressed in terms of such specialized definite integrals.

g(x) =

∫ bx

ax

t2 + ct2 − c

dt =

∫ 0

ax

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

= −∫ ax

0

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

(b) For the summand

bx∫

0

t2 + ct2 − c

dt, let u = bx. Then

ddx

bx∫

0

t2 + ct2 − c

dt =ddx

u∫

0

t2 + ct2 − c

dt

=d

du

u∫

0

t2 + ct2 − c

dt · dudx

=u2 + cu2 − c

· dudx

=u2 + cu2 − c

· b

=(bx)2 + c(bx)2 − c

· b


(c) For the summand∫ ax

0

t2 + ct2 − c

dt, let u = ax. Then, analogously to the preceding

step,ddx

∫ ax

0

t2 + ct2 − c

dt =(ax)2 + c(ax)2 − c

· a .

(d)

g′(x) =(bx)2 + c(bx)2 − c

· b − (ax)2 + c(ax)2 − c

· a .

4. [10 MARKS] Find a function f (x) such that

k +

∫ x

a

f (t)t2 dt = `

√x (24)

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 (24) holds. Then differentiation of both sides with respect tox yields

0 +f (x)x2 =

12· ` · 1√

x.

(b) We solve the preceding equation to obtain that f (x) = `2 x

32 .

(c) Substitution into equation (24) yields

k +`

2

∫ x

at−

12 dt = `

√x .

We know how to integrate powers of t:

k +`

cot2 · √t

]x

a= `√

x .

(d) The preceding reduces to k = `√

a, which may be solved to obtain

a =

(k`

)2

.


Thursday version

1. [5 MARKS] Evaluate the integral

πb∫

πa

sin t dt.

Solution:

(a) An antiderivative of sin t is − cos t.

(b)πb∫

πa

sin t dt = [− cos t]πbπa

(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 finalnumerical answer.

Solution:

(a) The interval 0 ≤ x ≤ 2 is divided by 3 points into 4 subintervals of length ∆x =24 = 1

2 .

(b) The point x∗i 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 is4∑

i=1

f (x∗i )∆x =12

4∑

i=1

(a − i2

4

).


of the function

bx∫

cos x

cos (tc) dt, where a, b, c are real numbers.

Solution:


bx∫

cos x

cos (tc) dt =

0∫

cos x

cos (tc) dt +

bx∫

0

cos (tc) dt


= −cos x∫

0

cos (tc) dt +

bx∫

0

cos (tc) dt

(b) For the summand

bx∫

0

cos (tc) dt, let u = bx. Then

ddx

bx∫

0

cos (tc) dt =ddx

u∫

0

cos (tc) dt

=d

du

u∫

0

cos (tc) dt · dudx

= cos (uc) · b= cos ((bx)c) · b

(c) For the summand

cos x∫

0

cos (tc) dt, let v = cos x.

ddx

cos x∫

0

cos (tc) dt =ddx

v∫

0

cos (tc) dt

=ddv

v∫

0

cos (tc) dt · dvdx

= cos (vc) · (− sin x)= cos (cosc x) · (− sin x)

(d)bx∫

cos x

cos (tc) dt = − cos (cosc x) · (− sin x) + cos ((bx)c) · b .

4. [10 MARKS] Let f (x) =

0 if x < 0x if 0 ≤ x ≤ a2a − x if a < x < 2a0 if x > 2a

and g(x) =

∫ x

0f (t) dt, where a is

a positive constant. Showing all your work, find a formula for the value of g(x) whena < x < 2a.


Solution:

(a) The interval where we seek a formula is the third interval into which the domainhas been broken. For x in this interval the integral can be decomposed into

∫ x

0f (t) dt =

∫ a

0f (t) dt +

∫ x

af (t) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, sincewe are not finding area under that portion of the curve; the same applies to theportion of the definition for x > 2a.

(b)∫ a

0f (t) dt =

∫ a

0t dt

=

[t2

2

]t=a

t=0=

a2

2.

(c)∫ x

af (t) dt =

∫ x

a(2a − t) dt

=

[2at − t2

2

]t=x

t=a

=

(2ax − x2

2

)−

(2a2 − a2

2

).

(d)

g(x) =a2

2+

(2ax − x2

2

)−

(2a2 − a2

2

)= 2ax − x2

2− a2 .

E.1.2 Draft Solutions to Quiz Q2

Distribution Date: Posted on the Web on 28 February, 2007Caveat lector! There could be misprints or errors in these draft solutions.




• No calculators!




Monday version

1. [5 MARKS] Evaluate the integral∫ (

xb + a +1

x2 + 1

)dx, (where a and b are given

positive integers).

Solution:

(a)∫

xb dx =xb+1

b + 1+ C1,

(b)∫

a dx = ax + C2

(c)∫

1x2 + 1

dx = arctan x + C3

(d)∫ (

xb + a +1

x2 + 1

)dx =

xb+1

b + 1+ ax + arctan x + C.

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)∫

t2 cos(a − t3

)dt =

∫13

cos(a − u) du

= −13

sin(a − u) + C

= −13

sin(a − t3) + C.

(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.)


3. [10 MARKS] Find the area of the region bounded by the parabola y = x2, the tangentline 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 withrespect to x.

Integrating with respect to y: (a) Since y′ = 2x, the tangent line through (a, a2) hasequation

y − a2 = 2a(x − a)⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabolaand 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

equation x =y

2a+

a2

.

(c) The area of the horizontal element of area at height y is(y + a2

2a− √y

)∆y.

(d) The area is the value of the integral∫ a2

0

(y + a2

2a− √y

)dy .

(e) Integration yields[

y2

4a+

ay2− 2

3y

32

]a2

0=

(14

+12− 2

3

)a3 =

112

a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Itsintercept with the x-axis is at x =

a2

.

(b) The area of the vertical element of area at horizontal position x ≤ a2

is(x2 − 0

)∆x.

(c) The area of the vertical element of area at horizontal position x ≥ a2 is

(x2 − (2ax − a2)

)dx =

(x − a)2 ∆x.(d) The area of the region is the sum

∫ a2

0x2 dx +

∫ a

a2

(x − a)2 dx .

(e) Integration yields [x3

3

] a2

0+

[−(x − a)3

3

]a

a2

=a3

12.


4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded bythe 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 theregion. 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 revolutionis a horizontal line, the element of area being rotated is vertical. For arbitraryx 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.(c) The volume of the “washer” is, therefore,

π(−(1 − x)2 + (1 − n√x)2

)∆x .

(d) Correctly evaluate the integral:

π

∫ 1

0

(−(1 − x)2 + (1 − n√x)2

)dx

= π

∫ 1

0

(−2x + x2 + 2x

1n − x

2n)

dx

= π

[−x2 +

x2

3+

2nn + 1

xn+1

n − nn + 2

xn+2

n

]1

0

= π

(−1 +

13

+2n

n + 1− n

n + 2

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) Find the intersections of the curves bound-ing 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 revolutionis a horizontal line, the element of area being rotated is also horizontal. Forarbitrary 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 fromthe axis of symmetry is 1 − y.

(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 − y) · (y − yn) · ∆y .


2π∫ 1

0(1 − y)(y − yn) dy


= 2π∫ 1

0

(−yn + yn+1 + y − y2

)dy

= 2π[− 1

n + 1yn+1 +

1n + 2

yn+2 +12

y2 − 13

y3]1

0

= 2π(− 1

n + 1+

1n + 2

+12− 1

3

)− 0

= 2π(16− 1

(n + 1)(n + 2)

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Tuesday version

1. [5 MARKS] Evaluate the integral∫

(a − t)(b + t2) dt.

Solution:

(a) Expand the product in the integrand:∫

(a − t)(b + t2) dt =

∫ (ab − bt + at2 − t3

)dt .

(b) Integrate term by term:∫ (

ab − bt + at2 − t3)

dt = ab · t − b2· t2 +

a3· t3 − 1

4· t4 + C .

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)∫

cosn x sin x dx = −∫

un du

= − un+1

n + 1+ C

= − 1n + 1

cosn+1 x + C


3. [10 MARKS] Find the area of the region bounded by the parabola x = y2, the tangentline 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 Prob-lem 3 of the Monday quiz.

4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded byy = 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 bound-ing 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 revo-lution is a vertical line, the element of area being rotated is horizontal. Forarbitrary y the farther endpoint on the element is (y

1n , y); the nearer end-

point is (yn, y). The distances of these points from the axis are, respectively1 + n√

y and 1 + yn.(c) The volume of the “washer” is, therefore,

π(−(1 + yn)2 + (1 + n

√y)2

)∆y .


π

∫ 1

0

(−(1 + y)2 + (1 + n

√y)2

)dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]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 curvesbounding 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(x, x

1n ); the lower endpoint is (x, xn). The length of the element is, there-

fore, x1n − 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,

2π(1 + x) ·(x

1n − xn

).


2π∫ 1

0(1 + x)(x

1n − xn) dx

= 2π∫ 1

0

(x

1n − xn + x

n+1n − xn+1

)dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 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 bound-ing 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 theword 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 willfollow 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 rev-olution is a vertical line, the element of area being rotated is horizontal.But there are two kinds of elements, depending on whether y is positiveor negative. For arbitrary, positive y the farther endpoint on the element is(y

1n , 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 thenearer endpoint on the element is (y

1n , y); the farther endpoint is (yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and1 + 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 .


π

∫ 1

−1

∣∣∣−(1 + y)2 + (1 + n√

y)2∣∣∣ dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy


+π

∫ 0

−1

(−2y

1n − y

2n + 2yn + y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

+π

[− 2n

n + 1y

n+1n − n

n + 2y

n+2n +

2n + 1

yn+1 +1

2n + 1y2n+1

]0

−1

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

+2(n − 1)(n2 + 3n + 1)π(n + 1)(n + 2)(2n + 1)

=4(n − 1)π

n + 1.


(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 el-ement is (x, x

1n ); the lower endpoint is (x, xn); for arbitrary, negative x the

bottom endpoint on the element is (x, x1n ); the upper endpoint is (x, xn).

The length of the element is, therefore,∣∣∣∣x 1

n − xn∣∣∣∣; the distance of the ele-

ment from the axis of rotation is 1 + x.(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·∣∣∣∣x 1

n − xn∣∣∣∣ .


2π∫ 1

−1(1 + x)

∣∣∣∣x 1n − xn

∣∣∣∣ dx

= 2π∫ 1

−1

∣∣∣∣x 1n − xn + x

n+1n − xn+1

∣∣∣∣ dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

+2π(

nn + 1

− 1n + 1

− n2n + 1

+1

n + 2

)

=4(n − 1)π

n + 1

Wednesday version


1. [5 MARKS] Showing all your work, evaluate the indefinite integral∫

sin 2tcos t

dt.

Solution:

(a) Apply a “double angle” formula:∫

sin 2tcos t

dt =

∫2 sin t cos t

cos tdt = 2

∫sin t dt .

(b) Complete the integration:

2∫

sin t dt = −2 cos t + C .

2. Using a substitution, evaluate the indefinite integral∫

ex

ex + adx, where a is a non-zero

real number.

Solution:

(a) [2 MARKS] Try the substitution u = ex + a, so du = ex dx.

(b) ∫ex

ex + adx =

∫duu

= ln |u| + C = ln |ex + a| + C .

(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 boundedby the curves y = ax2 and y = k into two regions with equal area, where a, k are givenpositive constants.

Solution:

(a) Determine the range of values for integration by finding the intersections of the

bounding curves: solving the equations yields the points

∓√

ka, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin byrepeating the calculation of the preceding part: the corner points have coordinates∓

√ba, b

. The area is

∫ √ ba

−√

ba

(b − ax2) dx = 2[bx − ax3

3

]√ ba

0=

43

b

√ba.


(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is43

k

√ka

.

(d) The condition of the problem is that

43

b

√ba

=12· 4

3k

√ka

which is equivalent to 4b3 = k3, and implies that the line should be placed whereb = 2−

23 k.

4. [10 MARKS] Use the method of cylindrical shells to find the volume generated by ro-tating 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 hasleft 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 theaxis of rotation is b − y.

(d) Set up the integral for the volume by cylindrical shells:

2π∫ √

a−1

0(b − y)

(a − (1 + y2)

)dy .

(e) Evaluate the integral

2π∫ √

a−1

0(b − y)

(a − (1 + y2)

)dy

= 2π∫ √

a−1

0

(b(a − 1) − (a − 1)y − by2 + y3

)dy

= 2π[b(a − 1)y − a − 1

2y2 − b

3y3 +

14

y4]√a−1

0

= 2π(b(a − 1)

32 − a − 1

2(a − 1) − b

3(a − 1)

32 +

14

(a − 1)2)

= 2π(a − 1)32

(23· b − 1

4

√a − 1

).



Thursday version

1. [5 MARKS] Showing all your work, evaluate the integral by making a substitution:∫

b(1 + ax)3 dx ,

where a, b are non-zero constants.

Solution:

(a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, sodx = 1

a du.

(b) ∫b

(1 + ax)3 dx =ba

∫duu3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

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) ∫seca x tan x dx =

∫ua−1 du =

ua

a+ C =

seca xa

+ C .

Some students may have integrated by sight.

3. [10 MARKS] Find the number b such that the line divides the region bounded by thecurves 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 Prob-lem 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, wesolve 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.(d) The volume is given by the integral

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

(e) Evaluating the integral:

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

= 2π∫ 2

1

(−x3 + 2x2 + x − 2

)dx

= 2π[−1

4x4 +

23

x3 +12

x2 − 2x]2

1

= 2π(−4 +

163

+ 2 − 4 +14− 2

3− 1

2+ 2

)

=5π6

Using the method of “washers” (a) To find the lowest point on the parabola, we solvex2 − 3x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be(32,

194

).

(b) The horizontal element generating the “washer” at height y extends betweenthe solutions in x to the equation y = x2 − 3x + 7; these are

x =3 ± √

4y − 192

.

(c) The volume of the “washer” at height y is, therefore,

π

1 +

3 +√

4y − 192

2

−1 +

3 − √4y − 192

2 ∆y

= 5π√

4y − 19 ∆y

(d) The volume is given by the integral 5π∫ 5

194

√4y − 19 dy.

(e) Evaluation of the integral:

5π∫ 5

194

√4y − 19 dy =

[5π · 2

3· 1

4(4y − 19)

32

]5

194

=5π6.



Distribution Date: Posted on the Web on 21 March, 2007Caveat lector! There could be misprints or errors in these draft solutions.



• No calculators!




Monday version


t3eat dt, where a is a non-zero constant.

Solution: This problem requires several consecutive applications of integration by partsto reduce the exponent of the power of t to 0:

u = t3, dv = eatdt ⇒ du = 3t2 dt, v =1a

eat

⇒∫

t3eat dt =1a· t3eat − 3

a

∫t2eat dt

U = t2, dV = eatdt ⇒ dU = 2t dt,V =1a

eat

⇒∫

t3eat dt =

(1a· t3 − 3

a2 t2)

eat +6a2

∫teat dt

u = t, dv = eatdt ⇒ du = dt, v =1a

eat

⇒∫

t3eat dt =

(1a· t3 − 3

a2 · t2 +6a3 t

)eat − 6

a3

∫eat dt

=

(1a· t3 − 3

a2 · t2 +6a3 t − 6

a4

)eat + C


the correctness of which integration may be verified by differentiation of the product onthe right.

2. [8 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:

In =

∫cosn ax dx

(b) Integration by parts:

u = cosn−1 ax ⇒ du = −a(n − 1) cosn−2 ax · sin ax dx

dv = cos ax dx ⇒ v =1a· sin ax

In =1a· cosn−1 ax · sin ax + (n − 1)

∫cosn−2 ax · sin2 ax dx

=1a· cosn−1 ax · sin ax + (n − 1)

∫cosn−2 ax ·

(1 − cos2 ax

)dx

=1a· cosn−1 ax · sin ax + (n − 1)In−2 − (n − 1)In .

(c) Solve the last equation for In

In =1

an· sin ax · cosn−1 ax +

n − 1n

In−2 + C .

3. [4 MARKS] Showing all your work, evaluate the integral∫

1

x2√

x2 − a2dx , where a is

a non-zero constant.

Solution: The following solution uses a trigonometric substitution; it is also possible tosolve this problem using a hyperbolic substitution.

As is customary, I will proceed mechanically, taking square roots where necessary with-out much attention to the sign choices; and then verify at the end by differentiation thatthis process has produced a valid indefinite integral. This procedure can be made rigor-ous by defining the new variable in terms of an inverse trigonometric function of x. Thisis a useful exercise, but becomes extremely complicated in this case, because we wouldhave to work with either the inverse cosine or the inverse secant, and the textbook we


are using chooses different domains for these two functions. So I will avoid the nicetiesand 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

∫1

x2√

x2 − a2dx =

1a2

∫cos θ dθ

=1a2 sin θ + C

=1a2 tan θ · cos θ + C

=sec2 θ − 1a2 sec θ

+ C

=

√x2 − a2

a2x+ C ,

the correctness of which integration may be verified by differentiation of the quotient onthe right.

4. [8 MARKS] Showing all your work, evaluate the integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b are distinct constants, c is a non-zero constant, and the limits of integrationk, ` 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 aremainder:

∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .

(b) Expand the fraction into partial fractions. Assume that

c(x − a)(x − b)

=A

x − a+

Bx − 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 succes-sive values x = a and x = b:

c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .

(d) The integration reduces to∫

x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· 1x − b

)dx .

(e) Complete the integration:∫

x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln |x − a| − ln |x − b|) + C

=x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣ + C ,

the correctness of which integration may be verified by differentiation of the func-tion on the right.

(f) Provided it is convergent the given definite integral can now be evaluated:

∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣]`

k

=`2 − k2

2+

ca − b

(ln

∣∣∣∣∣` − a` − b

∣∣∣∣∣ − ln∣∣∣∣∣k − ak − b

∣∣∣∣∣)

=`2 − k2

2+

ca − b

(ln

∣∣∣∣∣(` − a)(k − b)(` − b)(k − a)

∣∣∣∣∣)

(g) All of the preceding is based on the integral being convergent. In some of theversions the integral was divergent. This was because at least one of the roots ofthe polynomial which is the denominator of the integrand was contained in theinterval of integration. In such a case the integral can be seen to diverge.


1√3∫

−1

earctan y

1 + y2 dy . I have stated the problem with just one

pair of possible limits for the integral; the variations of the problem included severalpossible limits in each case, for each of which students should have been familiar withthe arctangent.


Solution: For simplicity, I work a specific instance of this problem. Use the substitution

u = arctan y, so du =dy

1 + y2 . Then an antiderivative can be obtained as follows

∫earctan y

1 + y2 dy =

∫eu du = eu + C = earctan y + C

so the definite integral is equal to

[earctan y

] 1√3

−1= e

π6 − e−

π4 .

Alternatively, the substitution may be executed in the definite integral, replacing thelower limit of −1 by arctan(−1) = −π4 , and the upper limit of 1√

3by arctan 1√

3= π

6 .

Tuesday version


x2 cos ax dx, where a is a

non-zero constant.

Solution: Two applications of integration by parts will be used to reduce the exponentof the power of x to 0.

(a)

u = x2 ⇒ du = 2x dx

dv = cos ax dx ⇒ v =1a

sin ax∫

x2 cos ax dx =x2

a· sin ax − 2

a

∫x · sin ax dx

(b)

U = x ⇒ dU = dx

dV = sin ax dx ⇒ V = −1a· cos ax

∫x2 cos ax dx =

x2

a· sin ax − 2

a

(− x

a· cos ax +

1a

∫cos ax dx

)

=x2

a· sin ax +

2a2 x · cos ax − 2

a2

∫cos ax dx

=x2

a· sin ax +

2a2 x · cos ax − 2

a3 sin ax + C


The integration can be checked by differentiation of the alleged antiderivative.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral∫

xneax dx,

where a is a non-zero constant.

Solution:

(a) Introduce a symbol for the general indefinite integral sought:

In =

∫xneax dx .


u = xn ⇒ du = nxn−1 dx

dv = eax dx ⇒ v =1a· eax

In =1a· eax − n

a

∫xn−1eax dx

=1a· eax − n

aIn−1.

which is the desired reduction formula.

3. [4 MARKS] Showing your work, evaluate the integral∫

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.

∫sin3 ax · cos2 ax dx =

∫sin2 ax · u2 · −du

a

=

∫ (1 − cos2 ax

)· u2 · −du

a

=

∫ (1 − u2

)· u2 · −du

a

=1a

∫ (u4 − u2

)du

=1a

(u5

5− u3

3

)+ C

=1a

(cos5 ax

5− cos3 ax

3

)+ C ,


which integration may be checked by differentiation. (Of course, there are other, equiv-alent ways of expressing this indefinite integral.)


x3

√x2 + a2

dx, where a is a

given non-zero constant.

Solution: To simplify the surd in the denominator one may use either a trigonometricor a hyperbolic substitution. For students in this course a trigonometric substitution isusually a better choice. To arrange that x = a tan u, we use a substitution

u = arctanxa, (25)

and dx = a sec2 u du. We may assume that a > 0. The interval of validity for substitution(25) is −π2 < x < π

2 , in which the secant is positive.∫

x3

√x2 + a2

dx =

∫a3 tan3 u|a sec u| · a sec2 u du

= a3∫

tan2 u · sec u tan u du

= a3∫ (

sec2 u − 1)· d

dusec u du

= a3(sec3 u

3− sec u

)+ C

effectively by substitution U = sec u

=13

((a tan u)2 − 2a2

) √(a tan u)2 + a2 + C

=13

(x2 − 2a2

) √x2 + a2 + C

which integration may be verified by differentiation.

5. [9 MARKS] (To simplify the exposition of the solution, I work a specific example here.)

Showing all your work, evaluate the integral∫

x + 21(x + 9)(x − 5)

dx.

Solution: Since the degree of the numerator is less than that of the denominator, we candispense with the first step of dividing denominator into numerator.

(a) We need to expand the integrand into a sum of partial fractions; fortunately thefactorization of the denominator has been given. Assuming there are constantsA, B such that

x + 21(x + 9)(x − 5)

=A

x + 9+

Bx − 5


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 powersof x, or by assigning to x successively the “convenient” values 5, -9: we obtain that

26 = 14B ⇒ B =137

12 = −14A ⇒ A = −67.

(c) We may now complete the integration:∫

x + 21(x + 9)(x − 5)

dx =17

∫ (− 6

x + 9+

13x − 5

)dx

= −17

(−6 ln(x + 9) + 13 ln(x − 5)) + C

which can also be expressed in other, equivalent ways. This integration may beverified by differentiation.

Wednesday version

1. [4 MARKS] Showing your work, evaluate

a∫

0

(x2 +1)e−x dx , where a is a given constant.

Solution: I will integrate by parts twice, in order to reduce the degree of the polynomialfactor of the integrand.

(a) First integration by parts:

u = x2 + 1 ⇒ du = 2x dxdv = e−x dx ⇒ v = −e−x

a∫

0

(x2 + 1)e−x dx =[(x2 + 1)

(−e−x)]a

0+

∫ a

02x · e−x dx .

(b) Second integration by parts:

U = 2x ⇒ dU = 2 dxdV = e−x dx ⇒ V = −e−x

a∫

0

(x2 + 1)e−x dx =[(x2 + 1 + 2x)

(−e−x)]a

0+

∫ a

02 · e−x dx .


(c) Completion of the integration:a∫

0

(x2 + 1)e−x dx =[(x2 + 2 + 2x)

(−e−x)]a

0

= −(a2 + 2a + 3)e−a + 3.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral

π2∫

0

sinn ax dx,

where a is a given integer.

Solution: Assume n is an integer greater than 1.

(a) Introduce a symbol for the definite integral sought:

In =

∫ π2

0sinn ax dx .


dv = sin ax dx⇒ v = −cos axa

dx

u = sinn−1 ax⇒ du = a(n − 1) sinn−2 ax · cos ax dx

In =

[−1

a· sinn−1 ax · cos ax

] π2

0+ (n − 1)

∫ π2

0sinn−2 ax · cos2 ax dx

(c) Decomposition of integral:∫ π

2

0sinn−2 ax · cos2 ax dx =

∫ π2

0sinn−2 ax ·

(1 − sin2 ax

)dx

=

∫ π2

0sinn−2 ax dx −

∫ π2

0sinn ax dx

= In−2 − In

(d) Solution of equation to obtain reduction formula

In =

[−1


] π2

0+ (n − 1) (In−2 − In)

⇒ nIn =

[−1


] π2

0+ (n − 1)In−2

⇒ In =1n

(0 − 0) +n − 1

nIn−2


Because of the choice of limits and the fact that a is an integer, the “net change” is0. Thus we obtain a very simple relationship, which can be solved. Students werenot asked to complete this part of the solution. For example, it is possible to proveby induction that, if n = 2m, an even, positive integer, then

I2m =2m − 1

2m· 2m − 3

2m − 2· . . . · 3

2I0

=(2m)!

4mm!m!· π

2.

(This is Exercise 44, page 481 in the textbook.)


x2

(a2 − x2) 3

2

dx, where a is a

non-zero constant.

Solution: Without limiting generality we take a > 0. A trigonometric substitution cansimplify this integral. One such substitution would have x = a sin u; more precisely,u = arcsin x

a , defined for −π2 ≤ x ≤ π2 , in which interval the cosine and secant are

positive. Then dx = a du√1−u2

.

∫x2

(a2 − x2) 3

2

=

∫a2 sin2 ua3 cos3 u

· a cos u du

=

∫ (tan2 u

)du =

∫ (sec2 u − 1

)du

= tan u − u + C

=sin ucos u

− u + C

=sin u√

1 − sin2 u− u + C

=

xa√

1 − x2

a2

− arcsinxa

+ C

=x√

a2 − x2− arcsin

xa

+ C

which may be verified by differentiation.

4. [9 MARKS] Showing all your work, evaluate the integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b, k, ` are distinct constants such that the integrand is defined throughout the


given interval, and c is a non-zero constant. (The integrand was not presented to studentsin factored form.)

Solution:


∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .


c(x − a)(x − b)

=A

x − a+

Bx − b

,


c = A · (x − b) + B · (x − a) .


c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .


x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· cx − b

)dx .


x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln(x − a) − ln(x − b)) + C

=x2

2+

ca − b

lnx − ax − b

+ C ,



(f) Evaluate the definite integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

(ln |x − a| − ln |x − b|)]`

k

=

[x2

2+

ca − b

lnx − ax − b

]`

k

=`2 − k2

2+

ca − b

ln∣∣∣∣∣(` − a)(k − b)(` − b)(k − a)

∣∣∣∣∣

5. [4 MARKS] Showing all work, evaluate the integral∫

sin3 ax dx, where a is a positive

integer.

Solution:∫

sin3 ax dx =

∫ (1 − cos2 ax

)· sin ax dx

under substitution u = cos ax, where du = −a sin ax dx

=

∫(1 − u2)(−1

a) du

= −1a

(u − u3

3

)+ C

= −cos axa

+cos3 ax

3a+ C .

Thursday version

1. [4 MARKS] Showing all your work, evaluate the integral∫ b

ae√

t dt.

Solution: Begin with a substitution√

t = u, so t = u2, dt = 2u du. When t = a, u =√

a,etc.: ∫

e√

t dt =

∫2ueu du .

Now apply integration by parts:

U = u ⇒ dU = dudV = 2eu du ⇒ V = 2eu∫

2ueu du = u · 2eu −∫

2eu du

= u · 2eu − 2eu + C = 2(u − 1)eu + C= 2(

√t − 1)e

√t + C


The definite integral given is then equal to[2(√

t − 1)e√

t]b

a.

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 ⇒ du = a dx

In =1a

∫(ln u)n du

(c) Integration by parts:

U = (ln u)n ⇒ dU = n(ln u)n−1 · 1u

dV = 1 du ⇒ V = u

In = u(ln u)n − n∫

(ln u)n−1 du

= (ax + 1)(ln(ax + 1))n − na∫

(ln(ax + 1))n−1 dx

= (ax + 1)(ln(ax + 1))n − na · In−1


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 .

∫cos4 at dt =

∫ (1 + cos 2at

2

)2

dt

=14

(∫cos2 2at dt + 2

∫cos 2at dt +

∫1 dt

)


=14

(∫1 + cos 4at

2dt + 2

∫cos 2at dt +

∫1 dt

)

=18· 1

4asin 4at +

12· 1

2asin 2at +

38

t + C

=1

32asin 4at +

14a

sin 2at +38

t + C

which may be verified by differentiation. (Of course, the integral may be expressed inother ways under transformation by trigonometric identities.)

4. [4 MARKS] Showing your work, evaluate the integral∫

dx

x√

x2 + a, where a is a given

positive integer, not a perfect square.

Solution: The surd in the denominator may be simplified by either a trigonometric or ahyperbolic substitution. For students in this course the trigonometric substitutions areusually easier.

x =√

a · tan θ ⇒ dx =√

a sec2 θ dθ∫dx

x√

x2 + a=

∫sec2 θ dθ

tan θ · √a · | sec θ| .

The actual substitution is given by θ = arctan x√a , valid for −π2 < x < π

2 . In that intervalthe secant function is positive, so the absolute signs may be dropped. The integral isequal to

1√a

∫csc θ dθ =

1√a

ln | csc θ − cot θ| + C

=1√a

ln∣∣∣∣∣sec θ − 1

tan θ

∣∣∣∣∣ + C

=1√a

ln

∣∣∣∣∣∣∣

√tan2 θ − 1

tan θ

∣∣∣∣∣∣∣ + C

=1√a

ln

∣∣∣∣∣∣∣

√x2 + a − √a

x

∣∣∣∣∣∣∣ + C

which can be verified by differentiation.

5. [9 MARKS] Showing all your work, evaluate the integral

`∫

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b, k, ` are distinct constants such that a, b are not contained in the interval whoseend-points are k, `, and c is a non-zero constant.

Solution: I will first determine an indefinite integral.



∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .


c(x − a)(x − b)

=A

x − a+

Bx − b

,


c = A · (x − b) + B · (x − a) .


c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .


x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· cx − b

)dx .


x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln |x − a| − ln |x − b|) + C

=x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣ + C ,


(f) ∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣]`

k



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. Eachtype of quiz was generated in multiple varieties for each of the tutorial sections. The order ofthe problems in the varieties was also randomly assigned. Each version of the quiz was gradedout 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


• No calculators!




Monday version

1. [6 MARKS] Showing all of your work, find the length of the following curve for theinterval 0 < a ≤ u ≤ b :

y = ln(eu + 1eu − 1

).

Solution:

(a)

dydx

=eu − 1eu + 1

(eu(eu − 1) − (eu + 1)eu

(eu − 1)2

)

= − 2eu

e2u − 1

1 +

(dydx

)2

=

(e2u + 1e2u − 1

)2

= coth2 u

(b)

arc length =

∫ b

a

√1 +

(dydx

)2

du =

∫ b

a| coth u| du


(c) Successful completion of the integration:∫ b

a| coth u| du =

∫ b

acoth u du

since a < b[ln sinh u]b

a

= lnsinh bsinh a

= ln(eb − e−b

ea − e−a

)

0.0

2.5

−1.0

0.5

1.51.00.5

3.0

2.0

−0.5

1.5

1.0

0.0

−1.5

Figure 1: The limacon r = 1 + 2 sin θ

2. [10 MARKS] (see Figure 1 on page 5071) The graph of the following curve is given.


Showing detailed work, find the area that is enclosed between the inner and the outerloops: 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 curvecrosses 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 overuniquely determined values. How then can the curve cross itself? This can happeneither

i. at points (r(θ1), θ1) and (r(θ1 + π), θ1 + π) where r(θ1 + π) = −r(θ1); orii. 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 crossitself only at the pole. That occurs where 1 + 2 sin θ = 0, i.e., where sin θ = − 1

2 .The values of θ satisfying this equation are 2nπ− π

6 and (2n+1)π+ π6 , where n is any

integer. The outer loop of this limacon is traced out, for example, for −π6 ≤ θ ≤ 7π6 .

The inner loop is traced for 7π6 ≤ θ ≤ 11π

6 .

(b) The area of the region bounded by the larger, outer loop is

a2∫ 7π

6

− π6

12· (1 + 2 sin θ)2 dθ

= a2∫ 7π

6

− π6

12·(1 + 4 sin θ + 4 sin2 θ

)dθ

= a2∫ 7π

6

− π6

12· (1 + 4 sin θ + 2 − 2 cos 2θ) dθ

=a2

2

∫ 7π6

− π6(3 + 4 sin θ − 2 cos 2θ) dθ

=a2

2[3θ − 4 cos θ − sin 2θ]

7π6− π6

=a2

2

(7π2− 4 cos

7π6− sin

7π3

)− a2

2

(−π

2− 4 cos

−π6− sin

−π3

)


= a2

2π +3√

32

.

(c) The area of the inner loop is

a2∫ 11π

6

7π6

12· (1 + 2 sin θ)2 dθ

=a2

2[3θ − 4 cos θ − sin 2θ]

11π6

7π6

=a2

2

(11π

2− 4 cos

11π6− sin

11π3

)− a2

2

(7π2− 4 cos

7π6− sin

7π3

)

= a2

π − 3√

32

.

(d) The area of the region between the loops is the excess of the area inside the outerloop over the area inside the inner loop, i.e.,

a2

2π +

3√

32

−π − 3

√3

2

= a2(π + 3

√3) .

Note that a cleaner way of solving this problem would have been to first integratefrom 0 to 2π, which would give the area between the loops plus twice the areainside 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 theperiodic terms contribute nothing.

This curve is discussed in Exercise 10.4.21, on page 683 of the textbook, and is solvedin the Student Solutions Manual and also on one of the CD-Roms which accompany thetextbook.

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:(dxdt

)2

+

(dydt

)2

=(et − e−t)2

+ 4

=(et + e−t)2

arc length =

∫ b

0

(et + e−t) dt

=[et − e−t]b

0 = eb − e−b = 2 sinh b .


4. [4 MARKS] Find the value of the limit for the sequence. If it diverges, prove that fact:arctan

(3n

3n + 1

).

Solution: As n → ∞, 3n3n+1 = 1 − 1

3n+1 → 1. Since the arctangent function is continuousat the point 1, the limit of the sequence is the arctangent of 1, i.e., π

4 .

5. [6 MARKS] The given curve is rotated about the y-axis. Find the area of the resultingsurface: x = 1

2√

2

(y2 − ln y

), (1 ≤ y ≤ a), where a is a real constant greater than 1.

Solution:

(a)

dxdy

=1

2√

2

(2y − 1

y

)

1 +

(dxdy

)2

= 1 +18

(4y2 +

1y2 − 4

)

=18

(4y2 +

1y2 + 4

)

=

(1

2√

2

(2y +

1y

))2

(b)

surface area =

∫ a

12πx

√1 +

(dxdy

)2

dy

=π

4

∫ a

1

(y2 − ln y

) (2y +

1y

)dy

=π

4

∫ a

1

(2y3 + y − 2y ln y − ln y

y

)dy

=π

4

[y4

2+

y2

2− 1

2(ln y)2

]a

1− π

2

∫ a

1y ln y dy

(c) One way to integrate∫

y ln y dy is by parts, with u = ln y, v′ = y: u′ = 1y , v =

y2

2 ,

∫y ln y dy =

y2

2· ln y −

∫y2

dy

=y2

4(2 ln y − 1) + C .


Another way is to use the substitution w = y2, so dw = 2y dy, ln w = 2 ln y:∫y ln y dy =

∫ln w

2· dw

2=

14

∫ln w dw =

14

w (ln w − 1) + C etc.

Thus the surface area isπ

4

[23

y3 + ln y − 12

(ln y)2 − y2

4(2 ln y − 1)

]a

1

=π

4

(a4

2+

a2

2− 1

2(ln a)2 − a2

4(2 ln a − 1) − 1

)

Tuesday version

1. [4 MARKS] Showing detailed work, find the arc length function for the curve y = ax32

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 withabscissa x.

Solution:

(a)dydx

= a · 32· √x

1 +

(dydx

)2

= 1 +9a2x

4

(b)

f (x) =

∫ x

1

√1 +

9a2t4

dt

=

23· 4

9a2

(1 +

9a2t4

) 32

x

1

=

(4 + 9a2x

) 32 −

(4 + 9a2

) 32

27a2 .

2. [6 MARKS] Showing detailed work, find the area of the surface obtained by rotating thefollowing curve about the x-axis:

y =x2

4− ln x

2(a ≤ x ≤ b)

where a, b are two positive real constants, a < b.

Solution:


(a)

dydx

=12

(x − 1

x

)

1 +

(dydx

)2

= 1 +14

(x2 +

1x2 − 2

)

=14

(x2 +

1x2 + 2

)

=

(12

(x +

1x

))2

(b)

surface area =

∫ b

a2πy

√1 +

(dydx

)2

dx

= π

∫ b

a

(x2

4− ln x

2

) (x +

1x

)dx

= π

∫ b

a

(x3

4+

x4− x ln x

2− ln x

2x

)dx

= π

[116

x4 +x2

8− 1

4(ln x)2

]b

a− π

∫ b

ax ln x dx

(c) One way to integrate∫

x ln x dx is by parts, with u = ln x, v′ = x: u′ = 1x , v = x2

2 ,

∫x ln x dx =

x2

2· ln x −

∫x2

dx

=x2

4(2 ln x − 1) + C .

Another way is to use the substitution w = x2, so dw = 2x dx, ln w = 2 ln x:∫

x ln x dx =

∫ln w

2· dw

2=

14

∫ln w dw =

14

w (ln w − 1) + C etc.

Thus the surface area is

π

[x4

16+

x3

8− 1

4(ln x)2 − x2

4(2 ln x − 1)

]b

a


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)

dxdt

= et(cos t − sin t)

dydt

= et(sin t + cos t)(dxdt

)2

+

(dydt

)2

= e2t(2 cos2 t + 2 sin2 t) = 2e2t

(b) The arc length is ∫ a

0

√2et dt =

√2(ea − 1) .

4. [10 MARKS] Working only with polar coordinates, find the area of the region that liesinside the first curve and outside the second curve: r = b sin θ, r = a, where a and b arepositive constants.

Solution:

(a) Both of these curves are circles; we need to determine the coordinates of the pointsof intersection. Solving the equations yields

r = a sin θ =ab.

One point of intersection will be (r, θ) =(a, arcsin a

b

). Another point of intersection

will be (r, θ) =(a, π − arcsin a

b

)— remember that the values of the arcsine function

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 preciselythe same points, albeit with different coordinates. If we attempt to replace theequation 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 ofintersection except the pole, which must be checked separately. But the pole cannotlie on r = a, since the pole has first coordinate 0 always. Thus we have, indeed,found all the points of intersection.


(b) The area bounded by the arcs can be considered to consist of the disk r = b sin θdiminished by a sector of the circle r = a for arcsin a

b ≤ 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

12

∫ π−arcsin ab

arcsin ab

((b sin θ)2 − a2

)dθ

=

∫ π2

arcsin ab

((b sin θ)2 − a2

)dθ

by symmetry around the line θ = π2

=

∫ π2

arcsin ba

(b2

(1 − cos 2θ

2

)− a2

)dθ

=

∫ π2

arcsin ba

((b2

2− a2

)− b2

2cos 2θ

)dθ

=

[(b2

2− a2

)θ − b2

4sin 2θ

] π2

arcsin ba

=b2 − 2a2

2

(π

2− arcsin

ab

)+

a2

√b2 − a2

5. [6 MARKS] Determine whether the series is convergent or divergent. If it is convergent,

find its sum. Otherwise prove that it is divergent:∞∑

n=1

b

n(n + a)

, where a, b are positive

integers.

Solution:

(a) Expand the general term into partial fractions: there exist constants A, B such that

bn(n + a)

=An

+B

n + a.

To determine the coefficients A, B we can proceed in several ways. If we take thefractions 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), soB = − b

a ; and, setting n = 0, we obtain A = ba , hence

bn(n + a)

=ba

(1n− 1

n + a

).


(b) For sufficiently large N the Nth partial sum is equal to

N∑

n=1

(b

n(n + a)

)

=ba

(11

+12

+ . . . +1

a − 1− 1

N + 1− 1

N + 2− . . . − 1

N + a

)

→ ba

(11

+12

+ . . . +1

a − 1− 0 − 0 − . . . − 0

)

as N → ∞. Hence the series converges to the sum

ba

(11

+12

+ . . . +1

a − 1

).

Convergence could be proved in other ways, thereby earning the student part marks. Forexample, 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 thefollowing 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)

dxdθ

= a (− sin θ − 2 cos θ(− sin θ))

= a(sin θ)(2 cos θ − 1)dydθ

= 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 dxdt .

(b) There will be a horizontal tangent at the point with parameter value t if dydx = 0,

i.e., ifdydθdxdθ

= 0, implying that dydθ must be 0, provided dx

dθ , 0 at the same value of t.


(This last requirement is subtle, and you weren’t expected to actually check it. It isbecause of this restriction that I explicitly excluded the origin from consideration.)The equations we have to solve are

cos θ = 1

andcos θ = −1

2.

They are not being solved simultaneously: we are looking for points t that satisfyat 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. Butthis parameter value corresponds always to the origin, which was excluded fromconsideration.

(d) The second is satisfied when θ is of the form

(2n + 1)π ± 13π .

(e) The functions defining this curve are all periodic with period 2π. Thus we canstudy the curve completely by examining its behavior for parameters θ chosen overan interval of length 2π, e.g. 0 ≤ θ < 2π. There are precisely 3 points here wheredydθ = 0:

(x(0), y(0)) = (0, 0)(x(2π3

), y

(2π3

))=

−a4,

3√

3a4

(x(4π3

), y

(4π3

))=

−a4,−3√

3a4

.

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 thefollowing curve which is in the first quadrant: x = a cos3 θ, y = a sin3 θ, where a is apositive constant.

Solution:

dxdθ

= −3a cos2 θ · sin θ

dydθ

= 3a sin2 θ · cos θ


⇒(dxdθ

)2

+

(dydθ

)2

= 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 isa cube of a sine or cosine, this means that the portion of the curve in the first quadrant isthat given by 0 ≤ θ ≤ π

2 . The length of the arc is

∫ π2

0

√9a2 sin2 θ cos2 θ dθ =

∫ π2

03a sin θ · cos θ dθ = 3a

[sin2 θ

2

] π2

0=

3a2.

3. [10 MARKS] Find the area of the region that lies inside both of the following curvesr = 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 curvescross. We begin by solving the two given equations, and find that

sin θ = −12⇒ θ =

7π6

or11π

6

or any angle obtained from these by adding an integer multiple of 2π. This yieldsthe points (

a − 1,7π6

),

(a − 1,

11π6

).

Students weren’t expected to pursue this question further. Strictly speaking, theyshould then have solved r = a + 2 sin θ, r = −(a− 1), which would have yielded nopoints; 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 alreadyfound.

(b) This problem could then be approached in several ways. To find the area “directly”would require finding the sum of the integrals

12

∫ 7π6

−pi6

(a − 1)2 dθ

and12

∫ 11π6

7pi6

(a + 2 sin θ)2 dθ .


The first of these is just 2/3 of the area of a disk of radius a − 1, i.e., 2π(a−1)2

3 . Thesecond is

12

∫ 11π6

7pi6

(a + 2 sin θ)2 dθ

=12

∫ 11π6

7pi6

(a2 + 4a sin θ + 2(1 − cos 2θ)

)dθ

=12

[a2θ − 4a cos θ + 2θ − sin 2θ

] 11π6

7π6

=1π3· a2 − 2

√3a +

2π3.

(Another way to solve this would be to find the area of the disk of radius a − 1 andsubtract from it the portion that is cut off.These integrals could have been slightly more efficiently computed by taking onlythe area up to the y-axis and doubling it.)Hence the net area of the region inside both of the curves is

5π9

a2 −(4π3

+ 2√

3)

+4π3.

4. [5 MARKS] Showing detailed work, determine whether the following series is conver-gent or divergent. If it is convergent, find its sum. Otherwise explain why it diverges:

∞∑

n=1

an + bn

(ab)n ,

where a, b are integers greater than 1.

Solution: The nth term is

an =

(1b

)n

+

(1a

)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 1in magnitude. Since the two separate partial sums approach a limit, the sum of thesesequences approaches as its limit the sum of the limits of the two sequences. In the caseof the series whose nth term is

(1b

)n, the limit of the partial sums is

1b

1 − 1b

=1

b − 1;


similarly, the second series sums to 1a−1 . Hence the given series sums to the sum of these

limits, i.e.1

b − 1+

1a − 1

=a + b − 2

(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 agiven positive integer, crosses itself at some point (x0, y0). Showing all your work, findthe point of crossing, and the equations of both tangents at the point. (In determiningthe point of crossing you are expected to investigate the parametric functions: it is notsufficient 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 ofthis length for t, and that will reveal all aspects of the behavior of this curve. (Moreprecisely, the tangent function has period π, and, while the cosine function hasperiod 2π, its square has period π.) So, without limiting generality, let’s consider−π2 ≤ t ≤ π

2 : we have to exclude both end points of this interval, since the tangentfunction is not defined at either of them.Suppose that the curve crosses itself at the points with parameter values t = t1 andt = t2; without limiting generality, we can assume that these parameter values havebeen 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) (26)

socos t1 = ± cos t2 . (27)

Since the y-coordinates must also coincide, we also have

(tan t1)(1 − 2 cos2 t1) = (tan t2)(1 − 2 cos2 t2) (28)

which implies that either

cos2 t1 = cos2 t2 =12, (29)

ortan t1 = tan t2 . (30)

In the interval we have chosen for t, the cosines are always positive; the only solu-tion to (29) is t1 = −π4 , t2 = +π

4 . In that same interval for t there will be no solutionsto (30), since the tangent function is increasing there. Thus the only possible cross-ing 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 asymptoticto the vertical line x = 1.

(b) Tangent at the point with parameter value t = −π4 :

dxdt

= 4a(cos t)(sin t) = 2a sin 2t = −2a

y = (tan t)(1 − 2 cos2 t) = tan t − sin 2tdydt

= sec2 t − 2 cos 2t = 2 − 0

dydx

=

dydtdxdt

= − 22a

= −1a,

and the tangent has equation y = − xa .

(c) Tangent at the point with parameter value t = π4 :

dxdt

= 4a(cos t)(sin t) = 2a sin 2t = 2a

dydt

= sec2 t − 2 cos 2t = 2 − 0

dydx

=

dydtdxdt

=22a

=1a,

and the tangent has equation y = xa .

2. [5 MARKS] Showing detailed work, find the surface area generated by rotating thefollowing curve about the y-axis.

x = at2 , y = bt3 , 0 ≤ t ≤ 5.

Solution:

dxdt

= 2at

dydt

= 3bt2

(dxdt

)2

+

(dydt

)2

= 4a2t2 + 9b2t4

area about y-axis = 2π∫ 5

0at2√

4a2t2 + 9b2t4 dt

= 2πa∫ 5

0t3√

4a2 + 9b2t2 dt


Under the substitution u = 4a2 + 9b2t2,

du = 18b2t dt

2πa∫ 5

0t3√

4a2 + 9b2t2 dt =aπ

81b4

∫ 4a2+225b2

4a2(u

32 − 4a2u

12 ) du

=πa

81b4

[25

u52 − 8a2

3u

32

]4a2+225b2

4a2

=2πa

(81)(15)b4

((4a2 + 225b2)

32 (2a2 + 675b2) − 16a5

)

= ...

3. [10 MARKS] There is a region in the first quadrant that is bounded by arcs of both ofthe 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 givenfunctions are periodic with period π. When, for either of these curves, we permitθ to range over an interval of length π, we will trace out the entire curves. Theintersections of the curves in the first quadrant will be where sin 2θ = cos 2θ ispositive: thus the only point we have found by this algebraic solution of the twoequations is at θ = π

8 .However there are other ways in which curves can intersect, since points haveinfinitely many different sets of polar coordinates. If we transform either of thegiven equations under the substitution (r, θ) → (−r, θ + π), we find that there isno change in the equation. Thus we haven’t missed any points because of theconvention that permits the first coordinate to be negative.But there is another situation that leads to multiple sets of coordinates; that is at thepole, where the second — angular — coordinate is totally arbitrary; the pole canlie on a curve simply because of the fact that its distance coordinate r = 0, with noreference to θ. To determine whether the pole lies on a curve we must investigatewhether the equation is satisfied by r = 0 with any value of θ. We find the curver2 = a sin 2θ does contain the pole: when r = 0 the equation is satisfied by any θsuch that sin 2θ = 0; so two solutions are θ = 0 and θ = π

2 . Similarly, the pole lieson the curve r2 = a cos 2θ with θ = π

4 . I have given only the coordinates in the firstquadrant. To summarize: there are 2 intersection points in the first quadrant:

(r, θ) =

(a

12 2−

14 ,π

4

)


where the point lies on both of the curves with the same pair of coordinates; andthe pole, which lies on the two curves with different sets of coordinates.

(b) We can find the area by joining the point( √

a4√2, π8

)to the pole and calculating the

sum of two integrals:

πa2

∫ π4

0sin 2θ dθ +

πa2

∫ π2

π4

cos 2θ, dθ

=πa4

[− cos 2θ]π40 +

πa4

[sin 2θ]π2π4

=πa4

+πa4

=πa2.

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 aspossible.

Solution: The repeating decimal is the sum of an infinite series(10a + b

100

)+

(10a + b

100

)1

100+

(10a + b

100

)1

1002 +

(10a + b

100

)1

1003 + . . .

= limN→∞

N∑

n=0

(10a + b

100

)1

100n

=

(10a + b

100

)lim

N→∞

N∑

n=0

(1

100

)n

=

(10a + b

100

)lim

N→∞

1 −(

1100

)N+1

1 − 1100

=

(10a + b

100

)1

1 − 1100

=

(10a + b

100

)10099

=10a + b

99


Release Date: Mounted on the Web on Friday, 01 February, 2008 (but subject to correction)




• No calculators!




Monday version

1. [5 MARKS] If

c∫

a

f (x) dx = k and

c∫

b

f (x) dx = `, find

b∫

a

f (x) dx. Show your work.

Solution:

(a)c∫

a

f (x) dx =

b∫

a

f (x) dx +

c∫

b

f (x) dx .

(b) Hencec∫

b

f (x) dx =

c∫

a

f (x) dx −b∫

a

f (x) dx .

(c)= k − ` .

2. [5 MARKS] Find an antiderivative of the integrand of the integral∫ a

0

√x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expectedto simplify your numerical answer, but no marks will be given unless all your work isclearly shown.


(a) One antiderivative of x12 is

112 + 1

· x 12 +1 =

23

x32 .

(b) ∫ a

0

√x dx =

[23

x32

]a

0=

23

(a

32 − 0

)=

23

a32 .

3. [10 MARKS] Showing all your work, differentiate the function g(x) =

x4∫

tan x

1√2 + t2

dt.

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

g(x) =

0∫

tan x

1√2 + t2

dt +

x4∫

0

1√2 + t2

dt .

(b) Then reverse the limits in the first summand and change its sign, so that the variablelimit is the upper one:

g(x) = −tan x∫

0

1√2 + t2

dt +

x4∫

0

1√2 + t2

dt .

(c) Denote the upper limit of the first integral by u = tan x. Then

ddx

tan x∫

0

1√2 + t2

dt =d

du

u∫

0

1√2 + t2

dt · dudx

=1√

2 + u2· sec2 x

=2 sec x · tan x√

2 + tan2 x=

2 sec x · tan x√1 + sec2 x

.

(d) Denote the upper limit of the second integral by v = x4. Then

ddx

x4∫

0

1√2 + t2

dt =ddv

v∫

0

1√2 + t2

dt · dvdx


=1√

2 + v2· 4x3

=4x3

√2 + x8

.

(e) Henceddx

g(x) = −2 sec x · tan x√2 + tan2 x

+4x3

√2 + x8

.

4. [10 MARKS] If F(x) =

∫ x

1f (t) dt, where f (t) =

∫ t2

1

a + ub

udu and a, b are constants,

find F′′(2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F′(x) = f (x) =

∫ x2

1

a + ub

udu .

(b) A second application of Part 1 of the Fundamental Theorem yields

F′′(x) = f ′(x) =ddx

∫ x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

ddx

∫ x2

1

a + ub

udu =

ddx

∫ v

1

a + ub

udu

=ddv

∫ v

1

a + ub

udu · dv

dx

=a + vb

v· dv

dx

=a + vb

v· 2x

=a + x2b

x2 · 2x

=2(a + x2b

)

x.


Tuesday version


g(x) =

∫ a

xb tan(t) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluateg(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 afunction of its upper index of integration. Here the variable is the lower index ofintegration.

ddx

∫ a

xb tan(t) dt

=ddx

(−

∫ x

ab tan(t) dt

)

= − ddx

∫ x

ab tan(t) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definiteintegral with respect to the lower index, and this should be accepted if work hasbeen shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

ddx

ln | sec x| =ddu

ln |u| · ddx

secx

=1u· sec x tan x

=1

sec x· sec x tan x

= tan x .

Henceg(x) = b ln | sec t|ax = b ln

∣∣∣∣∣sec asec x

∣∣∣∣∣ = b ln∣∣∣∣∣cos xcos a

∣∣∣∣∣ .


2. [5 MARKS] Evaluate the limit by first recognizing the sum as a Riemann sum for afunction defined on [0, 1]:

limn→∞

1n

√

7n

+

√14n

+

√21n

+ . . . +

√7nn

.

Solution:

(a) We are told to take the interval of integration to be [0, 1]; when this is divided into

n equal parts, each has length ∆x =1n

. Such a factor has been explicitly written inthe sum.

(b) The typical summand is — aside from the common factor 1n — of the form

√7in

.

Since the distance of the left end-point of the ith subinterval from 0 is i∆x = in , we

may interpret

√7in

=√

7x .

(c) Thus the limit must be equal to∫ 1

0

√7x dx =

√7 · 2

3· x 3

2

]1

0=

23

√7.


g(x) =

∫ √x

ab

cos tt

dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u(x) =√

x.

(b) Then

ddx

g(x) =ddx

∫ √x

ab

cos tt

dt

=ddx

∫ u(x)

ab

cos tt

dt

=d

du

∫ u(x)

ab

cos tt

dt · du(x)dx

= bcos u

u· du(x)

dx


= bcos u

u· 1

2√

x

= bcos√

x√x· 1

2√

x

= bcos√

x2x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y =x∫

0

11 + at + bt2 dt is concave downward, where a, b are constants.

Solution:

(a) By Part 1 of the Fundamental Theorem,

y′(x) =1

1 + ax + bx2 .

(b) Differentiating a second time yields

y′′(x) =ddx

(1

1 + ax + bx2

)

= − 1(1 + ax + bx2)2

· ddx

(1 + ax + bx2

)

= − a + 2bx(1 + ax + bx2)2 .

(c) The curve is concave downward where y′′ < 0:

− a + 2bx(1 + ax + bx2)2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive⇔ 2bx < −a

⇔

x < − a2b when b > 0

x > − a2b when b < 0

never concave upward when b = 0

Wednesday version



πb∫

πa

sin t dt.

Solution:

(a) An antiderivative of sin t is − cos t.

(b)πb∫

πa

sin t dt = [− cos t]πbπa

(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

sum for a function defined on [0, 1]: limn→∞

n∑

i=1

i8

n9 . A full solution is required — it is not

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

n equal parts, each has length ∆x =1n

.

(b) The typical summand is — aside from the common factor1n

— of the form( in

)8

.

Since the distance of the left end-point of the ith subinterval from 0 is i∆x =in

, we

may interpret( in

)8

= x8 .

(c) Thus the limit must be equal to

∫ 1

0x8 dx =

19

x9]1

0=

19.


of the function

bx∫

cos x

cos (tc) dt, where a, b, c are real numbers.

Solution:



bx∫

cos x

cos (tc) dt =

0∫

cos x

cos (tc) dt +

bx∫

0

cos (tc) dt

= −cos x∫

0

cos (tc) dt +

bx∫

0

cos (tc) dt

(b) For the summand

bx∫

0

cos (tc) dt, let u = bx. Then

ddx

bx∫

0

cos (tc) dt =ddx

u∫

0

cos (tc) dt

=d

du

u∫

0

cos (tc) dt · dudx

= cos (uc) · b= cos ((bx)c) · b

(c) For the summand

cos x∫

0

cos (tc) dt, let v = cos x.

ddx

cos x∫

0

cos (tc) dt =ddx

v∫

0

cos (tc) dt

=ddv

v∫

0

cos (tc) dt · dvdx

= cos (vc) · (− sin x)= cos (cosc x) · (− sin x)

(d)bx∫

cos x

cos (tc) dt = − cos (cosc x) · (− sin x) + cos ((bx)c) · b .


4. [10 MARKS] Let f (x) =

0 if x < 0x if 0 ≤ x ≤ a2a − x if a < x < 2a0 if x > 2a

and g(x) =

∫ x

0f (t) dt, where a is

a positive constant. Showing all your work, find a formula for the value of g(x) whena < x < 2a.

Solution:

(a) The interval where we seek a formula is the third interval into which the domainhas been broken. For x in this interval the integral can be decomposed into

∫ x

0f (t) dt =

∫ a

0f (t) dt +

∫ x

af (t) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, sincewe are not finding area under that portion of the curve; the same applies to theportion of the definition for x > 2a.

(b)∫ a

0f (t) dt =

∫ a

0t dt

=

[t2

2

]t=a

t=0=

a2

2.

(c)∫ x

af (t) dt =

∫ x

a(2a − t) dt

=

[2at − t2

2

]t=x

t=a

=

(2ax − x2

2

)−

(2a2 − a2

2

).

(d)

g(x) =a2

2+

(2ax − x2

2

)−

(2a2 − a2

2

)= 2ax − x2

2− a2 .

Thursday version


1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ∫

aπ

cos θ dθ, where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ.

(b)bπ∫

aπ

cos θ dθ = [sin θ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiplesof π is 0, so the value of the definite integral is 0.

2. [5 MARKS] Express limn→∞

n∑

i=1

axi sin xi ∆x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: ∫ c

bax sin x dx .


g(x) =

∫ bx

ax

t2 + ct2 − c

dt ,

where a, b, c are positive integers.

Solution:


g(x) =

∫ bx

ax

t2 + ct2 − c

dt =

∫ 0

ax

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

= −∫ ax

0

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt


(b) For the summand

bx∫

0

t2 + ct2 − c

dt, let u = bx. Then

ddx

bx∫

0

t2 + ct2 − c

dt =ddx

u∫

0

t2 + ct2 − c

dt

=d

du

u∫

0

t2 + ct2 − c

dt · dudx

=u2 + cu2 − c

· dudx

=u2 + cu2 − c

· b

=(bx)2 + c(bx)2 − c

· b

(c) For the summand∫ ax

0

t2 + ct2 − c

dt, let u = ax. Then, analogously to the preceding

step,ddx

∫ ax

0

t2 + ct2 − c

dt =(ax)2 + c(ax)2 − c

· a .

(d)

g′(x) =(bx)2 + c(bx)2 − c

· b − (ax)2 + c(ax)2 − c

· a .

4. [10 MARKS] Find the derivative of the function f (x) =

x3∫

√x

√t cos t dt.

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

f (x) =

0∫

√x

√t cos t dt +

x3∫

0

√t cos t dt


(b) Then reverse the limits in the first summand and change its sign, so that the variablelimit is the upper one:

f (x) = −√

x∫

0

√t cos t dt +

x3∫

0

√t cos t dt .

(c) Denote the upper limit of the first integral by u =√

x. Then

ddx

√x∫

0

√t cos t dt =

ddx

u∫

0

√t cos t dt

=d

du

u∫

0

√t cos t dt · du

dx

=√

u cos u · dudx

=

√√x cos

√x · 1

2√

x

=cos√

x

2x14

.

(d) Denote the upper limit of the second integral by v = x3. Then

ddx

x3∫

0

√t cos t dt =

ddx

v∫

0

√t cos t dt

=ddv

v∫

0

√t cos t dt · dv

dx

=√

v cos v · dvdx

=√

x3 cos(x3

)· 3x2

= 3x72 cos

(x3

)

(e) Henceddx

f (x) = −cos√

x

2x14

+ 3x72 cos

(x3

).



Distribution Date: Mounted on the Web on Monday, March 03rd, 2008Caveat lector! There could be misprints or errors in these draft solutions.



• No calculators!




Monday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotatingabout 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 nis even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-tions 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 thecase of odd n, to permit the second intersection point (x, y) = (−1,−1); thedecision was left to the individual TA’s. The remainder of this solution coversthe case of even n; for odd n this solution does not consider the solid generatedby 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 revolutionis a horizontal line, the element of area being rotated is vertical. For arbitraryx 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.


(e) The volume of the “washer” is, therefore,

π(−(1 − x)2 + (1 − n√x)2

)∆x .

(f) Correctly evaluate the integral:

π

∫ 1

0

(−(1 − x)2 + (1 − n√x)2

)dx

= π

∫ 1

0

(−2x + x2 + 2x

1n − x

2n)

dx

= π

[−x2 +

x2

3+

2nn + 1

xn+1

n − nn + 2

xn+2

n

]1

0

= π

(−1 +

13

+2n

n + 1− n

n + 2

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) The solution I am giving is for the casewhere n is even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-tions 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 thecase of odd n, to permit the second intersection point (x, y) = (−1,−1); thedecision was left to the individual TA’s. The remainder of this solution coversthe case of even n; for odd n this solution does not consider the solid generatedby 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 revolutionis a horizontal line, the element of area being rotated is also horizontal. Forarbitrary 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 fromthe 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:

2π∫ 1

0(1 − y)(y − yn) dy

= 2π∫ 1

0

(−yn + yn+1 + y − y2

)dy


= 2π[− 1

n + 1yn+1 +

1n + 2

yn+2 +12

y2 − 13

y3]1

0

= 2π(− 1

n + 1+

1n + 2

+12− 1

3

)− 0

= 2π(16− 1

(n + 1)(n + 2)

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)


(a − t)(b + t2) dt.

Solution:

(a) Expand the product in the integrand:∫

(a − t)(b + t2) dt =

∫ (ab − bt + at2 − t3

)dt .

(b) Integrate term by term:∫ (

ab − bt + at2 − t3)

dt = ab · t − b2· t2 +

a3· t3 − 1

4· t4 + C .

3. [10 MARKS] Showing all your work, determine a number b such that the line x = bdivides into two regions of equal area the region bounded by the curves x = ay2 andx = k.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 1 of the Tuesday quiz.

4. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral∫ex

ex + adx, where a is a non-zero real number.

Solution:

(a) Try the substitution u = ex + a, so du = ex dx.

(b) ∫ex

ex + adx =

∫duu

= ln |u| + C = ln |ex + a| + C .

(If the constant a is positive, then the absolute signs are not required.)


Tuesday version

1. [10 MARKS] Showing all your work, find a number b such that the line y = b dividesthe 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 intersections of the

bounding curves: solving the equations yields the points

∓√

ka, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin byrepeating the calculation of the preceding part: the corner points have coordinates∓

√ba, b

. The area is

∫ √ ba

−√

ba

(b − ax2) dx = 2[bx − ax3

3

]√ ba

0=

43

b

√ba.

(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is43

k

√ka

.

(d) The condition of the problem is that

43

b

√ba

=12· 4

3k

√ka

which is equivalent to 4b3 = k3, and implies that the line should be placed whereb = 2−

23 k.

2. [10 MARKS] The region bounded by the curves y = 5 and y = x2 − ax + b is rotatedabout 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, wesolve 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.


(d) The volume is given by the integral

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

(e) Evaluating the integral:

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

= 2π∫ 2

1

(−x3 + 2x2 + x − 2

)dx

= 2π[−1

4x4 +

23

x3 +12

x2 − 2x]2

1

= 2π(−4 +

163

+ 2 − 4 +14− 2

3− 1

2+ 2

)

=5π6

Using the method of “washers” (a) To find the lowest point on the parabola, we solvex2 − 3x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be(32,

194

).

(b) The horizontal element generating the “washer” at height y extends betweenthe solutions in x to the equation y = x2 − 3x + 7; these are

x =3 ± √

4y − 192

.

(c) The volume of the “washer” at height y is, therefore,

π

1 +

3 +√

4y − 192

2

−1 +

3 − √4y − 192

2 ∆y

= 5π√

4y − 19 ∆y

(d) The volume is given by the integral 5π∫ 5

194

√4y − 19 dy.

(e) Evaluation of the integral:

5π∫ 5

194

√4y − 19 dy =

[5π · 2

3· 1

4(4y − 19)

32

]5

194

=5π6.


3. [5 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)∫

cosn x sin x dx = −∫

un du

= − un+1

n + 1+ C

= − 1n + 1

cosn+1 x + C

4. [5 MARKS] Showing all your work, evaluate the integral∫ (

xb + a +1

x2 + 1

)dx, (where

a and b are given positive integers).

Solution:

(a)∫

xb dx =xb+1

b + 1+ C1,

(b)∫

a dx = ax + C2

(c)∫

1x2 + 1

dx = arctan x + C3

(d)∫ (

xb + a +1

x2 + 1

)dx =

xb+1

b + 1+ ax + arctan x + C.

Wednesday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotatingabout the line x = −1 the region bounded by y = xn and x = yn, where n is a givenpositive integer.

Solution:

Case I: n is even

Using the method of “washers”: (a) Find the intersections of the curves bound-ing 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 revo-lution is a vertical line, the element of area being rotated is horizontal. Forarbitrary y the farther endpoint on the element is (y

1n , y); the nearer end-

point is (yn, y). The distances of these points from the axis are, respectively1 + n√

y and 1 + yn.(c) The volume of the “washer” is, therefore,

π(−(1 + yn)2 + (1 + n

√y)2

)∆y .


π

∫ 1

0

(−(1 + y)2 + (1 + n

√y)2

)dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)


(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(x, x

1n ); the lower endpoint is (x, xn). The length of the element is, there-

fore, x1n − 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,

2π(1 + x) ·(x

1n − xn

).


2π∫ 1

0(1 + x)(x

1n − xn) dx

= 2π∫ 1

0

(x

1n − xn + x

n+1n − xn+1

)dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)


Case II: n is oddUsing the method of “washers”: (a) Find the intersections of the curves bound-

ing 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 theword 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 willfollow 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 rev-olution is a vertical line, the element of area being rotated is horizontal.But there are two kinds of elements, depending on whether y is positiveor negative. For arbitrary, positive y the farther endpoint on the element is(y

1n , 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 thenearer endpoint on the element is (y

1n , y); the farther endpoint is (yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and1 + 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 .


π

∫ 1

−1

∣∣∣−(1 + y)2 + (1 + n√

y)2∣∣∣ dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

+π

∫ 0

−1

(−2y

1n − y

2n + 2yn + y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

+π

[− 2n

n + 1y

n+1n − n

n + 2y

n+2n +

2n + 1

yn+1 +1

2n + 1y2n+1

]0

−1

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

+2(n − 1)(n2 + 3n + 1)π(n + 1)(n + 2)(2n + 1)

=4(n − 1)π

n + 1.


(0, 0), (±1,±1).


(b) Since the axis of revolution is a vertical line, the element of area beingrotated is also vertical. For arbitrary, positive x the top endpoint on the el-ement is (x, x

1n ); the lower endpoint is (x, xn); for arbitrary, negative x the

bottom endpoint on the element is (x, x1n ); the upper endpoint is (x, xn).

The length of the element is, therefore,∣∣∣∣x 1

n − xn∣∣∣∣; the distance of the ele-

ment from the axis of rotation is 1 + x.(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·∣∣∣∣x 1

n − xn∣∣∣∣ .


2π∫ 1

−1(1 + x)

∣∣∣∣x 1n − xn

∣∣∣∣ dx

= 2π∫ 1

−1

∣∣∣∣x 1n − xn + x

n+1n − xn+1

∣∣∣∣ dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

+2π(

nn + 1

− 1n + 1

− n2n + 1

+1

n + 2

)

=4(n − 1)π

n + 1


sin 2tcos t

dt.

Solution:

(a) Apply a “double angle” formula:∫

sin 2tcos t

dt =

∫2 sin t cos t

cos tdt = 2

∫sin t dt .

(b) Complete the integration:

2∫

sin t dt = −2 cos t + C .

3. [10 MARKS] Showing all your work, find the area of the region bounded by the parabolay = x2, the tangent line to this parabola at (a, a2), and the x-axis, (where a is a given realnumber).

Solution: This area can be computed by integrating either with respect to y or withrespect to x.


Integrating with respect to y: (a) Since y′ = 2x, the tangent line through (a, a2) hasequation

y − a2 = 2a(x − a)⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabolaand 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

equation x =y

2a+

a2

.

(c) The area of the horizontal element of area at height y is(y + a2

2a− √y

)∆y.

(d) The area is the value of the integral∫ a2

0

(y + a2

2a− √y

)dy .

(e) Integration yields[

y2

4a+

ay2− 2

3y

32

]a2

0=

(14

+12− 2

3

)a3 =

112

a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Itsintercept with the x-axis is at x =

a2

.

(b) The area of the vertical element of area at horizontal position x ≤ a2

is(x2 − 0

)∆x.

(c) The area of the vertical element of area at horizontal position x ≥ a2 is

(x2 − (2ax − a2)

)dx =

(x − a)2 ∆x.(d) The area of the region is the sum

∫ a2

0x2 dx +

∫ a

a2

(x − a)2 dx .

(e) Integration yields [x3

3

] a2

0+

[−(x − a)3

3

]a

a2

=a3

12.


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) ∫seca x tan x dx =

∫ua−1 du =

ua

a+ C =

seca xa

+ C .

Some students may have integrated by sight.

Thursday version

1. [10 MARKS] Showing all your work, use the method of cylindrical shells to find thevolume generated by rotating about the axis x = b the region bounded by the curvesy =√

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 hasleft 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 theaxis of rotation is b − y.

(d) Set up the integral for the volume by cylindrical shells:

∫ √a−1

0(b − y)

(a − (1 + y2)

)dy .

(e) Evaluate the integral

∫ √a−1

0(b − y)

(a − (1 + y2)

)dy

=

∫ √a−1

0

(b(a − 1) − (a − 1)y − by2 + y3

)dy

=

[b(a − 1)y − a − 1

2y2 − b

3y3 +

14

y4]√a−1

0

= b(a − 1)32 − a − 1

2(a − 1) − b

3(a − 1)

32 +

14

(a − 1)2

= (a − 1)32

(23· b − 1

4

√a − 1

).


2. [5 MARKS] Showing all your work, use a substitution to evaluate the integral∫

b(1+ax)3 dx ,

where a, b are non-zero constants.

Solution:

(a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, sodx = 1

a du.

(b) ∫b

(1 + ax)3 dx =ba

∫duu3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

3. [5 MARKS] Showing all your work, 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)∫

t2 cos(a − t3

)dt =

∫13

cos(a − u) du

= −13

sin(a − u) + C

= −13

sin(a − t3) + C.

(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 theparabola x = y2, the tangent line to this parabola at (a2, a), and the y-axis, where a is afixed, positive real number.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 3 of the Wednesday quiz.


Release Date: Mounted on the Web on 05 April, 2008These draft solutions could contain errors, and they must be subject to correction. Caveat

lector!




• No calculators!




Monday version


x5 ln(20x) dx.

Solution:

(a) I will integrate by parts, setting u = ln(20x), dv = x5 dx. Then du =dxx

, v =x6

6.

(b)∫

x5 ln(20x) dx = (ln(20x)) ·(

x6

6

)− 1

6

∫x5 dx

=x6 ln(20x)

6− x6

36+ C .


1 − cot2 xcsc2 x

dx.

Solution:∫

1 − cot2 xcsc2 x

dx =

∫sin2 x − cos2 xsin2 x · csc2 x

dx

=

∫ (sin2 x − cos2 x

)dx

=

∫(− cos 2x) dx

= −12

sin 2x + C .


3. [10 MARKS] Use a substitution to transform this integral into the integral of a rationalfunction; then integrate, and express your answer in terms of x:

∫1

e−3x − e−x dx

Solution:

(a) I would try the substitution u = e−x, so du = −e−x dx = −u dx.

(b) Then∫

1e−3x − e−x dx =

∫du

u2(1 − u2)du

=

∫ 1u2 +

12

1 + u+

12

1 − u

du

= −1u

+12

ln |1 + u| − 12

ln |1 − u| + C

= −1u− 1

2ln

∣∣∣∣∣1 − u1 + u

∣∣∣∣∣ + C

= −ex − ln

√∣∣∣∣∣1 − e−x

1 + e−x

∣∣∣∣∣ + C

There are other, equivalent ways in which this last class of antiderivatives can be ex-pressed. (For example, if we define K = eC, so C = ln K, we can bring the logarithmterms together.)


x2

(4 − x2) 3

2

dx.

Solution:

(a) Use a trigonometric substitution, e.g., x = 2 cos θ, i.e., θ = arccos x2 . Then dx =

−2 sin θ dθ.

(b)∫

x2

(4 − x2) 3

2

dx = −∫

cos2 θ sin θsin3 θ

dθ

= −∫

cot2 θ dθ

= −∫ (

csc2 θ − 1)

dθ


= cot θ + θ + C

=cos θsin θ

+ θ + C

=cos θ√

1 − cos2 θ+ arccos

x2

+ C

=

x2√

1 − x2

4

+ arccosx2

+ C

=x√

4 − x2+ arccos

x2

+ C .

Tuesday version


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. Twoapplications of integration by parts, with appropriate choices of functions, willyield an equation that can be solved for the value of the indefinite integral. For thefirst application take, for example, u = e5x and dv = cos 2x. Then du = 5e5x dx,and v = 1

2 sin 2x.

(b) ∫e5x cos(2x) dx = e5x · 1

2sin 2x − 5

2

∫e5x sin 2x dx . (31)

(c) Now we apply integration by parts to evaluate∫

e5x sin 2x dx, taking U = e5x anddV = sin 2x. Then dU = 5e5x dx, and V = − 1

2 cos 2x.

(d) ∫e5x sin(2x) dx = e5x ·

(−1

2cos 2x

)+

52

∫e5x cos 2x dx . (32)

(e) Combining equations (31), (32) yields∫

e5x cos(2x) dx = e5x · 12

sin 2x − 52

(−e5x · 1

2· cos 2x +

52

∫e5x sin 2x dx

)

= e5x ·(12· sin 2x +

54

cos 2x)− 25

4

∫e5x cos(2x) dx .


(f) Solving the last equation yields∫

e5x cos(2x) dx =1

29e5x · (2 sin 2x + 5 cos 2x) + C .

2. [10 MARKS] Showing all your work, use a substitution to change this integral into theintegral of a rational function; then integrate and express your solution in terms of x:

∫1 + 7ex

1 − ex dx .

Solution:

(a) Let u = ex, so du = ex dx.

(b) Then∫

1 + 7ex

1 − ex dx =

∫1 + 7u1 − u

· duu

and we proceed to expand the integrand into

a sum of partial fractions.

(c) Assuming that1 + 7u

u(1 − u)=

Au

+B

1 − u, and multiplying both sides by u(1 − u), we

obtain the identity in u 1 + 7u = A(1 − u) + Bu. Setting u = 0 and u = 1 yields theequations 1 = A and 8 = B. We can now continue integration.

(d)∫

1 + 7u1 − u

du =

∫ (1u

+8

1 − u

)du

= ln |u| − 8 ln |1 − u| + C= ln ex − 8 ln |1 − ex| + C= x − 8 ln |1 − ex| + C

which solution can be checked by differentiation.


dx(x2 + 8x + 17

)2 dx.

Solution:

(a) Completing the square of the polynomial in the denominator, we obtain

x2 + 8x + 17 =

(x +

82

)2

+

17 −(82

)2 = (x + 4)2 + 1 .

Accordingly, we can simplify the integral by taking u = x + 4, du = dx.


(b) The preceding substitution is not sufficient, however. All we obtain is∫

dx(x2 + 8x + 17

)2 dx =

∫du

(u2 + 1

)2 .

We can simplify this further by taking u = tan θ, i.e., by taking θ = arctan u, so

dθ =du

1 + u2 .

(c)∫

du(1 + u2)2 =

∫sec2 θ

sec4 θdθ

=

∫cos2 θ dθ

=

∫1 + cos 2θ

2dθ =

θ

2+

sin 2θ4

+ C

=12

arctan u +u

2(1 + u2)+ C

=12

(arctan(x + 4) +

x + 4x2 + 8x + 17

)+ C

which can be verified by differentiation. You should always verify this type ofintegration by differentiation, in order to locate silly algebra mistakes (or worse).

Wednesday version


x cos(18x) dx.

Solution:

(a) This can be solved using integration by parts. Define u = x, dv = cos(18x) dx, so

du = dx and v =sin(18x)

18.

(b)∫

x cos(18x) dx = x · sin(18x)18

−∫

sin(18x)18

=x sin(18x)

18+

cos(18x)182 + C ,

which can be verified by differentiation.



1 − sin xcos x

dx.

Solution:∫

1 − sin xcos x

dx =

∫(sec x − tan x) dx

= 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.


π3∫

π6

ln tan x(sin x) · (cos x)

dx.

Solution:

(a) In view of the complicated nature of the integrand, one would be advised to seeka substitution that could render it more amenable. But the integrand involves bothsines and cosines. However, note that

(sin x)(cos x) = (tan x)(cos2 x) =tan xsec2 x

=tan x

tan2 x + 1.

Taking u = tan x, we have du = sec2 x dx. When x =π

3, u =

√3; when x =

π

6,

u =1√3

.

(b)

π3∫

π6

ln tan x(sin x) · (cos x)

dx

=

√3∫

1√3

ln uu

u2+1

duu2 + 1

=

√3∫

1√3

ln uu

du


(c) Now the integral looks as though it could be simplified by a substitution v = ln u,

so dv =duu

. When u =√

3, v = ln 32 ; when u = 1√

3, v = − ln 3

2 .

(d)√

3∫

1√3

ln uu

du =

∫ ln 22

− ln 32

v dv

=

[v2

2

] ln 32

− ln 32

= 0.


et√

49 − e2t dt.

Solution:

(a) Clearly a substitution of the form u = et is indicated, in order to simplify theintegrand. We find that du = et dt.

(b) We obtain∫

et√

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:∫ √

49 − u2 du = 49∫

cos2 θ dθ.

(c)

49∫

cos2 θ dθ =492

∫(1 + cos 2θ) dθ

=492

(θ + sin θ · cos θ) + C

=492

arcsinu7

+u2

√49 − u2 + C

=492

arcsinet

7+

et√

49 − e2t

2+ C .

Thursday version


sin3 9x dx.


Solution: The integrand is an odd power of the sine function. I will substitute u = cos 9x,so du = −9 sin 9x dx.

∫sin3 9x dx =

∫ (1 − u2

) (19

)du

= −u9

+u3

27+ C

= −cos x9

+cos3 x

27+ C .


π2∫

π4

cot2 x dx. Your answer

should be simplified as much as possible; the instructors are aware that you do not havethe use of a calculator.

Solution:

(a) Recall that cot2 x = csc2 x − 1, and thatddx

cot x = − csc2 x.

(b)

π2∫

π4

cot2 x dx =

π2∫

π4

(csc2 x − 1

)dx

= [− cot x − x]π4π2.

(c) = −π2

+ 1 +π

4= 1 − π

4.

3. [10 MARKS] Showing all your work, evaluate the integral∫ √

9 − x2

xdx.

Solution:

(a) To simplify the square root, substitute x = 3 cos θ, i.e., θ = arccos x3 . Then dx =

−3 sin θ dθ.

(b)

∫ √9 − x2

xdx = −3

∫sin2 θ

cos θdθ


= −3∫

1 − cos2 θ

cos θdθ

= −3∫

(sec θ − cos θ) dθ

= −3 ln | sec θ + tan θ| + 3 sin θ + C

(c)

−3 ln | sec θ + tan θ| + 3 sin θ + C = −3 ln

∣∣∣∣∣∣∣∣∣

1 +

√1 − x2

9x3

∣∣∣∣∣∣∣∣∣+ 3

√1 − x2

9+ C

= −3 ln

∣∣∣∣∣∣∣3 +√

9 − x2

x

∣∣∣∣∣∣∣ +√

9 − x2 + C

4. [10 MARKS] Showing all your work, determine whether the following integral is con-vergent or divergent. Evaluate it if it is convergent; in such a case you are expected tosimplify your answer as much as is consistent with not having the use of a calculator:

∫ 1

06(ln 7x√

x

)dx

Solution:

(a) Let’s look first at the associated indefinite integral,∫

ln 7x√x

dx. The integrand

can be expressed as a product, in which one of the factors “simplifies” upon dif-ferentiation, while the other does not become significantly “more difficult” uponintegration. So we will integrate by parts, taking u = ln 7x, and dv = x−

12 dx. Then

du = 1x , and we may take v = 2

√x.

(b)∫

ln 7x√x

dx = (ln 7x)(2√

x) − 2∫

dx√x

= (ln 7x)(2√

x) − 4√

x + C .

(c) The integrand is not defined at x = 0 in the given interval of integration. By thedefinition of an improper integral, we have

∫ 1

0

ln 7x√x

dx = lima→0+

∫ 1

a

ln 7x√x

dx


= lima→0+

[(ln 7x)(2

√x) − 4

√x]1

a

= lima→0+

[(2 ln 7 − 4) − √a · (2 ln(7a) − 4)

]

= 2 ln 7 − 4 − lima→0+

(ln 7a)(2√

a)

(d) The limit can be expressed as that of a ratio where numerator and denominator bothbecome infinite. Thus l’Hospital’s Rule may be used:

lima→0+

(ln 7a)(2√

a) = 2 lima→0+

ln 7a

a−12

= 2 lima→0+

1a

− 12a−

32

= 2 lima→0+

(−2√

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, andthen evaluate the integral. ∫ 4

1

√x

x − 16dx .

Solution:

(a) Start by substituting u =√

x, so x = u2, dx = 2u du. When x = 1, u = 1; whenx = 4, u = 2.

(b)∫ 4

1

√x

x − 16dx =

∫ 2

1

2u2

u2 − 36du

=

∫ 2

1

(2 +

72u2 − 36

)du

=

∫ 2

1

(2 +

6u − 6

− 6u + 6

)du

=[u2 + 6 ln |u − 6| − 6 ln |u + 6|

]2

1

= 3 + 6 ln7

10.



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 in-tended that Teaching Assistants would consult these draft solutions when they graded their stu-dents’ quizzes, and would report any errors or omissions. As the Teaching Assistants may be-lieve 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 releasedwith the cautionary warning, Caveat lector! — Let the reader beware! Use them at your ownrisk.

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. Theorder of the problems in the varieties was also randomly assigned. All of the quizzes had aheading that included the instructions.


• No calculators!




Monday version

1. [10 MARKS] Showing all your work, find the length of the curve y =x5

30+

12x3 (1 ≤

x ≤ 2). Simplify your answer as much as possible; the instructors are aware that you donot have the use of a calculator.

Solution:

dydx

=x4

6− 3

2x4

⇒√

1 +

(dydx

)2

=

√1 +

x8

36+

94x8 −

12

=

√x8

36+

94x8 +

12

=

√(x4

6+

32x4

)2

=

∣∣∣∣∣∣x4

6+

32x4

∣∣∣∣∣∣ .


The absolute signs may be dropped, since the given square root is a sum of positivemultiples of even powers, and must be non-negative. The length is

∫ 2

1

(x4

6+

32x4

)dx =

[x5

30− 1

2x3

]2

1

=353240

.

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating

x =12

(y2 + 2

) 32 (7 ≤ y ≤ 10) about the x-axis. Simplify your answer as much as

possible; the instructors are aware that you do not have the use of a calculator.

Solution:dxdy

=13· 3

2·(y2 + 2

) 12 · 2y = y

√y2 + 2 .

Thus √1 +

(dxdy

)2

=√

1 + y2(2 + y2) = y2 + 2 .

The area of the surface is∫ 10

7

(1 + y2

)· (2πy) dy = 2π

[y2

2+

y4

4

]10

7

=7701π

2.

3. [10 MARKS] Showing all your work, find the area enclosed by the curve (in polarcoordinates) r = 7 + 2 sin 6θ.

Solution: The curve surrounds the pole, and is periodic with period 2π. The area may beexpressed as an integral over an interval of length 2π; for example, as

12

∫ 2π

0(7 + 2 sin 6θ)2 dθ =

12

∫ 2π

0(49 + 28 sin 6θ + 4 sin2 6θ) dθ

=12

∫ 2π

0(49 + 28 sin 6θ + 2(1 − cos 12θ)) dθ

=12

[49θ − 28

6cos 6θ + 2θ − 2

12sin 12θ

]2π

0

=12

(51)(2π) = 51π .


4. [10 MARKS] Showing all your work, find the exact length of the curve x = 6 + 3t2,y = 2 + 2t3 (0 ≤ t ≤ 1).

Solution:

dxdt

= 6t

dydt

= 6t2

dydx

=

dydtdxdt

= t

⇒√

1 +

(dydx

)2

=√

1 + t2

Arc Length =

∫ 1

0

√(dxdt

)2

+

(dydt

)2

=

∫ 1

0

√36t2 + 36t4 dt

= 6∫ 1

0t√

1 + t2 dt

= 6 · 12· 2

3

[(1 + t2

) 32]1

0= 2

[(1 + t2

) 32]1

0

= 2(2

32 − 1

)= 4√

2 − 2 .

5. [10 MARKS] Showing all your work, sum a series in order to express the followingnumber as a ratio of integers: 0.35 = 0.35353535 . . ..

Solution:

0.35 =35

100

(1 +

1100

+1

1002 +1

1003 + . . .

)

=

35100

1 − 1100

=3599

.

Tuesday version

1. [10 MARKS] Showing all your work, find the length of the curve x =13√

y · (y − 3)(49 ≤ y ≤ 64). Simplify your answer as much as possible; the instructors are aware thatyou do not have the use of a calculator.


Solution:

dxdy

=13· 3

2√

y − 33· 1

2y−

12 =

12

(√y − 1√

y

)

⇒√

1 +

(dxdy

)2

=

√1 +

14

(y +

1y− 2

)

=

√14

(y +

1y

+ 2)

=

√14

(√y +

1√y

)2

=12

∣∣∣∣∣∣√

y +1√y

∣∣∣∣∣∣ .

The absolute signs may be dropped, since the square root is non-negative. The length is

12

∫ 64

49

(√y +

1√y

)dy =

12

[23· y 3

2 + 2√

y]64

49=

[13· y 3

2 +√

y]64

49

=1723

.

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotatingthe curve x = 6 + 2y2 (0 ≤ y ≤ 3) about the x-axis.

Solution:

dxdy

= 4y

⇒√

1 +

(dxdy

)2

=√

1 + 16y2

⇒ Area =

∫ 3

0

√1 + 16y2 · 2πy dy

=π

24

[(1 + 16y2)

32]3

0

=π

24

((145)

32 − 1

).

3. [15 MARKS] Showing all your work, find the area of the region that lies inside the curver = 15 cos θ, and outside the curve r = 5 + 5 cos θ.


Solution: The first curve is a circle; the second is a cardioid whose axis of symmetryis 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

12

∫ π3

− π3

((15 cos θ)2 − (5 + 5 cos θ)2

)dθ

=252

∫ π3

− π3

(8 cos2 θ − 2 cos θ − 1

)dθ

=252

∫ π3

− π3(4(1 + cos 2θ) − 2 cos θ − 1) dθ

=252

∫ π3

− π3(3 + 4 cos 2θ − 2 cos θ) dθ

= 25 [3θ + 2 sin 2θ − 2 sin θ]π30

= 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:

dxdt

= 18 − 18t2

dydt

= 36t√(

dxdt

)2

+

(dydt

)2

=

√182 (

1 − t2)2+ 362t2 = 18

(1 + t2

).

We must determine where the curve crosses itself. The student was expected to show thatshe knew how to find this crossing point systematically, not just by guessing or exam-ining a rough graph. The crossing point can be found by solving for distinct parametervalues t1 , t2 the equations x = 18t1 − 8t3

1 = 18t2 − 8t32, y = 18t2

1 = 18t2. Collectingterms and factorizing yields the system of equations

(t1 − t2)(18 − 6

(t21 + t1t2 + t2

2

))= 0 ,

18 (t1 − t2) (t1 + t2) = 0 .


Since we are looking for a solution where t1 , t2, we may divide by t1− t2, which cannotequal 0, and obtain the system

3 −(t21 + t1t2 + t2

2

)= 0 ,

t1 + t2 = 0 .

From the second equation we see that t2 = −t1, and then the first equation yields 3 = t21,

so the solutions are t1 = −t2 = ±√3; we may take the loop as beginning with parametervalue −√3 and ending with parameter value +

√3. The length of the arc will be

∫ +√

3

−√318

(1 + t2

)dt

= 2∫ √

3

018

(1 + t2

)dt

since the integrand is even and the interval is symmetric around 0

= 36[t +

13

t3]√3

0= 72

√3 .

Wednesday version

1. [10 MARKS] Showing all your work, find the length of the curve y = ln sec x. Simplifyyour answer as much as possible.

Solution:

dydx

=1

sec x· (sec x tan x) = tan x

⇒√

1 +

(dydx

)2

=√

1 + tan2 x = | sec x| .

In the following integral I will drop the absolute signs because the secant is positive overthe entire interval of integration; the length is

∫ π4

− π4| sec x| dx =

∫ π4

− π4sec x dx

= [ln | sec x + tan x|] π4− π4= ln(

√2 + 1) − ln(

√2 − 1)

= ln

√2 + 1√2 − 1

= ln(√

2 + 1)2

2 − 1

= ln((√

2 + 1)2)

= 2 ln(√

2 + 1) .


It would be sufficient for a student to obtain the different of logarithms above. Thesubsequent steps simplify the argument, and would be useful if the user did not have theuse of a calculator.

2. [10 MARKS] The curve y = 3√

x (1 ≤ y ≤ 3) is rotated about the y-axis. Showing allyour 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 isneeded. Since the limits are given in terms of y, I will integrate with respect to y; it willbe convenient to rewrite the equation of the curve as x = y3.

dxdy

= 3y2

⇒√

1 +

(dxdy

)2

=√

1 + 9y4

Area =

∫ 3

1

√1 + 9y4 · 2πy3 dy

=

[19· 1

4· 2π · 2

3

(1 + 9y4

) 32

]3

1

=π

27

(730

32 − 10

32).

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the innerloop of the curve r = 9 + 18 sin θ.

Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve istraced out as θ passes through an interval of that length. If, for example, we consider theinterval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π

6 andat θ = 11π

6 . Between these values the smaller loop is traced out; the larger loop is tracedout, for example, for −π6 ≤ θ ≤ 7π

6 . We can find the area of the small loop by integratingbetween the appropriate limits; the area is

12

∫ 11π6

7π6

(9 + 18 sin θ)2 dθ

=812

∫ 11π6

7π6

(1 + 4 sin θ + 4 sin2 θ) dθ

=812

∫ 11π6

7π6

(1 + 4 sin θ + 2(1 − cos 2θ)) dθ

=812

∫ 11π6

7π6

(3 + 4 sin θ − 2 cos 2θ) dθ


=812

[3θ − 4 cos θ − sin 2θ]11π

67π6

=812

(11π

2− 4 cos

11π6− sin

11π3

)− 81

2

(7π2− 4 cos

7π6− sin

7π3

)

=812

11π2− 2√

3 − −√

32

− 812

7π2

+ 2√

3 −√

32

=812

(2π − 3√

3) = 81π − 243√

32

.

4. [10 MARKS] Showing all your work, find the area of the surface obtained by rotatingthe curve x = 3t − t3, y = 3t2 (0 ≤ t ≤ 4) about the x-axis.

Solution:

dxdt

= 3 − 3t2

dydt

= 6t√(

dxdt

)2

+

(dydt

)2

=

√32 (

1 − t2)2+ 62t2 = 3

(1 + t2

).

The area of the surface of revolution about the x-axis will be∫ 4

03(1 + t2

)· 2π

(3t2

)dt = 18π

∫ 4

0

(t2 + t4

)dt

= 18π[13

t3 +15

t5]4

0=

6 × 64 × 535

=20352

5.

5. [10 MARKS] Showing all your work, determine the value of c, if it is known that∞∑

n=2

(5+

c)−n = 2.

Solution: We are told that the geometric series converges; this implies that its commonratio 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(5+c)2 · 1

1− 15+c

= 1(5+c)(4+c) . Equating this to 2, we obtain c2 + 9c + 18 = 0, implying that

(c + 3)(c + 6) = 0, so c = −3,−6. Of these two values, −6 lies outside of the admissibleinterval, and would yield a divergent series. Thus c can be equal only to −3.


Thursday version

1. [10 MARKS] Showing all your work, find the exact length of the polar curve r = 7e4θ

(0 ≤ θ ≤ 2π).

Solution:

drdθ

= 28e4θ

√r2 +

(drdθ

)2

=

√(72 + 282) e8θ

= 7√

17e4θ .

The length of the arc is then

7√

17∫ 2π

0e4θ dθ =

7√

174

e4θ

2π

0

=7√

174

(e8π − 1

).

2. [10 MARKS] The curve y = 4 − x2 (1 ≤ x ≤ 3) is rotated about the y-axis. Showing allyour work, find the area of the resulting surface.

Solution:

dydx

= −2x⇒√

1 +

(dydx

)2

=√

1 + 4x2

Area =

∫ 3

1

√1 + 4x2 · 2πx dx

= 2π · 14· 1

2· 2

3·[(

1 + 4x2) 3

2]3

1

=π

6

(37

32 − 5

32).

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the outerloop 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 istraced out as θ passes through an interval of that length. If, for example, we consider theinterval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π

6 andat θ = 11π

6 . Between these values the smaller loop is traced out; the larger loop is traced


out, for example, for −π6 ≤ θ ≤ 7π6 . We can find the area of the large loop by integrating

between those limits, and the area will include the entire smaller loop. The area is

12

∫ 11π6

7π6

(9 + 18 sin θ)2 dθ

=812

∫ 7π6

−π6

(1 + 4 sin θ + 4 sin2 θ) dθ

=812

∫ 7π6

−π6

(1 + 4 sin θ + 2(1 − cos 2θ)) dθ

=812

∫ 7π6

−π6

(3 + 4 sin θ − 2 cos 2θ) dθ

=812

[3θ − 4 cos θ − sin 2θ]7π6− π6

=812

(7π2− 4 cos

7π6− sin

7π3

)− 81

2

(−π

2− 4 cos

π

6+ sin

π

3

)

=812

7π2

+ 2√

3 −√

32

− 812

−π2 − 2√

3 +

√3

2

= 162π +243√

32

.

4. [10 MARKS] Showing all your work, sum a series in order to express the followingnumber as a ratio of integers: 4.645 = 4.645454545 . . ..

Solution:

4.645 = 4.6 +110

45100

(1 +

1100

+1

1002 +1

1003 + . . .

)

= 4.6 +

451000

1 − 1100

=4610

+1

10· 45

99=

511110

.

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

dxdθ

= − sin θ + 7 cos 7θ


dydθ

= cos θ − 2 sin θ

dydx

=

dydθdxdθ

=cos θ − 2 sin θ− sin θ + 7 cos 7θ

=17

when θ = 0 .

The line through (x(0), y(0)) = (1, 1) with slope 17 has equation y − 1 =

17

(x − 1), i.e.,x − 7y = −6.

Problems prepared but not used

1. The curve x =√

c2 − y2 (0 ≤ y ≤ c2 ) is rotated about the y-axis. (c is a fixed real number.)

Showing all your work, find the area of the resulting surface.

Solution:

dxdy

=12· 1√

c2 − y2· (−2y) = − y√

c2 − y2√

1 +

(dxdy

)2

=

√1 +

y2

c2 − y2 =|c|√

c2 − y2

Area =

∫ c2

0

|c|√c2 − y2

· 2π√

c2 − y2 dy

= 2π|c|∫ c

2

0dy = πc2 .

2. Showing all your work, find the slope of the tangent line to the curve with equation in

polar coordinates r =12θ

, at the point corresponding to θ = π.

Solution:

drdθ

= −12θ2

dydx

=sin θ · dr

dθ + r · cos θ

cos θ · drdθ − r · sin θ

=− sin θ + θ cos θ− cos θ − θ sin θ

.

At the point θ = π this ratio is equal to −π.


3. Showing all your work, find the area of one of the regions bounded by the line θ = π2 and

the closed curve r = 8 + 6 sin θ.

Solution: (The actual wording of the problem referred to a figure which it is not conve-nient to include in these notes.) The region can be interpreted as being swept out by aradius vector from the pole moving between −π2 and +π

2 . The area is this

12

∫ + π2

− π2(8 + 6 sin θ)2 dθ =

∫ + π2

− π2

(32 + 48 sin θ + 18 sin2 θ

)dθ

=

∫ + π2

− π2(32 + 48 sin θ + 9(1 − cos 2θ)) dθ

=

∫ + π2

− π2(41 + 48 sin θ − 9 cos 2θ) dθ

=

[41θ − 48 cos θ − 9

2sin 2θ

]+ π2

− π2= 41π .

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 beexpressed as an integral over an interval of length 2π; for example, as

12

∫ 2π

0(9 + cos 2θ)2 dθ =

12

∫ 2π

0(81 + 18 cos 2θ + cos2 2θ) dθ

=12

∫ 2π

0(81 + 18 cos 2θ +

1 + cos 4θ2

) dθ

=12

[81θ + 9 sin 2θ +

12θ +

18

sin 4θ]2π

0

=1634· 2π =

163π2

.

5. Showing all your work, find the exact length of the polar curve r = 4θ2 (0 ≤ θ ≤ 2π).

Solution:

drdθ

= 8θ√

r2 +

(drdθ

)2

=√

16θ4 + 64θ2

= 4|θ|√

4 + θ2 .


Over the interval in question θ is positive, and the absolute signs may be dropped. Thelength of the arc is

∫ 2π

04θ√

4 + θs dθ =43

[(4 + θ2)

32]2π

0

=323

((1 + π2)

32 − 1

).

6. Showing all your work, find equations of the tangents to the curve x = 3t2 +4, y = 2t3 +3that pass through the point (7, 5).

Solution: We might, in error think that we need first to determine the parameter valueassociated with the given point. We would then solve the system of equations

3t2 + 4 = 72t3 + 3 = 5

to obtain t = +1. This would be an error. It happens that the given curve passes throughthe point (7, 5), but that is fortuitous: we want the tangents to pass through the point, notthe curve! And we can’t find the points of contact of the tangents directly. So let’s firstdetermine the general tangent to the curve, at the point with parameter value t.

dxdt

= 6t

dydt

= 6t2

dydx

=

dydtdxdt

=6t2

6t= t ,

so the slope of the tangent at the point (3t2 + 4, 2t3 + 3) on the curve is t; the equation ofthat tangent is y −

(2t3 + 3

)= t

(x −

(3t2 + 4

)), or

y = tx − t3 − 4t + 3 . (33)

We now impose the condition that this line pass through the point (x, y) = (7, 5), i.e., thatits equation be satisfied by (x, y) = (7, 5), obtaining t3 − 3t + 2 = 0, whose left memberfactorizes to (t − 1)2(t + 2) = 0, so the points of contact of the tangents are t = 1 andt = −2. The equations of the tangents through the given point are found by giving theparameter t these two values in equation (33):

y = x − 2 and y = −2x + 19 .


7. Showing all your work, use methods of polar coordinates to find the length of the polarcurve r = 15 sin θ

(0 ≤ θ ≤ 4π

15

).

Solution:

drdθ

= 15 cos θ√

r2 +

(drdθ

)2

= 15 .

The length of the arc is, therefore

∫ 4π15

015 dθ = 4π .


Instructions to students

1. Show all your work. Marks may not be given for answers not supported by a full solu-tion. For future reference, the form of your solutions should be similar to those shownin the textbook or Student Solutions Manual for similar problems.

2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH0.) 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 thederivative of the function

f (x) =

∫ √x

√3

sin(t)t5 dt

Solution:


• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the new

variable/function is called u = u(x) =√

x, then f ′(x) =ddu

u∫

√3

sin(t)t5 dt · du

dx

• application of the Fundamental Theorem

ddu

u∫

√3

sin(t)t5 dt =

sin(u)u5 .

• completionsin(√

x)√

x5 · 12

x−12 =

sin√

x2x3

2. [10 MARKS] Compute

(a)∫ 0

−π/38 sec2(x) dx

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

− π3• compute the final answer. Students should know the trigonometric functions

of simple submultiples of π.(b)

∫ 1

0

√x(5x2 + 4x − 5) dx

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:∫ (

5x52 + 4x

32 − 5x

12)

dx = 5 · 27

x72 + 4 · 2

5x

52 − 5 · 2

3x

32 + C

• indicate that the value of the integral is the net change,[5 · 2

7x

72 + 4 · 2

5x

52 − 5 · 2

3x

32

]1

0

• compute the final answer correctly = 107 + 8

5 − 103


Tuesday Versions


f (x) =

∫ ex

5

√7 + ln6 t

tdt

Solution:

• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the newvariable/function is called u = u(x) = ex, then

f ′(x) =d

du

∫ u

5

√7 + ln6 t

tdt · du

dx


ddu

∫ u

5

√7 + ln6 t

tdt =

√7 + ln6 u

u.

• completion √7 + (ln(ex))6

ex · ex =

√7 + x6

ex · ex =√

7 + x6 .

While you may use some judgment about how much simplification you expect, I don’tbelieve it would not be appropriate to accept a composition like ln(ex) not simplified.


(a)∫ 0

−π/6[8 sec(x) tan(x) + 7 cos(x)] dx

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− π6


• compute the final answer correctly.

(8 + 0) −(8 · 2√

3− 7 · 1

2

)=

92− 16√

3

(b)∫ 4

1

−3x−1 + 5x + 3√x

dx

Solution: [5 MARKS TOTAL]

• state one antiderivative, here the obvious method is to express as a sum offractional powers and to integrate each separately:

∫ (−3x−1 + 5x + 3√x

)dx = −3 ·

(−2

1

)x−

12 + 5 · 2

3x

32 + 3 · 2

1x

12 + C

• indicate that the value of the integral is the net change,

[−3 ·

(−2

1

)x−

12 + 5 · 2

3x

32 + 3 · 2

1x

12

]4

1

• compute the final answer correctly

=

(62

+103· 8 + 6 · 2

)−

(6 +

103

+ 6)

=973

Wednesday Versions


f (x) =

∫ ln x

5et√

1 + t2 dt

Solution:

• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the newvariable/function is called u = u(x) = ln x, then

f ′(x) =d

du

∫ u

5et√

1 + t2 dt · dudx



ddu

∫ u

5et√

1 + t2 dt = eu√

1 + u2 .

• completion

eln x√

1 + (ln x)2 · 1x

=√

1 + (ln x)2

It is essential that eln x be simplified to x for full marks in this part.

2. [10 MARKS] Compute∫ π

4

− π6f (x) dx, where

f (x) =

4 sin x if x ≤ 0

5 sec x tan x if 0 < x < π2

Solution:

• decompose the interval into subintervals matching the intervals where the 2 partsof the definition apply:

∫ π4

− π6f (x) dx =

∫ 0

− π6f (x) dx +

∫ π4

0f (x) dx

• matching the different functions to the appropriate subintervals:∫ 0

− π6f (x) dx +

∫ π4

0f (x) dx =

∫ 0

− π64 sin x dx +

∫ π4

05 sec x tan x dx

• shift the constants outside of the integration:∫ 0

− π64 sin x dx +

∫ π4

05 sec x tan x dx = 4

∫ 0

− π6sin x dx + 5

∫ π4

0sec x tan x dx

• 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,

4[− cos x]0− π6 + 5[sec x]

π40

• correctly complete the computations

4(− cos 0 + cos

π

6

)+5

(sec

π

4− sec 0

)= 4

−1 +

√3

2

+5(√

2−1) = 2√

3+5√

2−9


Thursday Versions


f (x) =

∫ 7x

3x

√8 + 9t2

tdt

Solution:

• The integral must be split into two, at a convenient place, each integral with onefixed and one variable limit; note that the point where the integral is split CANNOTBE 0, since the integrand is undefined there:

f (x) =

∫ 7x

3x

√8 + 9t2

tdt =

∫ 1

3x

√8 + 9t2

tdt +

∫ 7x

1

√8 + 9t2

tdt

• one integral must be reversed so that the dependence on x is in the upper limit:

f (x) = −∫ 3x

1

√8 + 9t2

tdt +

∫ 7x

1

√8 + 9t2

tdt

• differentiate each of the integrals separately, using the Fundamental Theorem, andmultiply by the factor of the form du

dx from the Chain Rule (see problems on earlierversions)

ddx

f (x) = −√

8 + 9(3x)2

3x· d(3x)

dx+

√8 + 9(7x)2

7x· d(7x)

dx

• completion

−√

8 + 9(3x)2

3x· 3 +

√8 + 9(7x)2

7x· 7 =

−√

8 + 81x2 +√

8 + 9(49)x2

x

2. [10 MARKS] Compute∫ √

3

0f (x) dx, where

f (x) =

3x if 0 ≤ x ≤ 16

1+x2 if x > 1

Solution:


• for decomposing the interval into subintervals matching the intervals where the 2parts of the definition apply:

∫ √3

0f (x) dx =

∫ 1

0f (x) dx +

∫ √3

1f (x) dx

• for matching the different functions to the appropriate subintervals:

∫ √3

0f (x) dx =

∫ 1

03x dx +

∫ √3

1

61 + x2 dx

• for shifting constants outside of the integration:

∫ 1

03x dx +

∫ √3

1

61 + x2 dx =

32

∫ 1

02x dx + 6

∫ √3

1

11 + x2 dx

• 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,

32

[x2]10 + 6[arctan x]

√3

1

• for correctly completing the computations

32

[x2]10+6[arctan x]

√3

1 =32

(1−0)+6(arctan

√3 − arctan 1

)=

32

+6(π

3− π

4

)=

32

+π

2




2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.All submissions should carry your name and student number.




Monday Versions


(a)∫

sec2 x1 + 6 tan x

dx

(b)∫ 9/2

1/2

e√

2x

√2x

dx

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 thenet change [2 MARKS].)

2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtainedby revolving the region bounded by the x-axis, the lines x =

π

6and x =

π

3, and the curve

y =√

4 cos x.

Solution: It was intended that students solve this problem using the “Method of Wash-ers”. A solution using Cylindrical Shells would certainly be acceptable, but would bemore difficult, as students do not yet know how to integrate arccos y, and may not havemastered integration by parts. If they complete part of such a solution, allocate the markssimilarly to the scheme for Washers.

• [4 MARKS] for the integrand


• [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 cylindricalshells.

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 fullantiderivative

• [1 MARK] for apparently completing the integration correctly

Tuesday Versions


(a)∫

sec x tan x−7 − 8 sec x

dx

(b)

∫ e4

e

cos(9 ln x)x

dx

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 obtainedby revolving the region bounded by the lines y = 1, x = −π

4, x =

π

4, and the curve

y = 3 sec x.

Solution: Same scheme as for Problem 2 on Monday Versions.


3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = 1and x = 5, and the curve y = ln(5x). Compute the volume of the solid of revolutionobtained by revolving the region R about the y-axis. Hint: Use the method of cylindricalshells.

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 fullantiderivative

• [1 MARK] for apparently completing the integration correctly

Wednesday Versions


(a)∫

9x + 2√9x2 + 4x

dx

(b)∫ 1

0

e3x

e6x + 1dx

Solution:



2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtainedby revolving the region bounded by the y-axis, the lines y = ln 3 and y = ln 4, and thecurve y = ln

(x2

).

Solution: Note that this is a solid of revolution about the y-axis. To use the Methodof Washers, which is intended, students will have to rewrite the equation of the curvein the form x = 2ey. A correct solution using the Method of Cylindrical Shells wouldcertainly be acceptable. Follow the same grading scheme as shown above for Question2 of Monday Versions.


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 limitsof 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– [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, andobtaining the final answer.

Thursday Versions


(a)∫

x3

√x2 + 4

dx

(b)∫ 1

0

x + 2x2 + 1

dx

Solution:




2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained

by revolving the region bounded by the lines x = 1 and y = 2 and the curve y =3x

.

Solution: Note that this is a solid of revolution about the y-axis; moreover, it has a holein the middle. It was intended that students solve this problem using the “Method ofWashers”. However, the Method of Cylindrical Shells would be acceptable, and is notdifficult.

• [4 MARKS] for the integrand

• [2 MARKS] for the limits of integration; note that the data given are partly in termsof 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 limitsof 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– [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, andobtaining the final answer.





2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH0.) All submissions should carry your name and student number.



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∫

x3ex2dx .

Solution:

(a) [4 MARKS]

• [2 MARKS] for a correct choice of u and dv and correctly determining du andv

• [2 MARKS] for correctly implementing the selection of u and v and complet-ing 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 complet-

ing the integration in terms of the new variable• [1 MARK] for expressing the final, correct answer in terms of the original

variable


2. [10 MARKS] Use a trigonometric substitution to compute∫

1

(√

25 − x2)3dx . Verify

your 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 inversesine or inverse cosine, but it is common practice not to make that step explicit, soone can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral interms 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 obtainingthe original integrand

item [10 MARKS] Find the arc length of the parameterized curve x(t) = e2t +e−2t, y(t) =

4t − 2 . for t between 0 and12

.

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∫

x sec2(5x) dx

(b) Make a substitution and then use integration by parts to compute the integral∫

x3 cos(x2) dx


Solution:

(a) [4 MARKS]

• [2 MARKS] for a correct choice of u and dv and correctly determining du andv

• [2 MARKS] for correctly implementing the selection of u and v and complet-ing 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 complet-

ing the integration in terms of the new variable• [1 MARK] for expressing the final, correct answer in terms of the original

variable

2. [10 MARKS] Use a trigonometric substitution to compute∫

x2

√4 − x2

dx . Verify your

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 inversesine or inverse cosine, but it is common practice not to make that step explicit, soone can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral interms 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. Thisantiderivative does not include any term with plus/or/minus: there are no ambigu-ities of signs! If a student shows an ambiguity, this means he has not properlycompleted the differentiation of the next part, since only one of the signs will yieldthe correct derivative.

• [2 MARKS] for correctly differentiating the antiderivative and thereby obtainingthe original integrand

3. [10 MARKS] Find the arc length of the parameterized curve x(t) = −3 + e2t cos t, y(t) =

5 + e2t sin t for t between 0 and12

.







Wednesday Versions

1. [10 MARKS] Compute the integral∫ e9π

1cos(ln(x)) dx. Hint: A solution by integration

by parts could begin from the observation that cos(ln(x)) = 1 · cos(ln(x)). You could alsoapply integration by parts after making a substitution.

Solution: There appear to be several ways of attacking this problem, but the attacks willrequire 2 applications of integration by parts, followed by the solving of an equation.

Applying Integration by Parts immediately: • [1 MARK] for a correct selectionof 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 thenrequire a second application of integration by parts

• [1 MARK] for a correct selection of U and dV for the second integration byparts

• [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 cor-rect substitution, and implementing that substitution correctly both in the inte-grand and the limits of integration, so that the integral is now written in a formwhere 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 thenrequire a second application of integration by parts


• [1 MARK] for a correct selection of U and dV for the second integration byparts

• [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

2. [10 MARKS] Compute∫

13(x + 3)(x2 + 4)

dx .

Solution:

• [2+2 MARKS] for correctly factorizing the denominator and expressing the needto expand the function into a sum of 2 partial fractions, one with a linear denomi-nator, the other having a general numerator of degree 1 and denominator of degree2. Reserve a full 2 MARKS for the numerator of the fraction with the quadraticdenominator.

• [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 thegraph of the parametric curve x(t) = 2t − 3, y(t) = t2 − 3t − 1, 0 ≤ t ≤ 3/2 about they-axis.






Thursday Versions

1. [10 MARKS] Use a substitution and then integration by parts to compute the integral∫ 6

3

e3/x

x3 dx .

Solution:


• [5 MARKS] for selecting an appropriate substitution to simplify the integral, andcorrectly changing the integrand and the limits of integration. Some students mayarrive at the ultimate substitution through several composed substitutions. If theydo not succeed in completing a substitution and thus cannot begin seriously inte-gration 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 completingthe integration in terms of the new variable

• [1 MARK] for expressing the final, correct answer in terms of the original variable

2. [10 MARKS] Compute∫

25x(x + 5)2 dx .

Solution:

• [2+2 MARKS] for correctly factorizing the denominator and expressing the needto expand the function into a sum of either 2 or 3 partial fractions:

– One expansion would have one partial fraction with a linear denominator, theother having a general numerator of degree 1 and denominator of degree 2consisting 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 havingin its denominator the first power of the other linear factor; and the last havingin 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 forthe numerator of the fraction with the quadratic denominator, or, alternatively, the2 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

graph of the parametric curve x(t) =23

(t + 4)3/2, y(t) = 2√

t + 4, −4 ≤ t ≤ 0 about they-axis.







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

F(x) =

∫ x4

x2sin√

t dt .

2. [4 MARKS] Evaluate∫ π

−π2

f (x) dx , where

f (x) =

cos x, −π

2 ≤ x ≤ π3

3πx + 1, π

3 < x ≤ π .

3. [7 MARKS] Evaluate∫

x sin3 x2 cos x2 dx .


(x5 + 4−x) dx .

5. [10 MARKS] Calculate the area of the region bounded by the curves x = y2 andx − y = 2 .

6. [10 MARKS] The region bounded by f (x) = 4x − x2 and the x-axis, betweenx = 1 and x = 4 , is rotated about the y-axis. Find the volume of the solid that isgenerated.


x ln x dx .


sin2 x cos5 x dx .

9. [6 MARKS] Determine the partial fraction decomposition of the following ratio of poly-nomials:

x5 + 2x2 − 1

.

10. [4 MARKS] Determine whether or not the following sequence converges as n→ ∞ .If it does, find the limit: (

1 +xn

)3n.


11. [4 MARKS] Determine the following limit, if it exists:

limx→0+

√x√

x + sin√

x.

12. [6 MARKS] Determine whether the series∞∑

k=2

ke−k2converges or diverges.

13. [6 MARKS] Test the following series for

(a) absolute convergence,

(b) conditional convergence.

∞∑

k=10

(−1)k

√k(k + 1)

.

14. [10 MARKS] Find the area of the region that consists of all points that lie within thecircle 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 =32− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolution whichis generated.


3. [10 MARKS] Find the length of the curve y =x2

2− ln 4√x from x = 1 to

x = 2 .

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 limx→2

x3 − 8x4 − 16

.

6. [5 MARKS] Showing all your work, evaluate limx→0+

xx .


x3e−x2dx .


x3 − 1x3 + x

dx .


x3

√1 − x2

dx , where |x| < 1 .

10. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2 cos θand outside the circle r = 2 .

11. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f (x) =

∫ x

0et(1−t)dt at x = 0 .

12. [5 MARKS] Showing all your work, determine whether the following infinite seriesconverges or diverges. If it converges, find its sum.

∞∑

n=0

3n − 2n

4n

13. [5 MARKS] Showing all your work, determine whether or not the following series con-verges:

∞∑

n=1

21n

n2

14. [5 MARKS] Showing all your work, determine whether the following series converges:

∞∑

n=1

1n · 2n


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 + 2and x + y = 4 .

(b) Determine the area of the region.

2. [10 MARKS] The triangular region bounded by the lines

y = x , y =32− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolution whichis generated.

3. [10 MARKS] Find the area of the surface of revolution generated by revolving the curve

y =12

(ex + e−x) (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 limx→2

x − 2 cos πxx2 − 4

.

6. [5 MARKS] Evaluate limx→∞

(cos

1x2

)x4

.


e2x

1 + e4x dx .


x2 cos x dx .


x3 − 1x3 + x

dx .

10. [10 MARKS] Evaluate∫ √

a2 − u2 du , where |u| < a.


11. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2 cos θand outside the circle r = 2 .

12. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f (x) =

∫ x

0es(1−s)ds at x = 0 .

13. [5 MARKS] Showing all your work, determine whether the following infinite seriesconverges or diverges. If it converges, find its sum.

∞∑

n=0

1 + 2n + 3n

5n

14. [5 MARKS] Showing all your work, determine whether or not the following series con-verges.

∞∑

n=1

ln nn

15. [5 MARKS] Showing all your work, determine whether the following series convereges.

∞∑

n=1

n2 + 1en(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 theline 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 = 0the region bounded by the curves

y = sin xy = −2x = 0

and x = 2π .

4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = x2 (0 ≤x ≤ √2) about the y-axis.


5. Define the function F by F(x) =

x∫

0

et3dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the followinginequalities are true.

e < F(e) < ee3+1 .

(b) [4 MARKS] Determine the value ofddx

F(x3) at each of thefollowing points:

i. at x = 0 .ii. at x = 2 .

6. [4 MARKS] Showing all your work, evaluate∫

sin3 πx dx .


x2e−x dx .


x − 1x3 − x2 − 2x

dx .


x3 + x2 + x − 1x2 + 2x + 2

dx .

10. [8 MARKS] Find the area of the region inside the curve r = 3 sin θ and outside the curver = 2 − cos θ.

11. Showing all your work, determine whether each of the following integrals is convergentor divergent:

(a) [4 MARKS]

∞∫

0

sin x dx .


(b) [4 MARKS]

2∫

0

dx1 − x2 .

12. Showing all your work, determine whether each of the following sequences is convergentor divergent.

(a) [4 MARKS]n sin

π

n

(b) [4 MARKS](2 n + 1) e−n

13. Showing all your work, determine whether each of the following infinite series is con-vergent or divergent:

(a) [4 MARKS]∞∑

n=1

14n3 .

(b) [4 MARKS]∞∑

n=1

(1n

+1n2

).

14. Showing all your work, determine whether each of the following series is convergent,divergent, conditionally convergent and/or absolutely convergent.

(a) [4 MARKS]∞∑

n=1

(−1)n n + 2n(n + 1)

.

(b) [4 MARKS]∞∑

n=1

(−1)n cos nn2 .

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 theline x = 0 the region bounded by the curve y = 4 − x2 and the lines x = 0 andy = 0 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0the region bounded by the curves

y = sin xy = 2x = 0

and x = 2π .


4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = −x2 (0 ≤x ≤ √2) about the y-axis.

5. Define the function F by F(x) =

x∫

0

sin10 t dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the followinginequalities are true.

0 < F(e) < e .

(b) [4 MARKS] Determine the value ofddx

F(x) at each of thefollowing points:

i. at x = 0 .ii. at x =

π

2.


x5e−x2dx .


x3 − x2 + x + 1x2 − 2x + 2

dx .

8. [8 MARKS] Find the area of the region inside the curve r = 6 sin θ and outside the curver = 4 − 2 sin θ.

9. Showing all your work, determine whether each of the following integrals is convergentor divergent:

(a) [4 MARKS]

∞∫

0

cos x dx .

(b) [4 MARKS]

4∫

0

dx4 − x2 .

10. Showing all your work, determine whether each of the following sequences is convergentor divergent.


(a) [4 MARKS]n sin

π

n

(b) [4 MARKS](2 n + 1) e−n

11. Showing all your work, determine whether each of the following infinite series is con-vergent or divergent:

(a) [4 MARKS]∞∑

n=1

14n5 .

(b) [4 MARKS]∞∑

n=1

(1n− 1

n3

).

F.6 Final Examination in Mathematics 189-141B (1999/2000)1. [11 MARKS] Find the area of the region bounded by the curves x = y2 and x =

−y2 + 12y − 16 .

2. [11 MARKS] Let C denote the arc of the curve y = cosh x for −1 ≤ x ≤ 1 . Findthe volume of the solid of revolution generated by revolving about the line x = −2

the region bounded by C and the line y =e2 + 1

2e.

3. (a) [5 MARKS] Showing all your work, evaluate

∫ 92

32

√6t − t2 dt .

(b) [6 MARKS] Showing all your work, evaluate

∫ 3π4

π4

√1 − sin u du .

4. (a) [7 MARKS] Showing all your work, determine a reduction formula which ex-

presses, for any integer n not less than 2, the value of∫

xn sin 2x dx in

terms of∫

xn−2 sin 2x dx.

(b) [4 MARKS] Use your reduction formula to determine the indefinite integral∫

x2 sin 2x dx.



8x2 − 21x + 6(x − 2)2(x + 2)

dx .

6. [11 MARKS] Find the area of the region inside the curver = 1 + cos θ and outside the curve r = 1 − cos θ .

7. [11 MARKS] Determine whether the following integral is convergent or divergent. If itis convergent, find its value. Show all your work.

∫ 3

0

1

(x − 1)45

dx

8. [11 MARKS] Showing all your work, determine whether the following infinite series is

convergent or divergent:∞∑

n=1

n! e−(n − 1)2.

9. Showing all your work, determine whether each of the following series is convergent,divergent, conditionally convergent and/or absolutely convergent.

(a) [6 MARKS]∞∑

n=1

(−1)n(√

n + 2 − √n)

.

(b) [6 MARKS]∞∑

n=1

(−1)n nln

(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 = x4 and

y = 2 − x2 .

2. [11 MARKS] Determine the volume of the solid generated by rotating the region boundedby the curves y = 2x2 and y2 = 4x around the x-axis.

3. Evaluate the integrals:

(a) [5 MARKS]∫

x7

√1 − x4

dx .

(b) [6 MARKS]∫

x2

√4 − x2

dx .


4. [11 MARKS] Showing all your work, find∫ π/2

0e2x sin 3x dx .

5. [11 MARKS] Determine∫

6x3 − 18x(x2 − 1)(x2 − 4)

dx .

6. [11 MARKS] Find the area of the region inside the curve r = 2+2 sin θ and outsider = 2 .

7. [11 MARKS] Determine whether the following improper integral converges:∫ 1

0

ln xx2 dx .

8. [11 MARKS] Showing all your work, determine whether the following infinite series

converges:∞∑

n=1

1√15n3 + 3

.

9. Showing all your work, determine, for each of the following series, whether it is conver-gent, divergent, conditionally convergent and/or absolutely convergent.

(a) [6 MARKS]∞∑

n=1

(−1)n ln nn

.

(b) [6 MARKS]∞∑

n=1

cos nπn

.

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.

(a) [3 MARKS]∞∑

i=1

nn3 + 1

.

(b) [3 MARKS]∞∑

n=1

ln( n3n + 1

).

(c) [6 MARKS]∞∑

n=2

(−1)n(3n + 1)4

5n .

2. [12 MARKS] Determine the volume of the solid of revolution generated by revolvingabout the y-axis the region bounded by the curves

y = e−x2,


y = 0 ,x = 0 ,x = 1 .

3. [12 MARKS] Determine the area of the surface of revolution generated by revolvingabout the x-axis the curve

y = cos x ,(0 ≤ x ≤ π

6

).

[Hint: You may wish to make use of the fact that

2∫

sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C .]

4. [12 MARKS] Find the area that is inside the circler = 3 cos θ and outside the curve r = 2 − cos θ .


x(x − 1)(x2 + 4)

dx .

6. For the curve given parametrically by x = t3 + t2 + 1 , y = 1 − t2 , determine

(a) [6 MARKS] The equation of the tangent line at the point(x, y) = (1, 0) , written in the form y = mx + b , where m and b are con-stants;

(b) [6 MARKS] the value ofd2ydx2 at the point (x, y) = (1, 0) .

7. (a) [10 MARKS] Use integration by parts to determine the value of∫

ex cos x dx .

(b) [4 MARKS] Evaluate∫ 0

−∞ex cos x dx .

8. [12 MARKS] Find the area of the region bounded by the curves y = x2 − 4 andy = −2x2 + 5x − 2 .


F.9 Supplemental/Deferred Examination in Mathematics 189-141B (2000/2001)1. (a) [6 MARKS] Showing all your work, find F′(1) when

F(t) =

∫ 2t

1

xx3 + x + 7

dx .

(b) [6 MARKS] Showing all your work, evaluate∫ 6

0|x − 2| dx .

2. Showing all of your work, evaluate each of the following integrals:

(a) [4 MARKS]∫

x + 1√9 − x2

dx;

(b) [4 MARKS]∫

12x3 + x

dx;

(c) [4 MARKS]∫

sin2 2x cos2 2x dx;

(d) [4 MARKS]∫

ln x dx

3. [15 MARKS] Showing all your work, find the area of the region bounded below by the

line y =12

, and above by the curve y =1

1 + x2 .

4. [15 MARKS] Showing all your work, find the volume generated by revolving about they-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 givenby r ≥ 0 , −π

2≤ θ ≤ π

2;

(c) [5 MARKS] find the coordinates of the points on the curve where the tangent lineis parallel to the line θ = 0 .

6. For each of the following integrals, determine whether it is convergent or divergent; if itis convergent, you are expected to determine its value. Show all your work.

(a) [7 MARKS]∫ 2

−1

1x3 dx ;

(b) [7 MARKS]∫ ∞

1xe−x2

dx .


7. Showing all your work, determine, for each of the following series, whether or not itconverges:

(a) [5 MARKS]∞∑

n=2

1n(ln n)2 ;

(b) [5 MARKS]∞∑

n=1

(−1)n

(n2 − 1n2 + 1

);

(c) [5 MARKS]∞∑

n=1

n + 13n .

F.10 Final Examination in Mathematics 189-141B (2001/2002)1. Showing all your work, evaluate each of the following indefinite integrals:

(a) [3 MARKS]∫

x3

√4 − x2

dx

(b) [3 MARKS]∫

1

y√

ln ydy

(c) [3 MARKS]∫

sec u1 + sec u

· tan u du

(d) [3 MARKS]∫

et

1 + e2t dt

2. Let K denote the curvey = x2 , (0 ≤ x ≤ 1) .

(a) [6 MARKS] Determine the area of the surface of revolution generated by revolvingK about the y-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolvingabout the line y = 0 the region bounded by K and the lines x = 1 andy = 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 inte-gral.

(b) [4 MARKS] Determine the area of the region subtended by C at the pole — i.e. ofthe region bounded by the arc C and the line θ = 0.


(c) [4 MARKS] The given curve can be represented in cartesian coordinates paramet-rically as x = θ2 cos θ, y = θ2 sin θ. Determine the slope of the tangent to this curve

at the point (x, y) =

(0,

(π2

)2).


40 − 16x2

(1 − 4x2) (1 + 2x)

dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded by thecurves y = arctan x and 4y = π x in the first quadrant.

6. (a) [4 MARKS] Showing all your work, determine the value of∫sin3 x cos2 x dx .

(b) [4 MARKS] Showing all your work, determine the value of∫tan4 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2∫

0

tan4 x dx .

7. [12 MARKS] Showing all your work, determine the value of

d2

dx2

∫ x

0

∫ e2t

1

√u + 1 du

dt

when x = 0.

8. Showing all your work, determine, for each of the following infinite series, whether it isabsolutely convergent, conditionally convergent, or divergent.

(a) [4 MARKS]∞∑

n=5

(−1)n n2 − 16n2 + 4

.

(b) [4 MARKS]∞∑

n=2

(−1)n

n(ln n)2 .

(c) [4 MARKS]∞∑

n=2

(−1)n(n+1)

2

2n .

(d) [4 MARKS]∞∑

n=0

n + 52n .


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:

(a) [3 MARKS]∫

ex sin x dx

(b) [3 MARKS]∫ 1

2

0

sin−1 y√1 − y2

dy

equivalently,∫ 1

2

0

arcsin y√1 − y2

dy

.

(c) [3 MARKS]∫

(u2 + 2u)e−u du

(d) [3 MARKS]∫

1 + cos tsin t

dt

2. Let K denote the curve

y =√

2x − x2 ,

(0 ≤ x ≤ 1

2

).

(a) [6 MARKS] Showing all your work, use an integral to determine the area of thesurface of revolution generated by revolving K about the x-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolvingabout the line y =

√3

2 the region bounded by K and the lines x = 0 andy =

√3

2 .

(You may assume that∫ √

2x − x2 dx =x − 1

2

√2x − x2 +

12

arccos(1 − x) .)

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 thetangent is horizontal.

(b) [6 MARKS] By determining the value ofd2ydx2 as a function of t, determine all

points (x, y) on C at which the ordinate (y-coordinate) is a (local) maximum, andall points at which the ordinate is a (local) minimum.



4x3

(x2 − 9

)(3x + 9)

dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded by thecurves x − 2y + 7 = 0 and y2 − 6y − x = 0 .

6. (a) [6 MARKS] Showing all your work, evaluate∫

sin4 x cos2 x dx .

(b) [4 MARKS] Showing all your work, evaluate∫

tan5 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2∫

π4

tan5 θ dθ .

7. [12 MARKS] Showing all your work, determine the value of

d2

dx2

∫ x

0

(∫ π3

−2t

√4 + sin(−2u) du

)dt

when x = π4 . Your answer should be simplified, if possible.


(a) [4 MARKS]∞∑

n=5

(−1)n 1√n + 1

.

(b) [4 MARKS]∞∑

n=2

(−1)2n

n(ln n)3 .

(c) [4 MARKS]∞∑

n=2

(2n

1 + 5n

)3n

.

(d) [4 MARKS]∞∑

n=1

sin(

1n

)

cos(

1n

) · 1n

.


9. [10 MARKS] Prove or disprove the following statement: The point with polar coordi-nates

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 thecurves.

F.12 Final Examination in MATH 141 2003 011. [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 nothave the use of a calculator.)

2. Showing all your work, evaluate each of the following indefinite integrals:

(a) [5 MARKS]∫

1x2 + 2x + 17

dx

(b) [5 MARKS]∫

ln(ln x)x

dx

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.

I1 =

∫ ∞

2

2(x2 − x + 1)(x − 1)(x2 + 1)

dx , I2 =

∫ 1

0

2(x2 − x + 1)(x − 1)(x2 + 1)

dx

4. Let Ω denote the region in the first quadrant bounded by the curves x =√

16 + y2, y = 0,x = 0, and y = 3.


(a) [5 MARKS] Showing all your work, determine the volume of the solid of revolu-tion obtained by rotating Ω about the y-axis.

(b) [7 MARKS] Showing all your work, determine the area of the surface of revolutionobtained by rotating the arc

x =√

16 + y2, (0 ≤ y ≤ 3)

about the y-axis. You may assume that

ddθ

(sec θ tan θ + ln |sec θ + tan θ|) = 2 sec3 θ .

.

5. Consider the arc C given parametrically by

x =

∫ t

0

√4(1 − cos θ)θ2 dθ

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

t =3π2

.

(b) [6 MARKS] Find the length of C.

6. Give, for each of the following statements, a specific example to show that the statementis not a theorem:

(a) [3 MARKS] If an∞n=0 is a sequence such that limn→∞

an = 0, then∞∑

n=0an converges.

(b) [3 MARKS] If the series∞∑

n=0an and

∞∑n=0

bn are both divergent, then∞∑

n=0(an + bn) is

divergent.

(c) [3 MARKS] If a series∞∑

n=0an converges, then

∞∑n=0

a2n converges.


(a) [4 MARKS]∞∑

n=0

(−1)n

4n2 + 1.


(b) [4 MARKS]∞∑

n=2

(n − 1

n

)n2

(c) [4 MARKS]∞∑

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 011. [10 MARKS] Find the area of the region bounded by the curves

y = ex − 1, y = x2 − x, x = 1.

2. Showing all your work, evaluate each of the following:

(a) [5 MARKS]∫

e1x

x2 dx

(b) [5 MARKS]∫ 5

2|x2 − 4x| dx

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 thatthe integral does not converge. (The examiners are aware that you do not haveaccess to a calculator.)

Show all your work.

I1 =

∫ 1

0

2x(x − 1)(x2 + 1)

dx , I2 =

∫ 3

2

2x(x − 1)(x2 + 1)

dx

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 revolu-tion obtained by rotating Ω about the y-axis.

(b) [6 MARKS] Showing all your work, determine the volume of the solid of revolu-tion obtained by rotating Ω about the x-axis.


5. Consider the arc C given parametrically by

x = 2t(t2 − 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 valuet = − 1

2 .

(b) [6 MARKS] Find the area of the surface obtained by rotating the curve about thex-axis.

6. Give, for each of the following statements, a specific example to show that the statementis not a theorem:

(a) [3 MARKS] If bn∞n=0 is a sequence such that limn→∞

bn = 1, then∞∑

n=0bn converges.

(b) [3 MARKS] If a series∞∑

n=0b2

n converges, then∞∑

n=0bn converges.

(c) [3 MARKS] If the series∞∑

n=0an and

∞∑n=0

bn are both divergent, then∞∑

n=0(anbn) is di-

vergent.


(a) [4 MARKS]∞∑

n=3

(−1)n

e1n

.

(b) [4 MARKS]∞∑

n=0

(1

(n + 5)(n + 6)

)

(c) [4 MARKS]∞∑

n=2

(−1)n−1 2√n − 1

.

8. [10 MARKS]Showing all your work, find the area of the region that lies inside the curver = 2 − 2 sin θ and outside the curve r = 3 .

F.14 Final Examination in MATH 141 2004 01One of several versions


1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits, sums, inte-grals. If it does not exist write “DIVERGENT”.

(a)∞∑

n=1

1 + 2n−1

3n =

ANSWER ONLY

(b)∫ ∞

−∞xex2

dx =

ANSWER ONLY

(c) The limit of the Riemann sum limn→∞

n∑i=1

2n

((3 + 2i

n

)2 − 6(3 + 2i

n

)5)

=

ANSWER ONLY

(d)∫ −∞

∞xe−x2

dx =

ANSWER ONLY

(e)∞∑

n=3

4(2n + 1)(2n + 3)

=

ANSWER ONLY

2. BRIEF SOLUTIONS

[3 MARKS EACH] Simplify your answers as much as possible.


(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

(c) A curve is given parametrically by x(t) =∫ t

0e−u2

du, y(t) =∫ 4

teu2

du. Find theslope 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

the curve x =y3

6+

12y

(12 ≤ y ≤ 1

)about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

(e) On the interval 0 ≤ x ≤ 4 the average value of the function

f (x) =

√1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4is


ANSWER ONLY

3. BRIEF SOLUTIONS

[3 MARKS EACH] Give the value of each of the following indefinite integrals:

(a)∫

x3x2 + 1

dx =

ANSWER ONLY

(b)∫

ex√

1 + ex dx =

ANSWER ONLY

(c)∫

(sin2 x − 3 cos2 x) dx =

ANSWER ONLY

(d)∫

tan2 3x dx =

ANSWER ONLY


(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.


[12 MARKS] Evaluate the definite integral

∫ 12

− 12

4x(2x2 − x − 2

)

(x2 + 1)(x2 − 1)dx .


(a) [4 MARKS] Show that, for any positive integer n,∫

(ln x)2n dx = x(ln x)2n − 2n∫

(ln x)2n−1 dx

(b) [7 MARKS] Evaluate the integral∫ 1

0

y√2y − y2

dy .


[4 MARKS EACH] Determine for each of the following series whether it

• converges absolutely;

• converges conditionally; or

• diverges.


(a)∞∑

n=1

(1 +

1n

)2

e−n

(b)∞∑

n=10

(−1)n√

2n1 + 2

√n

(c)∞∑

n=2

(−1)n ·√

n + 2 − √n − 1n

(d)∞∑

n=0

2π + cos n6n


[6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ.

Another version

1. BRIEF SOLUTIONS


(a)∞∑

n=1

1 + 3n−1

4n =

ANSWER ONLY

(b)∫ ∞

−∞yey2

dy =

ANSWER ONLY


n∑i=1

2n

((4 + 2i

n

)2 − 7(4 + 2i

n

)5)

=

ANSWER ONLY


(d)∫ −∞

∞ye−y2

dy =

ANSWER ONLY

(e)∞∑

n=3

4(2n − 1)(2n + 1)

=

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) On the interval 0 ≤ x ≤ 4 the average value of the function

f (x) =

√1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4is

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


(d) A curve is given parametrically by x(t) =∫ t

0e−v2

dv, y(t) =∫ 4

tev2

dv. Find theslope 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

the curve x =y3

6+

12y

(12 ≤ y ≤ 1

)about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

3. BRIEF SOLUTIONS


(a)∫

sec3 x tan3 x dx =

ANSWER ONLY

(b)∫

x5x2 + 1

dx =

ANSWER ONLY

(c)∫

ex√

1 + ex dx =


ANSWER ONLY

(d)∫

(3 sin2 x − cos2 x) dx =

ANSWER ONLY

(e)∫

tan2 4x dx =

ANSWER ONLY


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 x-axis.


[12 MARKS] Evaluate the definite integral

∫ 12

− 12

4x(4x2 − 2x − 4

)

(x2 + 1)(x2 − 1)dx .


(a) [4 MARKS] Show that, for any positive integer m,∫

(ln y)2m dy = y(ln y)2m − 2m∫

(ln y)2m−1 dy


(b) [7 MARKS] Evaluate the integral∫ 1

0

x√2x − x2

dx .





• diverges.

(a)∞∑

n=10

(−1)n√

2n1 + 2

√n

(b)∞∑

n=1

(1 +

1n

)2

e−n

(c)∞∑

n=0

(cos n) − 2π4n

(d)∞∑

n=2

(−1)n ·√

n + 2 − √n − 1n


[6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ.

F.15 Supplemental/Deferred Examination in MATH 141 2004 011. BRIEF SOLUTIONS


(a)∞∑

n=1

en − e−n

3n =

ANSWER ONLY


(b)∫ 0

−∞xex dx =

ANSWER ONLY


π

n

n∑

i=1

sin2( iπ

n

)=

ANSWER ONLY

(d)∞∑

n=0

4(2n + 1)(2n + 3)

=

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) For the point with polar coordinates (r, θ) =(− 10π

3 ,−π6)

give another set of polarcoordinates (r1, θ1) in which r1 > 0 and θ1 > 2 .

ANSWER ONLY

(b) Find all points — if there are any — where the curvesr = 1 − cos θ and r = −3

2 intersect.


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

f (x) =

sinh x if ln 1

2 ≤ x ≤ 0sin x if 0 < x ≤ π is

ANSWER ONLY

3. BRIEF SOLUTIONS


(a)∫

14x2 + 1

dx =

ANSWER ONLY


(b)∫

dx√x2 − 25

=

ANSWER ONLY

(c)∫

(cos x + 1)(cos x − 2) dx =

ANSWER ONLY

(d)∫

tan 3x dx =

ANSWER ONLY


Let C be the arc x =13

( √y2 + 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-axisthe region bounded by C, the coordinate axes, and the line y = −√2.


[12 MARKS] Evaluate the indefinite integral∫

x3 − 8x − 1(x2 − 1)(x + 1)

dx .



Simplify your answers as much as possible.

(a) [6 MARKS] Evaluate the indefinite integral∫ √

1 − 9x2 dx .

(b) [6 MARKS] Evaluate the definite integral∫ −√3

0arctan x dx .





• diverges.

(a)∞∑

n=2

(−1)n ln√

nln(n2)

(b)∞∑

n=1

1 · 3 · 5 · . . . · (2n − 1)3nn!

(c)∞∑

n=1

sin 2n1 + 2n

(d)∞∑

n=1

(−1)n√

n3n − 1


[12 MARKS] Use polar coordinates — no other method will be accepted — to find the

area of the region bounded by the curve r = 2 and the line r =1

cos θ, and containing the

pole.

F.16 Final Examination in MATH 141 2005 011. SHOW ALL YOUR WORK!

(a) [4 MARKS] Evaluate∫ 3

0|x − 1| dx .


(b) [3 MARKS] Evaluateddx

∫ 5

x

√4 + t2 dt .

(c) [3 MARKS] Evaluateddx

∫ x2

2πsec t dt .

(d) [3 MARKS] Evaluate∫

x5( 3√

x3 + 1)

dx .

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests for conver-gence or divergence and determine whether the series is convergent. In each case youmust 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]∞∑

n=2

2 − cos nn

(b) [4 MARKS]∞∑

n=0

n(−3)n

4n

(c) [4 MARKS]∞∑

n=2

1n ln n

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand limits specific to the given problems:

(a) [6 MARKS] The area of the region bounded by the parabola y = x2, the x-axis, andthe tangent to the parabola at the point (1, 1).

DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)


(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 4 theregion bounded by x = 0 and the curve x =

√sin y (0 ≤ y ≤ π).


(c) [3 MARKS] The area of the surface obtained by revolving about the y-axis thecurve y = ex, 1 ≤ y ≤ 2.


(d) [2 MARKS] The average value of the function2x

(1 + x2)2 over the interval 0 ≤ x ≤2.




2x3 + 3x2 + 3x2 + x − 12

dx .


(a) [9 MARKS] Use integration by parts to prove that, for integersm ≥ 2, ∫

cosm x dx =1m

cosm−1 x · sin x +m − 1

m

∫cosm−2 x dx


(b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate∫ π2

0cos6 x dx.


Consider the curve C defined by

x = 2 cos t − cos 2ty = 2 sin t − sin 2t .

(a) [8 MARKS] Determine the points where the arc of the curve given by

π

4≤ t ≤ 7π

4

has a vertical tangent.

(b) [4 MARKS] Determine the length of the arc of the curve given by

0 ≤ t ≤ 2π .


(a) [5 MARKS] Determine whether the following integral is convergent; if it is con-vergent, determine its value: ∫ 1

−1

dx√1 − x2

(b) [5 MARKS] Determine whether the following series is conditionally convergent,absolutely convergent, or divergent.

∞∑

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.


[12 MARKS] Find the area of the region bounded by the curves

r = 4 + 4 sin θr sin θ = 3

which does not contain the pole.


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; andPages 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 questionis printed. When that space is exhausted, you may write on the facing page. Anysolution may be continued on the last pages, or the back cover of the booklet, but youmust 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 youare expected to write the correct answer in the box provided; you are not asked to showyour 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 theinvigilator if you find that your booklet is defective.)

6. A TOTAL OF 85 MARKS ARE AVAILABLE ON THIS EXAMINATION.



−2|x2 − 1| dx .

(b) [3 MARKS] Evaluateddx

∫ 4

xet2 dt .


∫ x2

1

dt1 + t5 .

(d) [3 MARKS] Evaluate∫

x sin(x2) dx .

SHOW ALL YOUR WORK!






(a) [4 MARKS]∞∑

n=2

(2n

3n − 1

)n

(b) [4 MARKS]∞∑

n=2

(−1)n

√n − 1

(c) [4 MARKS]∞∑

n=2

((−4)n

3n + 2n

)

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 donot evaluate the integral(s). It is not enough to quote a general formula: your integralsmust have integrand and limits specific to the given problems:

(a) [6 MARKS] The area of the infinite region containing the point(0, 1

2

)bounded by

the curve y = ex, the x-axis, and the tangent to the curve at the point (1, e).


(b) [3 MARKS] The volume generated by rotating the region bounded by the curvesy =√

x − 1, y = 0, x = 5 about the line y = 3.



(c) [3 MARKS] The area of the surface obtained by revolving about the x-axis thecurve x = ln y, 1 ≤ x ≤ 3.




x3 − x2 − 2x − 2x(x2 + x + 1)

dx .


(a) [9 MARKS] Use integration by parts to prove that, for integersm , 1,

∫secm x dx =

1m − 1

secm−2 x · tan x +m − 2m − 1

∫secm−2 x dx

(b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate∫ π3

0sec3 x dx.


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 linewith equations x = −7t, y = 12t − 5.

(b) [4 MARKS] Determine a definite integral whose value is the length of the arc ofC between the points with parameter values t = 1 and t = 2. YOU ARE NOTEXPECTED TO EVALUATE THE INTEGRAL!


[12 MARKS] Find the area of the region between the inner loop and the outer loop ofthe curve r = 1 − 2 cos θ.


F.18 Final Examination in MATH 141 2006 01 (One version)Instructions


2. Do not tear pages from this book; all your writing — even rough work — must be handedin. 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 transla-tion 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 bedifferent from yours.


• 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 ques-tion 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 isprinted!

• Some of the questions on this paper require only BRIEF SOLUTIONS ; for theseyou are expected to write the correct answer in the box provided; you are not askedto show your work, and you should not expect partial marks for solutions that arenot correct.


You are advised to spend the first few minutes scanning the problems. (Please informthe invigilator if you find that your booklet is defective.)




−1|x| dx .

(b) [3 MARKS] Evaluate

e3∫

1

dt

t√

1 + ln t.



∫ x2

0et2 dt .

(d) [4 MARKS] Evaluate

limn→∞

1n

(0n

)7

+

(1n

)7

+

(2n

)7

+ . . . +

(n − 1

n

)7 .

SHOW ALL YOUR WORK!





(a) [4 MARKS]∞∑

n=1

1(tanh n)2 + 1

(b) [4 MARKS]∞∑

n=1

n2ne−n2

(c) [4 MARKS]∞∑

n=1

n2 − 85n + 12n(n + 6)2

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand 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 regionbounded by the parabola y2 = 2x + 6 and the line y = x − 1. An answer involvingintegration along the y-axis will not be accepted.

(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “washers”.


(c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “cylindrical shells”.

(d) [3 MARKS] The length of the curve whose equation is

x2

4+

y2

9= 1 .



x5 + xx4 − 16

dx .


Showing all your work, evaluate each of the following:

(a) [4 MARKS]∫

cos x · cosh x dx

(b) [5 MARKS]

1∫

−3

√x2 + 2x + 5 dx

(c) [4 MARKS]∫

sin2 x · cos2 x dx


Consider the curve C defined by

x = x(t) = 10 − 3t2

y = y(t) = t3 − 3t ,

where −∞ < t < +∞.

(a) [8 MARKS] Determine the value ofd2ydx2 at the points where the tangent is horizon-

tal.

(b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of thearc

(x(t), y(t)) : −√

3 ≤ t ≤ 0.



(a) [5 MARKS] Showing detailed work, determine whether the following integral isconvergent; if it is convergent, determine its value:

∫ 0

−1

dx

x23

.

(b) [5 MARKS] Determine whether the following series is conditionally convergent,absolutely convergent, or divergent.

∞∑

n=1

(−1)n

n − ln n

(c) [3 MARKS] Give an example of a sequence an with the property that limn→∞

an = 0

but∞∑

n=1

an = +∞. You are expected to give a formula for the general term an of

your sequence.


[12 MARKS] The arcr = 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 containsthe point (r, θ) =

(12 ,

π2

).

F.19 Supplemental/Deferred Examination in MATH 141 2006 01

Instructions



3. Calculators are not permitted. This is a closed book examination. Regular and translationdictionaries are permitted.


4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank.








Simplify your answers as much as possible.

(a) [4 MARKS] Evaluate∫ ee

e2

dtt ln t

.

(b) [4 MARKS] Evaluate∫ + π

2

− π2| cosh x| dx .

(c) [4 MARKS] Evaluate

limn→∞

1n

(4 +

0n

)2

+

(4 +

1n

)2

+

(4 +

2n

)2

+ . . . +

(4 +

n − 1n

)2 .

(d) [3 MARKS] Evaluateddx

∫ −3x

0

√1 + t2 dt when x = 1.

SHOW ALL YOUR WORK!


• [1 MARK] Name the test(s) that you are using.


• [1 MARK] Explain why the test(s) you have chosen is/are applicable to the givenseries.

• [2 MARKS] Use the test(s) to conclude whether or not the series is convergent.

(a) [4 MARKS]∞∑

n=1

(−1)n nn2 + 4

(b) [4 MARKS]∞∑

n=1

ln( n3n + 2

)

(c) [4 MARKS]∞∑

n=1

3n + 65n

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand 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 theregion bounded by the curves y = x and y = x2. For this question you are expectedto use only the method of “cylindrical shells”.

(c) [4 MARKS] The volume of the solid obtained by rotating about the line y = 2 theregion bounded by the curves y = x and y = x2. For this question you are expectedto use only the method of “washers”.



x3

(x2 + 4)(x − 2)dx .


Showing all your work, evaluate each of the following. Simplify your answers as muchas possible.


(a) [4 MARKS]∫

8x cos 2x dx

(b) [4 MARKS]

2−√2∫

0

1√4x − x2

dx

(c) [4 MARKS]∫

ex sin x dx


Consider the arc C defined by

x = x(t) = 3t − t3

y = y(t) = 3t2 ,

where 0 ≤ t ≤ 1.

(a) [6 MARKS] Determine the value ofd2ydx2 at the point with parameter value t =

12

.

(b) [6 MARKS] Determine the area of the surface of revolution of C about the x-axis.



∫ 6

−∞xe

x3 dx .

(b) [4 MARKS] Give an example of a series which is convergent but not absolutelyconvergent. Justify all of your statements.

(c) [4 MARKS] Give an example of 2 divergent sequences an, bn with the propertythat the sequence anbn is convergent. You are expected to give formulas for thegeneral terms an, bn of both of your sequences.


[12 MARKS] The curves r = 2 cos 2θ and r = 2 sin θ define a number of regions in theplane. Let R denote the region containing the point (r, θ) = (1, 0), bounded by arcs ofboth of the curves. Showing all your work, carefully find the area of R.


F.20 Final Examination in MATH 141 2007 01 (One version)Instructions



3. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank. Your neighbour’s version of this test may be differentfrom yours.








Your answers must be simplified as much as possible.


−1|x|2 dx .


0∫

1

t4dt√t5 + 1

.

(c) [3 MARKS] Determine the value of

1n

(0n

)3

+

(1n

)3

+

(2n

)3

+ . . . +

(n − 1

n

)3 .


(d) [3 MARKS] Suppose it is known that f ′(x) = 4 cosh x for all x. Showing all yourwork, determine the value of f (1)− f (−1), expressed in terms of the values of eitherexponentials or hyperbolic functions.

(e) [4 MARKS] Evaluateddx

∫ x2

12

ett dt when x = 1 .

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests for conver-gence or divergence to determine whether the series is absolutely convergent, condition-ally convergent, or divergent. All tests used must be named, and all statements must becarefully justified.

(a) [4 MARKS]∞∑

n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n

(b) [4 MARKS]∞∑

n=1

(−1)n+1 n!n22n

(c) [4 MARKS]∞∑

n=1

(−1)n sin1n

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 arenot 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 arcy = 3x (−1 ≤ x ≤ 2).

(a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sumof integrals whose value is the volume of the resulting solid.



(c) [3 MARKS] Express in terms of integrals — which you need not evaluate — theaverage length that R cuts off from the vertical lines which it meets.


(d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluatethe integral.


(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line y = −1; you need not evaluate the integral.



[12 MARKS] Evaluate the indefinite integral∫ x

(x2 − 4

)(x − 2) + 4

(x2 − 4

)(x − 2)

dx .




(a) [4 MARKS]∫

e−x · cos x dx

(b) [5 MARKS]

52∫

− 12

x√8 + 2x − x2

dx

(c) [4 MARKS]∫ (

cos2 x +1

cos2 x

)· tan2 x dx


Consider the arc C defined by

x = x(t) = cos t + t sin ty = y(t) = sin t − t cos t ,

where 0 ≤ t ≤ π2 .

(a) [6 MARKS] Determine as a function of t the value ofd2ydx2 .

(b) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.



∫ π

π2

sec x dx .

(b) [5 MARKS] Showing all your work, carefully determine whether the series∞∑

n=3

4n ln n

is convergent.(c) [3 MARKS] Showing all your work, determine whether the following sequence

converges; if it converges, find its limit:

a1 = 1.a2 = 1.23a3 = 1.2345a4 = 1.234545a5 = 1.23454545a6 = 1.2345454545


etc., where each term after a2 is obtained from its predecessor by the addition onthe right of the decimal digits 45.


[10 MARKS] The polar curves

r = 2 + 2 sin θ (0 ≤ θ ≤ 2π)and

r = 6 − 6 sin θ (0 ≤ θ ≤ 2π)

divide the plane into several regions. Showing all your work, carefully find the area ofthe region bounded by these curves which contains the point (r, θ) = (1, 0).

F.21 Supplemental/Deferred Examination in MATH 141 2007 01 (Oneversion)

Instructions



3. The use of calculators is not permitted. This is a closed book examination. Use of regularand translation dictionaries is permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may bedifferent from yours.


• 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 questionis printed. When that space is exhausted, you may write on the facing page. Anysolution may be continued on the last pages, or the back cover of the booklet, but youmust indicate any continuation clearly on the page where the question is printed!








− 1√2

1√1 − x2

dx .

(b) [2 MARKS] Evaluate∫

t3 cosh(t4)

dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral∫ x

−xtet2 dt.


∫ 1

sin x(ln | sec t + tan t|) dt when x =

π

4.

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determinewhether the series is convergent or divergent. All tests used must be named, and allstatements must be carefully justified.

(a) [4 MARKS]∞∑

n=3

1(n + 2)(n − 2)

(b) [4 MARKS]∞∑

n=1

∞∑

i=n

3−i

(c) [4 MARKS]∞∑

n=1

(n + 1

n

)n2


R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2; C is thearc x = t, y = e3t (−2 ≤ t ≤ 1).


(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sumof integrals whose value is the volume of the resulting solid.


(c) [3 MARKS] Let f (t) denote the vertical distance of the point(t, e3t

)from the x-

axis. Express in terms of integrals — which you need not evaluate — the averagevalue of f (t) over the interval −2 ≤ t ≤ 1.




(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line x = 1; you need not evaluate the integral.




[12 MARKS] Evaluate the indefinite integral

∫ x3(x4 − 4x2

)− 16

x4 − 4x2 dx .


(a) [4 MARKS] Showing all your work, evaluate∫

sin2(3x) cos2(3x) dx .


1

x2√

9x2 − 16dx .

(c) [5 MARKS] Assume that

f (x) =

∫ x

0sec100 t dt

is known. Showing all your work, express the value of∫ x

0sec102 t dt in terms of

f (x). (You are not expected to determine f (x) explicitly.)


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

value t =12

.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.



∫ ∞

−∞

xx2 + 4

dx .

(b) [7 MARKS] Showing all your work, carefully determine whether the series

∞∑

n=3

(−1)n

√ln nn

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of thelimacon 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 As-sistants were not available for grading purposes. The following additional instructions weredistributed with the examination.

VERSION nMcGILL UNIVERSITYFACULTY OF SCIENCEFINAL EXAMINATION

IMPORTANT ADDITIONAL INSTRUCTIONS

MATHEMATICS 141 2008 01CALCULUS 2EXAMINER: Professor W. G. Brown DATE: Monday, April 14th, 2008ASSOCIATE EXAMINER: Mr. S. Shahabi 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 ormore, 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 APERFECT PAPER. You may attempt as many problems as you wish.

All other instructions remain valid. Where a problem requires that all work be shown, thatremains the requirement; where a problem requires only that an answer be written in a boxwithout work being graded, that also remains the requirement.

Students are advised to spend time checking their work; for that purpose you could verifyyour answers by solving problems in more than one way. Remember that indefinite integralscan be checked by differentiation.

W. G. Brown, Examiner.

Instructions




4. This examination booklet consists of this cover, Pages 1 through 9 containing questions; andPages 10, 11, and 12, which are blank. Your neighbour’s version of this test may be differentfrom yours.


• 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 questionis 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, youmay write on the facing page. Any solution may be continued on the last pages, or theback cover of the booklet, but you must indicate any continuation clearly on the pagewhere the question is printed!








(a) [2 MARKS] Evaluate∫

4 − 6x1 + x2 dx .

ANSWER (SHOW YOUR WORK OUTSIDE THE BOX)


2∫

0

y2√

y3 + 1 dy .



(c) [3 MARKS] Evaluate∫

sin(18 θ) cos(30 θ) dθ .



(a) [3 MARKS] Simplifying your answer as much as possible, evaluateddx

∫ √3

−xearcsin z dz .



(b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for thefunction f (x) = 3 − x, where the sample points are the left end-point of each of nsubintervals of equal length.

ANSWER ONLY

(c) [4 MARKS] Determine the value of the preceding Riemann sum as a function ofn, simplifying your work as much as possible. (NOTE: You are being asked todetermine the value of the sum as a function of n, not the limit as n→ ∞.)




For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=3

(1

n√

ln n

)

(b) [4 MARKS]∞∑

n=1

(−1)n+1

√4n + 5

3n + 10

(c) [4 MARKS]∞∑

n=1

(cot−1

(1

n + 1

)− cot−1

(1n

))

4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by thecurves 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 sumof integrals whose value is the volume of the resulting solid.


(b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies insideR. Express in terms of integrals — which you need not evaluate — the average ofthe positive lengths L(a).



(c) [4 MARKS] Let C1 be the curve x(t) = t, y(t) = cosh t (0 ≤ t ≤ ln 2). Simplifyingyour answer as much as possible, find the length of C1.

ANSWER ONLY


(a) [8 MARKS] Evaluate the indefinite integral∫

36(x + 4)(x − 2)2 dx .

(b) [4 MARKS] Determine whether

∞∫

3

36(x + 4)(x − 2)2 dx converges.

If it converges, find its value.



(a) [4 MARKS]∫ √

e√

x dx



(b) [5 MARKS]

0∫

− 12

x√3 − 4x − 4x2

dx


(c) [4 MARKS]∫ π

0sin2 t cos4 t dt .




Consider the curve C2 defined by x = x(t) = 1 + e−t , y = y(t) = t + t2 .

(a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis.


(b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent ishorizontal.



(c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis.



(a) [5 MARKS] Showing all your work, determine whether the series∞∑

n=2

√n(√

n + 2 −√

n − 2)

is convergent or divergent.

(b) [5 MARKS] Showing all your work, determine whether the following sequenceconverges; if it converges, find its limit:

a1 = 3.a2 = 3.14a3 = 3.1414a4 = 3.141414a5 = 3.14141414a6 = 3.1414141414

etc., where each term after a2 is obtained from its predecessor by the addition onthe right of the decimal digits 14.



Curves C3 and C4, respectively represented by polar equations

r = 4 + 2 cos θ (0 ≤ θ ≤ 2π) (34)and

r = 4 cos θ + 5 (0 ≤ θ ≤ 2π) , (35)

divide the plane into several regions.

(a) [8 MARKS] Showing all your work, carefully find the area of the one region whichis bounded by C3 and C4 and contains the pole.

(b) [4 MARKS] Find another equation — call it (35*) — that also represents C4, andhas the property that there do not exist coordinates (r, θ) which satisfy equations(34) and (35*) simultaneously. You are expected to show that equations (34) and(35*) have no simultaneous solutions.

F.23 Supplemental/Deferred Examination in MATH 141 2008 01 (oneversion)

Instructions



3. The use of calculators is not permitted. This is a closed book examination. Use of regularand translation dictionaries is permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may bedifferent from yours.










− 1√2

1√1 − x2

dx .


t3 cosh(t4)

dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral∫ x

−xtet2 dt.


∫ 1

sin x(ln | sec t + tan t|) dt when x =

π

4.

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determinewhether the series is convergent or divergent. All tests used must be named, and allstatements must be carefully justified.

(a) [4 MARKS]∞∑

n=3

1(n + 2)(n − 2)

(b) [4 MARKS]∞∑

n=1

∞∑

i=n

3−i

(c) [4 MARKS]∞∑

n=1

(n + 1

n

)n2


R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2; C is thearc x = t, y = e3t (−2 ≤ t ≤ 1).


(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sumof integrals whose value is the volume of the resulting solid.


(c) [3 MARKS] Let f (t) denote the vertical distance of the point(t, e3t

)from the x-

axis. Express in terms of integrals — which you need not evaluate — the averagevalue of f (t) over the interval −2 ≤ t ≤ 1.




(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line x = 1; you need not evaluate the integral.




[12 MARKS] Evaluate the indefinite integral

∫ x3(x4 − 4x2

)− 16

x4 − 4x2 dx .


(a) [4 MARKS] Showing all your work, evaluate∫

sin2(3x) cos2(3x) dx .


1

x2√

9x2 − 16dx .

(c) [5 MARKS] Assume that

f (x) =

∫ x

0sec100 t dt

is known. Showing all your work, express the value of∫ x

0sec102 t dt in terms of

f (x). (You are not expected to determine f (x) explicitly.)


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

value t =12

.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.




∫ ∞

−∞

xx2 + 4

dx .

(b) [7 MARKS] Showing all your work, carefully determine whether the series∞∑

n=3

(−1)n

√ln nn

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of thelimacon r = 2 sin θ − 1.

F.24 Final Examination in MATH 141 2009 01 (one version)Instructions



3. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10 and 11, which are blank. A TOTAL OF 75 MARKS ARE AVAILABLE ON THISEXAMINATION.

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 answeralone will not be sufficient unless substantiated by your work. You may continue yoursolution on the facing page, or on the last pages, or the back cover of the booklet, butyou 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 partmust be deemed to be more than 50% correct. Most of these questions will requirethat 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.


• Some questions on this paper require only BRIEF SOLUTIONS ; for these you mustwrite the correct answer in the box provided; you are not asked to show your work, andyou should not expect partial marks for solutions that are not correct. Check your work!


(a) [4 MARKS] Evaluate∫

t3 cos t2 dt .


(b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivativeddt

∫ t2

0tanh x2 dx .





(a) [4 MARKS] Evaluate

12∫

1√2

dx√1 − x2 · arcsin x

.




2y√y2 − y + 1

dy


SHOW ALL YOUR WORK!

3. For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=1

(−1)n+1(cos n

2

)n

(b) [4 MARKS]∞∑

n=2

(−1)n 1

n√

ln n.

(c) [4 MARKS]∞∑

n=4

(−1)n 1n

ln (3n + 1)

4. [12 MARKS] SHOW ALL YOUR WORK!

(a) [3 MARKS] Evaluate limn→∞

1n·

n∑

r=1

cos2(rπ

n

) . (Hint: This could be a Riemann

sum.)

(b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an =

arctan(−2n) as n→ ∞.


(c) [3 MARKS] Showing all your work, prove divergence, or find the value of∞∑

n=1

∞∑

i=1

15i

n

.


(a) [8 MARKS] Evaluate the indefinite integral∫

t + 12t2 − t − 1

dt .

(b) [2 MARKS] Determine whether the following improper integral converges or di-verges; if it converges, find its value:

∞∫

4

t + 12t2 − t − 1

dt .


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 generatedwhen 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 thenormal to the curve is parallel to the line x + y = 8.


Curves C3 and C4, are respectively represented by polar equations

r = 3 + 3 cos θ (0 ≤ θ ≤ 2π) (36)and

r = 9 cos θ (0 ≤ θ ≤ π) . (37)

(a) [7 MARKS] Showing all your work, carefully find the area of the region lyinginside both of the curves.

(b) [3 MARKS] Determine the length of the curves which form the boundary of theregion whose area you have found.

8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves y =8x2 , x = y, x = 1.


(a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensionalsolid, 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).


(b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensionalsolid, 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 evaluatethe integral(s).


(c) [3 MARKS] Calculate the area of R.

ANSWER ONLY


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 isaccessible 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 programmedto 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 StatisticsDepartment, and is not accessible directly through the myMcGill Portal; your initial passwordwill 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 thispassword after you sign on to WeBWorK.36

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 thatyou prefer. You should endeavour to keep your e-mail address up to date, since the instructorsmay send messages to the entire class through this route.

We suggest that you use either your UEA37 or your po-box address. You may be able toforward 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.

36If you forget your password you will have to send a message to Professor Brown so that the system adminis-trator may be instructed to reset the password at its initial value.

37Uniform E-mail Address



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 youafter 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 ableto solve in this course?

The questions on the WeBWorK assignments (A1 through A6) are a sampling of some typesof problem you should be able to solve after successfully completing this course. Some typesof calculus problems do not lend themselves to this kind of treatment, and may not appear onthe WeBWorK assignments. Use of WeBWorK does not replace studying the textbook —including the worked examples, attending lectures and tutorials, and working exercisesfrom the textbook — using the Student Solutions Manual [3] to check your work. Studentsare cautioned not to draw conclusions from the presence, absence, or relative frequencies ofproblems of particular types, or from particular sections of the textbook. Certain sectionsof the textbook remain examination material even though no problems are included in theWeBWorK assignments.

G.1.6 May I assume that the distribution of topics on quizzes and final examinationswill parallel the distribution of topics in the WeBWorK assignments?

No! While the order of topics on WeBWorK assignments should conform to the order of thelectures, 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 yourtextbook. Also, remember that WeBWorK — which checks answer only — cannot ascertainwhether you are using a correct method for solving problems. But, if you write out a solutionto an odd-numbered textbook problem, you can compare it with the solution in the SolutionsManual; and, if in doubt, you can show your work to a Teaching Assistant at one of the manyoffice 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, whenyou submit your answer. You may wish to experiment with the different formats. The defaultis jsMath mode, which should look similar to the version that you print out (cf. next question).


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; ifthe Reader has not been installed on your computer38, you will find instructions for download-ing this (free) software in §1.6.5 of these notes. Most computers available to you on campusshould 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 imple-mented by McGill University. You may access the web page for this course, and WeBWorKthrough your WebCT account39, and WebCT will link you to the appropriate server for WeB-WorK. If you follow this route to WeBWorK, you will still have to log in when you reachthe WeBWorK site. At the present time we will be using WebCT primarily for the posting ofgrades, and as a convenient repository for links to notes and announcements in the course. Weare not planning to use the potential WebCT sites that exist for the tutorial sections: use onlythe 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 As-signments, each with a due date. Since there is no limit to the number of attempts at problemson P0 or the other “Practice” assignments, you may play with these assignments to learn howto use the WeBWorK software.

You may print out a copy of your assignment by clicking on “Get hard copy”. This isyour version of the assignment, and it will differ from the assignments of other students inthe 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 thesystem as you attempt several problems; but, in the beginning, there are likely to be situationswhere you cannot understand what the system finds wrong with some of your answers. It isuseful to click on the Preview Answers button to see how the system interprets an answer thatyou 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 Answersbutton shows that the system will be taking of your answer.

38At the time these notes were written, the latest version of the Reader was 9.0, but recent, earlier versionsshould also work properly.

39http://mycourses.mcgill.ca


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 tryassignment P0, which does not count. The system is self-instructive, so we will not burdenyou with a long list of instructions.

You will need to learn how to enter algebraic expressions into WeBWorK as it is coded toread what you type in a way that may different from what you expect. For example, the symbolˆ is used for writing exponents (powers). If you type 2ˆ3, WeBWorK will interpret this as23 = 8. However, if you type 2ˆ3+x, WeBWorKwill interpret it as 23 + x, i.e. as 8 + x; if youwish to write 23+x, you have to type 2ˆ(3+x). You may obtain more information from the Listof 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 onWeBWorK, or with the specific problems on your version of an assignment, you may sendan 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 ofresolving your difficulty is usually the Feedback . Please give as much information as youcan. (All of the instructors and tutors are able to view from within WeBWorK the answersthat you have submitted to questions.)

If your problem is mathematical, and you need help in solving a problem, you shouldconsult one of the tutors at their office hours; you may go to any tutor’s office hours, not onlyto 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 man-ager 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 duedate, 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! Tobenefit from the speed that the system can deliver under normal conditions, do not delay yourWeBWorK until the last possible day! If a systems failure interferes with the due date of anassignment, arrangements could be made to change that date, and an e-mail message could


be broadcast to all users (to the e-mail addresses on record), or a message could be posted onWeBCT or the WeBWorK sign-on screen.40

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, andpermit unlimited numbers of attempts; your grades on these “Practice” do not count in yourterm mark. For the problems on assignments A1 — A6 you will normally be permitted about 5tries: 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 individualproblems may vary in difficulty. Assignments A1 — A6 will count equally in the computationof 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 exam-

ination.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 onereason why we have made the WeBWorK assignments compulsory.

40But slowness of the system just before the due time will not normally be considered a systems failure.


H Contents of the DVD disks forLarson/Hostetler/Edwards

These excellent disks were produced to accompany the textbook, Calculus of a Single Vari-able: Early Transcendental Functions, 3rd Edition[28] (called LHE in the charts below). Thecorrespondence shown to sections of [7] are only approximate. (NOTE THAT THIS BOOKDOES NOT FOLLOW STEWART’S CONVENTIONS FOR INVERSE SECANT/COSECANT!)

[All references in this table are to the 5th edition of Stewart, [7].]

DVD LHE Stewart# Section Subject Minutes Section1 P Chapter P: Preparation for Calculus1 P.1 Graphs and Models 451 P.2 Linear Models and Rates of Change 27 A101 P.3 Functions and Their Graphs 48 1.11 P.4 Fitting Models to Data 21 1.21 P.5 Inverse Functions 48 1.61 P.6 Exponential and Logarithmic Functions 30 1.5

DVD LHE Stewart# Section Subject Minutes Section1 1 Chapter 1: Limits and Their Properties1 1.1 A Preview of Calculus 11 2.11 1.2 Finding Limits Graphically and Numerically 25 2.2, 2.41 1.3 Evaluating Limits Analytically 28 2.31 1.4 Continuity and One-Sided Limits 22 2.51 1.5 Infinite Limits 18 2.6

DVD LHE Stewart# Section Subject Minutes Section1 2 Chapter 2: Differentiation1 2.1 The Derivative and the Tangent Line Problem 68 2.11 2.2 Basic Differentiation Rules and Rates of

Change34 2.3

1 2.3 The Product and Quotient Rules and HigherOrder Derivatives

25 3.2, 3.7


DVD LHE Stewart# Section Subject Minutes Section2 2 Chapter 2 (continued): Differentiation2 2.4 The Chain Rule 44 3.52 2.5 Implicit Differentiation 50 3.62 2.6 Derivatives of Inverse Functions 17 3.5, 3.8, 3.92 2.7 Related Rates 34 3.102 2.8 Newton’s Method 26 4.9

DVD LHE Stewart# Section Subject Minutes Section2 3 Chapter 3: Applications of Differentiation2 3.1 Extrema on an Interval 41 4.12 3.2 Rolle’s Theorem and the Mean Value Theo-

rem15 4.2

2 3.3 Increasing and Decreasing Functions and theFirst Derivative Test

19 4.3

2 3.4 Concavity and the Second Derivative Test 24 4.32 3.5 Limits at Infinity 23 2.62 3.6 A Summary of Curve Sketching 43 4.52 3.7 Optimization Problems 37 4.72 3.8 Differentials 51 3.11

DVD LHE Stewart# Section Subject Minutes Section3 4 Chapter 4: Integration3 4.1 Antiderivatives and Indefinite Integration 40 4.10

DVD LHE Stewart# Section Subject Minutes Section4 7 Chapter 7: Integration by Parts Trigonomet-

ric Substitution Partial Fractions L’Hopital’sRule

4 7.7 Indeterminate Forms and L’Hopital’s Rule 22 4.4

(The coverage extends to part of the material for Math 141 as well.)

math 141 long document

Documents

d problem assignments

written assignment w1

written assignment w2

c supplementary notes

numbers of pages

relevant pages

specific pages

calculus textbooks