Cambridge HSC Maths [2U]

YEAR

12

2 UnitSecondEdition

CAMBRIDGEMathematics

BILL PENDER

DAVID SADLER

JULIA SHEA

DEREK WARD

COLOUR

VERSION WITH

STUDENT CD-ROM

Now in colour wit

h an

electronic version

of

the book on CD

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, So Paulo, Delhi Cambridge University Press 477 Williamstown Road, Port Melbourne, VIC 3207, Australia www.cambridge.edu.au Information on this title: www.cambridge.org/9780521177504 Bill Pender, David Sadler, Julia Shea, Derek Ward 2009 First edition 1999 Reprinted 2001, 2004 Second edition 2005 Colour version 2009 Cover design by Sylvia Witte Typeset by Aptara Corp. Printed in China by Printplus National Library of Australia Cataloguing in Publication data Bill Pender Cambridge mathematics 2 unit : year 12 / Bill Pender [et al.] . 2nd ed. 9780521177504 (pbk.) Includes index. For secondary school age. Mathematics. Mathematics--Problems, exercises, etc. Sadler, David. Shea, Julia. Ward, Derek. 510 ISBN 978-0-521-17750-4 paperback Reproduction and Communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 Email: [email protected] Reproduction and Communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Cambridge University Press has no responsibility for the persistence or accuracy of URLS for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work are correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. Student CD-ROM licence Please see the file 'licence.txt' on the Student CD-ROM that is packed with this book.

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vHow to Use This Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAbout the Authors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiChapter One Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1A Areas and the Denite Integral . . . . . . . . . . . . . . . . . . . . . . 11B The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . 61C The Denite Integral and its Properties . . . . . . . . . . . . . . . . . 121D The Indenite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 191E Finding Areas by Integration . . . . . . . . . . . . . . . . . . . . . . . 251F Areas of Compound Regions . . . . . . . . . . . . . . . . . . . . . . . . 331G Volumes of Solids of Revolution . . . . . . . . . . . . . . . . . . . . . . 391H The Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 471I Simpsons Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511J Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Chapter Two The Exponential Function . . . . . . . . . . . . . . . . . . . . . . . 602A Review of Exponential Functions . . . . . . . . . . . . . . . . . . . . . 602B The Exponential Function ex and the Denition of e . . . . . . . . . . 652C Dierentiation of Exponential Functions . . . . . . . . . . . . . . . . . 732D Applications of Dierentiation . . . . . . . . . . . . . . . . . . . . . . . 792E Integration of Exponential Functions . . . . . . . . . . . . . . . . . . . 842F Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 902G Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Chapter Three The Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . 993A Review of Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . 993B The Logarithmic Function Base e . . . . . . . . . . . . . . . . . . . . . 1053C Dierentiation of Logarithmic Functions . . . . . . . . . . . . . . . . . 1123D Applications of Dierentiation of log x . . . . . . . . . . . . . . . . . . 1163E Integration of the Reciprocal Function . . . . . . . . . . . . . . . . . . 1213F Applications of Integration of 1/x . . . . . . . . . . . . . . . . . . . . . 1283G Calculus with Other Bases . . . . . . . . . . . . . . . . . . . . . . . . . 1333H Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 139

iv Contents

Chapter Four The Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 1414A Radian Measure of Angle Size . . . . . . . . . . . . . . . . . . . . . . . 1414B Mensuration of Arcs, Sectors and Segments . . . . . . . . . . . . . . . 1474C Graphs of the Trigonometric Functions in Radians . . . . . . . . . . . 1534D The Behaviour of sinx Near the Origin . . . . . . . . . . . . . . . . . . 1594E The Derivatives of the Trigonometric Functions . . . . . . . . . . . . . 1644F Applications of Dierentiation . . . . . . . . . . . . . . . . . . . . . . . 1724G Integration of the Trigonometric Functions . . . . . . . . . . . . . . . . 1784H Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 1864I Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 192

Chapter Five Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1965A Average Velocity and Speed . . . . . . . . . . . . . . . . . . . . . . . . 1965B Velocity as a Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 2035C Integrating with Respect to Time . . . . . . . . . . . . . . . . . . . . . 2135D Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 220

Chapter Six Rates and Finance . . . . . . . . . . . . . . . . . . . . . . . . . . . 2236A Applications of APs and GPs . . . . . . . . . . . . . . . . . . . . . . . 2236B The Use of Logarithms with GPs . . . . . . . . . . . . . . . . . . . . . 2326C Simple and Compound Interest . . . . . . . . . . . . . . . . . . . . . . 2386D Investing Money by Regular Instalments . . . . . . . . . . . . . . . . . 2446E Paying O a Loan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2526F Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2606G Natural Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . 2686H Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 278

Chapter Seven Euclidean Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 2817A Points, Lines, Parallels and Angles . . . . . . . . . . . . . . . . . . . . 2827B Angles in Triangles and Polygons . . . . . . . . . . . . . . . . . . . . . 2917C Congruence and Special Triangles . . . . . . . . . . . . . . . . . . . . . 2987D Trapeziums and Parallelograms . . . . . . . . . . . . . . . . . . . . . . 3087E Rhombuses, Rectangles and Squares . . . . . . . . . . . . . . . . . . . 3117F Areas of Plane Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . 3187G Pythagoras Theorem and its Converse . . . . . . . . . . . . . . . . . . 3217H Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3247I Intercepts on Transversals . . . . . . . . . . . . . . . . . . . . . . . . . 3327J Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 336

Chapter Eight Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3418A Probability and Sample Spaces . . . . . . . . . . . . . . . . . . . . . . 3418B Sample Space Graphs and Tree Diagrams . . . . . . . . . . . . . . . . 3488C Sets and Venn Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 3528D Venn Diagrams and the Addition Theorem . . . . . . . . . . . . . . . . 3578E Multi-Stage Experiments and the Product Rule . . . . . . . . . . . . . 3618F Probability Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 3678G Chapter Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 372

Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413

Preface

This textbook has been written for students in Years 11 and 12 taking the 2 Unitcalculus course Mathematics for the NSW HSC. The book covers all the contentof the course at the level required for the HSC examination. There are twovolumes the present volume is roughly intended for Year 12, and the previousvolume for Year 11. Schools will, however, dier in their choices of order of topicsand in their rates of progress.

Although the Syllabus has not been rewritten for the new HSC, there has beena gradual shift of emphasis in recent examination papers. The interdependence of the course content has been emphasised. Graphs have been used much more freely in argument. Structured problem solving has been expanded. There has been more stress on explanation and proof.

This text addresses these new emphases, and the exercises contain a wide varietyof dierent types of questions.

There is an abundance of questions and problems in each exercise too manyfor any one student carefully grouped in three graded sets, so that with properselection the book can be used at all levels of ability in the 2 Unit course.

This new second edition has been thoroughly rewritten to make it more acces-sible to all students. The exercises now have more early drill questions to reinforceeach new skill, there are more worked exercises on each new algorithm, and somechapters and sections have been split into two so that ideas can be introducedmore gradually. We have also added a review exercise to each chapter.

We would like to thank our colleagues at Sydney Grammar School and NewingtonCollege for their invaluable help in advising us and commenting on the successivedrafts. We would also like to thank the Headmasters of our two schools fortheir encouragement of this project, and Peter Cribb, Sarah Buerckner and theteam at Cambridge University Press, Melbourne, for their support and help indiscussions. Finally, our thanks go to our families for encouraging us, despite thedistractions that the project has caused to family life.

Dr Bill PenderSubject Master in MathematicsSydney Grammar SchoolCollege StreetDarlinghurst NSW 2010

David SadlerMathematicsSydney Grammar School

Julia SheaDirector of CurriculumNewington College200 Stanmore RoadStanmore NSW 2048

Derek WardMathematicsSydney Grammar School

How to Use This Book

This book has been written so that it is suitable for the full range of 2 Unitstudents, whatever their abilities and ambitions.

The Exercises: No-one should try to do all the questions! We have written longexercises so that everyone will nd enough questions of a suitable standard each student will need to select from them, and there should be plenty left forrevision. The book provides a great variety of questions, and representatives ofall types should be attempted.

Each chapter is divided into a number of sections. Each of these sections has itsown substantial exercise, subdivided into three groups of questions:

Foundation: These questions are intended to drill the new content of the sec-tion at a reasonably straightforward level. There is little point in proceedingwithout mastery of this group.

Development: This group is usually the longest. It contains more substantialquestions, questions requiring proof or explanation, problems where the newcontent can be applied, and problems involving content from other sectionsand chapters to put the new ideas in a wider context.

Challenge: Many questions in recent 2 Unit HSC examinations have beenvery demanding, and this section is intended to match the standard of thoserecent examinations. Some questions are algebraically challenging, some re-quire more sophistication in logic, some establish more dicult connectionsbetween topics, and some complete proofs or give an alternative approach.

The Theory and the Worked Exercises: All the theory in the course has been properlydeveloped, but students and their teachers should feel free to choose how thor-oughly the theory is presented in any particular class. It can often be helpful tolearn a method rst and then return to the details of the proof and explanationwhen the point of it all has become clear.

The main formulae, methods, denitions and results have been boxed and num-bered consecutively through each chapter. They provide a bare summary only,and students are advised to make their own short summary of each chapter usingthe numbered boxes as a basis.

The worked examples have been chosen to illustrate the new methods introducedin the section. They should provide sucient preparation for the questions in thefollowing exercise, but they cannot possibly cover the variety of questions thatcan be asked.

viii How to Use This Book

The Chapter Review Exercises: A Chapter Review Exercise has been added to eachchapter of the second edition. These exercises are intended only as a basic reviewof the chapter for harder questions, students are advised to work through moreof the later questions in the exercises.

The Order of the Topics: We have presented the topics in the order that we havefound most satisfactory in our own teaching. There are, however, many eectiveorderings of the topics, and apart from questions that provide links betweentopics, the book allows all the exibility needed in the many dierent situationsthat apply in dierent schools.

The time needed for the Euclidean geometry in Chapter Seven and probabilityin Chapter Eight will depend on students experiences in Years 9 and 10.

We have left Euclidean geometry and probability until Year 12 for two reasons.First, we believe that functions and calculus should be developed as early aspossible because these are the fundamental ideas in the course. Secondly, thecourses in Years 9 and 10 already develop most of the work in Euclidean geometryand probability, at least in an intuitive fashion, so that revisiting them in Year 12,with a greater emphasis now on proof in geometry, seems an ideal arrangement.

The Structure of the Course: Recent examination papers have made the interconnec-tions amongst the various topics much clearer. Calculus is the backbone of thecourse, and the two processes of dierentiation and integration, inverses of eachother, are the basis of most of the topics. Both processes are introduced as ge-ometrical ideas dierentiation is dened using tangents and integration usingareas but the subsequent discussions, applications and exercises give manyother ways of understanding them.

Besides linear functions, three groups of functions dominate the course:

The Quadratic Functions: These functions are known from earlier years.They are algebraic representations of the parabola, and arise naturally whenareas are being considered or a constant acceleration is being applied. Theycan be studied without calculus, but calculus provides an alternative andsometimes quicker approach.

The Exponential and Logarithmic Functions: Calculus is essential forthe study of these functions. We have begun the topic with the exponentialfunction. This has the great advantage of emphasising the fundamental prop-erty that the exponential function with base e is its own derivative this isthe reason why it is essential for the study of natural growth and decay, andtherefore occurs in almost every application of mathematics. The logarithmicfunction, and its relationship with the rectangular hyperbola y = 1/x, hasbeen covered in a separate chapter.

The Trigonometric Functions: Calculus is also essential for the study ofthe trigonometric functions. Their denitions, like the associated denitionof , are based on the circle. The graphs of the sine and cosine functions arewaves, and they are essential for the study of all periodic phenomena.

Thus the three basic functions in the course, x2 , ex and sinx, and the relatednumbers e and , can all be developed from the three most basic degree-2 curves the parabola, the rectangular hyperbola and the circle. In this way, everything

How to Use This Book ix

in the course, whether in calculus, geometry, trigonometry, coordinate geometryor algebra, can easily be related to everything else.

Algebra and Graphs: One of the chief purposes of the course, stressed heavily in re-cent examinations, is to encourage arguments that relate a curve to its equation.Algebraic arguments are constantly used to investigate graphs of functions. Con-versely, graphs are constantly used to solve algebraic problems. We have drawnas many sketches in the book as space allowed, but as a matter of routine, stu-dents should draw diagrams for most of the problems they attempt. It is becausesketches can so easily be drawn that this type of mathematics is so satisfactoryfor study at school.

Theory and Applications: Although this course develops calculus in a purely mathe-matical way using geometry and algebra, its content is fundamental to all thesciences. In particular, the applications of calculus to maximisation, motion,rates of change and nance are all parts of the syllabus. The course thus allowsstudents to experience a double view of mathematics, as a system of pure logicon the one hand, and an essential part of modern technology on the other.

Limits, Continuity and the Real Numbers: This is a rst course in calculus, and rigorousarguments about limits, continuity or the real numbers would be quite inappro-priate. Any such ideas required in this course are not dicult to understandintuitively. Most arguments about limits need only the limit lim

x 1/x = 0 andoccasionally the sandwich principle. Introducing the tangent as the limit of thesecant is a dramatic new idea, clearly marking the beginning of calculus, and isquite accessible. The functions in the course are too well-behaved for continuityto be a real issue. The real numbers are dened geometrically as points on thenumber line, and any properties that are needed can be justied by appealing tointuitive ideas about lines and curves. Everything in the course apart from thesesubtle issues of foundations can be proven completely.

Technology: There is much discussion about what role technology should play in themathematics classroom and what calculators or software may be eective. Thisis a time for experimentation and diversity. We have therefore given only a fewspecic recommendations about technology, but we encourage such investigation,and to this new colour version we have added some optional technology resourceswhich can be accessed via the student CD in the back of the book. The graphs offunctions are at the centre of the course, and the more experience and intuitiveunderstanding students have, the better able they are to interpret the mathemat-ics correctly. A warning here is appropriate any machine drawing of a curveshould be accompanied by a clear understanding of why such a curve arises fromthe particular equation or situation.

About the Authors

Dr Bill Pender is Subject Master in Mathematics at Sydney Grammar School,where he has taught since 1975. He has an MSc and PhD in Pure Mathemat-ics from Sydney University and a BA(Hons) in Early English from MacquarieUniversity. In 197374, he studied at Bonn University in Germany, and he haslectured and tutored at Sydney University and at the University of NSW, wherehe was a Visiting Fellow in 1989. He has been involved in syllabus developmentsince the early 1990s he was a member of the NSW Syllabus Committee inMathematics for two years and of the subsequent Review Committee for the 1996Years 910 Advanced Syllabus. More recently he was involved in the writing ofthe new K10 Mathematics Syllabus. He is a regular presenter of inservice coursesfor AIS and MANSW, and plays piano and harpsichord.

David Sadler is Second Master in Mathematics at Sydney Grammar School, wherehe has taught since 1980. He has a BSc from the University of NSW and an MAin Pure Mathematics and a DipEd from Sydney University. In 1979, he taughtat Sydney Boys High School, and he was a Visiting Fellow at the Universityof NSW in 1991.

Julia Shea is now Director of Curriculum at Newington College, having beenappointed Head of Mathematics there in 1999. She has a BSc and DipEd fromthe University of Tasmania, she taught for six years at Rosny College, a StateSenior College in Hobart, and was a member of the Executive Committee of theMathematics Association of Tasmania for ve years. She then taught for veyears at Sydney Grammar School before moving to Newington College.

Derek Ward has taught Mathematics at Sydney Grammar School since 1991 andis Master in Charge of Statistics. He has an MSc in Applied Mathematics and aBScDipEd, both from the University of NSW, where he was subsequently SeniorTutor for three years. He has an AMusA in Flute, and is a lay clerk at St James,King Street, where he sings counter-tenor. He also does occasional solo work atvarious venues.

The Book of Nature is written in the language of Mathematics.

The seventeenth-century Italian scientist Galileo

It is more important to have beauty in ones equations than tohave them t experiment.

The twentieth-century English physicist Paul Dirac

Even if there is only one possible unied theory, it is just aset of rules and equations. What is it that breathes re intothe equations and makes a universe for them to describe? Theusual approach of science of constructing a mathematical modelcannot answer the questions of why there should be a universefor the model to describe.

Steven Hawking, A Brief History of Time

CHAPTER ONE

Integration

x

yy = 4 x2

2 2

4

The calculation of areas has so far been restricted to regionsbounded by straight lines or parts of circles. This chapterwill extend the study of areas to regions bounded by moregeneral curves. For example, it will be possible to calculatethe area of the shaded region in the diagram to the right,bounded by the parabola y = 4 x2 and the x-axis.The method developed in this chapter is called integration.The basis of this method is the fact that nding tangents andnding areas are inverse processes, so that integration is theinverse process of dierentiation. This result is called thefundamental theorem of calculus and it will greatly simplifycalculation of the required areas.

1 A Areas and the Denite IntegralAll area formulae and calculations of area are based on two principles:1. Area of a rectangle = length breadth.2. When a region is dissected, the area is unchanged.

A region bounded by straight lines, like a triangle or a trapezium, can be cut upand rearranged into a rectangle with a few well-chosen cuts. Dissecting a curvedregion into rectangles, however, requires an innite number of rectangles and somust be a limiting process, like dierentiation.

A New Symbol The Denite Integral: Some new notation is needed to reect thisprocess of innite dissection as it applies to functions and their graphs.

The diagram on the left below shows the region contained between a given curvey = f(x) and the x-axis, from x = a to x = b. The curve must be continuousand, for the moment, entirely above the x-axis.

x

y

a b x

y

a b x

y

a b

f x( )

x

x + x

2 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

In the middle diagram, the region has been dissected into a number of strips.Each strip is approximately a rectangle, but only roughly so, because the upperboundary is curved. The area of the region is the sum of the areas of all the strips.

The third diagram shows just one of the strips, above the value x on the x-axis.Its height at the left-hand end is f(x), and provided the strip is very thin, theheight is still about f(x) at the right-hand end. Let the width of the strip be x,where x is, as usual in calculus, thought of as being very small. Then, roughly,

area of strip = width height= f(x) x.

Adding up the areas of all the strips gives the following rough formula. We needsigma notation, based on the Greek upper-case letter

, meaning S for sum.

Area of shaded region =b

x=a

area of each strip

=b

x=a

f(x) x.

If, however, there were innitely many of these strips, each innitesimally thin,one can imagine that the inaccuracy would disappear. This involves taking thelimit so that the equality is exact:

area of shaded region = limx0

bx=a

f(x) x.

At this point, the width x is replaced by the symbol dx, which suggests aninnitesimal width, and an old form

of the letter S is used to suggest an

innite sum. The result is the strange-looking symbol ba

f(x) dx, invented by

Leibnitz. This symbol is now dened to denote the area of the shaded region: ba

f(x) dx = area of shaded region.

The Denite Integral: This new object ba

f(x) dx is called a denite integral. The rest

of the chapter is concerned with evaluating denite integrals and applying them.

1

THE DEFINITE INTEGRAL:Let f(x) be a function that is continuous in the interval a x b.

For the moment, suppose that f(x) is never negative in the interval.

The denite integral ba

f(x) dx is dened to be the area of the region between

the curve and the x-axis, from x = a to x = b.

The function f(x) is called the integrand and the values x = a and x = b are calledthe lower and upper bounds of the integral.

The name integration suggests putting many parts together to make a whole.The notation arises from building up the region from an innitely large number ofinnitesimally thin strips. Integration is making a whole from these thin slices.

CHAPTER 1: Integration 1A Areas and the Denite Integral 3

Evaluating Denite Integrals Using Area Formulae: When the function is linear or cir-cular, the denite integral can be calculated from the graph using well-known areaformulae, although a quicker method will be developed later for linear functions.

Here are the relevant area formulae:

2FOR A TRIANGLE: Area = 12 base heightFOR A TRAPEZIUM: Area = width average of parallel sidesFOR A CIRCLE: Area = r2

WORKED EXERCISE:Evaluate using a graph and area formulae:

(a) 41(x 1) dx (b)

42(x 1) dx

SOLUTION:(a) The graph of y = x 1 has gradient 1 and y-intercept 1.

The area represented by the integral is the shaded triangle,with base 4 1 = 3 and height 3.

Hence 41(x 1) dx = 12 base height

= 12 3 3 x

y

1 4

1

3

1= 412 .

(b) The function y = x 1 is the same as before.The area represented by the integral is the shaded trapezium,with width 4 2 = 2 and parallel sides of length 1 and 3.

Hence 42(x 1) dx = width average of parallel sides

= 2 1 + 32 x

y

1 2 4

1

3

1

= 4.

WORKED EXERCISE:Evaluate using a graph and area formulae:

(a) 22|x| dx (b)

55

25 x2 dx

SOLUTION:(a) The function y = |x| is a V-shape with vertex at the origin.

Each shaded triangle has base 2 and height 2.

Hence 22|x| dx = 2 ( 12 base height)

= 2 ( 12 2 2) x

y

2 2

2

= 4.


(b) The function y =

25 x2 is a semicirclewith centre at the origin and radius 5.

Hence 55

25 x2 dx = 12 r2

= 12 52 x

y

5 5

5

= 252 .

The Area of a Circle: In earlier years, the formula A = r2 for the area of a circle wasproven. Because the boundary is a curve, some limiting process had to be used inthat proof. For comparison with the notation for the denite integral explainedat the start of this section, here is the most common version of that argument a little rough in its logic, but very quick. It involves dissecting the circle intoinnitesimally thin sectors and then rearranging them into a rectangle.

r

r

r

r

r

The height of the rectangle in the lower diagram is r. Since the circumference 2ris divided equally between the top and bottom sides, the length of the rectangleis r. Hence the rectangle has area r2, which is therefore the area of the circle.

Exercise 1A1. Use area formulae to calculate the following integrals (sketches are given):

x

y

2

3

(a) 20

3 dx

x

y

4

3

(b) 30

4 dx

x

y

4

4

(c) 40

x dx

x

y

3

6

(d) 30

2x dx

CHAPTER 1: Integration 1A Areas and the Denite Integral 5

x

y

2

2

(e) 20(2 x) dx

x

y

5

5

(f) 50(5 x) dx

x

y

0

2

4

2

(g) 20(x + 2) dx

x

y

7

3

40

(h) 40(x + 3) dx

2. Use area formulae to calculate the following integrals (sketches are given):

x

y

2

1 3

(a) 31

2 dx

x

y

3 2

5

(b) 23

5 dx

( )1 6,

x

y

12

4

(c) 12

(2x + 4) dx

x

y

31

12

3

(d) 31

(3x + 3) dx

x

y

51

4

(5, 9)

(1, 3)

(e) 51

(x + 4) dx

x

y

( )2 4,( )2 8,6

(f) 22

(x + 6) dx

x

y

3 3

3

(g) 33|x| dx

x

y

2 2

4

(h) 22|2x| dx

x

y

1

1

3. The diagram to the right shows the graph of y = x2

from x = 0 to x = 1, drawn on graph paper.The scale is 20 little divisions to 1 unit. This meansthat 400 little squares make up 1 square unit.(a) Count how many little squares there are under the

graph from x = 0 to x = 1 (keeping reasonabletrack of fragments of squares), then divide by 400

to approximate 10

x2 dx.

(b) By counting the appropriate squares, approximate:

(i) 1

2

0x2 dx (ii)

112

x2 dx

Conrm that the sum of the answers to parts (i) and (ii) is the answer to part (a).


D E V E L O P M E N T

4. Sketch a graph of each denite integral, then use an area formula to calculate it:

(a) 30

5 dx

(b) 03

5 dx

(c) 41

5 dx

(d) 62

5 dx

(e) 05

(x + 5) dx

(f) 20(x + 5) dx

(g) 42(x + 5) dx

(h) 31

(x + 5) dx

(i) 84(x 4) dx

(j) 104

(x 4) dx

(k) 75(x 4) dx

(l) 106

(x 4) dx

(m) 22|x| dx

(n) 44|x| dx

(o) 50|x 5| dx

(p) 105

|x 5| dxC H A L L E N G E

5. [Technology] Questions 3 and 7 of this exercise involve counting squares under a curve.Many programs can do such things automatically, usually dividing the region under thecurve into thin strips rather than the squares used in questions 3 and 7. Steadily increasingthe number of strips should show the value converging to a limit, which can be checkedeither using mensuration formulae or using the exact value of the integral as calculated inthe next section.

6. Sketch a graph of each denite integral, then use an area formula to calculate it:

(a) 44

16 x2 dx (b)

05

25 x2 dx

7. The diagram to the right shows the quadrant

y =

1 x2 , from x = 0 to x = 1.As in question 3, the scale is 20 little divisions to 1 unit.

x

y

1

1

(a) Count how many little squares there are under thegraph from x = 0 to x = 1.

(b) Divide by 400 to approximate 10

1 x2 dx.

(c) Hence nd an approximation for .

1 B The Fundamental Theorem of CalculusThe fundamental theorem is a formula for evaluating denite integrals. Its proofis rather demanding, so only the algorithm is presented in this section, by meansof some worked examples. The proof is given in the appendix to this chapter.

Primitives: The formula of the fundamental theorem relies on primitives. Recall thatF (x) is called a primitive of a function f(x) if the derivative of F (x) is f(x):

F (x) is a primitive of f(x) if F (x) = f(x).

We will need the result established in the last section of the Year 11 volume:

3

FINDING PRIMITIVES: Suppose that n = 1.

Ifdy

dx= xn , then y =

xn+1

n + 1+ C, for some constant C.

Increase the index by 1 and divide by the new index.

CHAPTER 1: Integration 1B The Fundamental Theorem of Calculus 7

Statement of the Fundamental Theorem: The fundamental theorem says that a deniteintegral can be evaluated by writing down any primitive F (x) of f(x), thensubstituting the upper and lower bounds into it and subtracting.

4

THE FUNDAMENTAL THEOREM:Let f(x) be a function that is continuous in a closed interval a x b. Then

ba

f(x) dx = F (b) F (a),

where F (x) is any primitive of f(x).

Using the Fundamental Theorem to Evaluate an Integral: The conventional way to setout these calculations is to enclose the primitive in square brackets, writing theupper and lower bounds as superscript and subscript respectively.

WORKED EXERCISE:Evaluate the following denite integrals:

(a) 20

2x dx (b) 42(2x 3) dx

Then draw diagrams to show the regions that they represent.

SOLUTION:

(a) 20

2x dx =[x2]20

(x2 is a primitive of 2x.)

= 22 02 (Substitute 2, then substitute 0.)= 4

This value agrees with the area of x

yy x= 2

4

2

the triangle shaded in the diagram to the right.(Note that area of triangle = 12 base height

= 12 2 4= 4.)

(b) 42(2x 3) dx =

[x2 3x

]42

(Take the primitive of each term.)

= (16 12) (4 6) (Substitute 4, then substitute 2.)= 4 (2)= 6

Again, this value agrees with the area of x

y y x= 2 3

2 4

1

5

the trapezium shaded in the diagram to the right.(Note that area of trapezium = width average of parallel sides

= 2 1 + 52

= 2 3= 6.)

Note: Whenever there are two or more terms in the primitive, brackets areneeded when substituting the upper and lower bounds of integration. Misuse ofthese brackets is a common source of error.


WORKED EXERCISE:Evaluate the following denite integrals:

(a) 10

x2 dx (b) 22

(x3 + 8) dx

SOLUTION:

(a) 10

x2 dx =[x3

3

]10

(Increase the index 2 to 3, then divide by 3.)

= 13 0 (Substitute 1, then substitute 0.)= 13

This integral was approximated by counting squares in question 3 of Exercise 1A.

(b) 22

(x3 + 8) dx =[x4

4+ 8x

]22

(Take the primitive of each term.)

= (4 + 16) (4 16) (Substitute 2, then substitute 2.)= 20 (12)= 32

Expanding Brackets in the Integrand: As with dierentiation, it is often necessary toexpand the brackets in the integrand before nding a primitive.

WORKED EXERCISE:Expand the brackets, then evaluate these denite integrals:

(a) 61

x(x + 1) dx (b) 30(x 4)(x 6) dx

Note: Fractions arise very often in denite integrals because the standardforms for primitives involve fractions. Care is needed with the resulting com-mon denominators, mixed numerals and cancelling.

SOLUTION:

(a) 61

x(x + 1) dx = 61(x2 + x) dx

=[x3

3+

x2

2

]61

= (72 + 18) (13 + 12 )= 90 56= 8916

(b) 30(x 4)(x 6) dx =

30(x2 10x + 24) dx

=[x3

3 5x2 + 24x

]30

= (9 45 + 72) (0 0 + 0)= 36


Writing the Integrand as Two Separate Fractions: If the integrand is a fraction with twoterms in the numerator, it should normally be written as two separate fractions,as with dierentiation.

WORKED EXERCISE:Write each integrand as two separate fractions, then evaluate:

(a) 21

3x4 2x2x2

dx (b) 23

x3 2x4x3

dx

SOLUTION:

(a) 21

3x4 2x2x2

dx = 21

(3x2 2) dx (Divide both terms on the top by x2.)

=[x3 2x

]21

= (8 4) (1 2)= 4 (1)= 5

(b) 23

x3 2x4x3

dx = 23

(1 2x) dx (Divide both terms by x3 .)

=[x x2

]23

= (2 4) (3 9)= 6 (12)= 6 + 12= 6

Negative Indices: The fundamental theorem works just as well when the indices arenegative. The working, however, requires care when converting between negativepowers of x and fractions.

WORKED EXERCISE:Use negative indices to evaluate these denite integrals:

(a) 51

x2 dx (b) 21

1x4

dx

SOLUTION:

(a) 51

x2 dx =[x1

1]51

(Increase the index to 1 and divide by 1.)

=[ 1

x

]51

(Rewrite x1 as1x

before substitution.)

= 15 (1)= 15 + 1= 45


(b) 21

1x4

dx = 21

x4 dx (Rewrite1x4

as x4 before nding the primitive.)

=[x3

3]21

(Increase the index to 3 and divide by 3.)

=[ 1

3x3

]21

(Rewrite x3 as1x3

before substitution.)

= 124 ( 13 )= 124 + 824= 724

Exercise 1BTechnology: Many programs allow denite integrals to be calculated automatically.This allows not just quick checking of the answers, but experimentation with furtherdenite integrals. It would be helpful to generate screen sketches of the graphs and theregions involved in the integrals.

1. Evaluate the following denite integrals, using the fundamental theorem:

(a) 10

2x dx

(b) 41

2x dx

(c) 31

4x dx

(d) 52

8x dx

(e) 32

3x2 dx

(f) 30

5x4 dx

(g) 21

10x4 dx

(h) 10

12x5 dx

(i) 10

11x10 dx

2. (a) Evaluate the following denite integrals, using the fundamental theorem:

(i) 10

4 dx (ii) 72

5 dx (iii) 54

dx

(b) Check your answers by sketching the graph of the region involved.


(a) 63(2x + 1) dx

(b) 42(2x 3) dx

(c) 30(4x + 5) dx

(d) 32(3x2 1) dx

(e) 41(6x2 + 2) dx

(f) 10(3x2 + 2x) dx

(g) 21(4x3 + 3x2 + 1) dx

(h) 20(2x + 3x2 + 8x3) dx

(i) 53(3x2 6x + 5) dx

4. Evaluate the following denite integrals, using the fundamental theorem. You will needto take care when nding powers of negative numbers.

(a) 01

(1 2x) dx

(b) 01

(2x + 3) dx

(c) 12

3x2 dx

(d) 25

dx

(e) 21

(4x3 + 5) dx

(f) 22

(5x4 + 6x2) dx

(g) 26

3x2 dx

(h) 43

(12 2x) dx

(i) 12

(4x3 + 12x2 3) dx


5. Evaluate the following denite integrals, using the fundamental theorem. You will needto take care when adding and subtracting fractions.

(a) 30

x dx

(b) 41(x + 2) dx

(c) 31

x2 dx

(d) 20(x2 + x) dx

(e) 30(x + x2 + x3) dx

(f) 21

(3x + 5) dx

(g) 11

(x3 x + 1) dx

(h) 32

(2x2 3x + 1) dx

(i) 24

(16 x3 x) dxD E V E L O P M E N T

6. By expanding the brackets where necessary, evaluate the following denite integrals:

(a) 32

x(2 + 3x) dx

(b) 20(x + 1)(3x + 1) dx

(c) 10

3x(2 + x) dx

(d) 32

2x(x 1) dx

(e) 11

x2(5x2 + 1) dx

(f) 31(x + 2)2 dx

(g) 21

(x 3)2 dx

(h) 32

(4 3x)2 dx

(i) 01

x(x 1)(x + 1) dx

(j) 12

x(x 2)(x + 3) dx

(k) 01

(1 x2)2 dx

(l) 94

(x + 1

) (x 1) dx

7. By dividing each fraction through by the denominator, evaluate each integral:

(a) 31

3x3 + 4x2

xdx

(b) 21

4x4 xx

dx

(c) 32

5x2 + 9x4

x2dx

(d) 21

x3 + 4x2

xdx

(e) 31

x3 x2 + xx

dx

(f) 12

x3 2x5x2

dx

8. Evaluate the following denite integrals, using the fundamental theorem. You will needto take care when nding the powers of fractions.

(a) 1

2

0x2 dx (b)

123

(2x + 3x2) dx (c) 4

3

34

(6 4x) dx

9. (a) Evaluate the following denite integrals:

(i) 105

x2 dx (ii) 32

2x3 dx (iii) 1

12

4x5 dx

(b) By writing them with negative indices, evaluate the following denite integrals:

(i) 21

dx

x2(ii)

41

dx

x3(iii)

112

3x4

dx

10. (a) (i) Show that k2

3 dx = 3k 6.

(ii) Hence nd the value of k if k2

3 dx = 18.

(b) (i) Show that k0

x dx = 12 k2 .

(ii) Hence nd k if k > 0 and k0

x dx = 18.


11. Use area formulae to nd 40

f(x) dx in each sketch of f(x):

x

y

3

1

1 2 4

(a)

x

y

3

1

1 2 4

(b)

C H A L L E N G E

12. By dividing each fraction through by the denominator, evaluate each integral:

(a) 21

1 + x2

x2dx (b)

12

1 + 2xx3

dx (c) 13

1 x3 4x52x2

dx

13. Evaluate the following denite integrals:

(a) 31

(x +

1x

)2dx (b)

13

(x2 +

1x2

)2dx (c)

32

(x2 x)2 dx

14. (a) Explain why the function y =1x2

is never negative.

(b) Sketch the integrand and explain why the argument below is invalid: 11

dx

x2=[ 1

x

]11

= 1 1 = 2.

1 C The Denite Integral and its PropertiesThis section will rst extend the theory to functions with negative values. Thensome simple properties of the denite integral will be established using argumentsabout the dissection of the area associated with the integral.

Integrating Functions with Negative Values: When a function has negative values, itsgraph is below the x-axis, so the heights of the little rectangles in the dissec-tion are negative numbers. This means that any areas below the x-axis shouldcontribute negative values to the value of the nal integral.

x

y

a b

A

B

C

For example, in the diagram to the right,the region B is below the x-axis and so willcontribute a negative number to the de-nite integral: b

a

f(x) dx = area A area B + area C.

Because areas under the x-axis are countedas negative, the denite integral is some-times referred to as the signed area underthe curve, to distinguish it from area, whichis always positive.

CHAPTER 1: Integration 1C The Denite Integral and its Properties 13

5

THE DEFINITE INTEGRAL:Let f(x) be a function that is continuous in the interval a x b.

Suppose now that f(x) may take positive and negative values in the interval.

The denite integral ba

f(x) dx is the sum of the areas above the x-axis, from

x = a to x = b, minus the sum of the areas below the x-axis.

WORKED EXERCISE:Evaluate these denite integrals:

(a) 40(x 4) dx (b)

64(x 4) dx (c)

60(x 4) dx

Sketch the graph of y = x 4 and then shade the regions associated with theseintegrals. Then explain how each result is related to the shaded regions.

SOLUTION:

(a) 40(x 4) dx =

[12x

2 4x]40

= (8 16) (0 0)= 8

Triangle OAB has area 8 and is below the x-axis;this is why the value of the integral is 8.

(b) 64(x 4) dx =

[12x

2 4x]64

= (18 24) (8 16)

x

y

4

2

4 6O

A

B

C

M

= 6 (8)= 2

Triangle BMC has area 2 and is above the x-axis;this is why the value of the integral is 2.

(c) 60(x 4) dx =

[12x

2 4x]60

= (18 24) (0 0)= 6

This integral represents the area of BMC minus the area of OAB;this is why the value of the integral is 2 8 = 6.

Dissection of the Interval: When a region is dissected, its arearemains the same. We can always dissect the region by dis-secting the interval a x b of integration.Thus if f(x) is continuous in the interval a x b, and thenumber c lies in this interval, then:

6 DISSECTION: ba

f(x) dx = ca

f(x) dx+ bc

f(x) dxx

y

a bc


Odd and Even Functions: In the rst example below, the function y = x3 4x is anodd function, with point symmetry in the origin. Thus the area of each shadedhump is the same. Hence the whole integral from x = 2 to x = 2 is zero,because the equal humps above and below the x-axis cancel out.

In the second diagram, the function y = x2 + 1 is even, with line symmetry inthe y-axis. Thus the areas to the left and right of the y-axis are equal, so thereis a doubling instead of a cancelling.

7ODD FUNCTIONS: If f(x) is odd, then

aa

f(x) dx = 0.

EVEN FUNCTIONS: If f(x) is even, then aa

f(x) dx = 2 a0

f(x) dx.

WORKED EXERCISE:Sketch these integrals, then evaluate them using symmetry:

(a) 22

(x3 4x) dx (b) 22

(x2 + 1) dx

SOLUTION:

(a) 22

(x3 4x) dx = 0, since the integrand is odd.(Without this simplication, the calculation is: 22

(x3 4x) dx =[14x

4 2x2]22

= (4 8) (4 8)

x

y

22

= 0, as before.)

(b) Since the integrand is even, 22

(x2 + 1) dx = 2 20(x2 + 1) dx

= 2[13x

3 + x]20

= 2((223 + 2) (0 + 0)

)x

y

2 2

1

5

= 913 .

x

y

a

Intervals of ZeroWidth: Suppose that a function is integrated overan interval a x a of width zero. In this situation, theregion also has width zero and so the integral is zero.

8 INTERVALS OF ZERO WIDTH: aa

f(x) dx = 0

Running an Integral Backwards from Right to Left: A further small qualication mustbe made to the denition of the denite integral. Suppose that the bounds of theintegral are reversed, so that the integral runs backwards from right to left overthe interval. Then its value reverses in sign:


9

REVERSING THE INTERVAL: Let f(x) be continuous in a x b. Then ab

f(x) dx = ba

f(x) dx.

This agrees perfectly with the fundamental theorem, because

F (a) F (b) = (F (b) F (a)

).

WORKED EXERCISE:Evaluate and compare the two denite integrals:

(a) 42(x 1) dx (b)

24(x 1) dx

SOLUTION:

(a) 42(x 1) dx =

[x2

2 x

]42

= (8 4) (2 2)= 4,

which is positive, since the region is above the x-axis.

(b) 24(x 1) dx =

[x2

2 x

]24

= (2 2) (8 4) x

y

2 41

1

3

= 4,which is the opposite of part (a), because the integralruns backwards from right to left, from x = 4 to x = 2.

Sums of Functions: When two functions are added, the two regions are piled on topof each other, so that:

10 INTEGRAL OF A SUM: ba

(f(x) + g(x)

)dx =

ba

f(x) dx + ba

g(x) dx

WORKED EXERCISE:Evaluate these two expressions and show that they are equal:

(a) 10(x2 + x + 1) dx (b)

10

x2 dx + 10

x dx + 10

1 dx

SOLUTION:

(a) 10(x2 + x + 1) dx =

[x3

3+

x2

2+ x

]10

= (13 +12 + 1) (0 + 0 + 0)

= 156 .


(b) 10

x2 dx + 10

x dx + 10

1 dx =[

x3

3

]10

+[

x2

2

]10

+[

x

]10

= (13 0) + (12 0) + (1 0)= 156 , the same as in part (a).

Multiples of Functions: Similarly, when a function is multiplied by a constant, theregion is expanded vertically by that constant, so that:

11 INTEGRAL OF A MULTIPLE: ba

kf(x) dx = k ba

f(x) dx

WORKED EXERCISE:Evaluate these two expressions and show that they are equal:

(a) 31

10x3 dx (b) 10 31

x3 dx

SOLUTION:

(a) 31

10x3 dx =[

10x4

4

]31

=8104 10

4

=8004

= 200.

(b) 10 31

x3 dx = 10[

x4

4

]31

= 10(814 1

4

)

= 10 804

= 200.

Inequalities with Denite Integrals: Suppose that a curve y = f(x) is always under-neath another curve y = g(x) in an interval a x b. Then the area under thecurve y = f(x) from x = a to x = b must be less than the area under the curvey = g(x).

In the language of denite integrals:

12INEQUALITY: If f(x) g(x) in the interval a x b, then b

a

f(x) dx ba

g(x) dx.

WORKED EXERCISE:(a) Sketch the graph of f(x) = 4 x2, for 2 x 2.

(b) Explain why 0 22

(4 x2) dx 16.

x

y

2 2

4 y = 4

SOLUTION:(a) The parabola and line are sketched opposite.

(b) Clearly 0 4 x2 4 over the interval 2 x 2.Hence the region associated with the integral is insidethe square of side length 4 in the diagram opposite.


Exercise 1CTechnology: All the properties of the denite integral discussed in this section havebeen justied visually from sketches of the graphs. Screen sketches of the graphs in this ex-ercises would be helpful in reinforcing these explanations. Questions 6, 7, 8, 11, 12 and 13deal with these properties. The simplication of integrals of odd and even functions isparticularly important and is easily demonstrated visually by curve-sketching programs.


(a) 02

2x dx

(b) 12

6x dx

(c) 21

4x3 dx

(d) 11

6x5 dx

(e) 03

3x2 dx

(f) 03

x2 dx

(g) 41

10x4 dx

(h) 23

x3 dx

(i) 22

x7 dx


(a) 23

(1 + 4x) dx

(b) 02

(3x2 5) dx

(c) 11

(7 4x3) dx

(d) 20(2x 4x3) dx

(e) 11

(6x2 8x) dx

(f) 60(x2 6x) dx

(g) 11

(x3 x) dx

(h) 30(4x3 2x2) dx

(i) 102

(12 3x) dx

(j) 31(3x2 5x4 10x) dx

(k) 13

(1 x x2) dx

(l) 22

(7 2x + x4) dxD E V E L O P M E N T

3. By expanding the brackets where necessary, evaluate the following denite integrals:

(a) 31

3x(x 4) dx

(b) 11

(3x 1)(3x + 1) dx

(c) 02

x2(6x3 + 5x2 + 4x + 3) dx

(d) 20

x(1 x) dx

(e) 22

(2 x)(1 + x) dx

(f) 50

x(x + 1)(x 1) dx

4. By dividing through by the denominator, evaluate the following denite integrals:

(a) 12

2x2 5xx

dx (b) 13

3x3 + 7xx

dx (c) 32

x2 6x3x2

dx

5. Find the value of k if:

(a) 3k

2 dx = 4

(b) 8k

3 dx = 12

(c) 32(k 3) dx = 5

(d) k3

(x 3) dx = 0

(e) k1

(x + 1) dx = 6

(f) k1

(k + 3x) dx = 132


6. Evaluate each group of denite integrals and use the properties of the denite integral toexplain the relationships within each group:

(a) (i) 20(3x2 1) dx (ii)

02(3x2 1) dx

(b) (i) 10

20x3 dx (ii) 20 10

x3 dx

(c) (i) 41(4x + 5) dx (ii)

41

4x dx (iii) 41

5 dx

(d) (i) 20

12x3 dx (ii) 10

12x3 dx (iii) 21

12x3 dx

(e) (i) 33(4 3x2) dx (ii)

22

(4 3x2) dx

7. Without nding a primitive, use the properties of the denite integral to evaluate thefollowing, stating reasons:

(a) 33

9 x2 dx

(b) 44(x33x2 +5x7) dx

(c) 11

x3 dx

(d) 55

(x3 25x) dx

(e) 9090

sinx dx

(f) 22

x

1 + x2dx

8. (a) On one set of axes sketch y = x2 and y = x3, clearly showing the point of intersection.

(b) Hence explain why 0 < 10

x3 dx 0 (c) 12

1x

dx > 0 (d) 12

1x

dx > 0

CHAPTER 1: Integration 1D The Indenite Integral 19

1 D The Indenite IntegralNow that primitives have been established as the key to calculating denite inte-grals, this section turns again to the task of nding primitives. First, a new andconvenient notation for the primitive is introduced.

The Indenite Integral: Because of the close connection established by the fundamentaltheorem between primitives and denite integrals, the term indenite integral isoften used for the primitive. The usual notation for the primitive of a functionf(x) is an integral sign without any upper or lower bounds. For example, theprimitive or indenite integral of x2 + 1 is

(x2 + 1) dx =x3

3+ x + C, for some constant C.

The word indenite implies that the integral cannot be evaluated further becauseno bounds for the integral have yet been specied.

A denite integral ends up as a pure number. An indenite integral, on the otherhand, is a function of x the pronumeral x is carried across to the answer.The constant is called a constant of integration and is an important part of theanswer. Despite being a nuisance to write down every time, it must always beincluded. In most problems other than denite integrals, it will not be zero.

Standard Forms for Integration: The rules for nding primitives given in the last sec-tion of the Year 11 volume can now be restated in this new notation.

13

STANDARD FORMS FOR INTEGRATION: Suppose that n = 1. Thenxn dx =

xn+1

n + 1+ C, for some constant C.

(ax + b)n dx =

(ax + b)n+1

a(n + 1)+ C, for some constant C.

The word integration is commonly used to refer to both the nding of a primitiveand the evaluating of a denite integral.

Note: Strictly speaking, the words for some constant C or where C is aconstant should follow the rst mention of the new pronumeral C, because nopronumeral should be used without having been formally introduced. There isa limit to ones patience, however, and usually in this situation it is quite clearthat C is the constant of integration. If another pronumeral such as D is used,it would be wise to introduce it formally.

WORKED EXERCISE:

Use the standard form

xn dx =xn+1

n + 1+ C to nd:

(a)

9 dx (b)

12x3 dx

SOLUTION:

(a)

9 dx = 9x + C, for some constant C


Note: We know that 9x is the primitive of 9, becaused

dx(9x) = 9.

But the formula still gives the correct answer, because 9 = 9x0,and so increasing the index to 1 and dividing by this new index 1,

9x0 dx =9x1

1+ C, for some constant C

= 9x + C.

(b)

12x3 dx = 12 x4

4+ C, for some constant C

= 3x4 + C

WORKED EXERCISE:

Use the standard form

(ax + b)n dx =(ax + b)n+1

a(n + 1)+ C to nd:

(a)

(3x + 1)5 dx (b)

(5 2x)2 dx

SOLUTION:

(a)

(3x + 1)5 dx =(3x + 1)6

3 6 + C (Here a = 3 and b = 1.)= 118 (3x + 1)

6 + C

(b)

(5 2x)2 dx = (5 2x)3

(2) 3 + C (Here a = 2 and b = 5.)= 16 (5 2x)3 + C

Negative Indices: Both standard forms apply with negative indices as well as positiveindices, as in the following worked exercise.

WORKED EXERCISE:Use negative indices to nd the indenite integrals:

(a)

12x3

dx (b)

dx

(3x + 4)2

SOLUTION:

(a)

12x3

dx =

12x3 dx (Rewrite1x3

as x3 before integrating.)

= 12 x2

2 + C (Increase the index to 2 and then divide by 2.)

= 6x2

+ C (Rewrite x2 as1x2

.)

(b)

dx

(3x + 4)2=

(3x + 4)2 dx (Rewrite1

(3x + 4)2as (3x + 4)2 .)

=(3x + 4)1

3 (1) + C (Here a = 3 and b = 4.)

= 13(3x + 4)

+ C (Rewrite (3x + 4)1 as1

3x + 4.)


Special Expansions: In many integrals, brackets must be expanded before the in-denite integral can be found. The following worked exercises use the specialexpansions; the second also requires negative indices.

WORKED EXERCISE: Find these indenite integrals:

(a)

(x3 1)2 dx (b) (

3 1x2

)(3 +

1x2

)dx

SOLUTION:

(a)

(x3 1)2 dx =

(x6 2x3 + 1) dx (Use (A + B)2 = A2 + 2AB + B2 .)

=x7

7 x

4

2+ x + C

(b) (

3 1x2

)(3 +

1x2

)dx =

(9 1

x4

)dx (Use (AB)(A + B) = A2 B2 .)

= (

9 x4) dx (Use 1x4

= x4.)

= 9x x3

3 + C

= 9x +1

3x3+ C

Fractional Indices: The standard forms for nding primitives of powers also apply tofractional indices. These calculations require quick conversions between fractionalindices and surds.

WORKED EXERCISE: Use fractional and negative indices to evaluate:

(a) 41

x dx (b)

41

1x

dx

SOLUTION:

(a) 41

x dx =

41

x12 dx (Rewrite

x as x

12 before nding the primitive.)

= 23[x

32

]41

(Increase the index to 32 and divide by32 .)

= 23 (8 1) (Note that 432 = 23 = 8 and 1

32 = 1.)

= 423

(b) 41

1x

dx = 41

x12 dx (Rewrite

1x

as x12 before nding the primitive.)

= 21[x

12

]41

(Increase the index to 12 and divide by12 .)

= 2 (2 1) (Note that 412 =4 = 2 and 1

12 = 1.)

= 2


WORKED EXERCISE:

(a) Use index notation to express1

9 2x as a power of 9 2x.

(b) Hence nd the indenite integral

dx9 2x .

SOLUTION:

(a)1

9 2x = (9 2x) 12 . (Use

1u

= u12 .)

(b) Hence

19 2x =

(9 2x) 12 dx

=(9 2x) 122 12

+ C (Use


a(n + 1).)

= 9 2x + C (Use u 12 = u .)

Running a Chain-Rule Differentiation Backwards: Finding primitives is the reverse pro-cess of dierentiation. Thus once any dierentiation has been performed, theprocess can then be reversed to give a primitive.

WORKED EXERCISE: [These questions are always dicult.](a) Dierentiate (x2 + 1)4.(b) Hence nd a primitive of 8x(x2 + 1)3.

SOLUTION:(a) Let y = (x2 + 1)4 .

Thendy

dx=

dy

du du

dx

= 4(x2 + 1)3 2x= 8x(x2 + 1)3 .

Let u = x2 + 1.Then y = u4 .

Hencedu

dx= 2x

anddy

du= 4u3 .

(b) Henced

dx(x2 + 1)4 = 8x(x2 + 1)3 .

Reversing this,

8x(x2 + 1)3 dx = (x2 + 1)4 + C, for some constant C.

Note: Questions in the 2 Unit course would never ask for such an integralwithout rst asking for the appropriate derivative.

Exercise 1DTechnology: Many programs that can perform algebraic manipulation are also ableto deal with indenite integrals. They can be used to check the questions in this exerciseand to investigate the patterns arising in such calculations.

1. Find the following indenite integrals:

(a)

4 dx

(b)

1 dx

(c)

0 dx

(d)

(2) dx

(e)

x dx

(f)

x2 dx

(g)

x3 dx

(h)

x7 dx


2. Find the indenite integral of each function. Use the notation of the previous question.(a) 2x(b) 4x

(c) 3x2

(d) 4x3(e) 10x9

(f) 2x3(g) 4x5

(h) 3x8

3. Find the following indenite integrals:

(a)

(x + x2) dx

(b)

(x4 x3) dx

(c)

(x7 + x10) dx

(d)

(2x + 5x4) dx

(e)

(9x8 11) dx

(f)

(7x13 + 3x8) dx

(g)

(4 3x) dx

(h)

(1 x2 + x4) dx

(i)

(3x2 8x3 + 7x4) dx

4. Find the indenite integral of each function. (Leave negative indices in your answers.)(a) x2

(b) x3(c) x8

(d) 3x4(e) 9x10

(f) 10x6

5. Find the following indenite integrals. (Leave fractional indices in your answers.)

(a)

x12 dx

(b)

x13 dx

(c)

x14 dx

(d)

x23 dx

(e)

x12 dx

(f)

4x12 dx


6. By expanding brackets where necessary, nd the following indenite integrals:

(a)

x(x + 2) dx

(b)

x(4 x2) dx

(c)

x2(5 3x) dx

(d)

x3(x 5) dx

(e)

(x 3)2 dx

(f)

(2x + 1)2 dx

(g)

(1 x2)2 dx

(h)

(2 3x)(2 + 3x) dx

(i)

(x2 3)(1 2x) dx

7. By dividing through by the denominator, perform the following integrations:

(a)

x2 + 2xx

dx (b)

x7 + x8

x6dx (c)

2x3 x4

4xdx

8. Write each of these functions with negative indices and nd its indenite integral:

(a)1x2

(b)1x3

(c)1x5

(d)1

x10

(e)3x4

(f)5x6

(g)7x8

(h)1

3x2

(i)1

7x5

(j) 15x3

(k)1x2 1

x5

(l)1x3

+1x4

9. Write these functions with fractional indices and hence nd their indenite integrals:

(a)

x (b) 3

x (c)1x

(d) 3

x2

10. Use the indenite integrals of the previous question to evaluate:

(a) 90

x dx (b)

80

3

x dx (c) 4925

1x

dx (d) 10

3

x2 dx


11. By using the formula


a(n + 1)+ C, nd:

(a)

(x + 1)5 dx

(b)

(x + 2)3 dx

(c)

(4 x)4 dx

(d)

(3 x)2 dx

(e)

(3x + 1)4 dx

(f)

(4x 3)7 dx

(g)

(5 2x)6 dx

(h)

(1 5x)7 dx

(i)

(2x + 9)11 dx

(j)

3(2x 1)10 dx

(k)

4(5x 4)6 dx

(l)

7(3 2x)3 dx



a(n + 1)+ C, nd:

(a)

(13x 7)4 dx (b)

(14x 7)6 dx (c)

(1 15x)3 dx



a(n + 1)+ C, nd:

(a)

1(x + 1)3

dx

(b)

1(x 5)4 dx

(c)

1(3x 4)2 dx

(d)

1(2 x)5 dx

(e)

3(x 7)6 dx

(f)

8(4x + 1)5

dx

(g)

2(3 5x)4 dx

(h)

45(1 4x)2 dx

(i)

78(3x + 2)5

dx

14. By expanding the brackets, nd:

(a)

x(3

x x) dx (b) (

x 2) (x + 2) dx (c) (

2

x 1)2 dx15. (a) Evaluate the following denite integrals:

(i) 10

x12 dx (ii)

41

x12 dx (iii)

80

x13 dx

(b) By writing them with fractional indices, evaluate the following denite integrals:

(i) 40

x dx (ii)

91

x

x dx (iii) 91

dxx

16. By expanding the brackets where necessary, nd:

(a) 42

(2x ) (2 +x ) dx (b)

10

x(

x 4) dx (c) 94

(x 1)2 dx

C H A L L E N G E

17. Explain why the indenite integral

1x

dx cant be found in the usual way using the

standard form

xn dx =xn+1

n + 1+ C.

18. Find each of the following indenite integrals:

(a)

2x 1 dx

(b)

7 4x dx

(c)

34x 1 dx

(d)

13x + 5

dx

CHAPTER 1: Integration 1E Finding Areas by Integration 25

19. Evaluate the following:

(a) 20(x + 1)4 dx

(b) 32(2x 5)3 dx

(c) 22

(1 x)5 dx

(d) 50

(1 x

5

)4dx

(e) 10

9 8x dx

(f) 72

dxx + 2

(g) 02

3

x + 1 dx

(h) 51

3x + 1 dx

(i) 03

1 5x dx

20. (a) (i) Findd

dx(x2 + 1)5. (ii) Hence nd

10x(x2 + 1)4 dx.

(b) (i) Findd


12x2(x3 + 1)3 dx.

(c) (i) Findd

dx(x5 7)8. (ii) Hence nd

40x4(x5 7)7 dx.

(d) (i) Findd

dx(x2 + x)6. (ii) Hence nd

6(2x + 1)(x2 + x)5 dx.

21. (a) (i) Findd

dx(x2 3)5. (ii) Hence nd

x(x2 3)4 dx.

(b) (i) Findd


x2(x3 + 1)10 dx.

(c) (i) Findd


x3(x4 + 8)6 dx.

(d) (i) Findd

dx(x2 + 2x)3. (ii) Hence nd

(x + 1)(x2 + 2x)2 dx.

1 E Finding Areas by IntegrationThe aim of this section and the next is to use denite integrals to nd the areasof regions bounded by curves, lines and the coordinate axes.

Area and the Denite Integral: A denite integral is a pure number, which can bepositive or negative remember that a denite integral representing a regionbelow the x-axis is negative in value. An area has units (called square units oru2 in the absence of any physical interpretation) and cannot be negative.

Any problem on areas requires some care when nding the correct integral orcombination of integrals required. Some particular techniques are listed below,but the general rule is to draw a diagram rst to see which bits need to be addedor subtracted.

14

FINDING AN AREA: When using integrals to nd the area of a region:1. Draw a sketch of the curves, showing relevant intercepts and intersections.

2. Evaluate the necessary denite integral or denite integrals.

3. Write a conclusion, giving the required area in square units.

Areas Above the x-axis: When a region lies entirely above the x-axis, the relevantintegral will be positive and the area will be equal to the integral, apart fromneeding units.


WORKED EXERCISE:Find the area of the region bounded by the curve y = 4 x2 and the x-axis.(This was the example given on page 1 in the introduction to the chapter.)

SOLUTION:The curve meets the x-axis at (2, 0) and (2, 0).The region lies entirely above the x-axis and the relevant integral is 2

2(4 x2) dx =

[4x x

3

3

]22

= (8 83 ) (8 + 83 )= 513 (513 )= 1023 ,

which is positive because the region lies above the x-axis.x

yy = 4 x2

2 2

4

Hence the required area is 1023 square units.

Areas Below the x-axis: When a region lies entirely below the x-axis, the relevantintegral will be negative and the area will then be the opposite of this.

WORKED EXERCISE:Find the area of the region bounded by the curve y = x2 1 and the x-axis.SOLUTION:The curve meets the x-axis at (1, 0) and (1, 0).The region lies entirely below the x-axis and the relevant integral is 1

1(x2 1) dx =

[x3

3 x

]11

= (13 1) ( 13 + 1)= 23 23= 113 ,

which is negative, because the region lies below the x-axis.

x

y

1 1

1


Areas Above and Below the x-axis: When a curve crosses the x-axis, the area of theregion between the curve and the x-axis cannot usually be found by means ofa single integral. This is because integrals representing regions below the x-axishave negative values.

WORKED EXERCISE:(a) Sketch the cubic curve y = x(x + 1)(x 2), showing the x-intercepts.(b) Shade the region enclosed between the x-axis and the curve, and nd its area.

[Hint: The expansion of the function is y = x(x + 1)(x 2)= x(x2 x 2)= x3 x2 2x.]

(c) Find 21

x(x+1)(x 2) dx and explain why this integral does not representthe area of the region described in part (b).


SOLUTION:(a) The curve has x-intercepts x = 1, x = 0 and x = 2 and is graphed below.(b) For the region above the x-axis, 0

1(x3 x2 2x) dx =

[x4

4 x

3

3 x2

]01

= (0 0 0) (14 + 13 1) x

y

1 2

= 512 ,

and so area above = 512 square units.

For the region below the x-axis, 20(x3 x2 2x) dx =

[x4

4 x

3

3 x2

]20

= (4 223 4) (0 0 0)= 223 ,

and so area below = 223 square units.

Adding these, total area = 512 + 223

= 3 112 square units.

(c) 21

x(x + 1)(x 2) dx =[x4

4 x

3

3 x2

]21

= (4 223 4) (14 + 13 1)= 223 + 512= 214 .

This integral represents the area from x = 1 to x = 0 above the x-axis,minus, rather than plus, the area from x = 0 to x = 2 below the x-axis.

Areas Associated with Odd and Even Functions: As always, these calculations are oftenmuch easier if symmetries can be recognised.

WORKED EXERCISE:Find the area between the curve y = x3 x and the x-axis.SOLUTION:Factoring, y = x(x2 1)

x

y

11

= x(x 1)(x + 1),and so the x-intercepts are x = 1, x = 0 and x = 1.The two shaded regions have equal areas, since the function is odd.

First, 10(x3 x) dx =

[x4

4 x

2

2

]10

= (14 12 ) (0 0)= 14 ,

so area below the x-axis = 14 square units.

Doubling, total area = 12 square units.


Area Between a Graph and the y-axis: Integration with respect to y rather than x canoften give a result more quickly without the need for subtraction.

When x is a function of y, the denite integral with respect to y represents thearea of the region between the curve and the y-axis, except that areas of regions tothe left of the y-axis are subtracted rather than added. The limits of integrationare values of y rather than of x.

15

THE DEFINITE INTEGRAL AND INTEGRATION WITH RESPECT TO y:

Let x be a continuous function of y in some closed interval a y b.

Then the denite integral ba

x dy is the sum of the areas to the right of the y-axis,

from y = a to y = b, minus the sum of the areas to the left of the y-axis.

WORKED EXERCISE:(a) Sketch the lines y = x + 1 and y = 5 and shade the region between these

lines to the right of the y-axis.(b) Use integration with respect to y to nd the area of this region.(c) Conrm the result by mensuration.

SOLUTION:(a) The lines are sketched below. They meet at (4, 5).

(b) The given equation is y = x + 1.Solving for x, x = y 1,and the required integral is 5

1(y 1) dy =

[y2

2 y

]51

=(252 5

)(12 1

)

= 712 ( 12 )= 8,

x

y

1

5

which is positive, since the region is to the right of the y-axis.


(c) By mensuration, area = 12 base height= 12 4 4= 8 square units.

WORKED EXERCISE:The curve in the diagram below is the cubic y = x3. Use integration with respectto y to nd:(a) the areas of the shaded regions to the right and left of the y-axis,(b) the total area of the two shaded regions.


SOLUTION:(a) The given equation is y = x3 .

Solving for x, x3 = y

x = y13 .

For the region to the right of the y-axis, 80

y13 dy = 34

[y

43

]80

= 34 (16 0) (Note that 843 = 24 = 16.)

= 12,x

y

1

8

so area = 12 square units.

For the region to the left of the y-axis, 01

y13 dy = 34

[y

43

]01

= 34 (0 1) (Note that (1)43 = (1)4 = 1.)

= 34 ,so area = 34 square units.

(b) Adding these, total area = 1234 square units.

Exercise 1ETechnology: Any curve-sketching program will help in identifying the denite inte-grals that need to be evaluated to nd the area of a given region. Programs that canapproximate areas of regions on the screen graph can demonstrate how the nal area isbuilt up from the separate pieces.

1. Find the area of each shaded region below by evaluating the appropriate integral:

x

y

y x= 2

2

(a)

x

y

y = 3x2

1 3

(b)

x

y

y x= 4 3

3

(c)

x

y

y = 3x + 12

1 2

1

(d)

x

y

3

y = x2

(e)

x

y

2 4

y x= 2x2

(f)

x

y

16

y = x

(g)

x

y

5

5

31

y x= 5

(h)


x

y

1

y x= 3 x

(i)

x

y

3 4

y x= 12 2x

(j)

x

y

1 2

1

y x= 5 + 14(k)

x

y

1 27

y = x3(l)


x

y

x y= 25

(a)

x

y

x = 3y2

2

(b)

x

y

x y= 2 42

4

(c)

x

y

x = 27 3y2

273

3

(d)

x

y3

x y=

(e)

x

y

35

x y= 2 + 1

(f)

x

y9

x = y

(g)

x

y4

1 yx = 1

(h)


x

y

31

3 y x= 4x + 32

(a)

x

y

3y = 3x

(b)

x

y

3y x= 3

(c)

x

y1 31

y x= 1 4

(d)


x

y

1

4

x y= 1

(a)

x

y

8

2

4

x y= 6y + 82

(b)

x

y

1

8

x = y3(c)

x

y3

x y= 2

(d)



x

y

3 12

5. The sketch shows the line y = x + 1.(a) Copy the diagram and then shade the region bounded

by y = x+1, the x-axis and the lines x = 3 and x = 2.(b) By evaluating

21

(x+1) dx, nd the area of the shaded

region above the x-axis.

(c) By evaluating 13

(x+1) dx, nd the area of the shaded

region below the x-axis.(d) Hence nd the area of the entire shaded region.

(e) Find 23

(x + 1) dx, and explain why this integral does

not give the area of the shaded region.

x

y

23 1

6. The sketch shows the curve y = (x1)(x+3) = x2 +2x3.(a) Copy the diagram and shade the region bounded by the

curve y = (x 1)(x + 3), the x-axis and the line x = 2.(b) By evaluating

13

(x2 + 2x 3) dx, nd the area of theshaded region below the x-axis.

(c) By evaluating 21(x2 + 2x 3) dx, nd the area of the

shaded region above the x-axis.(d) Hence nd the area of the entire shaded region.

(e) Find 23

(x2 + 2x 3) dx, and explain why this integraldoes not give the area of the shaded region.

x

y

21

7. The sketch shows the curve y = x(x+1)(x2) = x3x22x.(a) Copy the diagram and shade the region bounded by the

curve and the x-axis.

(b) By evaluating 20(x3 x2 2x) dx, nd the area of the

shaded region below the x-axis.

(c) By evaluating 01

(x3 x2 2x) dx, nd the area of theshaded region above the x-axis.

(d) Hence nd the area of the entire region you have shaded.

(e) Find 21

(x3x22x) dx, and explain why this integraldoes not give the area of the shaded region.

8. In each part below, nd the area of the region bounded by the graph of the given functionand the x-axis between the specied values. You should draw a diagram for each part andcheck to see whether the region is above or below the x-axis.(a) y = x2, between x = 3 and x = 2(b) y = 2x3, between x = 4 and x = 1


(c) y = 3x(x 2), between x = 0 and x = 2(d) y = x 3, between x = 1 and x = 4(e) y = (x 1)(x + 3)(x 2), between x = 3 and x = 2(f) y = 2x(x + 1), between x = 2 and x = 2(g) y = x(3 x)2, between x = 0 and x = 3(h) y = x4 4x2, between x = 5 and x = 0

9. In each part below, nd the area of the region bounded by the graph of the given functionand the y-axis between the specied values. You should draw a diagram for each part andcheck whether the region is to the right or left of the y-axis.(a) x = y 5, between y = 0 and y = 6(b) x = 3 y, between y = 2 and y = 5(c) x = y2 , between y = 1 and y = 3(d) x = (y 1)(y + 1), between y = 3 and y = 0

10. In each part below you should draw a graph and look carefully for any symmetries thatwill simplify the calculation.(a) Find the area of the region bounded by the curve and the x-axis:

(i) y = x7, for 2 x 2(ii) y = x3 16x = x(x 4)(x + 4), for 4 x 4(iii) y = x4 9x2 = x2(x 3)(x + 3), for 3 x 3

(b) Find the area of the region bounded by the curve and the y-axis:(i) x = 2y, for 5 y 5(ii) x = y2 , for 3 y 3(iii) x = 4 y2 = (2 y)(2 + y), for 2 y 2

11. Find the area of the region bounded by y = |x + 2| and the x-axis, for 2 x 2.C H A L L E N G E

x

y12. The diagram shows a graph of y2 = 16(2 x).(a) Find the x-intercept and the y-intercepts.(b) Find the area of the shaded region:

(i) by considering the region between the curvey = 4

2 x and the x-axis,

(ii) by considering the region between the curvex = 2 116 y2 and the y-axis.

13. The gradient of a curve is y = x2 4x + 3 and the curve passes through the origin.(a) Find the equation of the curve.(b) Show that the curves turning points are (1, 113 ) and (3, 0), and sketch its graph.(c) Find the area of the region enclosed between the curve and the x-axis between the

two turning points.

14. Sketch y = x2 and mark the points A(a, a2), B(a, a2), P (a, 0) and Q(a, 0).(a) Show that

a0

x2 dx = 23 (areaOAP ).

(b) Show that aa

x2 dx = 13 (area of rectangle ABQP ).

CHAPTER 1: Integration 1F Areas of Compound Regions 33

1 F Areas of Compound RegionsWhen a region is bounded by two or more dierent curves, some dissection processis usually needed before integrals can be used to calculate its area.

Thus a preliminary sketch of the region becomes all the more important.

Areas of Regions Under a Combination of Curves: Some regions are bounded by dif-ferent curves in dierent parts of the x-axis.

WORKED EXERCISE:(a) Sketch the curves y = x2 and y = (x 2)2 on one set of axes.(b) Shade the region bounded by y = x2, y = (x 2)2 and the x-axis.(c) Find the area of this shaded region.

SOLUTION:(a) The two curves intersect at (1, 1), because it can be checked by substitution

that this point lies on both curves.

(b) The whole region is above the x-axis, but it will be necessary to nd sepa-rately the areas of the regions to the left and right of x = 1.

(c) First, 10

x2 dx =[x3

3

]10

= 13 .

Secondly, 21(x 2)2 dx =

[(x 2)3

3

]21

x

y

21

1

= 0 ( 13 )= 13 .

Combining these, area = 13 +13

= 23 square units.

Areas of Regions Between Curves: Suppose that one curve y = f(x) is always belowanother curve y = g(x) in an interval a x b. Then the area of the regionbetween the curves from x = a to x = b can be found by subtraction.

16

AREA BETWEEN CURVES: If f(x) g(x) in the interval a x b, then

area between the curves = ba

(g(x) f(x)

)dx.

That is, take the integral of the top curve minus the bottom curve.

The assumption that f(x) g(x) is important. If the curves cross each other,then separate integrals will need to be taken or else the areas of regions wheredierent curves are on top will begin to cancel each other out.


WORKED EXERCISE:(a) Find the two points where the curve y = (x 2)2 meets the line y = x.(b) Draw a sketch and shade the area of the region between these two graphs.(c) Find the shaded area.

SOLUTION:(a) Substituting y = x into y = (x 2)2 gives

(x 2)2 = xx2 4x + 4 = xx2 5x + 4 = 0

(x 1)(x 4) = 0,x = 1 or 4,

x

y

1 2 4

1

4

so the two graphs intersect at (1, 1) and (4, 4).

(b) The sketch is drawn to the right.

(c) In the shaded region, the line is above the parabola.

Hence area = 41

(x (x 2)2

)dx

= 41

(x (x2 4x + 4)

)dx

= 41(x2 + 5x 4) dx

=[ x

3

3+

5x2

2 4x

]41

= (2113 + 40 16) ( 13 + 212 4)= 223 + 1

56

= 412 square units.

Note: The formula (given in Box 16 on the previous page) for the area of theregion between two curves holds even if the region crosses the x-axis.

To illustrate this point, the next example is the previous example shifted down2 units so that the region between the line and the parabola crosses the x-axis.The area of course remains the same and notice how the formula still givesthe correct answer.

WORKED EXERCISE:(a) Find the two points where the curves y = x2 4x + 2 and y = x 2 meet.(b) Draw a sketch and nd the area of the region between these two curves.

SOLUTION:(a) Substituting y = x 2 into y = x2 4x + 2 gives

x2 4x + 2 = x 2x2 5x + 4 = 0

(x 1)(x 4) = 0x = 1 or 4. x

y

1 24

1

2

so the two graphs intersect at (1,1) and (4, 2).


(b) Again, the line is above the parabola.

Hence area = 41

((x 2) (x2 4x + 2)

)dx

= 41(x2 + 5x 4) dx

=[ x

3

3+

5x2

2 4x

]41

= (2113 + 40 16) ( 13 + 212 4)= 412 square units.

Areas of Regions Between Curves that Cross: Now suppose that one curve y = f(x)is sometimes above and sometimes below another curve y = g(x) in the relevantinterval. In this case, separate integrals will need to be calculated.

WORKED EXERCISE:The diagram below shows the curves

y = x2 + 4x 4 and y = x2 8x + 12meeting at the points (2, 0) and (4,4). Find the area of the shaded region.SOLUTION:In the left-hand region, the second curve is above the rst.

Hence area = 20

((x2 8x + 12) (x2 + 4x 4)

)dx

= 20(2x2 12x + 16) dx

=[2x3

3 6x2 + 16x

]20

= 513 24 + 32x

y12

4

24

6

= 1313 square units.

In the right-hand region, the rst curve is above the second.

Hence area = 42

((x2 + 4x 4) (x2 8x + 12)) dx

= 42(2x2 + 12x 16) dx

=[ 2x

3

3+ 6x2 16x

]42

= (4223 + 96 64) (513 + 24 32)= 1023 + 1313= 223 square units.

Hence total area = 1313 + 223

= 16 square units.


Exercise 1FTechnology: Screen graphing programs are particularly useful with compound regionsbecause they allow the separate parts of the region to be identied clearly.

1. Find the area of the shaded region in each diagram below.

( )1 1,

x

y

y = x2y x=

(a)

( )1 1,

x

y

y x= 3y x=

(b)

( )1 1,

x

y

y x=

y x= 4

(c)

x

y

y x= 2

y x= 3

(1,1)

(d)

( )1 1,

x

y

y x= 6y x= 4

(e)

x

y

y x= 3 + 4

y x= 2

(4,16)

(1,1)

(f)

x

y

y x= 9 2

y x= + 12

(2,5)

(4,17)

(g)

( )3 1,

( )2 6,

x

y

y = 10 2x

y = x + 4(h)

2. By considering regions between the curves and the y-axis, nd the area of the shadedregion in each diagram below.

( )4 2,x

y

x = y2

x y= 2

(a)

x

y

x y= 3 2x y= 2

(1,1)

(4,2)

(b)

x

y

1

4x y= 4

x y y= 5 42

(2,2)

(c)

( )3 1,

( )6 2,

x

y

x = y2 + 2

y = 4x

(d)

3. Find the areas of the shaded region in the diagrams below. In each case you will need tond two areas and subtract one from the other.

x

y

62

3

4

y x= 6 8x2

(a)

x

y

12

12

y x= 1 2

y x= 4 2(b)


4. Find the areas of the shaded regions in the diagrams below. In each case you will need tond two areas and add them.

y x= ( 2)2

y x= ( + 2)2

x

y

2 2

(a)

x

y

3

y x= 2

y x= ( 3)232

94( , )

(b)

5. (a) By solving the equations simultaneously, show that the curves y = x2+4 and y = x+6intersect at the points (1, 5) and (2, 8).

(b) Sketch the curves on the same diagram and shade the region enclosed between them.(c) Show that this region has area 2

1

((x + 6) (x2 + 4)) dx =

21

(x x2 + 2) dx

and evaluate the integral.

6. (a) By solving the equations simultaneously, show that the curves y = 3xx2 = x(3x)and y = x intersect at the points (0, 0) and (2, 2).


0(3x x2 x) dx =

20(2x x2) dx


7. (a) By solving the equations simultaneously, show that the curves y = (x 3)2 andy = 14 2x intersect at the points (1, 16) and (5, 4).


1

((14 2x) (x 3)2) dx =

51

(4x + 5 x2) dx



8. Solve simultaneously the equations of each pair of curves below to nd their points ofintersection. Sketch each pair of curves on the same diagram and shade the region enclosedbetween them. By evaluating the appropriate integral, nd the area of the shaded regionin each case.(a) y = x4 and y = x2

(b) y = 3x2 and y = 6x3

(c) y = 9 x2 and y = 3 x(d) y = x + 10 and y = (x 3)2 + 1

9. (a) By solving the equations simultaneously, show that the curves y = x2 + 2x 8 andy = 2x + 1 intersect at the points (3, 7) and (3,5).

(b) Sketch both curves on the same diagram and shade the region enclosed between them.


(c) Despite the fact that it crosses the x-axis, the region has area given by 33

((2x + 1) (x2 + 2x 8)

)dx =

33

(9 x2) dx.

Evaluate the integral and hence nd the area of the region enclosed between the curves.

10. (a) By solving the equations simultaneously, show that the curves y = x2 x 2 andy = x 2 intersect at the points (0,2) and (2, 0).

(b) Sketch both curves on the same diagram and shade the region enclosed between them.(c) Despite the fact that it is below the x-axis, the region has area given by 2

0

((x 2) (x2 x 2)) dx =

20(2x x2) dx.

Evaluate this integral and hence nd the area of the region between the curves.

11. Solve simultaneously the equations of each pair of curves below to nd their points ofintersection. Sketch each pair of curves on the same diagram and shade the region enclosedbetween them. By evaluating the appropriate integral, nd the area of the shaded regionin each case.(a) y = x2 6x + 5 and y = x 5(b) y = 3x and y = 4 x2(c) y = x2 1 and y = 7 x2(d) y = x and y = x3

12. Find the area bounded by the lines y = 14x and y = 12x between x = 1 and x = 4.13. (a) On the same number plane, sketch the graphs of the functions y = x2 and x = y2

Cambridge HSC Maths [2U]

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