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YEAR 12 2 Unit Second Edition CAMBRIDGE Mathematics BILL PENDER DAVID SADLER JULIA SHEA DEREK WARD COLOUR VERSION WITH STUDENT CD-ROM Now in colour with an electronic version of the book on CD

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São 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.] . 2 nd 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: info@copyright.com.au 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 URL S 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 .......................................... v How to Use This Book .................................. vii About the Authors .................................... xi Chapter One — Integration ............................... 1 1A AreasandtheDeﬁniteIntegral ...................... 1 1B The Fundamental Theorem of Calculus.................. 6 1C TheDeﬁniteIntegralanditsProperties ................. 12 1D The Indeﬁnite Integral........................... 19 1E FindingAreasbyIntegration ....................... 25 1F AreasofCompoundRegions ........................ 33 1G VolumesofSolidsofRevolution ...................... 39 1H TheTrapezoidalRule ........................... 47 1I Simpson’sRule ............................... 51 1J ChapterReviewExercise .......................... 55 Chapter Two — The Exponential Function ....................... 60 2A ReviewofExponentialFunctions ..................... 60 2B The Exponential Functione xand the Deﬁnition ofe.......... 65 2C DiﬀerentiationofExponentialFunctions ................. 73 2D ApplicationsofDiﬀerentiation ....................... 79 2E IntegrationofExponentialFunctions ................... 84 2F ApplicationsofIntegration ......................... 90 2G ChapterReviewExercise .......................... 96 Chapter Three — The Logarithmic Function ...................... 99 3A Review of Logarithmic Functions . .................... 99 3B The Logarithmic Function Basee.....................105 3C Diﬀerentiation of Logarithmic Functions . ................112 3D Applications of Diﬀerentiation of logx..................116 3E IntegrationoftheReciprocalFunction ..................121 3F Applications of Integration of 1/x.....................128 3G CalculuswithOtherBases .........................133 3H ChapterReviewExercise ..........................139

iv Contents Chapter Four — The Trigonometric Functions .....................141 4A RadianMeasureofAngleSize .......................141 4B Mensuration of Arcs, Sectors and Segments ...............147 4C GraphsoftheTrigonometricFunctionsinRadians ...........153 4D The Behaviour of sinxNeartheOrigin ..................159 4E TheDerivativesoftheTrigonometricFunctions .............164 4F ApplicationsofDiﬀerentiation .......................172 4G IntegrationoftheTrigonometricFunctions ................178 4H ApplicationsofIntegration .........................186 4I ChapterReviewExercise ..........................192 Chapter Five — Motion .................................196 5A AverageVelocityandSpeed ........................196 5B VelocityasaDerivative ..........................203 5C IntegratingwithRespecttoTime .....................213 5D ChapterReviewExercise ..........................220 Chapter Six — Rates and Finance ...........................223 6A ApplicationsofAPsandGPs .......................223 6B The Use of Logarithms with GPs.....................232 6C SimpleandCompoundInterest ......................238 6D InvestingMoneybyRegularInstalments .................244 6E PayingOﬀaLoan .............................252 6F RatesofChange ..............................260 6G NaturalGrowthandDecay ........................268 6H ChapterReviewExercise ..........................278 Chapter Seven — Euclidean Geometry .........................281 7A Points,Lines,ParallelsandAngles ....................282 7B AnglesinTrianglesandPolygons .....................291 7C CongruenceandSpecialTriangles .....................298 7D TrapeziumsandParallelograms ......................308 7E Rhombuses, Rectangles and Squares...................311 7F AreasofPlaneFigures ...........................318 7G Pythagoras’ Theorem and its Converse..................321 7H Similarity ..................................324 7I InterceptsonTransversals .........................332 7J ChapterReviewExercise ..........................336 Chapter Eight — Probability ...............................341 8A Probability and Sample Spaces......................341 8B SampleSpaceGraphsandTreeDiagrams ................348 8C SetsandVennDiagrams ..........................352 8D VennDiagramsandtheAdditionTheorem................357 8E Multi-StageExperimentsandtheProductRule .............361 8F Probability Tree Diagrams . . .......................367 8G ChapterReviewExercise ..........................372 Answers to Exercises ..................................374 Index ...........................................413

Preface This textbook has been written for students in Years 11 and 12 taking the 2 Unit calculus course ‘Mathematics’ for the NSW HSC. The book covers all the content of the course at the level required for the HSC examination. There are two volumes — the present volume is roughly intended for Year 12, and the previous volume for Year 11. Schools will, however, diﬀer in their choices of order of topics and in their rates of progress. Although the Syllabus has not been rewritten for the new HSC, there has been a 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 variety of diﬀerent types of questions. There is an abundance of questions and problems in each exercise — too many for any one student — carefully grouped in three graded sets, so that with proper selection the book can be used at all levels of ability in the 2 Unit course. This new second editionhas been thoroughly rewritten to make it more acces- sible to all students. The exercises now have more early drill questions to reinforce each new skill, there are more worked exercises on each new algorithm, and some chapters and sections have been split into two so that ideas can be introduced more gradually. We have also added a review exercise to each chapter. We would like to thank our colleagues at Sydney Grammar School and Newington College for their invaluable help in advising us and commenting on the successive drafts. We would also like to thank the Headmasters of our two schools for their encouragement of this pro ject, and Peter Cribb, Sarah Buerckner and the team at Cambridge University Press, Melbourne, for their support and help in discussions. Finally, our thanks go to our families for encouraging us, despite the distractions that the pro ject has caused to family life. Dr Bill Pender Sub ject Master in Mathematics Sydney Grammar School College Street Darlinghurst NSW 2010 David Sadler Mathematics Sydney Grammar SchoolJulia Shea Director of Curriculum Newington College 200 Stanmore Road Stanmore NSW 2048 Derek Ward Mathematics Sydney Grammar School

How to Use This Book This book has been written so that it is suitable for the full range of 2 Unit students, whatever their abilities and ambitions. The Exercises: No-one should try to do all the questions!We have written long exercises so that everyone will ﬁnd enough questions of a suitable standard — each student will need to select from them, and there should be plenty left for revision. The book provides a great variety of questions, and representatives of all types should be attempted. Each chapter is divided into a number of sections. Each of these sections has its own 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 proceeding without mastery of this group. Development:This group is usually the longest. It contains more substantial questions, questions requiring proof or explanation, problems where the new content can be applied, and problems involving content from other sections and chapters to put the new ideas in a wider context. Challenge:Many questions in recent 2 Unit HSC examinations have been very demanding, and this section is intended to match the standard of those recent examinations. Some questions are algebraically challenging, some re- quire more sophistication in logic, some establish more diﬃcult connections between topics, and some complete proofs or give an alternative approach. The Theory and the Worked Exercises: All the theory in the course has been properly developed, 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 to learn a method ﬁrst and then return to the details of the proof and explanation when the point of it all has become clear. The main formulae, methods, deﬁnitions 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 using the numbered boxes as a basis. The worked examples have been chosen to illustrate the new methods introduced in the section. They should provide suﬃcient preparation for the questions in the following exercise, but they cannot possibly cover the variety of questions that can be asked.

viii How to Use This Book The Chapter Review Exercises: A Chapter Review Exercise has been added to each chapter of the second edition. These exercises are intended only as a basic review of the chapter — for harder questions, students are advised to work through more of the later questions in the exercises. The Order of the Topics: We have presented the topics in the order that we have found most satisfactory in our own teaching. There are, however, many eﬀective orderings of the topics, and apart from questions that provide links between topics, the book allows all the ﬂexibility needed in the many diﬀerent situations that apply in diﬀerent schools. The time needed for the Euclidean geometry in Chapter Seven and probability in 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 as possible because these are the fundamental ideas in the course. Secondly, the courses in Years 9 and 10 already develop most of the work in Euclidean geometry and 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 the course, and the two processes of diﬀerentiation and integration, inverses of each other, are the basis of most of the topics. Both processes are introduced as ge- ometrical ideas — diﬀerentiation is deﬁned using tangents and integration using areas — but the subsequent discussions, applications and exercises give many other 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 when areas are being considered or a constant acceleration is being applied. They can be studied without calculus, but calculus provides an alternative and sometimes quicker approach. The Exponential and Logarithmic Functions:Calculus is essential for the study of these functions. We have begun the topic with the exponential function. This has the great advantage of emphasising the fundamental prop- erty that the exponential function with baseeis its own derivative — this is the reason why it is essential for the study of natural growth and decay, and therefore occurs in almost every application of mathematics. The logarithmic function, and its relationship with the rectangular hyperbolay=1/x,has been covered in a separate chapter. The Trigonometric Functions:Calculus is also essential for the study of the trigonometric functions. Their deﬁnitions, like the associated deﬁnition ofπ, are based on the circle. The graphs of the sine and cosine functions are waves, and they are essential for the study of all periodic phenomena. Thus the three basic functions in the course,x 2,e xand sinx, and the related numb erseandπ, 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 geometry or 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 drawn as 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 because sketches can so easily be drawn that this type of mathematics is so satisfactory for 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 the sciences. In particular, the applications of calculus to maximisation, motion, rates of change and ﬁnance are all parts of the syllabus. The course thus allows students to experience a double view of mathematics, as a system of pure logic on the one hand, and an essential part of modern technology on the other. Limits, Continuity and the Real Numbers: This is a ﬁrst course in calculus, and rigorous arguments about limits, continuity or the real numbers would be quite inappro- priate. Any such ideas required in this course are not diﬃcult to understand intuitively. Most arguments about limits need only the limit lim x→∞ 1/x=0and occasionally the sandwich principle. Introducing the tangent as the limit of the secant is a dramatic new idea, clearly marking the beginning of calculus, and is quite accessible. The functions in the course are too well-behaved for continuity to be a real issue. The real numbers are deﬁned geometrically as points on the number line, and any properties that are needed can be justiﬁed by appealing to intuitive ideas about lines and curves. Everything in the course apart from these subtle issues of ‘foundations’ can be proven completely. Technology: There is much discussion about what role technology should play in the mathematics classroom and what calculators or software may be eﬀective. This is a time for experimentation and diversity. We have therefore given only a few speciﬁc recommendations about technology, but we encourage such investigation, and to this new colour version we have added some optional technology resources which can be accessed via the student CD in the back of the book. The graphs of functions are at the centre of the course, and the more experience and intuitive understanding students have, the better able they are to interpret the mathemat- ics correctly. A warning here is appropriate — any machine drawing of a curve should be accompanied by a clear understanding of why such a curve arises from the particular equation or situation.

About the Authors Dr Bill Pender is Sub ject 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 Macquarie University. In 1973–74, he studied at Bonn University in Germany, and he has lectured and tutored at Sydney University and at the University of NSW, where he was a Visiting Fellow in 1989. He has been involved in syllabus development since the early 1990s — he was a member of the NSW Syllabus Committee in Mathematics for two years and of the subsequent Review Committee for the 1996 Years 9–10 Advanced Syllabus. More recently he was involved in the writing of the new K–10 Mathematics Syllabus. He is a regular presenter of inservice courses for AIS and MANSW, and plays piano and harpsichord. David Sadler is Second Master in Mathematics at Sydney Grammar School, where he has taught since 1980. He has a BSc from the University of NSW and an MA in Pure Mathematics and a DipEd from Sydney University. In 1979, he taught at Sydney Boys’ High School, and he was a Visiting Fellow at the University of NSW in 1991. Julia Shea is now Director of Curriculum at Newington College, having been appointed Head of Mathematics there in 1999. She has a BSc and DipEd from the University of Tasmania, she taught for six years at Rosny College, a State Senior College in Hobart, and was a member of the Executive Committee of the Mathematics Association of Tasmania for ﬁve years. She then taught for ﬁve years at Sydney Grammar School before moving to Newington College. Derek Ward has taught Mathematics at Sydney Grammar School since 1991 and is Master in Charge of Statistics. He has an MSc in Applied Mathematics and a BScDipEd, both from the University of NSW, where he was subsequently Senior Tutor 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 at various venues.

ﬀ 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 one’s equations than to have them ﬁt experiment. — The twentieth-century English physicist Paul Dirac Even if there is only one possible uniﬁed theory, it is just a set of rules and equations. What is it that breathes ﬁre into the equations and makes a universe for them to describe? The usual approach of science of constructing a mathematical model cannot answer the questions of why there should be a universe for the model to describe. — Steven Hawking,A Brief History of Time

CHAPTERONE Integration x y y= 4 −x 2 −2 24 The calculation of areas has so far been restricted to regions bounded by straight lines or parts of circles. This chapter will extend the study of areas to regions bounded by more general curves. For example, it will be possible to calculate the area of the shaded region in the diagram to the right, bounded by the parabolay=4−x 2and thex-axis. The method developed in this chapter is calledintegration. The basis of this method is the fact that ﬁnding tangents and ﬁnding areas are inverse processes, so that integration is the inverse process of diﬀerentiation. This result is called the fundamental theorem of calculusand it will greatly simplify calculation of the required areas. 1AAreas and the Deﬁnite Integral All 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 up and rearranged into a rectangle with a few well-chosen cuts. Dissecting a curved region into rectangles, however, requires an inﬁnite number of rectangles and so must be a limiting process, like diﬀerentiation. A New Symbol — The Deﬁnite Integral: Some new notation is needed to reﬂect this process of inﬁnite dissection as it applies to functions and their graphs. The diagram on the left below shows the region contained between a given curve y=f(x)andthex-axis, fromx=atox=b. The curve must be continuous and, for the moment, entirely above thex-axis. x y ab x y ab x y ab fx() x x+x δ

2 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT 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 upper boundary 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 valuexon thex-axis. Its height at the left-hand end isf(x), and provided the strip is very thin, the height is still aboutf(x) at the right-hand end. Let the width of the strip beδx, whereδxis, as usual in calculus, thought of as being very small. Then, roughly, area of strip = width×height =f(x)δx. Adding up the areas of all the strips gives the following rough formula. We need sigma 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)δx. If, however, there were inﬁnitely many of these strips, each inﬁnitesimally thin, one can imagine that the inaccuracy would disappear. This involves taking the limit so that the equality is exact: area of shaded region = lim δx→0b x=a f(x)δx. At this point, the widthδxis replaced by the symboldx, which suggests an inﬁnitesimal width, and an old form of the letter S is used to suggest an inﬁnite sum. The result is the strange-looking symbol b a f(x)dx, invented by Leibnitz. This symbol is nowdeﬁnedto denote the area of the shaded region: b a f(x)dx= area of shaded region. The Deﬁnite Integral: This new ob ject b a f(x)dxis called adeﬁnite integral. The rest of the chapter is concerned with evaluating deﬁnite integrals and applying them. 1 THE DEFINITE INTEGRAL : Letf(x) be a function that is continuous in the intervala≤x≤b. For the moment, suppose thatf(x) is never negative in the interval. Thedeﬁnite integral b a f(x)dxis deﬁned to be the area of the region between the curve and thex-axis, fromx=atox=b. The functionf(x) is called theintegrandand the valuesx=aandx=bare called thelower and upper boundsof the integral. The name ‘integration’ suggests putting many parts together to make a whole. The notation arises from building up the region from an inﬁnitely large number of inﬁnitesimally thin strips. Integration is ‘making a whole’ from these thin slices.

CHAPTER 1: Integration 1A Areas and the Deﬁnite Integral 3 Evaluating Deﬁnite Integrals Using Area Formulae: When the function is linear or cir- cular, the deﬁnite integral can be calculated from the graph using well-known area formulae, although a quicker method will be developed later for linear functions. Here are the relevant area formulae: 2 FOR A TRIANGLE :Area= 12×base×height FOR A TRAPEZIUM :Area=width×average of parallel sides FOR A CIRCLE :Area=πr 2 WORKED EXERCISE : Evaluate using a graph and area formulae: (a) 4 1 (x−1)dx(b) 4 2 (x−1)dx SOLUTION : (a) The graph ofy=x−1 has gradient 1 andy-intercept−1. The area represented by the integral is the shaded triangle, with base 4−1 = 3 and height 3. Hence 4 1 (x−1)dx= 12×base×height = 12×3×3 x y 14 1 3 −1 =4 12. (b) The functiony=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 4 2 (x−1)dx=width×average of parallel sides =2×1+3 2 x y 12 4 1 3 −1 =4. WORKED EXERCISE : Evaluate using a graph and area formulae: (a) 2 −2 |x|dx(b) 5 −5 25−x 2dx SOLUTION : (a) The functiony=|x|is a V-shape with vertex at the origin. Each shaded triangle has base 2 and height 2. Hence 2 −2 |x|dx=2× 12×base×height =2× 12×2×2 x y −2 22 =4.

4 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 (b) The functiony= 25−x 2is a semicircle with centre at the origin and radius 5. Hence 5 −5 25−x 2dx= 12×πr 2 = 12×5 2×π x y −5 55 = 25π2 . The Area of a Circle: In earlier years, the formulaA=πr 2for the area of a circle was proven. Because the boundary is a curve, some limiting process had to be used in that proof. For comparison with the notation for the deﬁnite integral explained at 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 into inﬁnitesimally thin sectors and then rearranging them into a rectangle. r r r πr πr The height of the rectangle in the lower diagram isr. Since the circumference 2πr is divided equally between the top and bottom sides, the length of the rectangle isπr. Hence the rectangle has areaπr 2, which is therefore the area of the circle. Exercise1A 1.Use area formulae to calculate the following integrals (sketches are given): x y 2 3 (a) 2 0 3dx x y 4 3 (b) 3 0 4dx x y 4 4 (c) 4 0 xdx x y 3 6 (d) 3 0 2xdx

CHAPTER 1: Integration 1A Areas and the Deﬁnite Integral 5 x y 2 2 (e) 2 0 (2−x)dx x y 5 5 (f ) 5 0 (5−x)dx x y 0 2 4 2 (g) 2 0 (x+2)dx x y 7 3 4 0 (h) 4 0 (x+3)dx 2.Use area formulae to calculate the following integrals (sketches are given): x y 2 −1 3 (a) 3 −1 2dx x y −3 2 5 (b) 2 −3 5dx ()1 6, x y 1 −24 (c) 1 −2 (2x+4)dx x y 3 −112 3 (d) 3 −1 (3x+3)dx x y 5 −14(5, 9) (−1, 3) (e) 5 −1 (x+4)dx x y ()−2 4,()2 8, 6 (f ) 2 −2 (x+6)dx x y −3 33 (g) 3 −3 |x|dx x y −2 24 (h) 2 −2 |2x|dx x y 1 1 3.The diagram to the right shows the graph ofy=x 2 fromx=0 tox= 1, drawn on graph paper. The scale is 20 little divisions to 1 unit. This means that 400 little squares make up 1 square unit. (a)Count how many little squares there are under the graph fromx=0tox= 1 (keeping reasonable track of fragments of squares), then divide by 400 to approximate 1 0 x2dx. (b)By counting the appropriate squares, approximate: (i) 12 0 x2dx(ii) 1 12 x2dx Conﬁrm that the sum of the answers to parts (i) and (ii) is the answer to part (a).

6 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 DEVELOPMENT 4.Sketch a graph of each deﬁnite integral, then use an area formula to calculate it: (a) 3 0 5dx (b) 0 −3 5dx (c) 4 −1 5dx (d) 6 −2 5dx(e) 0 −5 (x+5)dx (f ) 2 0 (x+5)dx (g) 4 2 (x+5)dx (h) 3 −1 (x+5)dx(i) 8 4 (x−4)dx (j) 10 4 (x−4)dx (k) 7 5 (x−4)dx (l) 10 6 (x−4)dx(m) 2 −2 |x|dx (n) 4 −4 |x|dx (o) 5 0 |x−5|dx (p) 10 5 |x−5|dx CHALLENGE 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 the curve into thin strips rather than the squares used in questions 3 and 7. Steadily increasing the number of strips should show the value converging to a limit, which can be checked either using mensuration formulae or using the exact value of the integral as calculated in the next section. 6.Sketch a graph of each deﬁnite integral, then use an area formula to calculate it: (a) 4 −4 16−x 2dx(b) 0 −5 25−x 2dx 7.The diagram to the right shows the quadrant y= 1−x 2,fromx=0 tox=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 the graph fromx=0 tox=1. (b) Divide by 400 to approximate 1 0 1−x 2dx. (c) Hence ﬁnd an approximation forπ. 1BThe Fundamental Theorem of Calculus The fundamental theorem is a formula for evaluating deﬁnite integrals. Its proof is rather demanding, so only the algorithm is presented in this section, by means of some worked examples. The proof is given in the appendix to this chapter. Primitives: The formula of the fundamental theorem relies on primitives. Recall that F(x) is called aprimitiveof a functionf(x) if the derivative ofF(x)isf(x): F(x) is a primitive off(x)ifF (x)=f(x). We will need the result established in the last section of the Year 11 volume: 3 FINDING PRIMITIVES : Suppose thatn =−1. Ifdy dx=x n,theny=x n+1 n+1+C,for some constantC. ‘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 deﬁnite integral can be evaluated by writing down any primitiveF(x)off(x), then substituting the upper and lower bounds into it and subtracting. 4 THE FUNDAMENTAL THEOREM : Letf(x) be a function that is continuous in a closed intervala≤x≤b.Then b a f(x)dx=F(b)−F(a), whereF(x) is any primitive off(x). Using the Fundamental Theorem to Evaluate an Integral: The conventional way to set out these calculations is to enclose the primitive in square brackets, writing the upper and lower bounds as superscript and subscript respectively. WORKED EXERCISE : Evaluate the following deﬁnite integrals: (a) 2 0 2xdx(b) 4 2 (2x−3)dx Then draw diagrams to show the regions that they represent. SOLUTION : (a) 2 0 2xdx= x 2 2 0 (x 2is a primitive of 2x.) =2 2−0 2 (Substitute 2, then substitute 0.) =4 This value agrees with the area of x y yx= 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) 4 2 (2x−3)dx= x 2−3x 4 2 (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 yx= 2 − 3 24 1 5 the trapezium shaded in the diagram to the right. (Note that area of trapezium = width×average of parallel sides =2×1+5 2 =2×3 =6.) Note:Whenever there are two or more terms in the primitive, brackets are needed when substituting the upper and lower bounds of integration. Misuse of these brackets is a common source of error.

8 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 WORKED EXERCISE : Evaluate the following deﬁnite integrals: (a) 1 0 x2dx(b) 2 −2 (x 3+8)dx SOLUTION : (a) 1 0 x2dx= x 3 3 1 0 (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) 2 −2 (x 3+8)dx= x 4 4+8x 2 −2 (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 diﬀerentiation, it is often necessary to expand the brackets in the integrand before ﬁnding a primitive. WORKED EXERCISE : Expand the brackets, then evaluate these deﬁnite integrals: (a) 6 1 x(x+1)dx(b) 3 0 (x−4)(x−6)dx Note:Fractions arise very often in deﬁnite integrals because the standard forms for primitives involve fractions. Care is needed with the resulting com- mon denominators, mixed numerals and cancelling. SOLUTION : (a) 6 1 x(x+1)dx= 6 1 (x 2+x)dx = x 3 3+x 2 2 6 1 = (72 + 18)−( 13+ 12) =90− 56 =89 16 (b) 3 0 (x−4)(x−6)dx= 3 0 (x 2−10x+ 24)dx = x 3 3−5x 2+24x 3 0 =(9−45 + 72)−(0−0+0) =36

CHAPTER 1: Integration 1B The Fundamental Theorem of Calculus 9 Writing the Integrand as Two Separate Fractions: If the integrand is a fraction with two terms in the numerator, it should normally be written as two separate fractions, as with diﬀerentiation. WORKED EXERCISE : Write each integrand as two separate fractions, then evaluate: (a) 2 13x 4−2x 2 x2 dx(b) −2 −3 x3−2x 4 x3 dx SOLUTION : (a) 2 13x 4−2x 2 x2 dx= 2 1 3x 2−2 dx(Divide both terms on the top byx 2.) = x 3−2x 2 1 =(8−4)−(1−2) =4−(−1) =5 (b) −2 −3 x3−2x 4 x3 dx= −2 −3 (1−2x)dx(Divide both terms byx 3.) = x−x 2 −2 −3 =(−2−4)−(−3−9) =−6−(−12) =−6+12 =6 Negative Indices: The fundamental theorem works just as well when the indices are negative. The working, however, requires care when converting between negative powers ofxand fractions. WORKED EXERCISE : Use negative indices to evaluate these deﬁnite integrals: (a) 5 1 x−2 dx(b) 2 1 1 x4dx SOLUTION : (a) 5 1 x−2 dx= x −1 −1 5 1 (Increase the index to−1 and divide by−1.) = −1 x 5 1 (Rewritex −1 as1 xbefore substitution.) =− 15−(−1) =− 15+1 = 45

10 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 (b) 2 1 1 x4dx= 2 1 x−4 dx(Rewrite1 x4asx −4 before ﬁnding the primitive.) = x −3 −3 2 1 (Increase the index to−3 and divide by−3.) = −1 3x 3 2 1 (Rewritex −3 as1 x3before substitution.) =− 124 −(− 13) =− 124 + 824 = 724 Exercise1B Technology:Many programs allow deﬁnite integrals to be calculated automatically. This allows not just quick checking of the answers, but experimentation with further deﬁnite integrals. It would be helpful to generate screen sketches of the graphs and the regions involved in the integrals. 1.Evaluate the following deﬁnite integrals, using the fundamental theorem: (a) 1 0 2xdx (b) 4 1 2xdx (c) 3 1 4xdx(d) 5 2 8xdx (e) 3 2 3x 2dx (f ) 3 0 5x 4dx(g) 2 1 10x 4dx (h) 1 0 12x 5dx (i) 1 0 11x 10 dx 2.(a)Evaluate the following deﬁnite integrals, using the fundamental theorem: (i) 1 0 4dx(ii) 7 2 5dx(iii) 5 4 dx (b)Check your answers by sketching the graph of the region involved. 3.Evaluate the following deﬁnite integrals, using the fundamental theorem: (a) 6 3 (2x+1)dx (b) 4 2 (2x−3)dx (c) 3 0 (4x+5)dx(d) 3 2 (3x 2−1)dx (e) 4 1 (6x 2+2)dx (f ) 1 0 (3x 2+2x)dx(g) 2 1 (4x 3+3x 2+1)dx (h) 2 0 (2x+3x 2+8x 3)dx (i) 5 3 (3x 2−6x+5)dx 4.Evaluate the following deﬁnite integrals, using the fundamental theorem. You will need to take care when ﬁnding powers of negative numbers. (a) 0 −1 (1−2x)dx (b) 0 −1 (2x+3)dx (c) 1 −2 3x 2dx(d) 2 −5 dx (e) 2 −1 (4x 3+5)dx (f ) 2 −2 (5x 4+6x 2)dx(g) −2 −6 3x 2dx (h) 4 −3 (12−2x)dx (i) −1 −2 (4x 3+12x 2−3)dx

CHAPTER 1: Integration 1B The Fundamental Theorem of Calculus 11 5.Evaluate the following deﬁnite integrals, using the fundamental theorem. You will need to take care when adding and subtracting fractions. (a) 3 0 xdx (b) 4 1 (x+2)dx (c) 3 1 x2dx(d) 2 0 (x 2+x)dx (e) 3 0 (x+x 2+x 3)dx (f ) 2 −1 (3x+5)dx(g) 1 −1 (x 3−x+1)dx (h) 3 −2 (2x 2−3x+1)dx (i) −2 −4 (16−x 3−x)dx DEVELOPMENT 6.By expanding the brackets where necessary, evaluate the following deﬁnite integrals: (a) 3 2 x(2 + 3x)dx (b) 2 0 (x+ 1)(3x+1)dx (c) 1 0 3x(2 +x)dx (d) 3 2 2x(x−1)dx(e) 1 −1 x2(5x 2+1)dx (f ) 3 1 (x+2) 2dx (g) 2 −1 (x−3) 2dx (h) 3 −2 (4−3x) 2dx(i) 0 −1 x(x−1)(x+1)dx (j) −1 −2 x(x−2)(x+3)dx (k) 0 −1 (1−x 2)2dx (l) 9 4 √ x+1 √ x−1 dx 7.By dividing each fraction through by the denominator, evaluate each integral: (a) 3 13x 3+4x 2 xdx (b) 2 14x 4−x xdx(c) 3 25x 2+9x 4 x2 dx (d) 2 1x3+4x 2 xdx(e) 3 1x3−x 2+x xdx (f ) −1 −2 x3−2x 5 x2 dx 8.Evaluate the following deﬁnite integrals, using the fundamental theorem. You will need to take care when ﬁnding the powers of fractions. (a) 12 0 x2dx(b) 1 23 (2x+3x 2)dx (c) 43 3 4 (6−4x)dx 9.(a)Evaluate the following deﬁnite integrals: (i) 10 5 x−2 dx(ii) 3 2 2x −3 dx(iii) 1 12 4x −5 dx (b)By writing them with negative indices, evaluate the following deﬁnite integrals: (i) 2 1dx x2 (ii) 4 1dx x3 (iii) 1 12 3 x4dx 10.(a)(i) Show that k 2 3dx=3k−6. (ii) Hence ﬁnd the value ofkif k 2 3dx= 18. (b)(i) Show that k 0 xdx= 12k2. (ii) Hence ﬁndkifk>0and k 0 xdx= 18.

12 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 11.Use area formulae to ﬁnd 4 0 f(x)dxin each sketch off(x): x y 3 1 12 4 (a) x y 3 1 12 4 (b) CHALLENGE 12.By dividing each fraction through by the denominator, evaluate each integral: (a) 2 11+x 2 x2 dx(b) −1 −2 1+2x x3 dx(c) −1 −3 1−x 3−4x 5 2x 2 dx 13.Evaluate the following deﬁnite integrals: (a) 3 1 x+1 x 2 dx(b) −1 −3 x 2+1 x2 2 dx(c) 3 −2 (x 2−x) 2dx 14.(a) Explain why the functiony=1 x2is never negative. (b) Sketch the integrand and explain why the argument below is invalid: 1 −1dx x2= −1 x 1 −1 =−1−1=−2. 1CThe Deﬁnite Integral and its Properties This section will ﬁrst extend the theory to functions with negative values. Then some simple properties of the deﬁnite integral will be established using arguments about the dissection of the area associated with the integral. Integrating Functions with Negative Values: When a function has negative values, its graph is below thex-axis, so the ‘heights’ of the little rectangles in the dissec- tion are negative numbers. This means that any areas below thex-axis should contribute negative values to the value of the ﬁnal integral. x y abA BC For example, in the diagram to the right, the regionBis below thex-axis and so will contribute a negative number to the deﬁ- nite integral: b a f(x)dx=areaA−areaB+areaC. Because areas under thex-axis are counted as negative, the deﬁnite integral is some- times referred to as thesigned areaunder the curve, to distinguish it from area, which is always positive.

CHAPTER 1: Integration 1C The Deﬁnite Integral and its Properties 13 5 THE DEFINITE INTEGRAL : Letf(x) be a function that is continuous in the intervala≤x≤b. Suppose now thatf(x) may take positive and negative values in the interval. Thedeﬁnite integral b a f(x)dxis the sum of the areas above thex-axis, from x=atox=b, minus the sum of the areas below thex-axis. WORKED EXERCISE : Evaluate these deﬁnite integrals: (a) 4 0 (x−4)dx(b) 6 4 (x−4)dx(c) 6 0 (x−4)dx Sketch the graph ofy=x−4 and then shade the regions associated with these integrals. Then explain how each result is related to the shaded regions. SOLUTION : (a) 4 0 (x−4)dx= 12x2−4x 4 0 =(8−16)−(0−0) =−8 TriangleOABhas area 8 and is below thex-axis; this is why the value of the integral is−8. (b) 6 4 (x−4)dx= 12x2−4x 6 4 =(18−24)−(8−16) x y −42 46 O ABC M =−6−(−8) =2 TriangleBM Chas area 2 and is above thex-axis; this is why the value of the integral is 2. (c) 6 0 (x−4)dx= 12x2−4x 6 0 =(18−24)−(0−0) =−6 This integral represents the area ofBM Cminus the area ofOAB; this is why the value of the integral is 2−8=−6. Dissection of the Interval: When a region is dissected, its area remains the same. We can always dissect the region by dis- secting the intervala≤x≤bof integration. Thus iff(x) is continuous in the intervala≤x≤b,andthe numb erclies in this interval, then: 6 DISSECTION : b a f(x)dx= c a f(x)dx+ b c f(x)dx x y abc

14 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 Odd and Even Functions: In the ﬁrst example below, the functiony=x 3−4xis an odd function, with point symmetry in the origin. Thus the area of each shaded hump is the same. Hence the whole integral fromx=−2tox= 2 is zero, because the equal humps above and below thex-axis cancel out. In the second diagram, the functiony=x 2+ 1 is even, with line symmetry in they-axis. Thus the areas to the left and right of they-axis are equal, so there is a doubling instead of a cancelling. 7 ODD FUNCTIONS :Iff(x) is odd, then a −a f(x)dx=0. EVEN FUNCTIONS :Iff(x)iseven,then a −a f(x)dx=2 a 0 f(x)dx. WORKED EXERCISE : Sketch these integrals, then evaluate them using symmetry: (a) 2 −2 (x 3−4x)dx(b) 2 −2 (x 2+1)dx SOLUTION : (a) 2 −2 (x 3−4x)dx=0,since the integrand is odd. (Without this simpliﬁcation, the calculation is: 2 −2 (x 3−4x)dx= 14x4−2x 2 2 −2 =(4−8)−(4−8) x y −2 2 =0,as before.) (b) Since the integrand is even, 2 −2 (x 2+1)dx=2 2 0 (x 2+1)dx =2 13x3+x 2 0 =2 (2 23+2)−(0 + 0) x y −2 2 1 5 =9 13. x y a Intervals of Zero Width: Suppose that a function is integrated over an intervala≤x≤aof width zero. In this situation, the region also has width zero and so the integral is zero. 8 INTERVALS OF ZERO WIDTH : a a f(x)dx=0 Running an Integral Backwards from Right to Left: A further small qualiﬁcation must be made to the deﬁnition of the deﬁnite integral. Suppose that the bounds of the integral are reversed, so that the integral ‘runs backwards’ from right to left over the interval. Then its value reverses in sign:

CHAPTER 1: Integration 1C The Deﬁnite Integral and its Properties 15 9 REVERSING THE INTERVAL :Letf(x) be continuous ina≤x≤b.Then a b f(x)dx=− b a 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 deﬁnite integrals: (a) 4 2 (x−1)dx(b) 2 4 (x−1)dx SOLUTION : (a) 4 2 (x−1)dx= x 2 2−x 4 2 =(8−4)−(2−2) =4, which is positive, since the region is above thex-axis. (b) 2 4 (x−1)dx= x 2 2−x 2 4 =(2−2)−(8−4) x y 24 −11 3 =−4, which is the opposite of part (a), because the integral runs backwards from right to left, fromx=4 tox=2. Sums of Functions: When two functions are added, the two regions are piled on top of each other, so that: 10 INTEGRAL OF A SUM : b a f(x)+g(x) dx= b a f(x)dx+ b a g(x)dx WORKED EXERCISE : Evaluate these two expressions and show that they are equal: (a) 1 0 (x 2+x+1)dx(b) 1 0 x2dx+ 1 0 xdx+ 1 0 1dx SOLUTION : (a) 1 0 (x 2+x+1)dx= x 3 3+x 2 2+x 1 0 =( 13+ 12+1)−(0+0+0) =1 56.

16 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 (b) 1 0 x2dx+ 1 0 xdx+ 1 0 1dx= x 3 3 1 0 + x 2 2 1 0 + x 1 0 =( 13−0) + ( 12−0) + (1−0) =1 56,the same as in part (a). Multiples of Functions: Similarly, when a function is multiplied by a constant, the region is expanded vertically by that constant, so that: 11 INTEGRAL OF A MULTIPLE : b a kf(x)dx=k b a f(x)dx WORKED EXERCISE : Evaluate these two expressions and show that they are equal: (a) 3 1 10x 3dx(b)10 3 1 x3dx SOLUTION : (a) 3 1 10x 3dx= 10x 4 4 3 1 =810 4−10 4 =800 4 = 200.(b) 10 3 1 x3dx=10× x 4 4 3 1 =10× 81 4−1 4 =10×80 4 = 200. Inequalities with Deﬁnite Integrals: Suppose that a curvey=f(x) is always under- neath another curvey=g(x)inanintervala≤x≤b. Then the area under the curvey=f(x)fromx=atox=bmust be less than the area under the curve y=g(x). In the language of deﬁnite integrals: 12 INEQUALITY :Iff(x)≤g(x)intheintervala≤x≤b,then b a f(x)dx≤ b a g(x)dx. WORKED EXERCISE : (a)Sketch the graph off(x)=4−x 2,for−2≤x≤2. (b)Explain why 0≤ 2 −2 (4−x 2)dx≤16. x y −2 24 y= 4 SOLUTION : (a) The parabola and line are sketched opposite. (b) Clearly 0≤4−x 2≤4overtheinterval−2≤x≤2. Hence the region associated with the integral is inside the square of side length 4 in the diagram opposite.

CHAPTER 1: Integration 1C The Deﬁnite Integral and its Properties 17 Exercise1C Technology:All the properties of the deﬁnite integral discussed in this section have been justiﬁed 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 13 deal with these properties. The simpliﬁcation of integrals of odd and even functions is particularly important and is easily demonstrated visually by curve-sketching programs. 1.Evaluate the following deﬁnite integrals, using the fundamental theorem: (a) 0 −2 2xdx (b) 1 −2 6xdx (c) 2 −1 4x 3dx(d) 1 −1 6x 5dx (e) 0 −3 3x 2dx (f ) 0 −3 x2dx(g) 4 −1 10x 4dx (h) −2 −3 x3dx (i) 2 −2 x7dx 2.Evaluate the following deﬁnite integrals, using the fundamental theorem: (a) 2 −3 (1 + 4x)dx (b) 0 −2 (3x 2−5)dx (c) 1 −1 (7−4x 3)dx (d) 2 0 (2x−4x 3)dx(e) 1 −1 (6x 2−8x)dx (f ) 6 0 (x 2−6x)dx (g) 1 −1 (x 3−x)dx (h) 3 0 (4x 3−2x 2)dx(i) 10 −2 (12−3x)dx (j) 3 1 (3x 2−5x 4−10x)dx (k) −1 −3 (1−x−x 2)dx (l) 2 −2 (7−2x+x 4)dx DEVELOPMENT 3.By expanding the brackets where necessary, evaluate the following deﬁnite integrals: (a) 3 1 3x(x−4)dx (b) 1 −1 (3x−1)(3x+1)dx (c) 0 −2 x2(6x 3+5x 2+4x+3)dx(d) 2 0 x(1−x)dx (e) 2 −2 (2−x)(1 +x)dx (f ) 5 0 x(x+1)(x−1)dx 4.By dividing through by the denominator, evaluate the following deﬁnite integrals: (a) −1 −2 2x 2−5x xdx(b) −1 −3 3x 3+7x xdx(c) 3 2x2−6x 3 x2 dx 5.Find the value ofkif: (a) 3 k 2dx=4 (b) 8 k 3dx=12 (c) 3 2 (k−3)dx=5(d) k 3 (x−3)dx=0 (e) k 1 (x+1)dx=6 (f ) k 1 (k+3x)dx= 132

18 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 6.Evaluate each group of deﬁnite integrals and use the properties of the deﬁnite integral to explain the relationships within each group: (a)(i) 2 0 (3x 2−1)dx(ii) 0 2 (3x 2−1)dx (b)(i) 1 0 20x 3dx(ii) 20 1 0 x3dx (c)(i) 4 1 (4x+5)dx(ii) 4 1 4xdx(iii) 4 1 5dx (d)(i) 2 0 12x 3dx(ii) 1 0 12x 3dx(iii) 2 1 12x 3dx (e)(i) 3 3 (4−3x 2)dx(ii) −2 −2 (4−3x 2)dx 7.Without ﬁnding a primitive, use the properties of the deﬁnite integral to evaluate the following, stating reasons: (a) 3 3 9−x 2dx (b) 4 4 (x 3−3x 2+5x−7)dx(c) 1 −1 x3dx (d) 5 −5 (x 3−25x)dx(e) 90◦ −90 ◦sinxdx (f ) 2 −2 x 1+x 2dx 8.(a)On one set of axes sketchy=x 2andy=x 3, clearly showing the point of intersection. (b)Hence explain why 00

CHAPTER 1: Integration1D The Indeﬁnite Integral 19 1DThe Indeﬁnite Integral Now that primitives have been established as the key to calculating deﬁnite inte- grals, this section turns again to the task of ﬁnding primitives. First, a new and convenient notation for the primitive is introduced. The Indeﬁnite Integral: Because of the close connection established by the fundamental theorem between primitives and deﬁnite integrals, the termindeﬁnite integralis often used for the primitive. The usual notation for the primitive of a function f(x) is an integral sign without any upper or lower bounds. For example, the primitive or indeﬁnite integral ofx 2+1 is (x 2+1)dx=x 3 3+x+C,for some constantC. The word ‘indeﬁnite’ implies that the integral cannot be evaluated further because no bounds for the integral have yet been speciﬁed. A deﬁnite integral ends up as a pure number. An indeﬁnite integral, on the other hand, is a function ofx— the pronumeralxis carried across to the answer. The constant is called a ‘constant of integration’ and is an important part of the answer. Despite being a nuisance to write down every time, it must always be included. In most problems other than deﬁnite integrals, it will not be zero. Standard Forms for Integration: The rules for ﬁnding 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 thatn =−1. Then x ndx=x n+1 n+1+C,for some constantC. (ax+b) ndx=(ax+b) n+1 a(n+1)+C,for some constantC. The word ‘integration’ is commonly used to refer to both the ﬁnding of a primitive and the evaluating of a deﬁnite integral. Note:Strictly speaking, the words ‘for some constantC’or‘whereCis a constant’ should follow the ﬁrst mention of the new pronumeralC, because no pronumeral should be used without having been formally introduced. There is a limit to one’s patience, however, and usually in this situation it is quite clear thatCis the constant of integration. If another pronumeral such asDis used, itwouldbewisetointroduceitformally. WORKED EXERCISE : Use the standard form x ndx=x n+1 n+1+Cto ﬁnd: (a) 9dx(b) 12x 3dx SOLUTION : (a) 9dx=9x+C,for some constantC

20 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 Note:We know that 9xis the primitive of 9, becaused dx(9x)=9. But the formula still gives the correct answer, because 9 = 9x 0, and so increasing the index to 1 and dividing by this new index 1, 9x 0dx=9x 1 1+C,for some constantC =9x+C. (b) 12x 3dx=12×x 4 4+C,for some constantC =3x 4 +C WORKED EXERCISE : Use the standard form (ax+b) ndx=(ax+b) n+1 a(n+1)+Cto ﬁnd: (a) (3x+1) 5dx(b) (5−2x) 2dx SOLUTION : (a) (3x+1) 5dx=(3x+1) 6 3×6+C(Herea=3 andb=1.) = 118(3x+1) 6+C (b) (5−2x) 2dx=(5−2x) 3 (−2)×3+C(Herea=−2andb=5.) =− 16(5−2x) 3+C Negative Indices: Both standard forms apply with negative indices as well as positive indices, as in the following worked exercise. WORKED EXERCISE : Use negative indices to ﬁnd the indeﬁnite integrals: (a) 12 x3dx (b) dx (3x+4) 2 SOLUTION : (a) 12 x3dx= 12x −3 dx(Rewrite1 x3asx −3 before integrating.) =12×x −2 −2+C(Increase the index to−2 and then divide by−2.) =−6 x2 +C(Rewritex −2 as1 x2.) (b) dx (3x+4) 2= (3x+4) −2 dx(Rewrite1 (3x+4) 2as (3x+4) −2 .) =(3x+4) −1 3×(−1)+C(Herea=3 andb=4.) =−1 3(3x+4)+C(Rewrite (3x+4) −1 as1 3x+4.)

CHAPTER 1: Integration1D The Indeﬁnite Integral 21 Special Expansions: In many integrals, brackets must be expanded before the in- deﬁnite integral can be found. The following worked exercises use the special expansions; the second also requires negative indices. WORKED EXERCISE :Find these indeﬁnite integrals: (a) (x 3−1) 2dx (b) 3−1 x2 3+1 x2 dx SOLUTION : (a) (x 3−1) 2dx= (x 6−2x 3+1)dx(Use (A+B) 2=A 2+2AB+B 2.) =x 7 7−x 4 2+x+C (b) 3−1 x2 3+1 x2 dx= 9−1 x4 dx(Use (A−B)(A+B)=A 2−B 2.) = 9−x −4 dx(Use1 x4=x −4 .) =9x−x −3 −3+C =9x+1 3x 3 +C Fractional Indices: The standard forms for ﬁnding primitives of powers also apply to fractional indices. These calculations require quick conversions between fractional indices and surds. WORKED EXERCISE :Use fractional and negative indices to evaluate: (a) 4 1√xdx(b) 4 1 1√xdx SOLUTION : (a) 4 1√xdx= 4 1 x12dx(Rewrite√ xasx 12before ﬁnding the primitive.) = 23 x 32 4 1 (Increase the index to 32and divide by 32.) = 23×(8−1) (Note that 4 32=2 3=8 and 1 32=1.) =4 23 (b) 4 1 1√xdx= 4 1 x−12dx(Rewrite1 √xasx −12before ﬁnding the primitive.) = 21 x 12 4 1 (Increase the index to 12and divide by 12.) =2×(2−1) (Note that 4 12=√ 4=2 and 1 12=1.) =2

22 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 WORKED EXERCISE : (a)Use index notation to express1 √9−2xas a power of 9−2x. (b)Hence ﬁnd the indeﬁnite integral dx √9−2x. SOLUTION : (a)1 √9−2x=(9−2x) −12.(Use1 √u=u −12.) (b) Hence 1 √9−2x= (9−2x) −12dx =(9−2x) 12 −2× 12 +C(Use (ax+b) ndx=(ax+b) n+1 a(n+1).) =−√ 9−2x+C(Useu 12=√ u.) Running a Chain-Rule Differentiation Backwards: Finding primitives is the reverse pro- cess of diﬀerentiation. Thus once any diﬀerentiation has been performed, the process can then be reversed to give a primitive. WORKED EXERCISE :[These questions are always diﬃcult.] (a)Diﬀerentiate (x 2+1) 4. (b)Hence ﬁnd a primitive of 8x(x 2+1) 3. SOLUTION : (a) Lety=(x 2+1) 4. Thendy dx=dy du×du dx =4(x 2+1) 3×2x =8x(x 2+1) 3. Letu=x 2+1. Theny=u 4. Hencedu dx=2x anddy du=4u 3. (b) Henced dx(x 2+1) 4=8x(x 2+1) 3. Reversing this, 8x(x 2+1) 3dx=(x 2+1) 4+C,for some constantC. Note:Questions in the 2 Unit course would never ask for such an integral without ﬁrst asking for the appropriate derivative. Exercise1D Technology:Many programs that can perform algebraic manipulation are also able to deal with indeﬁnite integrals. They can be used to check the questions in this exercise and to investigate the patterns arising in such calculations. 1.Find the following indeﬁnite integrals: (a) 4dx (b) 1dx(c) 0dx (d) (−2)dx(e) xdx (f ) x 2dx(g) x 3dx (h) x 7dx

CHAPTER 1: Integration1D The Indeﬁnite Integral 23 2.Find the indeﬁnite integral of each function. Use the notation of the previous question. (a)2x (b)4x(c)3x 2 (d)4x 3 (e)10x 9 (f )2x 3 (g)4x 5 (h)3x 8 3.Find the following indeﬁnite integrals: (a) (x+x 2)dx (b) (x 4−x 3)dx (c) (x 7+x 10)dx(d) (2x+5x 4)dx (e) (9x 8−11)dx (f ) (7x 13 +3x 8)dx(g) (4−3x)dx (h) (1−x 2+x 4)dx (i) (3x 2−8x 3+7x 4)dx 4.Find the indeﬁnite integral of each function. (Leave negative indices in your answers.) (a)x −2 (b)x −3 (c)x −8 (d)3x −4 (e)9x −10 (f )10x −6 5.Find the following indeﬁnite integrals. (Leave fractional indices in your answers.) (a) x 12dx (b) x 13dx(c) x 14dx (d) x 23dx(e) x −12dx (f ) 4x 12dx DEVELOPMENT 6.By expanding brackets where necessary, ﬁnd the following indeﬁnite integrals: (a) x(x+2)dx (b) x(4−x 2)dx (c) x 2(5−3x)dx(d) x 3(x−5)dx (e) (x−3) 2dx (f ) (2x+1) 2dx(g) (1−x 2)2dx (h) (2−3x)(2 + 3x)dx (i) (x 2−3)(1−2x)dx 7.By dividing through by the denominator, perform the following integrations: (a) x 2+2x xdx(b) x 7+x 8 x6 dx(c) 2x 3−x 4 4xdx 8.Write each of these functions with negative indices and ﬁnd its indeﬁnite integral: (a)1 x2 (b)1 x3 (c)1 x5 (d)1 x10 (e)3 x4 (f )5 x6 (g)7 x8 (h)1 3x 2 (i)1 7x 5 (j)−1 5x 3 (k)1 x2−1 x5 (l)1 x3+1 x4 9.Write these functions with fractional indices and hence ﬁnd their indeﬁnite integrals: (a)√ x(b) 3√x(c)1 √x(d) 3√ x2 10.Use the indeﬁnite integrals of the previous question to evaluate: (a) 9 0√xdx(b) 8 03√xdx(c) 49 25 1√xdx(d) 1 03√ x2dx

24 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 11.By using the formula (ax+b) ndx=(ax+b) n+1 a(n+1)+C, ﬁnd: (a) (x+1) 5dx (b) (x+2) 3dx (c) (4−x) 4dx (d) (3−x) 2dx(e) (3x+1) 4dx (f ) (4x−3) 7dx (g) (5−2x) 6dx (h) (1−5x) 7dx(i) (2x+9) 11 dx (j) 3(2x−1) 10 dx (k) 4(5x−4) 6dx (l) 7(3−2x) 3dx 12.By using the formula (ax+b) ndx=(ax+b) n+1 a(n+1)+C, ﬁnd: (a) ( 13x−7) 4dx(b) ( 14x−7) 6dx(c) (1− 15x) 3dx 13.By using the formula (ax+b) ndx=(ax+b) n+1 a(n+1)+C, ﬁnd: (a) 1 (x+1) 3dx (b) 1 (x−5) 4dx (c) 1 (3x−4) 2dx(d) 1 (2−x) 5dx (e) 3 (x−7) 6dx (f ) 8 (4x+1) 5dx(g) 2 (3−5x) 4dx (h) 4 5(1−4x) 2dx (i) 7 8(3x+2) 5dx 14.By expanding the brackets, ﬁnd: (a) √ x 3√ x−x dx(b) √ x−2 √ x+2 dx(c) 2√ x−1 2dx 15.(a)Evaluate the following deﬁnite integrals: (i) 1 0 x12dx(ii) 4 1 x−12dx(iii) 8 0 x13dx (b)By writing them with fractional indices, evaluate the following deﬁnite integrals: (i) 4 0√xdx(ii) 9 1 x√ xdx(iii) 9 1dx√x 16.By expanding the brackets where necessary, ﬁnd: (a) 4 2 2−√ x 2+√ x dx(b) 1 0√x √ x−4 dx(c) 9 4 √ x−1 2dx CHALLENGE 17.Explain why the indeﬁnite integral 1 xdxcan’t be found in the usual way using the standard form x ndx=x n+1 n+1+C. 18.Find each of the following indeﬁnite integrals: (a) √ 2x−1dx (b) √ 7−4xdx(c) 3√4x−1dx (d) 1 √3x+5dx

CHAPTER 1: Integration 1E Finding Areas by Integration 25 19.Evaluate the following: (a) 2 0 (x+1) 4dx (b) 3 2 (2x−5) 3dx (c) 2 −2 (1−x) 5dx(d) 5 0 1−x 5 4 dx (e) 1 0√9−8xdx (f ) 7 2 dx √x+2(g) 0 −2 3√x+1dx (h) 5 1√3x+1dx (i) 0 −3√1−5xdx 20.(a) (i) Findd dx(x 2+1) 5.(ii) Hence ﬁnd 10x(x 2+1) 4dx. (b) (i) Findd dx(x 3+1) 4.(ii) Hence ﬁnd 12x 2(x 3+1) 3dx. (c) (i) Findd dx(x 5−7) 8.(ii) Hence ﬁnd 40x 4(x 5−7) 7dx. (d) (i) Findd dx(x 2+x) 6.(ii) Hence ﬁnd 6(2x+1)(x 2+x) 5dx. 21.(a) (i) Findd dx(x 2−3) 5.(ii) Hence ﬁnd x(x 2−3) 4dx. (b) (i) Findd dx(x 3+1) 11.(ii) Hence ﬁnd x 2(x 3+1) 10 dx. (c) (i) Findd dx(x 4+8) 7.(ii) Hence ﬁnd x 3(x 4+8) 6dx. (d) (i) Findd dx(x 2+2x) 3.(ii) Hence ﬁnd (x+1)(x 2+2x) 2dx. 1EFinding Areas by Integration The aim of this section and the next is to use deﬁnite integrals to ﬁnd the areas of regions bounded by curves, lines and the coordinate axes. Area and the Deﬁnite Integral: A deﬁnite integral is a pure number, which can be positive or negative — remember that a deﬁnite integral representing a region below thex-axis is negative in value. An area has units (called ‘square units’ or u 2in the absence of any physical interpretation) and cannot be negative. Any problem on areas requires some care when ﬁnding the correct integral or combination of integrals required. Some particular techniques are listed below, but the general rule is to draw a diagram ﬁrst to see which bits need to be added or subtracted. 14 FINDING AN AREA : When using integrals to ﬁnd the area of a region: 1. Draw a sketch of the curves, showing relevant intercepts and intersections. 2. Evaluate the necessary deﬁnite integral or deﬁnite integrals. 3. Write a conclusion, giving the required area in square units. Areas Above the x-axis: When a region lies entirely above thex-axis, the relevant integral will be positive and the area will be equal to the integral, apart from needing units.

26 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 WORKED EXERCISE : Find the area of the region bounded by the curvey=4−x 2and thex-axis. (This was the example given on page 1 in the introduction to the chapter.) SOLUTION : The curve meets thex-axis at (2,0) and (−2,0). The region lies entirely above thex-axis and the relevant integral is 2 −2 (4−x 2)dx= 4x−x 3 3 2 −2 =(8− 83)−(−8+ 83) =5 13−(−5 13) =10 23, which is positive because the region lies above thex-axis. x y y= 4 −x 2 −2 24 Hence the required area is 10 23square units. Areas Below the x-axis: When a region lies entirely below thex-axis, the relevant integral will be negative and the area will then be the opposite of this. WORKED EXERCISE : Find the area of the region bounded by the curvey=x 2−1andthex-axis. SOLUTION : The curve meets thex-axis at (1,0) and (−1,0). The region lies entirely below thex-axis and the relevant integral is 1 −1 (x 2−1)dx= x 3 3−x 1 −1 =( 13−1)−(− 13+1) =− 23− 23 =−1 13, which is negative, because the region lies below thex-axis. x y −1 1 −1 Hence the required area is 1 13square units. Areas Above and Below the x-axis: When a curve crosses thex-axis, the area of the region between the curve and thex-axis cannot usually be found by means of a single integral. This is because integrals representing regions below thex-axis have negative values. WORKED EXERCISE : (a)Sketch the cubic curvey=x(x+1)(x−2), showing thex-intercepts. (b)Shade the region enclosed between thex-axis and the curve, and ﬁnd its area. [Hint:The expansion of the function isy=x(x+1)(x−2) =x(x 2−x−2) =x 3−x 2−2x.] (c)Find 2 −1 x(x+1)(x−2)dxand explain why this integral does not represent the area of the region described in part (b).

CHAPTER 1: Integration 1E Finding Areas by Integration 27 SOLUTION : (a) The curve hasx-interceptsx=−1,x=0 andx=2andisgraphedbelow. (b) For the region above thex-axis, 0 −1 (x 3−x 2−2x)dx= x 4 4−x 3 3−x 2 0 −1 =(0−0−0)−( 14+ 13−1) x y −1 2 = 512, and so area above = 512 square units. For the region below thex-axis, 2 0 (x 3−x 2−2x)dx= x 4 4−x 3 3−x 2 2 0 =(4−2 23−4)−(0−0−0) =−2 23, and so area below = 2 23square units. Adding these, total area = 512 +2 23 =3 112 square units. (c) 2 −1 x(x+1)(x−2)dx= x 4 4−x 3 3−x 2 2 −1 =(4−2 23−4)−( 14+ 13−1) =−2 23+ 512 =−2 14. This integral represents the area fromx=−1tox=0 above thex-axis, minus, rather than plus, the area fromx=0 tox=2 below thex-axis. Areas Associated with Odd and Even Functions: As always, these calculations are often much easier if symmetries can be recognised. WORKED EXERCISE : Find the area between the curvey=x 3−xand thex-axis. SOLUTION : Factoring,y=x(x 2−1) x y −1 1 =x(x−1)(x+1), and so thex-intercepts arex=−1,x=0 andx=1. The two shaded regions have equal areas, since the function is odd. First, 1 0 (x 3−x)dx= x 4 4−x 2 2 1 0 =( 14− 12)−(0−0) =− 14, so area below thex-axis = 14square units. Doubling, total area = 12square units.

28 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 Area Between a Graph and the y-axis: Integration with respect toyrather thanxcan often give a result more quickly without the need for subtraction. Whenxis a function ofy, the deﬁnite integral with respect toyrepresents the area of the region between the curve and they-axis, except that areas of regions to the left of they-axis are subtracted rather than added. The limits of integration are values ofyrather than ofx. 15 THE DEFINITE INTEGRAL AND INTEGRATION WITH RESPECT TO y: Letxbe a continuous function ofyin some closed intervala≤y≤b. Then the deﬁnite integral b a xdyis the sum of the areas to the right of they-axis, fromy=atoy=b, minus the sum of the areas to the left of they-axis. WORKED EXERCISE : (a)Sketch the linesy=x+1 andy= 5 and shade the region between these lines to the right of they-axis. (b)Use integration with respect toyto ﬁnd the area of this region. (c)Conﬁrm the result by mensuration. SOLUTION : (a) The lines are sketched below. They meet at (4,5). (b) The given equation isy=x+1. Solving forx,x=y−1, and the required integral is 5 1 (y−1)dy= y 2 2−y 5 1 = 25 2−5 − 1 2−1 =7 12−(− 12) =8, x y 1 5 which is positive, since the region is to the right of they-axis. Hence the required area is 8 square units. (c) By mensuration, area = 12×base×height = 12×4×4 = 8 square units. WORKED EXERCISE : The curve in the diagram below is the cubicy=x 3. Use integration with respect toyto ﬁnd: (a)the areas of the shaded regions to the right and left of they-axis, (b)the total area of the two shaded regions.

CHAPTER 1: Integration 1E Finding Areas by Integration 29 SOLUTION : (a) The given equation isy=x 3. Solving forx,x 3=y x=y 13. For the region to the right of they-axis, 8 0 y13dy= 34 y 43 8 0 = 34×(16−0) (Note that 8 43=2 4= 16.) =12, x y −1 8 so area = 12 square units. For the region to the left of they-axis, 0 −1 y13dy= 34 y 43 0 −1 = 34×(0−1) (Note that (−1) 43=(−1) 4=1.) =− 34, so area = 34square units. (b) Adding these, total area = 12 34square units. Exercise1E Technology:Any curve-sketching program will help in identifying the deﬁnite inte- grals that need to be evaluated to ﬁnd the area of a given region. Programs that can approximate areas of regions on the screen graph can demonstrate how the ﬁnal area is built up from the separate pieces. 1.Find the area of each shaded region below by evaluating the appropriate integral: x y yx= 2 2 (a) x y y= 3x 2 13 (b) x y yx= 4 3 3 (c) x y y= 3x+ 1 2 −1 21 (d) x y 3 y=x 2 (e) x y 24 yx=− 2x 2 (f ) x y 16 y= √x (g) x y 5 5 3 1yx= 5 − (h)

30 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 x y −1 yx=− 3 x (i) x y 3 −4 yx= 12 −− 2x (j) x y −1 21 yx= 5 + 1 4 (k) x y 127 y= √x 3 (l) 2.Find the area of each shaded region below by evaluating the appropriate integral: x y xy= 2 5 (a) x y x= 3y 2 −2 (b) x y xy= 2 − 4 2 4 (c) x y x= 27 − 3y 2 27 3 −3 (d) x y 3 xy= (e) x y 3 5 xy= 2+ 1 (f ) x y 9 x= √y (g) x y 4 1 √y x= 1 (h) 3.Find the area of each shaded region below by evaluating the appropriate integral: x y 3 1 3 yx=− 4x+ 3 2 (a) x y −3 y= 3x (b) x y −3yx= 3 (c) x y 13 1 yx= 1 − 4 (d) 4.Find the area of each shaded region below by evaluating the appropriate integral: x y 1 4 xy= 1 − (a) x y 8 2 4 xy=− 6y+ 8 2 (b) x y −1 −8 x= √y 3 (c) x y 3 xy= − 2 (d)

CHAPTER 1: Integration 1E Finding Areas by Integration 31 DEVELOPMENT x y −3 −1 2 5.The sketch shows the liney=x+1. (a)Copy the diagram and then shade the region bounded byy=x+1, thex-axis and the linesx=−3andx=2. (b)By evaluating 2 −1 (x+1)dx, ﬁnd the area of the shaded region above thex-axis. (c)By evaluating −1 −3 (x+1)dx, ﬁnd the area of the shaded region below thex-axis. (d)Hence ﬁnd the area of the entire shaded region. (e)Find 2 −3 (x+1)dx, and explain why this integral does not give the area of the shaded region. x y 2 −3 1 6.The sketch shows the curvey=(x−1)(x+3) =x 2+2x−3. (a)Copy the diagram and shade the region bounded by the curvey=(x−1)(x+ 3), thex-axis and the linex=2. (b)By evaluating 1 −3 (x 2+2x−3)dx, ﬁnd the area of the shaded region below thex-axis. (c)By evaluating 2 1 (x 2+2x−3)dx, ﬁnd the area of the shaded region above thex-axis. (d)Hence ﬁnd the area of the entire shaded region. (e)Find 2 −3 (x 2+2x−3)dx, and explain why this integral does not give the area of the shaded region. x y 2 −1 7.The sketch shows the curvey=x(x+1)(x−2) =x 3−x 2−2x. (a)Copy the diagram and shade the region bounded by the curve and thex-axis. (b)By evaluating 2 0 (x 3−x 2−2x)dx, ﬁnd the area of the shaded region below thex-axis. (c)By evaluating 0 −1 (x 3−x 2−2x)dx, ﬁnd the area of the shaded region above thex-axis. (d)Hence ﬁnd the area of the entire region you have shaded. (e)Find 2 −1 (x 3−x 2−2x)dx, and explain why this integral does not give the area of the shaded region. 8.In each part below, ﬁnd the area of the region bounded by the graph of the given function and thex-axis between the speciﬁed values. You should draw a diagram for each part and check to see whether the region is above or below thex-axis. (a)y=x 2, betweenx=−3andx=2 (b)y=2x 3, betweenx=−4andx=1

32 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 (c)y=3x(x−2), betweenx=0 andx=2 (d)y=x−3, betweenx=−1andx=4 (e)y=(x−1)(x+3)(x−2), betweenx=−3andx=2 (f )y=−2x(x+ 1), betweenx=−2andx=2 (g)y=x(3−x) 2, betweenx=0 andx=3 (h)y=x 4−4x 2, betweenx=−5andx=0 9.In each part below, ﬁnd the area of the region bounded by the graph of the given function and they-axis between the speciﬁed values. You should draw a diagram for each part and check whether the region is to the right or left of they-axis. (a)x=y−5, betweeny=0 andy=6 (b)x=3−y, betweeny=2 andy=5 (c)x=y 2, betweeny=−1andy=3 (d)x=(y−1)(y+ 1), betweeny=3 andy=0 10.In each part below you should draw a graph and look carefully for any symmetries that will simplify the calculation. (a)Find the area of the region bounded by the curve and thex-axis: (i)y=x 7,for−2≤x≤2 (ii)y=x 3−16x=x(x−4)(x+4), for−4≤x≤4 (iii)y=x 4−9x 2=x 2(x−3)(x+3), for−3≤x≤3 (b)Find the area of the region bounded by the curve and they-axis: (i)x=2y,for−5≤y≤5 (ii)x=y 2,for−3≤y≤3 (iii)x=4−y 2=(2−y)(2 +y), for−2≤y≤2 11.Find the area of the region bounded byy=|x+2|and thex-axis, for−2≤x≤2. CHALLENGE x y 12.The diagram shows a graph ofy 2= 16(2−x). (a) Find thex-intercept and they-intercepts. (b) Find the area of the shaded region: (i) by considering the region between the curve y=4√ 2−xand thex-axis, (ii) by considering the region between the curve x=2− 116y2and they-axis. 13.The gradient of a curve isy =x 2−4x+ 3 and the curve passes through the origin. (a) Find the equation of the curve. (b) Show that the curve’s turning points are (1,1 13)and(3,0), and sketch its graph. (c) Find the area of the region enclosed between the curve and thex-axis between the two turning points. 14.Sketchy=x 2and mark the pointsA(a, a 2),B(−a, a 2),P(a,0) andQ(−a,0). (a) Show that a 0 x2dx= 23(area ΔOAP). (b) Show that a −a x2dx= 13(area of rectangleAB QP).

CHAPTER 1: Integration 1F Areas of Compound Regions 33 1FAreas of Compound Regions When a region is bounded by two or more diﬀerent curves, some dissection process is 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 diﬀerent parts of thex-axis. WORKED EXERCISE : (a)Sketch the curvesy=x 2andy=(x−2) 2on one set of axes. (b)Shade the region bounded byy=x 2,y=(x−2) 2and thex-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 thex-axis, but it will be necessary to ﬁnd sepa- rately the areas of the regions to the left and right ofx=1. (c) First, 1 0 x2dx= x 3 3 1 0 = 13. Secondly, 2 1 (x−2) 2dx= (x−2) 3 3 2 1 x y 2 1 1 =0−(− 13) = 13. Combining these, area = 13+ 13 = 23square units. Areas of Regions Between Curves: Suppose that one curvey=f(x) is always b elow another curvey=g(x)inanintervala≤x≤b. Then the area of the region between the curves fromx=atox=bcan be found by subtraction. 16 AREA BETWEEN CURVES :Iff(x)≤g(x)intheintervala≤x≤b,then area between the curves = b a g(x)−f(x) dx. That is, take the integral of the top curve minus the bottom curve. The assumption thatf(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 where diﬀerent curves are on top will begin to cancel each other out.

34 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 WORKED EXERCISE : (a)Find the two points where the curvey=(x−2) 2meets the liney=x. (b)Draw a sketch and shade the area of the region between these two graphs. (c)Find the shaded area. SOLUTION : (a) Substitutingy=xintoy=(x−2) 2gives (x−2) 2=x x 2−4x+4 =x x 2−5x+4 = 0 (x−1)(x−4) = 0, x=1 or 4, x y 12 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 = 4 1 x−(x−2) 2 dx = 4 1 x−(x 2−4x+4) dx = 4 1 (−x 2+5x−4)dx = −x 3 3+5x 2 2−4x 4 1 =(−21 13+40−16)−(− 13+2 12−4) =2 23+1 56 =4 12square units. Note:The formula (given in Box 16 on the previous page) for the area of the region between two curves holds even if the region crosses thex-axis. To illustrate this point, the next example is the previous example shifted down 2 units so that the region between the line and the parabola crosses thex-axis. The area of course remains the same — and notice how the formula still gives the correct answer. WORKED EXERCISE : (a)Find the two points where the curvesy=x 2−4x+2 andy=x−2 meet. (b)Draw a sketch and ﬁnd the area of the region between these two curves. SOLUTION : (a) Substitutingy=x−2intoy=x 2−4x+2 gives x 2−4x+2 =x−2 x 2−5x+4 = 0 (x−1)(x−4) = 0 x=1 or 4. x y 12 4 −12 so the two graphs intersect at (1,−1) and (4,2).

CHAPTER 1: Integration 1F Areas of Compound Regions 35 (b) Again, the line is above the parabola. Hence area = 4 1 (x−2)−(x 2−4x+2) dx = 4 1 (−x 2+5x−4)dx = −x 3 3+5x 2 2−4x 4 1 =(−21 13+40−16)−(− 13+2 12−4) =4 12square units. Areas of Regions Between Curves that Cross: Now suppose that one curvey=f(x) is sometimes above and sometimes below another curvey=g(x)intherelevant interval. In this case, separate integrals will need to be calculated. WORKED EXERCISE : The diagram below shows the curves y=−x 2+4x−4andy=x 2−8x+12 meeting 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 ﬁrst. Hence area = 2 0 (x 2−8x+ 12)−(−x 2+4x−4) dx = 2 0 (2x 2−12x+ 16)dx = 2x 3 3−6x 2+16x 2 0 =5 13−24 + 32 x y 12 −424 6 =13 13square units. In the right-hand region, the ?rst curve is above the second. Hence area = 4 2 (−x 2+4x−4)−(x 2−8x+ 12) dx = 4 2 (−2x 2+12x−16)dx = −2x 3 3+6x 2−16x 4 2 =(−42 23+96−64)−(−5 13+24−32) =−10 23+13 13 =2 23square units. Hence total area = 13 13+2 23 = 16 square units.

36 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 Exercise1F Technology:Screen graphing programs are particularly useful with compound regions because they allow the separate parts of the region to be identiﬁed clearly. 1.Find the area of the shaded region in each diagram below. ()1 1, x y y=x 2 yx= (a) ()1 1, x y yx= 3 yx= (b) ()1 1, x y yx= yx= 4 (c) x y yx= 2 yx= 3 (1,1) (d) ()1 1, x y yx= 6 yx= 4 (e) x y yx= 3 + 4 yx= 2 (4,16) (−1,1) (f ) x y yx= 9 − 2 yx=+ 1 2 (2,5) (−4,17) (g) ()−3 1,()2 6, x y y= 10 − 2xy=x+ 4 (h) 2.By considering regions between the curves and they-axis, ﬁnd the area of the shaded region in each diagram below. ()4 2, x y x=y 2 xy= 2 (a) x y xy= 3 − 2 xy= 2 (1,1)(4,2) (b) x y 1 4 xy= 4 − xyy= 5 − − 4 2 (2,2) (c) ()3 −1,()6 2, x y x=y 2+ 2 y=− 4x (d) 3.Find the areas of the shaded region in the diagrams below. In each case you will need to ﬁnd two areas and subtract one from the other. x y 6 2 3 4 yx= 6 − − 8x 2 (a) x y −1 −2 12 yx= 1 − 2 yx= 4 − 2 (b)

CHAPTER 1: Integration 1F Areas of Compound Regions 37 4.Find the areas of the shaded regions in the diagrams below. In each case you will need to ﬁnd two areas and add them. yx= ( − 2) 2 yx= ( + 2) 2 x y −2 2 (a) x y 3 yx= 2 yx= ( − 3) 2 3 29 4 ( , ) (b) 5.(a)By solving the equations simultaneously, show that the curvesy=x 2+4 andy=x+6 intersect 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)−(x 2+4) dx= 2 −1 (x−x 2+2)dx and evaluate the integral. 6.(a)By solving the equations simultaneously, show that the curvesy=3x−x 2=x(3−x) andy=xintersect at the points (0,0) and (2,2). (b)Sketch the curves on the same diagram and shade the region enclosed between them. (c)Show that this region has area 2 0 (3x−x 2−x)dx= 2 0 (2x−x 2)dx and evaluate the integral. 7.(a)By solving the equations simultaneously, show that the curvesy=(x−3) 2 and y=14−2xintersect at the points (−1,16) and (5,4). (b)Sketch the curves on the same diagram and shade the region enclosed between them. (c)Show that this region has area 5 −1 (14−2x)−(x−3) 2 dx= 5 −1 (4x+5−x 2)dx and evaluate the integral. DEVELOPMENT 8.Solve simultaneously the equations of each pair of curves below to ﬁnd their points of intersection. Sketch each pair of curves on the same diagram and shade the region enclosed between them. By evaluating the appropriate integral, ﬁnd the area of the shaded region in each case. (a)y=x 4 andy=x 2 (b)y=3x 2 andy=6x 3 (c)y=9−x 2 andy=3−x (d)y=x+10 andy=(x−3) 2+1 9.(a)By solving the equations simultaneously, show that the curvesy=x 2+2x−8and y=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.

38 CHAPTER 1: Integration CAMBRIDGE MATHEMATICS 2U NIT YEAR 12 (c)Despite the fact that it crosses thex-axis, the region has area given by 3 −3 (2x+1)−(x 2+2x−8) dx= 3 −3 (9−x 2)dx. Evaluate the integral and hence ﬁnd the area of the region enclosed between the curves. 10.(a)By solving the equations simultaneously, show that the curvesy=x 2−x−2and y=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 thex-axis, the region has area given by 2 0 (x−2)−(x 2−x−2) dx= 2 0 (2x−x 2)dx. Evaluate this integral and hence ﬁnd the area of the region between the curves. 11.Solve simultaneously the equations of each pair of curves below to ﬁnd their points of intersection. Sketch each pair of curves on the same diagram and shade the region enclosed between them. By evaluating the appropriate integral, ﬁnd the area of the shaded region in each case. (a)y=x 2−6x+5 andy=x−5 (b)y=−3xandy=4−x 2 (c)y=x 2−1andy=7−x 2 (d)y=xandy=x 3 12.Find the area bounded by the linesy= 14xandy=− 12xbetweenx=1 andx=4. 13.(a)On the same number plane, sketch the graphs of the functionsy=x 2andx=y 2, clearly indicating their points of intersection. Shade the region enclosed between them. (b)Explain why the area of this region is given by 1 0 √ x−x 2 dx. (c)Find the area of the region bounded by the two curves. CHALLENGE 14.Consider the functionx 2=8y. Tangents are drawn at the pointsA(4,2) andB(−4,2) and intersect on they-axis. (a) Draw a diagram of the situation and note the symmetry about they-axis. (b) Find the equation of the tangent at the pointA. (c) Find the area of the region bounded by the curve and the tangents. 15.(a) Show that the tangent toy=x 3at the point wherex=2 isy−12x+16 = 0. (b) Show, by substituting the point into each equation, that the tangent and the curve meet again at the point (−4,−64). (c) Find the area of the region enclosed between the curve and the tangent.

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