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Printed in China by Printplus National Library of Australia Cataloguing in Publication data Bill Pender Cambridge mathematics 2 unit year 12 Bill Pender CambridgeMATHS Extension 1 Year 11 covers all syllabus dotpoints for Year 11 of the Mathematics Extension

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Description

CAMBRIDGE Mathematics COLOUR WITH N O I S R E V M T CD-RO n N E D'U T S with a colour Now in ion of nic vers D'o r t c'le e k on C the boo

BILL PENDER DAVID SADLER JULIA SHEA DEREK WARD

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

au Information on this title: www

org/9780521177504 © Bill Pender,

David Sadler,

Julia Shea,

Derek Ward 2009 First edition 1999 Reprinted 2001,

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

For secondary school age

Mathematics

Mathematics--Problems,

Sadler,

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,

criticism or review) no part of this publication may be reproduced,

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,

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

How to Use This Book

xi Chapter One — Integration

Areas and the Definite Integral

The Fundamental Theorem of Calculus The Definite Integral and its Properties The Indefinite Integral

Finding Areas by Integration

Areas of Compound Regions

Volumes of Solids of Revolution

The Trapezoidal Rule

Simpson’s Rule

Chapter Review Exercise

Chapter Two — The Exponential Function

Review of Exponential Functions

The Exponential Function ex and the Definition of e Differentiation of Exponential Functions

Applications of Differentiation

Integration of Exponential Functions

Applications of Integration

Chapter Review Exercise

Chapter Three — The Logarithmic Function

Review of Logarithmic Functions

The Logarithmic Function Base e

Differentiation of Logarithmic Functions Applications of Differentiation of log x

Integration of the Reciprocal Function

Applications of Integration of 1/x

Calculus with Other Bases

Chapter Review Exercise

Contents

Chapter Four — The Trigonometric Functions

Radian Measure of Angle Size

Mensuration of Arcs,

Sectors and Segments

Graphs of the Trigonometric Functions in Radians The Behaviour of sin x Near the Origin

The Derivatives of the Trigonometric Functions

Applications of Differentiation

Integration of the Trigonometric Functions

Applications of Integration

Chapter Review Exercise

Chapter Five — Motion

Average Velocity and Speed

Velocity as a Derivative

Integrating with Respect to Time Chapter Review Exercise

Applications of APs and GPs

The Use of Logarithms with GPs

Simple and Compound Interest

Investing Money by Regular Instalments Paying Off a Loan

Rates of Change

Natural Growth and Decay

Chapter Review Exercise

Chapter Six — Rates and Finance 6A 6B 6C 6D 6E 6F 6G 6H

Chapter Seven — Euclidean Geometry

Points,

Parallels and Angles

Angles in Triangles and Polygons

Congruence and Special Triangles

Trapeziums and Parallelograms

Rhombuses,

Rectangles and Squares

Areas of Plane Figures

Pythagoras’ Theorem and its Converse Similarity

Intercepts on Transversals

Chapter Review Exercise

Chapter Eight — Probability

Probability and Sample Spaces

Sample Space Graphs and Tree Diagrams

Sets and Venn Diagrams

Venn Diagrams and the Addition Theorem

Multi-Stage Experiments and the Product Rule Probability Tree Diagrams

Chapter Review Exercise

Answers to Exercises

374 Index

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,

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 edition has been thoroughly rewritten to make it more accessible 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 project,

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 project has caused to family life

Dr Bill Pender Subject Master in Mathematics Sydney Grammar School College Street Darlinghurst NSW 2010

Julia Shea Director of Curriculum Newington College 200 Stanmore Road Stanmore NSW 2048

David Sadler Mathematics Sydney Grammar School

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 section 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 require 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 thoroughly 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,

definitions and results have been boxed and numbered 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

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,

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

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,

The Structure of the Course: Recent examination papers have made the interconnections amongst the various topics much clearer

Calculus is the backbone of the course,

and the two processes of differentiation and integration,

are the basis of most of the topics

Both processes are introduced as geometrical 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 property that the exponential function with base e is 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 hyperbola y = 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 of π,

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,

and the related numbers e and π,

can all be developed from the three most basic degree-2 curves — the parabola,

the rectangular hyperbola and the circle

In this way,

How to Use This Book

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 recent examinations,

is to encourage arguments that relate a curve to its equation

Algebraic arguments are constantly used to investigate graphs of functions

Conversely,

graphs are constantly used to solve algebraic problems

We have drawn as many sketches in the book as space allowed,

students 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 mathematical way using geometry and algebra,

its content is fundamental to all the sciences

In particular,

the applications of calculus to maximisation,

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 inappropriate

Any such ideas required in this course are not difficult to understand intuitively

Most arguments about limits need only the limit lim 1/x = 0 and x→∞ occasionally the sandwich principle

Introducing the tangent as the limit of the secant is a dramatic new idea,

clearly marking the beginning of calculus,

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 mathematics 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 Subject Master in Mathematics at Sydney Grammar School,

where he has taught since 1975

He has an MSc and PhD in Pure Mathematics 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,

He also does occasional solo work at various 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 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

A Brief History of Time

CHAPTER ONE

Integration 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 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 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 calculus and it will greatly simplify calculation of the required areas

Area of a rectangle = length × breadth

When a region is dissected,

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,

requires an infinite number of rectangles and so must be a limiting process,

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) and the x-axis,

The curve must be continuous and,

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,

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 value x on the x-axis

Its height at the left-hand end is f (x),

and provided the strip is very thin,

the height is still about f (x) at the right-hand end

Let the width of the strip be δx,

thought of as being very small

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 ,

Area of shaded region =

there were infinitely many of these strips,

one can imagine that the inaccuracy would disappear

This involves taking the limit so that the equality is exact: area of shaded region = lim

δ x→0

At this point,

the width δx is replaced  by the symbol dx,

which suggests an infinitesimal width,

and an old form of the letter S is used to suggest an  b infinite sum

The result is the strange-looking symbol f (x) dx,

Leibnitz

This symbol is now defined to denote the area of the shaded region:  b f (x) dx = area of shaded region

The Definite Integral: This new object

f (x) dx is called a definite integral

The rest

of the chapter is concerned with evaluating definite integrals and applying them

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

 b f (x) dx is defined to be the area of the region between The definite integral a

The function f (x) is called the integrand and the values x = a and x = b are called the 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 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

Evaluating Definite Integrals Using Area Formulae: When the function is linear or circular,

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:

FOR A TRIANGLE: 2

× base × height

FOR A TRAPEZIUM: Area = width × average of parallel sides FOR A CIRCLE:

Area = πr2

WORKED EXERCISE: Evaluate using a graph and area formulae:  4  4 (x − 1) dx (b) (x − 1) dx (a) 1

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

 4 (x − 1) dx = 12 × base × height Hence 1

3 1 −1

(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

 4 (x − 1) dx = width × average of parallel sides Hence 2

WORKED EXERCISE: Evaluate using a graph and area formulae:   2 |x| dx (b) (a) −2

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

Each shaded triangle has base 2 and height 2

 2   Hence |x| dx = 2 × 12 × base × height −2   = 2 × 12 × 2 × 2 = 4

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

 (b) The function y = 25 − x2 is a semicircle with centre at the origin and radius 5

 5  Hence 25 − x2 dx = 12 × π r2 −5

The Area of a Circle: In earlier years,

the formula A = πr2 for 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,

It involves dissecting the circle into infinitesimally thin sectors and then rearranging them into a rectangle

The height of the rectangle in the lower diagram is r

Since the circumference 2πr is divided equally between the top and bottom sides,

the length of the rectangle is πr

Hence the rectangle has area πr2 ,

which is therefore the area of the circle

Exercise 1A 1

Use area formulae to calculate the following integrals (sketches are given):  2  3  4  3 (a) 3 dx (b) 4 dx (c) x dx (d) 2x dx 0

CHAPTER 1: Integration

1A Areas and the Definite Integral

Use area formulae to calculate the following integrals (sketches are given):  3  2  1  3 (a) 2 dx (b) 5 dx (c) (2x + 4) dx (d) (3x + 3) dx −1

1 3 −3

The diagram to the right shows the graph of y = x2 from x = 0 to x = 1,

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 from x = 0 to x = 1 (keeping reasonable track of fragments of squares),

3) 4 −1

(b) By counting the appropriate squares,

approximate:  1  12 2 x2 dx (ii) x dx (i) 0

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

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

DEVELOPMENT

Sketch a graph of each definite integral,

then use  0  3 (e) 5 dx (x + 5) dx (a) (i) −5 0 0  2 (b) (j) 5 dx (x + 5) dx (f) −3 0  4  4 (k) 5 dx (x + 5) dx (c) (g) −1 2  3  6 (l) (h) (x + 5) dx 5 dx (d) −1

an area formula to calculate it:  2  8 (m) (x − 4) dx |x| dx −2 4 10  4 (x − 4) dx |x| dx (n) 4 −4  7  5 (x − 4) dx |x − 5| dx (o) 0 5 10  10 (x − 4) dx (p) |x − 5| dx 6

CHALLENGE

[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

Sketch a graph of each definite integral,

then use an area formula to calculate it:  0   4  2 16 − x dx (b) 25 − x2 dx (a) y −4 −5 7

The diagram to the right shows the quadrant  y = 1 − x2 ,

As in question 3,

the scale is 20 little divisions to 1 unit

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

(b) Divide by 400 to approximate 0

(c) Hence find an approximation for π

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 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:

FINDING PRIMITIVES: Suppose that n = −1

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

CHAPTER 1: Integration

1B The Fundamental Theorem of Calculus

Statement of the Fundamental Theorem: The fundamental theorem says that a definite integral can be evaluated by writing down any primitive F (x) of f (x),

then substituting the upper and lower bounds into it and subtracting

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

Then  b 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 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:  4  2 2x dx (b) (2x − 3) dx (a) 0

Then draw diagrams to show the regions that they represent

SOLUTION:  2 2 2x dx = x2 (a) 0

) =4 This value agrees with the area of the triangle shaded in the diagram to the right

(Note that area of triangle = 12 × base × height = 12 × 2 × 4 = 4

= (16 − 12) − (4 − 6) (Substitute 4,

this value agrees with the area of the trapezium shaded in the diagram to the right

(Note that area of trapezium = width × average of parallel sides 1+5 =2× 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

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

WORKED EXERCISE: Evaluate the following definite integrals:   1 2 x dx (b) (a)

SOLUTION: 3 1  1 x x2 dx = (a) 3 0 0 =

1 3 1 3

(Increase the index 2 to 3,

(Substitute 1,

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

) (b) 4 −2 −2 = (4 + 16) − (4 − 16)

(Substitute 2,

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:  3  6 x(x + 1) dx (b) (x − 4)(x − 6) dx (a) 1

Note: Fractions arise very often in definite integrals because the standard forms for primitives involve fractions

Care is needed with the resulting common denominators,

SOLUTION:  6  6 x(x + 1) dx = (x2 + x) dx (a) 1

= (72 + 18) − ( 13 + 12 ) = 90 − = 89 16

CHAPTER 1: Integration

1B The Fundamental Theorem of Calculus

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,

WORKED EXERCISE: Write each integrand as two separate fractions,

then evaluate:  −2 3  2 4 3x − 2x2 x − 2x4 dx (b) dx (a) x2 x3 1 −3

SOLUTION:  2  2 4   2 3x − 2x2 dx = (a) 3x − 2 dx (Divide both terms on the top by x2

= (8 − 4) − (1 − 2) = 4 − (−1)  (b)

(Divide both terms by x3

= (−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,

requires care when converting between negative powers of x and fractions

WORKED EXERCISE: Use negative indices to evaluate these definite integrals:  2  5 1 −2 x dx (b) dx (a) 4 x 1 1

SOLUTION: −1 5  5 x −2 x dx = (a) −1 1 1 5 1 = − x 1 = − 15 − (−1) = − 15 + 1 =

(Increase the index to −1 and divide by −1

CHAPTER 1: Integration

1 dx = x4

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

(Rewrite

(Increase the index to −3 and divide by −3

Exercise 1B 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

Evaluate the following definite integrals,

using the fundamental theorem:  5  2  1 (d) (g) 2x dx 8x dx 10x4 dx (a) 

5x dx 4

12x5 dx

11x10 dx

(a) Evaluate the following definite integrals,

using the fundamental theorem:  7  5  1 (i) 4 dx (ii) 5 dx (iii) dx 0

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

Evaluate the following definite integrals,

using the fundamental theorem:  3  2  6 2 (d) (g) (2x + 1) dx (3x − 1) dx (4x3 + 3x2 + 1) dx (a) 

Evaluate the following definite integrals,

You will need to take care when finding powers of negative numbers

 2  −2  0 (d) (g) (1 − 2x) dx dx 3x2 dx (a)  (b) (c)

CHAPTER 1: Integration

1B The Fundamental Theorem of Calculus

Evaluate the following definite integrals,

You will need to take care when adding and subtracting fractions

 2  1  3 (d) (g) x dx (x2 + x) dx (x3 − x + 1) dx (a) 0

DEVELOPMENT

By expanding the brackets  3 (a) x(2 + 3x) dx 2  2 (b) (x + 1)(3x + 1) dx 0 1 (c) 3x(2 + x) dx 0  3 (d) 2x(x − 1) dx 2

evaluate the following definite integrals:  0  1 2 2 (i) x (5x + 1) dx x(x − 1)(x + 1) dx (e) −1 −1  3  −1 2 (x + 2) dx x(x − 2)(x + 3) dx (f) (j) 1 −2  2  0 (g) (x − 3)2 dx (1 − x2 )2 dx (k) −1 −1  3  9  √  √ (h) (4 − 3x)2 dx (l) x+1 x − 1 dx −2

By dividing each fraction through by the denominator,

evaluate each integral:  3 3  3 2  3 3 3x + 4x2 5x + 9x4 x − x2 + x dx dx (a) dx (c) (e) x x2 x 1 1 2 4  2 2 3 −1 3 4x − x x + 4x2 x − 2x5 (b) (f) (d) dx dx dx x x x2 1 1 −2 8

Evaluate the following definite integrals,

You will need to take care when finding the powers of fractions

 1  43  12 2 2 (b) (2x + 3x ) dx (c) x dx (6 − 4x) dx (a) 2 3

(a) Evaluate the following definite integrals:  3  10 −2 x dx (ii) 2x−3 dx (i) 5

4x−5 dx

(b) By writing them with negative indices,

evaluate the following definite integrals:  2  4  1 dx dx 3 (i) (ii) (iii) dx 2 3 4 1 x x x 1 1 2  k 3 dx = 3k − 6

(a) (i) Show that 2  k 3 dx = 18

(ii) Hence find the value of k if 2  k x dx = 12 k 2

(b) (i) Show that 0  k x dx = 18

(ii) Hence find k if k > 0 and 0

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

Use area formulae to find

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

CHALLENGE

By dividing each fraction through by the denominator,

evaluate each integral:  −1  −1  2 1 + x2 1 + 2x 1 − x3 − 4x5 dx (b) dx (c) dx (a) x2 x3 2x2 1 −2 −3 13

Evaluate the following definite integrals: 2 2  −1  3 1 1 2 x+ x + 2 dx (b) dx (a) x x 1 −3

x2 (b) Sketch the integrand and explain why the argument below is invalid: 1  1 dx 1 = − = −1 − 1 = −2

(a) Explain why the function y =

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 the x-axis,

so the ‘heights’ of the little rectangles in the dissection are negative numbers

This means that any areas below the x-axis should contribute negative values to the value of the final integral

For example,

the region B is below the x-axis and so will contribute a negative number to the definite integral:  b f (x) dx = area A − area B + area C

Because areas under the x-axis are counted as negative,

the definite integral is sometimes referred to as the signed area under the curve,

CHAPTER 1: Integration

1C The Definite Integral and its Properties

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

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

minus the sum of the areas below the x-axis

WORKED EXERCISE: Evaluate these definite integrals:  6  4 (x − 4) dx (b) (x − 4) dx (a) 0

Sketch the graph of y = x − 4 and then shade the regions associated with these integrals

Then explain how each result is related to the shaded regions

SOLUTION:  4

= (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

= (18 − 24) − (8 − 16) = −6 − (−8) =2 Triangle BM C has area 2 and is above the x-axis

this is why the value of the integral is 2

= (18 − 24) − (0 − 0) = −6 This integral represents the area of BM C 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,

We can always dissect the region by dissecting the interval a ≤ x ≤ b of integration

Thus if f (x) is continuous in the interval a ≤ x ≤ b,

and the number c'lies in this interval,

then:  b  c' b 6 DISSECTION: f (x) dx = f (x) dx + f (x) dx a

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

Odd and Even Functions: In the first example below,

the function y = x3 − 4x is an odd function,

with point symmetry in the origin

Thus the area of each shaded hump 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 in the y-axis

Thus the areas to the left and right of the y-axis are equal,

so there is a doubling instead of a cancelling

 a ODD FUNCTIONS: If f (x) is odd,

−a  a  a 7 EVEN FUNCTIONS: If f (x) is even,

WORKED EXERCISE: Sketch these integrals,

then evaluate them using symmetry:  2  2 3 (x − 4x) dx (b) (x2 + 1) dx (a) −2

SOLUTION:  2 (x3 − 4x) dx = 0,

(Without this simplification,

= (4 − 8) − (4 − 8) = 0,

(b) Since the integrand is even,

 2  2 2 (x + 1) dx = 2 (x2 + 1) dx −2 0

Intervals of Zero Width: Suppose that a function is integrated over

an interval a ≤ x ≤ a of width zero

In this situation,

the region also has width zero and so the integral is zero

 a 8 INTERVALS OF ZERO WIDTH: f (x) dx = 0 a

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

REVERSING THE INTERVAL: Let f (x) be continuous in a ≤ x ≤ b

Then  b  a f (x) dx = − 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:  2  4 (x − 1) dx (b) (x − 1) dx (a) 2

SOLUTION: 4 2  4 x −x (a) (x − 1) dx = 2 2 2 = (8 − 4) − (2 − 2) = 4,

since the region is above the x-axis

which is the opposite of part (a),

because the integral runs backwards from right to left,

Sums of Functions: When two functions are added,

the two regions are piled on top of each other,

WORKED EXERCISE: Evaluate these two expressions and show that they are equal:  1  1   1 (x2 + x + 1) dx (b) x2 dx + x dx + (a) 0

SOLUTION: 1 3  1 x2 x 2 (a) + + x (x + x + 1) dx = 3 2 0 0 = ( 13 + =

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

= ( 13 − 0) + ( 12 − 0) + (1 − 0) = 1 56 ,

Multiples of Functions: Similarly,

when a function is multiplied by a constant,

the region is expanded vertically by that constant,

WORKED EXERCISE: Evaluate these two expressions and show that they are equal:  3  3 10x3 dx (b) 10 x3 dx (a) 1

SOLUTION: 3  3 10x4 3 10x dx = (a) 4 1 1 810 10 − = 4 4 800 = 4 = 200

Inequalities with Definite 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 the curve y = f (x) from x = a to x = b must be less than the area under the curve y = g(x)

In the language of definite integrals:

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

then  b  b 12 f (x) dx ≤ g(x) dx

WORKED EXERCISE:

(a) Sketch the graph of f (x) = 4 − x2 ,

 2 (b) Explain why 0 ≤ (4 − x2 ) dx ≤ 16

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 inside the square of side length 4 in the diagram opposite

CHAPTER 1: Integration

1C The Definite Integral and its Properties

Exercise 1C 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 exercises would be helpful in reinforcing these explanations

Questions 6,

The simplification of integrals of odd and even functions is particularly important and is easily demonstrated visually by curve-sketching programs

Evaluate the following definite integrals,

using the fundamental theorem:  1  4  0 5 2x dx 6x dx 10x4 dx (d) (g) (a) −2 −1 −1  1  0  −2 2 6x dx 3x dx x3 dx (b) (e) (h) −2 −3 −3  2  0  2 4x3 dx x2 dx x7 dx (c) (f) (i) −1

Evaluate the following definite integrals,

using the fundamental theorem:  2  1  10 2 (i) (a) (e) (1 + 4x) dx (6x − 8x) dx (12 − 3x) dx −3 −1 −2  0  6  3 2 2 (b) (f) (j) (3x − 5) dx (x − 6x) dx (3x2 − 5x4 − 10x) dx −2 0 1  1  −1  1 (g) (k) (x3 − x) dx (1 − x − x2 ) dx (7 − 4x3 ) dx (c) −1 −3 −1  3  2  2 (4x3 − 2x2 ) dx (7 − 2x + x4 ) dx (h) (l) (2x − 4x3 ) dx (d) 0

DEVELOPMENT

By expanding the brackets where necessary,

 3 (a) 3x(x − 4) dx 1  1 (b) (3x − 1)(3x + 1) dx −1  0 x2 (6x3 + 5x2 + 4x + 3) dx (c) −2

evaluate the following definite integrals:  2 x(1 − x) dx (d) 0  2 (e) (2 − x)(1 + x) dx −2  5 x(x + 1)(x − 1) dx (f) 0

By dividing through by the denominator,

evaluate the following definite integrals:  −1 2  −1 3  3 2 2x − 5x 3x + 7x x − 6x3 (a) dx dx (b) dx (c) x x x2 −2 −3 2 5

Find the value of k if:  3 (a) 2 dx = 4 k 8 (b) 3 dx = 12 k  3 (c) (k − 3) dx = 5 2

(x − 3) dx = 0 (x + 1) dx = 6 (k + 3x) dx =

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

Evaluate each group of definite integrals and use the properties of the definite integral to explain the relationships within each group:  0  2 (3x2 − 1) dx (ii) (3x2 − 1) dx (a) (i) 2  1 0 1 3 20x dx (ii) 20 x3 dx (b) (i) 0 0  4  4  4 (4x + 5) dx (ii) 4x dx (iii) 5 dx (c) (i) 1 1 1  1  2  2 (d) (i) 12x3 dx (ii) 12x3 dx (iii) 12x3 dx 1 0 −2 0 3 2 2 (4 − 3x ) dx (ii) (4 − 3x ) dx (e) (i) −2

Without finding a primitive,

use the properties of the definite integral to evaluate the following,

stating reasons:  1  3  90 ◦ 3 2 (c) 9 − x dx x dx (a) sin x dx (e) 3 −1 −90 ◦  4  5  2 x 3 2 3 (x − 3x + 5x − 7) dx (d) (b) (x − 25x) dx dx (f) 2 4 −5 −2 1 + x 8

(a) On one set of axes sketch y = x2 and y = x3 ,

clearly showing the point of intersection

(c) Check the inequality in part (b) by evaluating each integral

given the following sketches of f (x): 9

Use area formulae to find 0

2 1 −1

By dividing through by the denominator,

evaluate the following definite integrals:  −1 5  2 2  4 3 x −3 x −2 2x − 3x + 1 (a) dx (b) dx (c) dx 3 3 x x x4 1 −2 1 11

Use the results of the previous question to write down the values of these definite integrals:  1 3  −2 5  1 2 x −3 x −2 2x − 3x + 1 (a) dx (b) dx (c) dx 3 3 x x x4 4 −1 2 12

Sketch a graph of each integral and hence determine whether each statement is true or false:  1  1  2  −1 1 1 (a) dx > 0 (d) dx > 0 2x dx = 0 (b) 3x > 0 (c) x x −1 0 −2 2

CHAPTER 1: Integration

1D The Indefinite Integral

this section turns again to the task of finding primitives

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 term indefinite integral is 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 of x2 + 1 is  x3 + x + C,

(x2 + 1) dx = 3 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,

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 the answer

Despite being a nuisance to write down every time,

In most problems other than definite integrals,

Standard Forms for Integration: The rules for finding primitives given in the last section of the Year 11 volume can now be restated in this new notation

STANDARD FORMS FOR INTEGRATION: Suppose that n = −1

Then  xn +1 + C,

xn dx = 13 n+1  (ax + b)n +1 (ax + b)n dx = + C,

a(n + 1) 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 constant C’ or ‘where C is a constant’ should follow the first mention of the new pronumeral C,

because no pronumeral should be used without having been formally introduced

There is a limit to one’s patience,

and usually in this situation it is quite clear that 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  (a) 9 dx

xn +1 + C to find: n+1  (b) 12x3 dx

SOLUTION:  (a) 9 dx = 9x + C,

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

dx But the formula still gives the correct answer,

and so increasing the index to 1 and dividing by this new index 1,

for some constant C 9x0 dx = 1 = 9x + C

We know that 9x is the primitive of 9,

x4 + C,

for some constant C 4 = 3x4 + C

WORKED EXERCISE: Use the standard form  (a) (3x + 1)5 dx

(ax + b)n +1 + C to find: a(n + 1) (b)

SOLUTION:  (3x + 1)6 (a) (3x + 1)5 dx = +C (