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- ge 2 Unit Year 12

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

- 2 Unit Second Edition

BILL PENDER DAVID SADLER JULIA SHEA DEREK WARD

Cape Town,

São Paulo,

Delhi Cambridge University Press 477 Williamstown Road,

Australia www

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org/9780521177504 © Bill Pender,

David Sadler,

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

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- vii About the Authors

xi Chapter One — Integration

- 1A 1B 1C 1D 1E 1F 1G 1H 1I 1J

Areas and the Deﬁnite Integral

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

The Trapezoidal Rule

Chapter Review Exercise

- 1 1 6 12 19 25 33 39 47 51 55

- 60 2A 2B 2C 2D 2E 2F 2G

Review of Exponential Functions

Applications of Diﬀerentiation

Applications of Integration

Chapter Review Exercise

- 60 65 73 79 84 90 96

- 99 3A 3B 3C 3D 3E 3F 3G 3H

Integration of the Reciprocal Function

Chapter Review Exercise

Contents

- 141 4A 4B 4C 4D 4E 4F 4G 4H 4I

Integration of the Trigonometric Functions

Chapter Five — Motion

- 196 5A 5B 5C 5D

Velocity as a Derivative

Applications of APs and GPs

The Use of Logarithms with GPs

Investing Money by Regular Instalments Paying Oﬀ a Loan

Rates of Change

Chapter Review Exercise

- 281 7A 7B 7C 7D 7E 7F 7G 7H 7I 7J

Points,

Intercepts on Transversals

Chapter Eight — Probability

- 341 8A 8B 8C 8D 8E 8F 8G

Sample Space Graphs and Tree Diagrams

Chapter Review Exercise

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

and the previous volume for Year 11

- 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

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

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,

- and Peter Cribb,

Sarah Buerckner and the team at Cambridge University Press,

for their support and help in discussions

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

David Sadler 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

and representatives of all types should be attempted

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

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

and this section is intended to match the standard of those recent examinations

some require more sophistication in logic,

some establish more diﬃcult connections between topics,

and some complete proofs or give an alternative approach

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

- methods,

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

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

The Chapter Review Exercises: A Chapter Review Exercise has been added to each chapter of the second edition

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

- 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

we believe that functions and calculus should be developed as early as possible because these are the fundamental ideas in the course

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 interconnections amongst the various topics much clearer

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 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

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,

- are based on the circle

and they are essential for the study of all periodic phenomena

Thus the three basic functions in the course,

- ex and sin x,

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

- everything

How to Use This Book

- in the course,
- whether in calculus,
- geometry,
- trigonometry,

coordinate geometry or algebra,

can easily be related to everything else

stressed heavily in recent examinations,

is to encourage arguments that relate a curve to its equation

graphs are constantly used to solve algebraic problems

- but as a matter of routine,

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

the applications of calculus to maximisation,

- motion,

rates of change and ﬁnance are all parts of the syllabus

as a system of pure logic on the one hand,

and an essential part of modern technology on the other

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

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,

- and is quite accessible

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

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

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

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

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

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

he taught at Sydney Boys’ High School,

and he was a Visiting Fellow at the University of NSW in 1991

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

both from the University of NSW,

where he was subsequently Senior Tutor for three years

and is a lay clerk at St James’,

King Street,

- where he sings counter-tenor

The Book of Nature is written in the language of Mathematics

— The seventeenth-century Italian scientist Galileo

— 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

? 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

Integration The calculation of areas has so far been restricted to regions bounded by straight lines or parts of circles

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

- y = 4 − x2
- 1 A Areas and the Deﬁnite Integral All area formulae and calculations of area are based on two principles: 1

Area of a rectangle = length × breadth

- the area is unchanged

like a triangle or a trapezium,

can be cut up and rearranged into a rectangle with a few well-chosen cuts

- 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) and the x-axis,

- from x = a to x = b

The curve must be continuous and,

- for the moment,
- entirely above the x-axis
- y x + δx

the region has been dissected into a number of strips

- but only roughly so,

because the upper boundary is curved

The third diagram shows just one of the strips,

above the value x on the x-axis

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,

- where δx is,
- as usual in calculus,

thought of as being very small

- roughly,

area of strip = width × height = f (x) δx

We need sigma notation,

based on the Greek upper-case letter ,

- meaning S for sum

Area of shaded region =

- area of each strip
- f (x) δx
- however,

there were inﬁnitely many of these strips,

- each inﬁnitesimally thin,

one can imagine that the inaccuracy would disappear

- f (x) δx

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

- invented by a

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

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

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

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

FOR A TRIANGLE: 2

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

with width 4 − 2 = 2 and parallel sides of length 1 and 3

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

- 1+3 =2× 2 = 4
- y 3 1 −1

- 25 − x2 dx

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

(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

- 2 1 2 ×5 25π 2

the formula A = πr2 for the area of a circle was proven

some limiting process had to be used in that proof

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

the length of the rectangle is πr

which is therefore the area of the circle

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

- (2 − x) dx
- (5 − x) dx
- (x + 2) dx
- (x + 3) dx

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

- −2 −3
- (x + 4) dx
- (x + 6) dx

1 3 −3

- (−2,4)
- |2x| dx y 4

- drawn on graph paper

(a) Count how many little squares there are under the graph from x = 0 to x = 1 (keeping reasonable track of fragments of squares),

- then divide by 400 1 x2 dx
- to approximate

3) 4 −1

(b) By counting the appropriate squares,

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

DEVELOPMENT

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

[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

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

- from x = 0 to x = 1

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

- 1 1 − x2 dx

(b) Divide by 400 to approximate 0

(c) Hence ﬁnd an approximation for π

- 1 B The Fundamental Theorem of Calculus The fundamental theorem is a formula for evaluating deﬁnite integrals

so only the algorithm is presented in this section,

by means of some worked examples

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)

- xn +1 dy = xn ,
- then y = + C,
- for some constant C

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

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

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

- (x is a primitive of 2x
- = 22 − 02 (Substitute 2,
- then substitute 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

- 4 (2x − 3) dx = x2 − 3x (Take the primitive of each term
- ) (b) 2

= (16 − 12) − (4 − 6) (Substitute 4,

- then substitute 2
- ) = 4 − (−2) =6 1 Again,

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

- y = 2x − 3

Note: Whenever there are two or more terms in the primitive,

brackets are needed when substituting the upper and lower bounds of integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

1 3 1 3

- (x3 + 8) dx

- then divide by 3

- then substitute 0

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

- 2 4 2 x 3 + 8x (x + 8) dx = (Take the primitive of each term

) (b) 4 −2 −2 = (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

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,

- mixed numerals and cancelling

- x3 x2 = + 3 2

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

- (x − 4)(x − 6) dx =
- (x2 − 10x + 24) dx
- 3 x3 2 = − 5x + 24x 3 0 = (9 − 45 + 72) − (0 − 0 + 0) = 36

CHAPTER 1: Integration

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

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

- ) 2 x 1 1
- 2 = x3 − 2x 1

= (8 − 4) − (1 − 2) = 4 − (−1) (b)

- =5 −2 x3 − 2x4 dx = (1 − 2x) dx x3 −3
- −2 = x − x2

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

- however,

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 =

- ) (Rewrite x−1 as
- 1 before substitution

1 dx = x4

- x−4 dx

- 2 x−3 = −3 1 2 1 = − 3 3x 1
- 1 as x−4 before ﬁnding the primitive

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

- ) (Rewrite x−3 as
- 1 before substitution
- 1 = − 24 − (− 13 ) 1 = − 24 +

but experimentation with further deﬁnite integrals

It would be helpful to generate screen sketches of the graphs and the regions involved in the integrals

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

- 4x dx

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

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

- (2x − 3) dx
- (4x + 5) dx
- (6x2 + 2) dx
- (3x + 2x) dx 2
- (2x + 3x2 + 8x3 ) dx
- (3x2 − 6x + 5) dx

- using the fundamental theorem

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)

- −1 1 −2
- (2x + 3) dx
- −1 2 −2
- (4x3 + 5) dx
- (5x4 + 6x2 ) dx
- (12 − 2x) dx
- −3 −1 −2
- (4x3 + 12x2 − 3) dx

- using the fundamental theorem

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

- (x + 2) dx
- (x + x2 + x3 ) dx
- (3x + 5) dx
- x2 dx
- (2x2 − 3x + 1) dx
- −2 −2 −4
- (16 − x3 − x) dx

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

- where necessary,

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

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

- using the fundamental theorem

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

- (iii) 1 2

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

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

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 − x)2 dx
- 1 is never negative

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

- 2 x −1 −1 x

(a) Explain why the function y =

- 1 C The 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

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

- in the diagram to the right,

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,

- to distinguish it from area,
- which is always positive

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

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

- from The deﬁnite integral a
- x = a to x = b,

minus the sum of the areas below the x-axis

- (x − 4) dx

Then explain how each result is related to the shaded regions

- 4 (x − 4) dx = 12 x2 − 4x (a) 0

= (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

- 6 (x − 4) dx = 12 x2 − 4x (b)

= (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

- 6 (x − 4) dx = 12 x2 − 4x (c)

= (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,

- its area remains the same

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

and the number c'lies in this interval,

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

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

because the equal humps above and below the x-axis cancel out

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,

- then f (x) dx = 0

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

- then f (x) dx = 2 f (x) dx

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

- since the integrand is odd

(Without this simpliﬁcation,

- the calculation is: 2
- 2 (x3 − 4x) dx = 14 x4 − 2x2 −2

= (4 − 8) − (4 − 8) = 0,

- as before

(b) Since the integrand is even,

2 2 2 (x + 1) dx = 2 (x2 + 1) dx −2 0

- 2 = 2 13 x3 + x 0 = 2 (2 23 + 2) − (0 + 0)

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

an interval a ≤ x ≤ a of width zero

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

CHAPTER 1: Integration

1C The Deﬁnite Integral and its Properties

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

- which is positive,

since the region is above the x-axis

- 2 2 2 x −x (b) (x − 1) dx = 2 4 4
- = (2 − 2) − (8 − 4)
- = −4,

which is the opposite of part (a),

because the integral runs backwards from right to left,

- from x = 4 to x = 2

Sums of Functions: When two functions are added,

the two regions are piled on top of each other,

- so that:
- 10 INTEGRAL OF A SUM:
- f (x) + g(x) dx =
- f (x) dx +
- g(x) dx

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

- + 1) − (0 + 0 + 0)

CHAPTER 1: Integration

- x dx + 2

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

- the same as in part (a)

when a function is multiplied by a constant,

the region is expanded vertically by that constant,

- so that:
- 11 INTEGRAL OF A MULTIPLE:
- kf (x) dx = k
- f (x) dx

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

- 3 x4 x dx = 10 × 4 1
- 81 1 − = 10 × 4 4 80 = 10 × 4 = 200
- (b) 10 1

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)

then b b 12 f (x) dx ≤ g(x) dx

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

- for −2 ≤ x ≤ 2

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

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

- 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

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

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

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

- (e) (f) 1

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

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

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

- 1 1 3 x dx < x2 dx < 1
- (b) Hence explain why 0 < 0

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

- 4 f (x) dx,

given the following sketches of f (x): 9

2 1 −1

- −1 CHALLENGE

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

- 1 D'The Indeﬁnite Integral Now that primitives have been established as the key to calculating deﬁnite integrals,

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

the primitive or indeﬁnite integral of x2 + 1 is x3 + x + C,

- for some constant 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

An indeﬁnite integral,

- on the other hand,

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,

- it must always be included

- it will not be zero

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

- for some constant C

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

- for some constant 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

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

- however,

and usually in this situation it is quite clear that C is the constant of integration

it would be wise to introduce it formally

- xn dx =

xn +1 + C to ﬁnd: n+1 (b) 12x3 dx

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

- for some constant C

CHAPTER 1: Integration

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 12

- d (9x) = 9

dx But the formula still gives the correct answer,

- because 9 = 9x0 ,

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

- 9x1 + C,

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

- because

x4 + C,

for some constant C 4 = 3x4 + C

- 12x3 dx = 12 ×

- (ax + b)n dx =

(ax + b)n +1 + C to ﬁnd: a(n + 1) (b)

- (5 − 2x)2 dx

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

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