An Analytical Calculus Volume I … · AN ANALYTICAL CALCULUS VOLUME I. AN ANALYTICAL CALCULUS FOR SCHOOL AND UNIVERSITY BY E. A. MAXWELL Fellow of Queens' College, Cambridge VOLUME

AN ANALYTICAL CALCULUS

VOLUME I

ANANALYTICAL CALCULUS

FOR SCHOOL AND UNIVERSITY

BY

E. A. MAXWELLFellow of Queens' College, Cambridge

VOLUME I

CAMBRIDGEAT THE UNIVERSITY PRESS

1966

CAMBRIDGE UNIVERSITY PRESSCambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

Published in the United States of America by Cambridge University Press, New York

www.cambridge.orgInformation on this title: www.cambridge.org/9780521056960

© Cambridge University Press 1954

This publication is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements,no reproduction of any part may take place without the written

permission of Cambridge University Press.

First published 1954Reprinted 1966

This digitally printed version 2008

A catalogue record for this publication is available from the British Library

ISBN 978-0-521-05696-0 hardbackISBN 978-0-521-09035-3 paperback

DEDICATEDTO

MY CHILDREN

PHYLLIS, SHEILA, DENNIS, JEAN

CONTENTS

PREFACE

INTRODUCTION

CHAPTER I:

1. Functions

THE IDEA

PAGE1

2. Discrete and continuous vari-ables

3. Notation4. Limits5. Continuity

345

12

OF DIFFERENTIATION

6. Rate of change7. Gradient8. The differential coefficient9. Notation

10. Tangent and normal

page IX

xi

PAGE1416192123

CHAPTER II: THE EVALUATION OF DIFFERENTIALCOEFFICIENTS

1. Some theorems on limits 252. The differential coefficient of a

sum of functions 263. The differential coefficient of

the product of two functions 264. The differential coefficient of a

'function of a function* 275. The differential coefficient of

xn is nxn~x 306. The limit as x ->• 0 of sin x\x is

unity 32

7. The differential coefficient ofsin a; is coscc

8. The differential coefficient ofcosa; is —sina?

9. Differential coefficients ofhigher order

10. Some standard forms11. The inverse circular functions12. Differentials13. The differentiation of deter-

minants

CHAPTER I I I : APPLICATIONS OF D I F F E R E N T I A T I O N

1. Illustration from dynamics;velocity, acceleration

2. Maxima and minima illus-trated

3. The determination of maximaand minima

4. Increasing and decreasing func-tions

5. The second differential co-efficient and concavity

6. Points of inflexion

R E V I S I O N E X A M P L E S I , ' A L T E R N A T I V E O R D I N A R Y ' L E V E L

CHAPTER IV: THE IDEA OF INTEGRATION1. The area 'under' a curve 772. The integral 813. Notation 834. The sign of the area 845. Definite and indefinite integrals 85

6. The evaluation of an integral7. The link with differentials8. The evaluation of a definite

integral9. Some simple standard forms

33

34

35363842

45

48

49

51

52

5455

7. Discrimination between max-ima and minima

8. The sketching of simplecurves

9. Rolle's theorem10. The mean value theorem11. The real roots of the equa-

tions/^) = 0ff'{x) = 012. Mean value theorems for two

functions13. Application to certain limits

56

596061

63

6668

70

8689

90

V1U C O N T E N T S

CHAPTEE V: DEVICES IN INTEGRATION

PAGK PAGE1. Substitution, or change of vari- 3. Integration by parts 103

able; indefinite integrals 94 4. Formulae of reduction 1062. Substitution, or change of vari-

able; definite integrals 98

CHAPTER VI: APPLICATIONS OF INTEGRATION

1. Use in dynamics2. Area; Cartesian coordinates3. The area of a sector, in polar

coordinates4. Centre of gravity, or centroid5. The moment of inertia of a

lamina6. Volume of revolution7. The centre of gravity of a

uniform solid of revolution 127

R E V I S I O N EXAMPLES I I , 'ALTERNATIVE O R D I N A R Y ' L E V E L 144

ANSWERS TO EXAMPLES 153

I N D E X 165

111113

114118

123125

8. First theorem of Pappus9. The moment of inertia of a

uniform solid of revolutionabout its axis

10. The area of a surface ofrevolution

11. Approximate integration12. Simpson's rule13. Mean values

128

129

130134135138

PREFACE

I HAVE been fortunate in the help received during the preparationof this work. The manuscript was read with great thoroughness byDr Sheila M. Edmonds, of Newnham College, Cambridge, whosecriticisms and suggestions were of great value and kept me firmlyin the paths of rigour. Dr J. W. S. Cassels, of Trinity College,Cambridge, read the proofs and drew attention to a number ofslips. To both I would express my sincere thanks.

A number of pupils helped me in the preparation of the answers.Special mention must be made of Mr J. E. Wallington andMr P. A. Wallington, who, acting almost as a committee, providedme with a complete set of checked answers; any slips that remainmust be due to my own carelessness in transcription. I am deeplyindebted to them for a very substantial piece of work.

As on former occasions, I have been greatly helped by the staffof the Cambridge University Press, and I should like to place onrecord how much I owe to their skilled interpretation of themanuscript.

The Examples come from many sources—the Oxford andCambridge Schools Examination Board, Scholarship examinationsin the University of Cambridge, and Degree Examinations in theUniversities of Cambridge and London. I am grateful forpermission to reproduce them.

E. A. M.

QUEENS' COLLEGE CAMBRIDGE

June, 1953

NOTE ON THE THIRD IMPRESSION

A number of corrections have been made, mainly small. I amindebted to Mr L. E. Clarke of the University College of Ghanafor a very helpful list covering this and the later volumes.

INTRODUCTION

T H E AIM of these volumes is that they shall together form acomplete course in Calculus from its beginnings up to the pointwhere it joins with the subject usually known as analysis. Thewhole conception is based on considerable dissatisfaction withmuch that seems rough-and-ready in the basic ideas with whichpupils reach the universities, so that almost anything seemsacceptable for 'proof which is superficially plausible. Of coursethe early work cannot be treated with the rigour appropriate tomore mature judgement; but I have tried here, however unsuccess-fully, to present the subject in such a way that the more exacttreatment, when it comes, can follow by natural development,without being forced to return to a fresh beginning which is oftenfelt to be both unnecessary and even pointless. (How manystudents lose the thread of analysis just because they do not seeany reason for the first few lectures and therefore do not givethem serious attention?)

The first volume deals with the basic ideas of differentiationand integration. Graphical methods are used freely, but, it ishoped, in such a way that the essential logical development isnever far away. The examples at this stage are mainly verysimple, and beginners should have no difficulty in acquiring afluent technique. Integration appears from the start as area andsummation, the method of calculation by inverse differentiationbeing deduced. All the usual elementary functions are treated,but the logarithmic and exponential functions are postponed.

In Volume II, more advanced parts of the theory make theirappearance. The logarithmic and exponential functions aretreated in some detail, followed by Taylor's series and thehyperbolic functions. The treatment of curves seems somewhatdifferent from that usually adopted—for example, the formulatan if; = rddjdr is derived without the help of the 'elementarytriangle', and other ideas often left to intuition are developed ona logical basis. There is a lengthy account of complex numbers,and the volume concludes with 'infinite* integrals and systematicintegration. The examples include many that are simple, but goup to scholarship, or early university, level.

Xll INTRODUCTION

Volume III contains a treatment of the functions of severalvariables, including both partial differentiation and multipleintegration, and also a chapter on curve-tracing. We are nowdefinitely at sixth form, or first-year university, level. Volume IVdeals with more advanced work, such as differential equations,Fourier series and similar topics. For these two volumes, theexamples are mainly of university standard.

It is hoped that a reader will find material for a continuousstudy of the subject from the start until he leaves it at the endfor more advanced study, or perhaps earlier in order to apply itto other sections of his work. Although I have tried to keep thestandard of discussion at a level which the mathematical specialistwill appreciate, I have also tried to remember the needs of othersand given much attention to keeping the actual exposition assimple and clear as is possible. In particular, I hope that thescientist and the engineer will find all their basic needs in suitablydigestible form. But I ought to state explicitly that, although anumber of practical applications appear as illustrations, I havemade no attempt at all to write a specifically 'applied' calculus.My own feeKng is that the underlying foundations should bemade firm by a study of calculus in its own right; the applicationscan then be made by others according to their own individualrequirements. To attempt the two things at once may lead toconfusion.

CHAPTER I

THE IDEA OF D I F F E R E N T I A T I O N

1. Functions. If a stone is thrown vertically upwards fromthe ground, it gets slower and slower till, at a certain height, itstops and immediately begins to fall down again. When the speedof projection is u feet per second, the height s feet at time t secondsis given approximately by the equation

This formula expresses a relation between the three quantitiess,u,t\ when two of them are known, the third can be calculated.

(i) If u> t are given, then s is determined uniquely; the heightcan be calculated at a given time for a definite speed of projection.

(ii) If $,t are given, then u is determined uniquely; the speedof projection can be calculated for a given height at a given time.

(iii) If s, u are given, the equation for t is the quadratic

16t2-ut + s = 0,

so that there are two values of t; the particle, projected with givenspeed, is at a given height twice (provided, of course, that it getsthere at all), once going up and once coming down.

We say that there is a functional relation connecting s, u, t, orthat each of them is a function of the other two. The magnitudesconnected by the relationship are called variables) those to whichwe assign values of our own choosing are called independentvariables and those whose values are then restricted (perhaps evendetermined, as in the example quoted) are called dependent. Thusthe value of a dependent variable is governed by that of theindependent variables.

In the example just given, the functional relationship wasexpressed by means of a precise algebraic formula. It may behelpful to point out at once that variables may be functions ofeach other even when such a formula is lacking; the essentialthing is that there should be some rule whereby the dependentvariable may be found when the independent is given. Suppose,for example, that y is the first prime number greater than x.

2 THE IDEA OF DIFFERENTIATION

When x is given, y is completely determined, but there is noformula connecting them. All the same, y is a definite function of x.

EXAMPLES I

1. Write down the functional relation connecting x and y forthe following data:

(i) The product of x2 and y exceeds yz by 4.(ii) The sum of x2 and y2 exceeds the square root of y by 2.

2. State the functional relation connecting the sides a, 6, c of atriangle ABC in which the angle at A is a right angle. Takinga, 6 as independent variables, express c in terms of them.

Let us return to the formula

Each of the variables is a function of the other two, but in aparticular problem any one of them may remain fixed in value.The most natural example would be the motion of a particle withgiven speed of projection u. Then we should regard u as a constantand the formula as a relation between TWO variables s,t, eachbeing a function of the other.

A number such as u, which enters into the formula but remainsunchanged throughout the problem, is called a parameter. Itspresence influences the 'size' of the genuine variables.

There is one difficulty to which we should refer at once. Considerthe two functions / 1 . o

*' s inW>where the symbol ° is used to denote measurement in degrees.Each has a definite value for every value of x, except when x is zero.

(i) The function - assumes the form - when x is zero, but therex 0

exists no such number.

(ii) The value of sin I-I always lies between — 1 and + 1, but

nothing at all can be said when x is zero. We may, for example,

evaluate sin (-) for x = -1, -01, -001,... (i.e. sin 10°, sin 100°,

sin 1000°,...), but we are no nearer to a definite value for x = 0.

FUNCTIONS 3

In what follows, we shall assume that a function has a definiteunique value for each relevant value of the independent variable,except perhaps in cases which we shall indicate explicitly.

EXAMPLES II

1. For what values, if any, of the independent variable x havethe following functions no definite value ?

I l l 1x-V x2-V

2. For what values of x have the functions

1since

no definite value?

tana;

3. For what values of x have the functions

1cos - ,

no definite value?

2. Discrete and continuous variables. Consider the follow-ing examples for a function y of an independent variable x:

(i) y is defined in terms of x by means of the relation y = 2x + 3;(ii) y is defined to be the first positive integer greater than x\

(iii) y is the average height in inches of the first x men namedin the Cambridge Telephone Directory for 1952;

(iv) y is the number of children of the #th man named in theCambridge Telephone Directory for 1952.

There are important differences in the allowable values of thevariables. Let us take the examples in order:

(i) The values of x and y are both unrestricted; each can takeany value, positive or negative, integral or fractional.

(ii) The value of x is unrestricted, but y must be a positiveinteger. That is, the value of y moves by 'jumps'.

(iii) The value of x is necessarily a positive integer; the valueof y may, by chance, be an integer, but is more likely to be afraction, probably between 54 and 78.

(iv) The values of #, y are both positive integers, with y possiblyzero.


A variable whose values proceed by steps is called discrete] avariable which can take all values (at any rate within limitsrelevant to the problem in hand) is called continuous.

3. Notation. We now direct our attention to functions of asingle continuous variable. A function of the independent variablex will be denoted by a symbol such as f(x)ig(x)9F(x), and so on.Thus, if the function were 2#2 — 3, we should write

where the sign ' = ' is used to mean 'is identically equal to5, or'stands for'. For example,

/(4) = 2 .4 2 -3 = 29, /(0)==2.02-3 = - 3 .

When more than one function is considered, we use separatesymbols for each; for example, we might have the three functions

These functions are actually re-lated, being connected by theidentity

It is sometimes convenient toomit the reference to the inde-pendent variable when no con-fusion can arise, and to use theshorter symbols/, g, F, and so on,for the functions. The identityjust given then appears morecompactly in the form

In practice, too, we often use Fi£- *•a SINGLE LETTER (usually y) todenote a function of the independent variable x. Thus, we mightwrite 2

to mean 'y is the function x2 — 5'. By convention, however, theordinary sign for equality is usually employed in this context,and we write 9 „

y = xl — 5.

NOTATION 5

This notation is linked up with the common representation ofa function by means of a graph. The diagram (Fig. 1) shows thegraph y = x2 — 5 representing the function just quoted.

A function may on occasion receive the alternative names y orf(x) according to convenience, and the graph y = f(x) is thencalled the graph of the function.

We assume that the reader has already had practice in thedrawing of graphs. The following revision examples are typical.

EXAMPLES IIIDraw the following graphs, choosing your own scales and ranges

of values for x:

1. y = z+3. 5. y = (x- 1) (x-2).

2. y= 2x2-3. 6. y = (x-l){x-2)(x-Z).

3. y = sin x. 7. y = x\x — 1).

4. y = xcosx. 8.* y = ^x.

4. Limits. Suppose that a stone, initially at rest, is droppedfrom some point at a certain height above the ground. As it falls,it is subject to resistance from the air, and some idea of themotion may be found by making the simplifying assumption thatthe resistance is proportional to the speed—say h times the speedper unit mass. It may then be proved that, if g (approximatelyequal to 32) is the usual constant of gravitation, the speed aftert seconds is

9 9_k he1*'

where e is a number about which we shall have much to say later,but which for the present we may take as having a value nearto 2-7.

As t increases from zero, ekt increases steadily, and the termgl(kekt) gets less and less, becoming almost negligible for largevalues of t. Hence as time goes on the speed of the stone (if it hasfar enough to fall before reaching the ground) approaches moreand more closely to the value g/k given by the first term. Thisvalue is, in fact, called the terminal speed of the stone.

* Here and elsewhere the symbol împlies the positive square root unlessthe contrary is stated explicitly.

6 THE IDEA OF DIFFEBENTIATION

It follows that the expression for the speed of the stone is afunction of the time which approaches more and more nearly tothe value gjk as t becomes larger and larger. We say that thefunction tends to the limit g\]c for large values of t.

We have begun by explaining the idea of a limit with the helpof a physical example, in which that limit is approached for largevalues of the independent variable. More generally, we mustconsider the limit approached by a function as the independentvariable tends towards any given finite value. In many cases theanswer is obvious; for example, the limit of 1 + x as x approaches 2

x2— 1is 3. But it is not so obvious what we are to say about

x — i.as x approaches 1, or about as x approaches 0. This is theproblem which we now begin to examine.

Suppose that f(x) is a given function of x, and that we wish todiscuss what happens to f(x) as x tends to a certain value a.Suppose, too, that there exists a number L (which, in practice,may require some skill to determine) such that f(x) is very nearto L whenever x is very near to a. We say that L is the limit off(x) as x tends to a, and we write this statement in the form

We must, of course, have some criterion to apply to thewords Very near'—that is, we must devise a 'test of nearness'.For this purpose we choose any small positive number, which itis customary to call €. We might take € to be *1, -01, '00001, andso on, according to the degree of accuracy which we propose toadopt. The point is that e is an arbitrary number of our ownselection, and that it may be taken as small as ever we please.In order to say that f(x) tends to the limit L, we are to insistthat the numerical value of the difference between f(x) and L,written \f(x)-L\,

is less than this number € for all values of x sufficiently near to a;and this must be true however small € is selected.

We require, in fact, that when e is given there can be found acertain positive number rj (whose value will depend on e) suchthat the difference \f(x) — L\ really is less than e whenever xdiffers from a by less than rj.

LIMITS

DEFINITION. A function f(x) tends to the limit L as x tends tothe value a if, when € is a given positive number (however small) apositive number r\ can be found, depending on e, such that

whenever 0 < | a; — a |

/w

L

L-e

0

I

im t

{'

i n X

Fig. 2.

The definition may be illustrated graphically. The diagramportrays the graph y = f(x), in which y = L when x = a, thecorresponding point on the graph being P. The band between thedotted lines at the levels L — *, L + c encloses that part of the curvefor which f(x) lies between L — €, £ + e; the points where the curvemeets the dotted lines are U, F, giving values m, n for x. If wetake rj to be any number less than both a — m and n — a, then thevalue of f(x) is between L—€ and L+€ whenever # lies betweena — rj and a + 77; that is, \f(x) — L\ is less than c whenever | a; —a |is less than rj. Hence lim f(x) = L.

Note. The function may never actually TAKE the value towardswhich it TENDS. For example, the relation

x-l


is true whenever x is not equal to 1, and this is so however closeto 1 it may be. For example, if x = 1-0000001, then

( z 2 - l ) / ( z - l ) = 2-0000001.

x2— 1Hence Km = 2.

But the function itself has no value when x = 1, since numeratorand denominator are both zero.

The modification when the independent variable becomesindefinitely large is easily supplied (compare the physical example,p. 5). The idea is that, if f(x) tends to a limit L for large x,then, whatever positive number € we take, we can ensure that\f(x) — L | is less than e by taking x sufficiently large; in fact, wemust ensure the existence of a number JV" (whose value willdepend on e), such that the difference \f(x) — L\ is less than ewhenever x is greater than N.

DEFINITION. A function f(x) tends to the limit L as x becomeslarger and larger (or, as we say, ltends to infinity') if, when e is agiven positive number, however small, a number N can be found,depending on e, such that

\f(x)-L\<€

whenever x > N.

The symbol coo' is often used for 'infinity', and the statement

'f(x) tends to L as x tends to infinity*

can be written in the form

'lim f(x) = V.

ILLUSTRATION 1. In order to show what is implied by thesedefinitions of a limit, we consider the two examples

v x + 5 x+5lim -, lim

+ 29

In each case the limiting value itself is easily obtained. Wedemonstrate how the value fits in with our formal definitions.

LIMITS

(i) We should naturally expect the solution

hm x + 5 _ 1 + 5x + 2~ 1 + 2

= 2.

For detailed proof, we turn to the definition of a limit, puttingf(x) = (a? + 5)1 (x + 2), L = 2. Then

x + 2 '

We have to show that, if € is any given positive small number,then we can establish the existence of a positive number rj (depend -

-x+1ing on e) with the property that

x 4-2is less than e whenever

x lies between l—rj and 1 +rj.Since we are concerned only with values of x near to 1, we may

take x + 2 as positive. The argument divides itself into two parts,according as x> 1 or x < 1.

If x > 1, we write x— 1 + S, where S is positive. Then

and |# + 2|

The condition is therefore

8

3+S.

3 + S

or

[When Clearing fractions' for an inequality, it is essential tobe sure that the denominator is positive. This is the point of theremark that rr+2 may be taken as positive.]

The inequality is certainly true if we choose 8 so that


If x < 1, we write x = 1 — S', where 8' is positive. Then

and | x + 2 | = 3 - 8'. The condition is therefore

8'

or, the denominator being positive,

or (l + «-)S'<3e,

If we choose S and S' so that both are less than 3€/(l + e), theneach of the inequalities

is satisfied. In other words, if we take

3e

then the expression (x + 5)/(x + 2) differs from 2 by less than € forall values of x in the range

Hence, by the formal definition of a limit,

hm ^ = 2.*->l* + 2

(ii) Consider next the limit

,. x + 5lim

We write the expression (x + 5)j(x+2) in the form

•©"

LIMITS 11

As x becomes very large, 5/x and 2\x become very small, so that

x + 5 1 + 0lim

1 + 0

= 1.

Now turn to the corresponding formal definition, and putf(x) = (x + 5)l(x + 2)9L = 1. Then

x + 2

3

!/(*)--&! =

We have to show that, if e is any given positive small number,then we can establish the existence of a number N (depending

3

x + 5

on e) with the property thatx + 2

is less than € whenever x is

greater than N.As we are concerned with large values of x9 we need only

consider x to be positive, so that

x + 2 x + 2'

and the condition is

z + 2 " '

or, the denominator being positive,

3 < ex + 2e,

or ex > 3 — 2e,

3 - 2 cor, finally,

If we write

x>

3-2€

then the expression (x + 5)/(x + 2) will differ from 1 by less than€ for all values of x greater than N. Hence, by the formal definitionof a limit,

hm = 1.+ 2


EXAMPLES IV

Evaluate the following limits:

1. lim 2. lim

4. lim

l+xl-x

3

5. lim

7.

6. limrr2 —4

x+l8. lim -2-5x + 5'

5. Continuity. The treatment of the preceding paragraphmust sometimes be modified. Anyone looking at the diagram(Fig. 3) would agree instinctively that the function f(x) repre-sented there is 'contin-uous' for negative values w\of x and 'continuous' forpositive values of x, but'discontinuous' when x iszero. We must examinethe idea of discontinuitymore closely.

We can, if necessary,split up the work of thepreceding paragraphs todefine TWO limits at x = a:

(i) The limit as x tends " Oto a 'from above', wherewe replace the conditions0 < | a; — a | < 77 in the de-finition by the conditions0<x — a<r], making x greater than a. This corresponds to anapproach to the point L2 in the diagram.

(ii) The limit as x tends to a 'from below', where we replacethe conditions 0<\x — a\<rj in the definition by the conditions0<a — x<7j, making x less than a. This corresponds to anapproach to the point Lx in the diagram.

We shall not have much to say about these two limits, exceptto note the definition of continuity wliich they imply:

Fig. 3.

CONTINUITY 13

DEFINITION. The function f(x) is CONTINUOUS when x = a iff(x) tends to a limit L as x tends to a from above and to the samelimit L as x tends to a from below, while f(x) = L when x = a.

Continuity is therefore a LOCAL property, as the diagram indi-cates. Continuity at one point by no means implies continuityat any other.

One of the commonest cases of discontinuity which occurs inpractice is when f(x) assumes the form of a 'fraction with a zerodenominator'. The simplest illustration is the function

and the diagram (Fig. 4) shows the graph_ 1

y ~~ x

Fig. 4.

O

Fig. 5.

For very small values of x, we see that y is large; but y jumpsfrom very large NEGATIVE values to very large POSITIVE valuesas x moves across the value zero. There is therefore a dis-continuity at x = 0; indeed, the formula does not define f(x) atall when x is zero.

Similarly a function such as

x+lx-l

has a discontinuity at x = 1.We add that a function such as

y = \x\

(i.e. y = the numerical value of x), whose graph we show (Fig. 5),is NOT discontinuous at x = 0.


EXAMPLES V

Name any points at which the folio wing functions are dis-

continuous:

A

7. tana;.

101-sma;

2.

5.

8.

11.

1

x2-V1

smx

sin a;

x-V1+a;2

4 —cos a;*

3.

6.

9.

12.

a : 2 -9 '

1cos a;

cos a:a;2-9*

1 + a^1 + 2 cos x

13. Prove that sin a; is continuous (i) when x = 0, (ii) whenx = JTT, (iii) when a: has any general value x0. [Remember thatsin (a; + h) — sin x = 2 cos (a; + |A) sin

14. Prove that cos a; is continuous (i) when x = 0, (ii) whenx = \TT, (iii) when x has any general value x0.

6. Rate of change. It is familiar that, under suitable con-ditions, the pressure p of a gas and its volume v are related by alaw of the form , ,

pv = constant,say pv = 6,

so that p is the function of v given by the relation

p = - .v

Suppose now that the volume of the gas is altered slightly;the pressure will change in sympathy. A standard notation is todenote the small change in the value of v by the symbol

8v [read 'delta v']

which may, of course, be positive or negative. As a result of thisvariation, the pressure sufiFers a change, which we denote similarlyby the symbol 8p.

BATE OF CHANGE 15

[The signs '8v' and 'S^' are composite symbols for these changesand stand for single algebraic variables. The (8\ V and *p' arenot separable entities.]

It is an obvious use of ordinary language to think of the ratio

8p8v

as measuring the rate at which p varies with v.We now form an estimate for this rate when v has a certain

given value which we may conveniently denote by vx. When thevolume has been increased by the amount 8vx (positive or negative)to the value v1+8v1, the pressure may be written as jPi + Sî,where px is the value of p corresponding to vv These numbersare connected by the relation

But

so that, by

Pi + &Pl

Pi

subtraction,

b

b

Hence

The rate of change of # with v is obviously best measured bytaking the values of the ratio 8p/8v for very small changes in v.In fact, if the ratio 8pj8v tends to a limit as the variation 8v istaken smaller and smaller, that limit will suit admirably to definethe rate of change. Now, vx itself being given, we have the result

lim ( 7

and so the rate of change of pressure, at an instant when the


volume is vlt is measured by the formula

Note. Since b is essentially positive, the rate of change isnegative. This agrees with the common-sense observation that pdecreases as v increases.

7. Gradient. Closely related to the idea of the rate of changeof a function is that of the gradient of a curve. We choose analternative illustration.

The electrical resistance of a metal may be expected to varywith the temperature, and experiment shows that for platinumthe resistance R when the temperature is 6° C. is given by therelation _

where Ro is the resistance at 0° C. and a, j8 are positive constants.

R

The relation between R and 8 may be illustrated by means ofthe accompanying diagram (Fig. 6).

Suppose that the resistance is Rl9R2 when the temperature is6lf 92, the corresponding points of the graph being Pl9 P2. Thevariation of resistance with temperature may suitably be measured

GRADIENT 17

by considering the ratio

resistance increase R2 — Rx

temperature increase 02 —1

In the diagram, draw the straight line P^^ and also the lineperpendicular to the ordinate through P2. Then, if the line

P1P2 makes an angle i[*f with the axis of 0, the value of tan?//,being the ratio (R2 — R1)l(02— #1) just indicated, gives a measure ofthe rate of change of R with 0 for the range of temperatures 0X to 62.

It is important to have an expression for the rate of change ofR with 0 at the value 0± itself. To do this, we take P2 progressivelynearer to P±; the chord PXP2 tends to take up a limiting positioncalled the tangent to the curve at Pl9 and I/J' assumes a limitingvalue r.

DEFINITION. The rate of change of R with 0 at the point 9X ismeasured by tan I/J, which is called the GRADIENT of the curve at Px.

We can find an expression for the gradient by a method similarto that used in the preceding paragraph. Suppose that the increaseof 0 from 0X to 01 + 801 causes an increase of resistance from Rx

to Rx + Si^. Then, by the formula for R,

± = RO{1 + ot{ei + S0±)

Subtracting, we have

Now R

"2— "l = = ^^1>

so that tan</r' = -^-~

ou1

= RQ{C

By definition, the gradient istan I/J = lim ' 1

80! J


Hence the gradient of the graph {the rate of change of R with 9) atthe value 0, is1

More generally, suppose that

is a given curve (Fig. 7) and P,P' two close points upon it given

x O

Fig. 7. Fig. 8.

by x = xvx1 + 8x1 and y = y1,y1 + 8y1 respectively. Then thechord PP' makes an angle <// with the #-axis, where

tanf = M

If P' is taken progressively nearer to P (Pig. 8), the chord PP'assumes (in ordinary cases) a limiting position, the tangent at P,and j/f' assumes a limiting value ip, where

= lim ~

This limiting value of tan j/r is called the gradient of the curvey = /(#) at the point for which x has the value xv

GBADIENT 19

EXAMPLES VI

1. Find the gradient of the straight line y = 3x + 4 at the pointswhere x = — 2,0,2,4 respectively.

2. Sketch the curve y = #2, and find the gradient at the pointswhere x = — 1,0,1, 2 respectively.

3. Prove that the gradient of the curve y = Xs, at the pointwhere x — xl9 is 3#f.

4. Prove that the gradient of the curve y = x2 + 5# + 4, at thepoint where # = xl9 is 2#x + 5.

5. Prove that the gradient of the curve y = x3 — x, at the pointwhere x = xl9 is Zx\ — 1.

6. Determine the range of values of x for which the gradientof the curve y = 2x3 — 9x2 + \2x is negative.

7. Prove that the gradient of the curve y = x2 — 4 is positivefor all positive values of x.

8. Find the equation of the tangent to the curve y = x2 at eachof the points (1,1), (-2,4), (0,0), (3,9).

9. Find the equation of the tangent to the curve y = x2 -f 2xat each of the points (1,3), ( - 1 , - 1), (0,0).

8. The differential coefficient. We now gather together(with some repetition) the ideas of the last two paragraphs.

Let f(x) be a given function, and xx a certain value of theindependent variable x. Suppose that x1 receives a small incre-ment; in the notation of the preceding paragraphs this would becalled Sa , but it is now more convenient to use the single letter h.The values of f(x) corresponding to the values x^x^h of theindependent variable are f(x1),f(x1 + h), and the rate of change off(x) between the values xv xx + h is

h

The rate of change off(x) at the point x1 is thus

h~* 0


and this limit (if it exists) is called the differential coefficient off(x) with respect to x at the value xv The phrase derivative of f(x)is also used.

Thus the differential coefficient of bjv with respect to v at thevalue vx is (p. 16) — b/vf, and the differential coefficient ofR0(l + ad + fid2) with respect to 0 at the value 6X is (p. 18)

ILLUSTRATION 2. To evaluate the differential coefficient of 3x2 —with respect to x when x = 4.

then /(4 + h) = 3(4 + h)2 - 4(4 + h)

= 3(16 + 8h + h2) -4(4 + h)

= 32+20A + 3A2,

and /(4) = 3 . 4 2 - 4 . 4 = 32.

Hence /(4 + h) - /(4) = 20A + 3A2,

and

The differential coefficient is the limiting value of this functionas h tends to zero, and so its value is 20.

In practice, it is customary to write the value of the differentialcoefficient so as to refer to a general value x; thus the limit

gives the differential coefficient of f(x) at the point x. But itshould always be kept in mind that a definite value of x is implied.

ILLUSTRATION 3. To find the differential coefficient of x2.The differential coefficient is

r (x + h)2-x2 .. 2hx + h2 .. , .hm 1 = lim T = hni (2x + A)h-+Q " hÔ h h~>Q

= 2x.

Hence the differential coefficient of x2 is 2x.

THE DIFFERENTIAL COEFFICIENT 21

ILLUSTRATION 4. The function x* has no differential coefficientat the point x = 0.

If f(x) = x\

As /& becomes smaller and smaller, 1/%* becomes larger and larger,

a n d

does not exist.

ILLUSTRATION 5. TAe differential coefficient of \\x is — 1/x2.The differential coefficient is

1 1,. x + h x .. x— (x + h)l i m = l i m ^ '

= l i m •=—, 7-r = — l i m — —

hôhx(x + h) h-*ox(x + h)_ 1

It is understood that o;#=O, otherwise there is no differentialcoefficient—indeed, the function is itself undefined.

9. Notation. Several notations are in use for the differentialcoefficient.

If f(x) is a given function of x, its differential coefficient withrespect to x is denoted by the symbol

Alternatively, if the function is denoted (as, for example, ingraphical work) by the letter y, so that

then the differential coefficient of y with respect to x is denotedby the symbol ,

dx*


Here we do NOT mean a process of division. The notation is anappeal to the eye based on the fact that, if 8y is written forf{x+8z)-f(x), then

f(x+8x)-f(x) 8y8x 8x

and the limit of this quotient is -j-.

The notations y', -=- {/(#)}CbX

are also in common use.When we wish to specify precisely that the differential coefficient

is evaluated at, say, x = a, we use one of the forms

\axj a

or other obvious modifications.The process of finding the differential coefficient is known as

differentiation, or differentiating the function.In illustration of the notation, consider the function x2 whose

differential coefficient (p. 20) is 2x.

(i) If f(x) = x\

then f\x) = 2x.

(ii) If y = x\

then - - = 2x,

dx

or y' = 2x.

If we evaluate the differential coefficient when x = 5, then

or (2) 5= 1 0 - or 2/5=10.

NOTATION 23

EXAMPLES VII

1. Prove that, if f(x) = x2 + x, then f(x) = 2x+l and that/'(0) = 1.

2. Prove that, if y = 4a:2, then -^ = 8a?.CLX

3. Prove that, if y = 5a:2, then y£ = 30.

4. Prove that, if/(z) = a;3, then/'(a:) = 3a:2.

5. Prove that, if y = x, then 2/' = 1.

6. Prove that, if y = x2- 8, then ^ = 3aA

7. Prove that, if 2/ = 5, then -^ = 0.

8. Prove that, if ?/ = 4a;-f 3, then y' = 4.

9. Evaluate/'(2) for each of the functions 4a;, 5a:2, a;3.

10. Prove that the function has no differential coefficientx-l

when x = 1.

10. Tangent and normal. We explained in § 7 (p. 18) whatis meant by the tangent to the curve

If P is the point (xv yx) of the curve (Fig. 9), then the tangent

O

Fig. 9.

at P makes with the a>axis an angle I/J such that

=f'(x±)= 2/1.


where f(x^) = y[ is the differential coefficient of f(x) evaluatedat xv Hence, by elementary analytical geometry, the equation ofthe tangent at P is

V -2/i= (*-*i)/'K)>

or y-Vi= (#-Zi)2/I.

DEFINITION. The normal to the curve at P is defined to be theline through P perpendicular to the tangent. Hence the equationof the normal at P is

or (y

ILLUSTRATION 6. To find the tangent and normal to the curve

y = 3x2- ±xat the point (4, 32).

We have proved (p. 20) that

/'(4) = 20.

Hence the equation of the tangent is

y- 32 = 20(a-4),

or 20x-y = 80-32

= 48,

and the equation of the normal is

or z+202/= 644.

EXAMPLES VIII

[Compare exx. 3, 4, 6 on p. 23.]1. Find the equation of the tangent and of the normal to the

curve y = 5x2 at each of the points (3,45), ( - 2,20), (0,0).

2. Find the equation of the tangent and of the normal to thecurve y = x* at each of the points (2, 8), (— 1, — 1), (0,0).

3. Find the equation of the tangent and of the normal to thecurve y = xz — 8 at the points where it crosses (i) the x-axis,(ii) the i/-axis.

CHAPTER I I

THE EVALUATION OF D I F F E R E N T I A LCOEFFICIENTS

1. Some theorems on limits. The direct evaluation of a dif-ferential coefficient from its definition is often a troublesomematter, and we must now devise a number of rules to shortenour labours. First of all, we 'borrow' from Pure Mathematicssome theorems about limits which seem obviously true but whichare not too easy to prove rigorously.

(i) The limit of the sum of two functions is the sum of theirindividual limits. Thus

lim {f(x) + g(x)} = limf(x) + lim g(x).

This result can be extended to any number of functions.For example, we can prove t ha t

1 — x2 1 — xz 1 — x*lim = 2, lim = 3, lim = 4;

Lx l x l x

A r ( l - z ) + ( l - a ; ) + ( l - : s ) ftand so hm — 1— ' = 9.

(ii) The limit of the product of two functions is the product oftheir individual limits. Thus

lim {f{x)g(x)} = lim/(a?) lim g (x).x-*-a x-+a x->a

This result can be extended to any number of functions.Properties (i), (ii) can be combined in obvious ways. For

example,

K (!-»*) ( l - ^ + q - ^ C l - ^ + q-a!*) (!-««)

= 4.3 + 3.2 + 2.4 = 26.

26 EVALUATION OF DIFFERENTIAL COEFFICIENTS

(iii) The limit of the quotient of two functions is the quotient oftheir limits, PROVIDED that the limit of the denominator is notzero. Thus

limf(x)

provided that Km g (x) #= 0.

For example,

Our next step is to establish three general theorems, after whichwe shall derive some standard formulae for differential coefficients.

2. The differential coefficient of a sum of functions.It is an immediate consequence of § 1 (i), p. 25, that

ddx{J

so that the differential coefficient of the sum of a number of functionsis equal to the sum of their differential coefficients.

The proof is similar to, but easier than, the correspondingtheorem for a product of functions given in the following para-graph, and is therefore left as an exercise for the reader.

3. The differential coefficient of the product of twofunctions. We prove that, if u, v are two functions of x, then

d dv dudx dx dx9

or (in other notation) (uv)f = uv' -

provided, of course, that the limits of -^-, ^- exist.ox ox

If u, v are the values of the functions when the independentvariable has the value x, and u+8u,v+8v the values for x + So;,then, by elementary algebra,

(u + 8u) (v + Sv) — uv Sv Su hu Sv-\ = u^- + v1E- + - r - .ox ox ox ox

COEFFICIENT

Since the

exist,

ddx(UV)

limits

— l im Û—— 1 1 i i _ i

= lim lu

dvdx

dv= U-J- + V

dx

OF

I

+ 8n

£)•dudx

dudx9

PRODUCT

8ulim —,

0 (v + 8v) — i8x

du--Z— l i m ov

OF TWO

r Svhm j -

FUNCTIONS

IV

— [definition; p. 19]

1 + lim I

[limit

[limit

u \— 8v\x 1of sum; p. 25]

of product; p. 25]

27

since lim Sv must be zero for lim (SvjSx) to exist.dx - • 0 ^ - > 0

The theorem is therefore proved.

COROLLARY. By repeated application of this theorem, we have

(uvw)' = (uv)'w+ (uv)wf

*= (u'v-\-uv')w+{uv)wf

In the same way, for any number of functions u, v, w,...,

(uvw...)f = w/t;t^...

4. The differential coefficient of a 'function of a func-tion*. We prove that, if u =f(x) is a given function of x, andy = g(u) a given function of u, so that y can be expressed as afunction of x in the form y == g{f(x)}, then

dy dy dudx "~ dudx*

Suppose that when x assumes the value x + 8x, the value of ubecomes u -f 8u, so that the value of y is y + hy. Then (but seethe Note below)


SO that

-Y- = lim -~ [definition; p. 19]dx sx->o$%

,. 8y8u= hm -s^ —

88

8y 8u= lim ~ lim — [product of limits; p. 25].

8x^0OU$x^0OX

(JuBut 8u->0 as 8x^0 (assuming that -=- exists, as is implicit in

dxthe enunciation). Hence

dy 8y 8u•f- = hm -f- hm •F-ax $u-+o oUfa-t.Q ox

dydu~~ du dx'

COROLLARY. The differential coefficient of the quotient u/v is

vu' — uv'

Let

Then, as above,

But (p. 21)

so that

dydx

dydx

y =

dudx

du"~ dx

d(%

V2

ujv = i

y—1 -\-11 •

V'1 -f U •

dv

Iduvdx

u dvv2dx

uv~\

dx

d(v~idv

1V*'

vuf

)

)dvdx'

— uv1

v2

TLT TTTi . 8y 8y 8u

Note. When we write -^ = -^---,8x 8u8x

we have in mind that the increment 8u is not zero. But we knowthat u is the given function f(x), and so there will usually exist

COEFFICIENT OF 'FUNCTION OF A FUNCTION* 29

(isolated) values of x at which/'(x) = 0; that is to say, the limitof Su -s- Sx is zero at such points, and so the initial step of dividing8y by Su is open to the suspicion of being division by a zerodenominator. To avoid this difficulty, we proceed as follows:

If Su = 0, we have

Sy = g(u + 8u)-g{u) _ g(u)-g(u)Sx Sx Sx

= 0,

and so -r—> 0Sx

as Sx -> 0 through values such that Su is zero.

But, by hypothesis, we are examining the case when — = 0,dx

and so we have the two relationsdy ^ du

o - 0dx dx

Hence it is still true that

dy __ dy dudx du dx*

ILLUSTRATION 1. To find the differential coefficient of

Write

duso that -7- = 5.

dx

We have proved (p. 20) that

dx du dx

= 2(5*+3). 5

« 10(53 + 3).


5. The differential coefficient of xn is /IJC"-1.

(This result is often proved by using the binomial series for(x + h)n. There is, however, the danger of argument in a circle ifthe proof of that series depends on the result known as Maclaurin'sexpansion, which comes later in a normal calculus course. Thereare, of course, other derivations of the series not open to thisobjection.)

(i) Suppose that n is a positive integer. If we assume that, forsome definite value of n9 we have

—(XX

then we can prove the general result by induction. For

(p. 26)

•= (n+l)xn.

If the result is true for any particular value of n, then it is truefor all subsequent values. But it is true for n = 0, since

dxK ' dxK '

dx d(xn)dx dx

Hence it is true for n = 1; hence for n = 2; hence for n = 3;and so on.

(ii) Suppose that n is a negative integer. Write n — — m, so thatm is a positive integer. Then

4- (x~™.xm) = 4~ (#-m+w) = i- (3°) = i- (1)dxx ' dxy ' dxK ' dxK

= 0.

[The reader may wish to remind himself from a text-book onalgebra about the rules for indices. The basic rule is that

THE DIFFERENTIAL COEFFICIENT 31

Hence a r ^ + ^ p = 0,ax ax

d(x\or x-™. mxm~^ + xm —^—- = 0,

dx

d(x~m)or mx~x + xm

~^T— = 0,dx

or —~—- = —mx~m-l>dxd(xn)

or, finally, = nxn~\ax

(iii) Suppose that n is a rational fraction. Suppose, that is, thatn is of the form pjq, where p,q are integers; we can regard q aspositive, but allow p to have either sign. Then

Differentiating by means of the formula for a 'function of afunction' and the rule already proved for integers, we have

Hidx

so that -=- (xp/q) = - x{p-dx q

= ?. xP-i~p+p/q

Hence -=- (xn) = nxn~x.dx

(iv) Suppose that n is irrational (for example, an unending, non-recurring decimal). We propose to regard it as obvious that nmay be approximated as closely as we please by a rational fraction,so that the theorem is true to as high a degree of accuracy as wedesire. But this seemingly innocent remark covers a number ofpoints of considerable difficulty, and a more advanced text-bookmust be consulted later when details are required.


EXAMPLES I

Find the differential coefficients of the following functions:

1. x*. 2. (z+3)4.

4. x2(x2+l). 5.

7.

10.

13.

16.

19.

9 9

1x'

1(2x + 3)5'

V(2^+3).

(x + l)~K

X2

8.

11.

14.

17.

20.

93

1

(3a

xK

(5a

< * •

i; - 5 ) 7

-3)Vl

3. (2a? + 3)4.

6. (a;-fl

19.

(z + 2)3 '

112.

15.

18. x-K

x

- 1 A3*

21.

24.

x+1'

6. The limit as JC->0 of sinx/jc is unity. In the diagram(Fig. 10), AOB is a triangle, right-angled at A, in which OA is ofunit length and the magnitudeof L BOA is x radians, wherex is small. (It should be re-membered carefully that, inwork of this kind, angles arealways measured in RADIANS.)

The line AC is drawn per- (

pendicular to OB, and an arc Yig. 10.of a circle of unit radius, withits centre at 0, passes through A and cuts OB in P. ThenP lies between B and C, and we propose to regard it as obviousby intuition thatJ AC<a,TcAP<AB.

Now AC = sin a:, arc.4P = x, AB = OB sin x, and so

sin x < x < OB sin x,

or 1< xsince

<OB.

THE LIMIT OF SIN XJX 33

Now let x-+Q] then OB tends to equality with OA, so that

lim OB = 1.

orIt follows that lim ——, which lies between 1 and Km OB, itself

smx

lim —— = 1.

has the value 1, and so

It is customary to write this result in the inverted form

sin#lim = 1.

mn irNote. Although lim exists, the value of the function

x—> Q X X

is indeterminate when x is actually zero.

7. The differential coeflBcient of sin x is cos x. For

sin (x + h) — sin x __ 2 cos (x + \h) sin \hh " h

= cos (x + \h). —fjr—•

Now, as ^ -> 0, cos (x + \h) -> cos a;

and -^f-*1 (§6)'(For the first limit, we have

cos (x + \h) — cos a; = — 2 sin (# + \h) sin JA.

But sin(# + |^) lies between —1 and 1, and sinJA->0, so thattheir product also tends to zero. That is,

cos (x + \h) — cos x -> 0.)

Consequently, since the limit of the product is the product ofthe limits,

.. sin (x + h) — sin xlim =^ = cos#.l,

d . . xor -y- sm x) = cos x.dx


8. The differential coefficient of cos x is - sin JC.

Let

and write

Then

Hence

and also

It follows that

y = cos a;,

U = \TT + X.

y = cos (u — \TT) = sin u.

dy-~- = cos it:du

dudx ~~

dy dy dudx du dx

= cosu

= cos (\TT-\-X)

= —sin x.

Note. Since 180° = TT radians, the differential coefficient ofsin#° is (77/180) cos #°, and the differential coefficient of cos#° is— (TT/180)sin#o. [For sino:0 = sin (TT^/180) in radian measure.]

EXAMPLES II

(It is most important that the student should acquire completemastery of the rules so far derived, and examples such as theseshould be practised regularly.)

Differentiate the following functions:

1. sin 2x. 2. cos 3x. 3.5 sin 5x. 4. x sin 2x.

5. x2 cosx. 6. 3x cos 3x. 7. (x + I)2sin 7a?. 8. sin (3x + 5).

9. (2z+l)3. 10. ll(x + 2)\ 11. z/sinz. 12.

13. sin2 a;. 14. cos2 a;. 15. sin3 a;. 16. cos3

17. a;sin2a;. 18. a:2cos2a;. 19. z2cos22z. 20.

21. cos (x- JTT). 22. sin2 (x + JTT). 23. Jx. 24. l/Jx.

25. a;*sin2a;. 26. ty(sinx). 27. xj(mnx). 28. x2j(sin

29. coseca:. 30. sec a:. 31. tana;. 32. cot a;.

33. cos 2a;°. 34. tan 3#°. 35. x sin x°. 36. sin2 x°.

DIFFERENTIAL COEFFICIENTS OF HIGHER ORDER 35

9. Differential coefficients of higher order. If f(x) is agiven function of x, its differential coefficient f'(x) is anotherfunction of x, having in general its own differential coefficient.This is called the second differential coefficient of f(x)> and isdenoted by the symbol »,,. x

Alternatively, if y = f(x), the second differential coefficient of yis written in one or other of the forms

„dtf' y •

In the same way, the differential coefficient of/"(#) is the thirddifferential coefficient of f(x); and so on. In this way we form asequence which it is customary to denote by the notations

dy d2y dzy d*y dryVy dx' d^' dx*' d^' "' dtf' "•

y, y', y'\ y'"> y™, .... y{r\ ....

For example, if y = x3,

then y' = 3#2, y" = 6s, y"' = 6, y™ = 0.

In the case of sin x, cos x, we can obtain a convenient expressionfor these coefficients:

Let y = sin x;

then y' = cos a;.

But, by elementary trigonometry,

cos x = sin {\IT + #),

so that y' = sin (|TT + a;).

Thus the effect of differentiating is to add \TT to the independentvariable. Proceeding in this way, we have

y" = sin(7r + #),

2/'" = sin (§77 +a*),

= s inl-


In the same way, if

y = cos#,

then yW = cos I - TT + x J.

EXAMPLES III

P i n d ^ d*y d3y t.. C i i l L L 5 7 O J 7 o I <

aa; aar aar1. a:5.

4. #s in# .

7. sin 2a;.

10. I/a:.

Dr each

2.

5.

8.

11.

of the following

a;3.

a:2 cos x.

cos 4a;.

l /(2s-3).

functions:

3.

6.

9.

12.

X.

x sin2 a;.

sin2 a;.

I/a:3.

13. Prove that, if f(x) is a cubic polynomial in x, then/(iv)(x) = 0.

14. Prove that, if f(x) is a polynomial in x of degree w, thenf{n)(x) is a constant independent of x.

15. Prove that, if y = sin mo;, then

16. Prove that, if y — #sin#, then

x2y" - 2xy' + (x2 + 2)y = 0.

17. Prove that ^- a ;^ = x ^ + ^f.

18. Prove that

10. Some standard forms. There are one or two basicformulae which follow directly from results already obtained, andwhich the reader should commit to memory. For convenience,we gather together the standard formulae of differentiation intothis one paragraph.

SOME STANDARD FORMS 37

I. GENERAL RULES.

,.., d . x dt; du . _„.( U ) ( W ) ^ + | ; ^ ( p - 2 6 ) -

(p. 28).

,. v dy dyduiv) -* = -* — (p. 27).

II. PARTICULAR FUNCTIONS.

( i )^ (z* ) = ^ - i (p. 30).

d(ii) -7- (sin a;) = cos a; (p. 33).

doc

d(iii) — (coso;) = — sin a; (p. 34).

CiOC

d(iv) -7- (sec a;) = sec x tan a;,

(tX

(v) -7- (tan a;) = sec2 x.(tX

(vi) -7- (cosec a;) = — cosec a; cot a;.ca7

(vii) — (cot a;) = —cosec2 a;.QJX

We prove the last four of these results (II, iv-vii):

(iv) Let y = sec x.

Then y = (cos a;)"1,

so that j - = ( - 1) (cos #)-2 -7- (cos x)

= sec x tan x.

4


(v) Let y =

Then

so that

(vi) Let

Then

so that, by

dydx

reasoning

y = sin x sec x,

cos x. sec x + sin x. sec x tan x

1+tan2^sec2 x.

y = cosec x.

y=(sinx)-\

similar to that given for sec#,

-^- = — (sin x)~2 cos xdx

= — cosec x cot x.

(vii) Let y = cot #.

Then i/ = cos # cosec x,

so that -p = (— sin x) cosec x + cos a; (— cosec x cot a:)

= - l ~ c o t 2 a ;

= — cosec2 x.

EXAMPLES IV"

Differentiate the following functions:

1. sec 2x. 2. sec2 2x. 3. tan2 2^.

4. tan3 3x. 5. # cosec #. 6. x2 cot 2x.

7. sec#tan#. 8. <J(secx). 9. cosec32x.

11.* The inverse circular functions.I. THE INVERSE SINE. The relation

v = sinw

serves to define v in the ordinary way as a function of u. But itcan also be used to define u as a function of v, and we use thenotation . ,

u = sm"1 v* This paragraph may be postponed, if desired.

THE INVERSE CIRCULAR FUNCTIONS 39

to mean that u is the function whose sine is v. The notation

u = arcsinvis also used.

It is familiar from elementary trigonometry that, if

sin u = sin a,

then

where n is a positive or negative integer. Hence the relationu = sin™1 v does not define u as a single-valued function of v.The graph shown in the diagram (Fig. 11) implies this; when u isgiven, v is determined uniquely, but, when v is given (lyingbetween — 1 and + 1) there are infinitely many values of u.

477

Fig. 11.

Our immediate problem is to evaluate the differential co-efficients, with respect to x, of the function sin"1 x.

We writey = sin"1 x9

so that, by the definition,

x = sin y.

Differentiate with respect to x. Then

dy

dy=

dx cos y1


Now inspection of the curve x = sin y, shown in the diagramflu

(Fig. 12), reveals that the gradient -~- is positive when y liesdx

between — - , - and, more generally, between2 2

~, -t

where n is any positive or negative integer; on the other hand,

the gradient is negative when y lies between ~^,-jr and, more2 2

generally, between

(2U+ 1)TT-| , \

Hencedy_ 1dx~~±

yJ(l-x2)'

with positive sign if the angle sin"1 x is between 2UTT — , 2mr + —,2 2

and with negative sign if sin"1 x is between (2w+l)7r —--,_ 2

In particular, if sin-1 a; is an ACUTE angle, then

dx " ^ ( l - * 2 ) '

II. THE INVERSE COSINE. Similar con-

siderations guide us in finding ~- when

P4TT

or

Differentiate

so that

y

X

with

dydx"

= cos"1 a;,= cosy,

respect to x. Then

dysin y

1siny

- _|_

(

0

2TT

)

-I*

Fig. 12.

THE INVERSE CIRCULAR FUNCTIONS 41

To determine the choice of sign, we appeal to the curve x = cos y

shown in the diagram (Fig. 13), and see that the gradient -^- is

positive when y lies between 77, 2TT and, more generally, between

the gradient is negative when y lies between 0,77 and, moregenerally, between

dy 1Hence - / = ± -r-—

with positive sign if the angle cos"1 x is between (2n 4-1) 77, (2n 4- 2) 77,and with negative sign if cos"1 a: is between 2nn, (2n + 1) n.

In particular, if cos"1:*; is an ACUTE angle, then

dy = 1dx ^(1—ju-) /

III. THE INVERSE TANGENT. The evaluation, 4TT - •

of -~ when

y = tan"1 x,

or x = tan y, 27r"

is much simpler. Differentiate with respect to x.

Then 1 - sec2y^, Oax

so thatsec2 y

1l+xr Fig. 13.

EXAMPLES V

dii1. Evaluate -^ for each of the functions sin"1 a;, cos"1 x. tan-1 x

dxwhen the angle has the value

, . . 77 577 1377 , . 5TT 777 . . . 1177

(!) 3, (u) T , (m) - r , (iv) T , (v) T , (V1) — .


Taking the angle to be acute, find -~- for each of the functions:

2. a; tan"1 a;. 3. sin"1 (a;2). 4.

5. x3 sin-1 (2a?). 6. x2 cos"1 (a;2). 7. (tan"1 a;)2.

8. sec"1 x. 9. cosec"1^. 10. cot"1^.

11. (sin-1^)2. 12. ^(cos"1^)3. 13. a:2 sec"1 a;.

14. ajcosec-1^. 15. a^sin-1 a:)2. 16. l/fsinâ;).

12.* Differentials. It follows, from the definition of thedifferential coefficient f'(x) of a function y=f(x), that the incre-ment 8y in the value of y when x receives a small increment Sa; is(in ordinary cases) not very different from f\x) 8x, so that

8y=f(x)8xapproximately.

For many purposes, however, it is desirable to work insteadwith an exact relation, and this leads to the idea of a differential,which we must now explain.

When the increment Sx is given an arbitrary value, not neces-sarily small, the expression

where f'(x) is the differential coefficient of the function y~f(x),has a definite value, and is denoted by the notation dy9 so that

dy=f(x)8x.

The expression dy} or f{x) 8x, is called the differential of y corre-sponding to the increment 8x. The relation dy = f'(x) 8x is EXACT,without any approximation; it is an automatic consequence ofthe definition of the differential. The differential dy and the incre-ment 8x are proportional, the coefficient of proportionality being

[Note that the increment S^ is not usually equal to the differ-ential dy, although they are approximately equal when Sa; issmall.]

* This paragraph may be postponed, if desired.

DIFFERENTIALS 43

In particular, we can assign a meaning to the symbol dx, whichis the differential of the function x itself corresponding to theincrement 8x; for the differential coefficient of x is unity, so that

dx = 1. 8x

= 8x.

It follows that, if dy, dx are the differentials corresponding tothe (arbitrary) increment 8x, then

dy = f'(x) 8x, dx = 8x.

Hence the differentials dy, dx satisfy the fundamental exact relation

dy — f'{x)dx.

The notation dy,dx for differentials is, of course, designed to

agree with the composite symbol -j- for the differential coefficient.dx

In fact, if we divide the relation just given by dx, we obtain theequation ,

or dy — -~ dx,ax

and obtain ~- as the actual coefficient of the differential dx.ax

In practice, especially in physical applications, it is customaryto use the symbols dy, dx when the increments 8y, 8x are reallyintended.

With this incorrect usage, the equation

dy=f'(x)dx

is often written where 8y ^f'(x)8xis meant.

The difference between dy, 8y may be illustrated graphically,as in the diagram (Fig. 14). Let

P(x,y),P'(x+8x,

be two points of the curve


Draw PM perpendicular to the ordinate through P' , and let thetangent at P meet that ordinate in Q. Then

PM = Sx = dx9

MPf = Sy.

Moreover,

Hence

Fig. 14.

| | = tan MPQ

= tan (angle between PQ, Ox)

= f{x) (p. 23).

y9

by definition.Thus MP' represents hy, while ikTQ represents dy. When Sx is

small, the difference between MP',MQ is very small indeed; infact, QP' is then small in comparison with S#.

ILLUSTRATION 2. A wire is pulled out so that its length is increasedby 1 per cent. Assuming that the wire can be treated as a cylinderof small cross-section and that the volume remains constant, by whatpercentage is its diameter decreased!

DIFFERENTIALS 45

Suppose that the constant volume is V and that the lengthis x. Then the area of a cross-section is Vjx, so that the radiusr is given by the relation

r =

Hence the differentials dr, dx are connected by the relation

so thatdr—r

Idx-—.2 x

For small variations, the differentials are approximately in thesame ratio as the increments, so that, if

IT1

Too'

then _200'

and the decrease in radius (or diameter) is approximately J percent.

EXAMPLES VI

1. If y — 5#3, find the percentage increase in y correspondingto an increase of 2 per cent in x.

2. If y = l/#4, find the percentage decrease in y correspondingto an increase of \ per cent in x.

3. If y = 2x2, find the approximate increase in y when x increasesfrom 5 to 5-01.

4. If y = 1/jx, find the approximate decrease in y when xincreases from 16 to 16-03.

13.* The differentiation of determinants. The process ofdifferentiation when applied to determinants may be illustratedby the special case when the order of the determinant is 4. Wethen have

wx w2 ws w4

Pi P2 P3 Pi* This paragraph may be postponed, if desired.

46 EVALUATION OF DIFFEBENTIAL COEFFICIENTS

where ul9 u2i ">,Pz,P& are all functions of the independent variablex. By definition,

A = S( + UiVjWjtpJ),

where the sign + or — must be taken according as the numberof interchanges required to bring the four distinct integers i, j , k, Ito the order 1, 2, 3, 4 is even or odd.

It follows that, if dashes denote differentiations with respect to x,

A' ^yL{±ufivjwkpl)

+ y£(±uiv'j wkpl)

+ S(±«< vf w'kPl)

+ S(± ui Vj wk p[).

Hence

wx w2 wz w±

Pi Pz P* P*

h u2 uz

Wi U>2 WZ W4

Pi P% Pz PA

IX u2 u3

w2 w'z W'I

Pz Pz Pi

ul U2 UZ U*

w1 w3

Pi P2 Pz n

Thus A' is the sum of the four determinants each obtained by differ-entiating one row of A and leaving the rest unaltered.

Similarly, A' is the sum of the four determinants each obtained bydifferentiating one column of A and leaving the rest unaltered.

DIFFERENTIATION OF DETERMINANTS

EXAMPLES VII

1. Prove that, if

x* (x+1)*

As

then

A'= 3

x2

x

1

2. Prove that, if

then

(z + 2)3

(a:+2)

1

( a : - a ) 3 (a; - a ) 2 1

{x-bf (x-b)2 1

(x-cf (x-cf 1

(a; — a ) 3 ic —a

{x-bf x-b

(# — c)3 a; — c

(a+ 2)

1

CHAPTER III

APPLICATIONS OF DIFFERENTIATION

1. Illustration from dynamics; velocity, acceleration.Suppose that a particle is moving in a straight line so that itsdistance at time t from a fixed point 0 of the line (Fig. 15) is x,

P PFig. 15.

and at time t+8t is x + Sx. The particle describes the distanceSx in the time 8t, and so its average speed in that time is

SxSf

We thus obtain an expression for the velocity at time t in the form

,. Sxlim £-

dx

TfLet us denote this velocity by u, where

dx

and proceed to find an expression for the rate of change ofvelocity with time. This rate is, in accordance with our usualprinciples, §u

lim -x-,st->o ot

where u + hu is the velocity at time t + St. This limit, called theacceleration at time t, is

du

d2xor (p. ) 35 w .

ILLUSTRATION FROM DYNAMICS 49

Note. Differentiations with respect to time are often indicatedby dots on top of the dependent variable. Thus the velocity andacceleration are x, x respectively.

Suppose, for example, that a particle moves in a straight lineso that its distance at time t from a fixed origin 0 is a sin nt; suchmotion is called simple harmonic motion. We have

x = a sin nt,

so that x = an cos nt,

and x = — an2 sin nt

= -n2x.

Thus the velocity at time t is an cos nt, and the acceleration is— an2 sin nt. The acceleration, being also expressed in the form— n2x, is directed TOWARDS the origin and is in magnitude pro-portional to the distance of the particle from the origin.

EXAMPLES I

Find the velocity and acceleration of a particle moving in astraight line so that its distance x at time t from a fixed pointis given by the following relations:

1 . X — COS7Tt.

3. x= 5t-

5. x =

7. x= (*-32)(*-64).

2. x = sinîrt.

4. x = 32*2.

6. x = t2 cos \nt.

8. x = tsin2^7rt.

2. Maxima and minima illustrated.ILLUSTRATION 1. A log of wood is in

the form of a cylinder of radius a. It isrequired to cut as strong a beam as possiblehaving rectangular section, on the assump-tion that strength is proportional to widthand to the square of height (Fig. 16).

If x, y denote the width and heightrespectively, and s the strength, then

8 = kxy2,

where & is a constant depending on the nature of the wood.

50 APPLICATIONS OF DIFFERENTIATION

By the theorem of Pythagoras,

x2 -f y2 = 4a2,

so that s = kx(4a2 - x2).

We have to find the greatest value of s.For this purpose, draw a graph of s against x (Fig. 17). It is

easy to obtain the shape shown in the diagram, where, of course,only the part between the values x = 0, x = 2a is relevant.

M

Fig. 17.

A glance at the diagram (Fig. 17) shows that the greatest valueof s occurs where x = OM and s — MP. Now the characteristicquality of the curve, from which we shall be able to calculatethese values of x and s, resides in the fact that the curve has, asit were, 'stopped rising' at P, so that the tangent at P is parallelto Ox. In the language of Chapter I, § 7, p. 16, the gradient ofthe curve is zero at P; that is, the greatest value of s occurs at apoint where

dxWe have the relation

so that

Hencedsdx"

0 when

s =

ds _dx

kx(4a2 — x2)4a2kx — kx3,

x- = %a\

or

MAXIMA AND MINIMA ILLUSTRATED 51

The relevant value of x is positive, and so we see that the beamof greatest strength has width fa^/3, height fa^/6, and thereforestrength ^ a 3 ^3.

EXAMPLES II

1. Taking k = J, a = 1, draw the graph s = £#(4 — x2), plottingthe points which arise from x = — 3, — 2^, — 2,..., 2,2\y 3.

2. By means of a graph, find the greatest value of the functiony = 2x — x2.

3. Find the least value of the function x2 — 4x.

3. The determination of maxima and minima. The dia-gram (Fig. 18) illustrates the graph of a function

with the properties that the value of f(x) at P exceeds that atany point NEAR IT, while the value at Q is less than that at anypoint NEAR IT. The function f(x) has a maximum at P and aminimum at Q.

O 0

Fig. 18. Fig. 19.

It is important to realize that the words maximum and minimumrefer to the parts of the curve near P and Q respectively. Clearlythe values at P and Q need not be the greatest or the least valueso(f(x) over the curve as a whole.

To find the positions of the maxima and minima, we followprecisely the argument of Illustration 1 (p. 49). The criterionis that , ,.

the gradient of the curve is zero


at a maximum or minimum. In other words, if the function f{x)has a maximum or a minimum value where x = a9 then

f'(a) = 0.

The converse result is NOT necessarily true At the point R inthe diagram (Fig. 19) the gradient is zero, although the functionhas neither a maximum nor a minimum there.

EXAMPLES III

Find the values of x at which each of the following functionshas a maximum or minimum value, and illustrate by sketchingthe curve y = f(x).

1. x2+l. 2. x2-2x. 3. x*-3x-4:. 4. x*-

5. sin#. 6. sin2#. 7. cos a;. 8. eos

4. Increasing and decreasing functions. We should saythat, from its appearance, the function f(x), whose graph isshown in the diagram (Fig. 20), is a steadily increasing functionof x. We must now consider how this feature is to be interpretedin terms of the differential coefficient of/(#).

If x undergoes a small positive variation 8x, then f(x), bydefinition, must increase, so that

f(x+8x)-f(x)

is positive. It follows that the quo-tient

f(x+8x)-f(x)Sx

is essentially positive, and so

f(z+8x)-f{x) Fig>

/w

o

A;

CX

c+Sx x

is positive. Hence the differential coefficient of a function is positivein an interval where the function is increasing.

Similarly, the differential coefficient of a function is negative inan interval where the function is decreasing.

The converses of these two results are also true.

INCREASING AND DECREASING FUNCTIONS 53

ILLUSTRATION 2. To prove that, if x is positive, the value ofsin# lies between

xz xz X5

* _ _ and a - g ^ .

X2

Write u = &\xix — x + —:.o!

X2

Then u' = cos x — 1 + ^r,

u" = — sinaj + o;,

The differential coefficient of the function u" is 1 — cosrr, whichis necessarily positive; hence u" is an increasing function of x.But u"(0) = 0, and so u" increases steadily from zero. That is,u" is positive when x is positive.

The differential coefficient of the function u' is u", which wehave just proved positive; hence u' is an increasing function of #.But ^'(0) = 0, and so u' increases steadily from zero. That is, u'is positive when x is positive.

Now consider u itself. Its differential coefficient u' has justbeen proved positive; hence u is an increasing function of x. Butw(0) = 0, and so u increases steadily from zero. That is, u ispositive when x is positive.

x3

Hence sin x exceeds x — -r for positive values of x.

In the same way, by considering the function

we may prove that sin# is less than x — 7- + —r for positive values

of #.The result is therefore established.

EXAMPLES IV

1. Prove that, if x > 0, then

, X2 , X2 X*


2. Prove that, if x lies between 0 and 7r, then

sin x > x cos x.

3. Prove that, if x lies between 0 and TT, then

4. Find the ranges of values of x for which the followingexpressions are increasing functions, and illustrate your answersgraphically:

(i) x2-x. (ii) xz + x-2. (iii) x*-3x + 2.

(iv) a 2 + l . (v) x*-2x2. (vi) 2x*-

5. The second differential coefficient and concavity.The diagram (Fig. 21) represents a curve where, in the ordinarysense of the phrase, 'the concavity is upwards'. The curve, asusual, is the graph of a function

2/=/(*).

oFig. 21. Fig. 22.

As x increases, the corresponding point describes the curvebetween A and B, moving in the sense indicated by the arrow.The gradient is negative to the left of C and positive to theright, but the important thing is that it INCREASES STEADILY

with x. At A the gradient has a fairly high negative value;between A and G it remains negative but decreases in numericalvalue so that (being negative) it actually increases; at C it hasincreased to the value zero. After C, the gradient remains positive,and continues to increase.

But the gradient of the function is/'(#), and the condition thatf'(x) should increase is that ITS differential coefficient/"^) should

THE SECOND DIFFERENTIAL COEFFICIENT 55

be positive. Hence the condition for the concavity of the curvey = f(x) to be 'upwards' during an interval a^x^b is that f"(x)should be positive in that interval.

In the same way, the diagram (Fig. 22) represents a curvewhose concavity is 'downwards' between E and F, The gradientis positive at E, decreases steadily to zero at O, and thereaftercontinues to decrease as its value becomes greater and greaterwith negative sign. Hence the condition for the concavity of thecurve y = f(x) to be 'downwards9 during an interval e^x^f is thatf"{x) should be negative in that interval.

6, Points of inflexion. The diagrams in § 5 (Figs. 21, 22)illustrated curves in which the concavity was always upwards, oralways downwards, in the interval considered. The present dia-gram (Fig. 23) illustrates a curve

Vi

in which the concavity is 'down-wards' between U and TF, but'upwards' between W and F,changing sense at the point Witself.

The value of f"{x) is negativeto the left of W and positive tothe right. At W, the value off"(x) is zero.

It is instructive to trace the _progress of the gradient as thecurve is traced by a point mov- Fig. 23.ing from U, through W, to V.At U the gradient is positive, but it decreases steadily as thepoint approaches W; after W, however, the gradient begins toincrease again, so that it has a minimum value at W. Hence thedifferential coefficient of the gradient vanishes at W; that is,

f"(x) = 0at W.

A point such as IF, where the concavity changes sense, is calleda point of inflexion of the curve.


The condition f"{x) = 0 is necessary for a point of inflexion,but not sufficient to ensure it, as the first two examples belowillustrate.

EXAMPLES v

1. Sketch the curve y = \xz for values of x between — 2,2.Verify that the curve has a point of inflexion at the origin.

2. Sketch the curve y = \x^ for values of x between — 2,2.Verify that the curve has NOT a point of inflexion at the origin

d2yalthough -z-| vanishes there.

ax

3. Sketch the curve y = \{x + xz) for values of # between — 3,3.

4. Sketch the curve y = |(4# — xz) for values of x between-4 ,4 .

7. Discrimination between maxima and minima. We arenow able to devise a method, which can be stated in two alterna-tive forms, for deciding whether a turning value of a functionf(x) is a maximum or a minimum. The diagram (Fig. 24) showsthe curve

with a minimum at P and amaximum at Q (see p. 51).

FORM I. Consider the pointP. The gradient there is zero,and the characteristic featureassociated with the minimumis that the gradient passes fromnegative, through zero, to positiveas x passes, in the increasingsense, through its value at P. **&• 24b-Thus:

/ / f'(x) is negative for values of x just less than xv zero at xx

itself, and positive for values of x just greater than xv then f(x) hasa minimum at x = xv

Similarly, {/ f'{x) is positive for values of x just less than xl9

zero at xx itself, and negative for values of x just greater than xl9

then f(x) has a maximum at x = xv

DISCRIMINATION BETWEEN MAXIMA AND MINIMA 57

FORM II. The first form just given may be reworded to statethat the gradient of the curve is an increasing function of a; at aminimum and a decreasing function at a maximum. Hence iff"{x) is positive at a turning point, the function has a minimumthere; and, if f"(x) is negative at a turning point, the function hasa maximum there.

Note. It is often better to use the first form rather than thesecond, especially if (as often happens) the second differentialcoefficient is awkward to manipulate.

ILLUSTRATION 3. The potential energy of a uniform rod.Suppose that OA (Fig. 25) is a uniform rod, of length 2a and

weight W, suspended from one end 0. If it makes ah angle 6with the downward vertical, the potential energy F is defined tobe the function of 0 given by the relation

V = const. — Wa cos 0.

HencedV-— = Wa sin 0,do

= Wa cos 6.

Fig. 25.

The significantly distinct values of 0 which giverise to the turning values of the potential energy arederived from the equation

so that 0 = 0 or TT.To distinguish between the cases 0 = 0, TT:FORM I.

(i) 0 = 0 (the rod hanging down).dV

When 0 is just less than zero, -j^ is negative; and, when 0 isdV

just greater than zero, -=j is positive. Hence the potential energy

is a minimum.(ii) 0 = TT (the rod 'standing up9).

dVWhen 0 is just less than TT, sin0 is positive, so that - ^ is

ddpositive; and, when 0 is just greater than TT, sin# is negative, so

dVthat -j-r is negative. Hence the potential energy is a maximum.


FORM II.

(i) 9 = 0.d2V

When 0 = 0, -j^ = Way so that V is a minimum.

(ii) 0 = TT.

d2VWhen 6 = IT, -j^ = — Wa, so that V is a maximum.

EXAMPLES VI

The positions of the maxima and minima of the followingfunctions were found in Examples III (p. 52). Use both formsof test to distinguish between maxima and minima.

1. x2+l. 2. x2-2x.

3. xz-3x-4:. 4. x*-2x2 + 7.

5. sin x. 6. sin 2x.

7. cos a;. 8. cosja;.

9. Prove that the curve

y = z 3 - 3x2- 9x

has a maximum where x = — 1, a minimum where x = 3, and apoint of inflexion where x = 1. Sketch the curve, and check thatthe concavity in your sketch agrees with the results of § 5 (p. 54).

10. Find the results analogous to those of Ex. 9 for the curves

(i) y = x*-12x,

(ii) y = 2x* + 3x2.

11. Find the maxima and minima of the function

sin 3# + 3 sin x9

and distinguish between them.

12. For the function y = 2xz- 9x2 + 12#,

(i) find the maxima and minima, and distinguish betweenthem,

(ii) find the values of x for which y is (a) an increasingfunction, (b) a decreasing function,

(iii) sketch the curve.

THE SKETCHING OF SIMPLE CUBVES 59

8. The sketching of simple curves. The ability to sketchquickly the principal features of a curve

is important for many purposes. The following properties areimportant:

(i) passage through characteristic points;

(ii) symmetry;(iii) the sign oif(x) for various ranges of values of x;

(iv) the gradient;

(v) maximum or minimum values of f(x);

(vi) the concavity.

Common sense and experience must decide which of theseproperties seem likely to be most helpful for a particular curve.The illustration which follows is typical.

ILLUSTRATION 4. To sketch the curve

- 3 ) ,

By differentiation,

y' = 3 3 2 -

y" = 63-12 = 6(3-2).

The critical features may be expected to be:

(a) where y = 0, but it is not easy to find such points for thisparticular curve;

(b) where y' = 0, that is, where x = 1,3;

(c) where y" = 0, that is, where 3 = 2.

We may therefore construct a table as follows:

X

y

y'

y"

x<\

?

+

-

1

3

0

- 6

1 <x<2

?

-

-

2

1

— 3

0

2<#<3

?

-

+

3

- 1

0

+ 6

x>Z

?

+

+


The concavity is 'downwards' for x < 2 and 'upwards' for x > 2;there is a point of inflexion at (2,1), with gradient —3.

There is a maximum at (1,3) and a minimum at (3, — 1).These features are all gathered together in the diagram (Fig. 26).

EXAMPLES VIISketch the curves given by the

following equations:

1. y = x2.

2. y = x*.

3. y = x2-3x + 2.

4. y=(x-l)(x-2)(x-3).

5. y — xz — 3x.

6. y = x* -2x2.

See also Examples V (p. 56).Fig. 26.

9. Rollers theorem. To verify that, iff(x) is a function of xcontinuous between two points x = a,x = b, at each of which itvanishes, the curve y — f(x) having a tangent at all points betweena, b, then there exists at least one value of x between a, 6 at whichf\x) vanishes.

The diagram (Fig. 27) is typical of the form which the graphy =f(x) must take, cutting the #-axis at the points A,B corre-sponding to x = a, b.

We propose to regard it as obviousfrom the diagram that there existsbetween A,B at least one pointwhere the tangent is parallel tothe #-axis. For a rigorous proof,the reader should consult a moreadvanced text-book.

O

Fig. 27.

EXAMPLES VIII

Verify RoUe's theorem by actual calculation of /'(#), and alsoby drawing a graph, for each of the following functions:

1. f(x) = x2-3z + 2; a = 1,6 = 2.

ROLLE'S THEOREM 61

2. f(x) = (x-l)(x-2)(x-3);a= 1,6 = 2.

3. f(x)==x2(x-l); a = 0,6 = 1.

4. f(x) = x(x-l)2;a = 0,b = 1.

5. /(#)== sin #; a = 77,6 = 2TT.

6. f(x) = cos2 a;; a = £77,6 = f7r.

10. The mean value theorem. To verify that, if f(x) is afunction of x (continuous and) having a differential coefficient* ateach point between x = a,x = 6, then there exists a value £ of xbetween a, b with the property that

In the diagram (Fig. 28), let A,B be the points of the curvey = f(x) for which x = a, 6, and draw AP perpendicular to theordinate through B. Then

f(b)-f(a) PBb — a AP

= tan PAB.

But the graph shows thatthere must be at least onepoint X between A,B wherethe tangent is parallel to AB.If this point is given by x = £,then

so that

f(b)-f(a)b — a =/m

This is equivalent to the required result.

ALITER. The following proof is instructive, and worthy of closestudy. It is the basis of the proof of a generalization of the meanvalue theorem to be given later (Vol. II, p. 44).

* Having a differential coefficient ensures that the function is continuous.It should be noted, however, that the converse need not be true—a functionmay be continuous but not have a differential coefficient. For example,if y = + x when x is positive and — x when x is negative, then (i) y iscontinuous at x — 0; (ii) yf = 4-1 when x is positive; (iii) y' = - 1 when xis negative; but (iv) y/ does not exist when x is zero.


Define a function u(x) as follows:

Since f(x) is continuous between a, 6, so also is u(x). Further, itis easy to see that u(a) = 0,^(6) = 0, so that u(x) satisfies theconditions of Rolle's theorem (p. 60). Hence there exists a value£ of x between a9 b such that

*{£) = 0.

Now u'(x) = —/

and so / ' ( I ) — r $b — a

as required.ILLUSTRATION 5. To prove that, iff(x) is a continuous function

of x such that f'(x) = 0 throughout a given interval, then f(x) isconstant in that interval.

The result is, of course, obvious graphically, but the followingproof is of interest.

If xl9 x2 are any two values of x in the interval, then, by themean value theorem, there exists at least one value of £ betweenxl9 x2 such that

But f'(£) = 0,

and so f(x2)=f(x1).

Since this is true for all values of xl9 x2 in the interval, f(x) mustbe constant.

EXAMPLES IX

1. Prove that, if f(x) is a continuous function of x such thatf\x) = 2 throughout a given interval, then f(x) = 2x + b in thatinterval, where 6 is some constant.

2. Verify the mean value theorem by calculating | for each ofthe following functions, and illustrate your results graphically.

(i) f{x) = x*;a = 0,b=:2.

(ii) f(x)==x2 + x] a = 1,6 = 3.

(iii) f(x) = xz + 3x; a = -1 ,6 = 2.

THE REAL ROOTS OF THE EQUATIONS 63

11. The real roots of the equations/(JC) = 0,/'(x) = 0.Suppose that we have to solve an equation f(x) = 0, and that thediagram (Pig. 29) represents the curve

Fig. 29.

We restrict ourselves for the moment to the case when the rootsare ALL DISTINCT. This restriction, the purpose of which is toensure that f'(x) and f(x) are not both zero together, is essentialfor the work which follows.It is assumed, too, thatthe curve is continuous,with a continuously turn-ing tangent.

Our aim is to discuss,without the analytical de-tail required for a fullstudy, those propertieswhich may be assertedwith reasonable confidencefrom a study of the graph.

(i) The sign of f(x). If x is imagined to increase from — oo to+ oo, the value of f(x) changes sign every time that x passesthrough a point where f(x) = 0. Hence the number of (real) rootsis equal to the number of changes in the sign of f(x) as x increasesfrom —oo to -t-oo.

If f(a)J(b) have the SAME signs, then there is an even number(or zero) of roots between a, b.

If f(a)}f(b) have OPPOSITE signs, then there is an odd number ofroots between a, b.

In particular, if

f(x) = aoxn + a^-1 +...+an

is a polynomial in x of degree n, and if n is odd, then, for largenegative values of x, f(x) has sign opposite to a0, and for largepositive values of x, f(x) has the same sign as a0. Hence if f(x) isa polynomial of ODD degree, the equation f(x) — 0 has at least onereal root,

(ii) The sign off(x). Suppose, for the sake of explanation, thatthe sign of f'(x) is positive at a point f where f(x) = 0. Then f(x)is an increasing function of x at x = £, so that the curve y = f(x)


goes 'above' the axis as x increases from £. As x, still increasing,approaches the next root x = y, the curve 'drops' towards thea;-axis, so that f'(x) is negative there. Hence the differential co-efficient f'{x) assumes alternate signs at the successive roots of theequation f(x) = 0.

Applying the result (i) above, we see too that the equationf'(x) = 0 has an odd number of roots (at least one) between con-secutive roots of the equation f(x) = 0.

(iii) Since f'(x) has constant sign between successive roots ofthe equation f'(x) = 0, the value of f(x) either rises or, alterna-tively, falls ALL THE WAY between successive maxima andminima. Hence the equation f(x) = 0 has either one root or noroots between consecutive roots of the equation f {x) = 0.

y

\

\oV ^/ A \

Fig. 30a.

y

\

\ oX X

Fig. 306.

When the roots of the equation f(x) = 0 are not all distinct, caremust be taken in applying the rules. Thus the equation

whose graph 2/ = f(x)is shown in Fig. 30a, has a double root x = £, corresponding tothe point A; and in Fig. 306 it has a triple root x = y correspondingto the point B. The function f(x) has the same sign for values of xon either side of £, but opposite signs for values on either side of rj.

It is probably wiser (at any rate, for the present) to treat eachcase on its merits as it arises rather than to seek a more elaborateset of rules.

ILLUSTRATION 6. / /

THE REAL ROOTS OF THE EQUATIONS 65

prove that the equation Pn(x) = 0 has no real roots when n is evenand exactly one real root when n is odd.

This is a typical problem. Assume that the result is true forn = 1,2,..., JV. We prove that it is then true for n = N+ 1.

(i) Suppose that N is even. By the assumption, the equation

has no real roots.Consider the equation

By direct differentiation, we have the relation

so that PH+1(X) is never zero. Hence the polynomial PN+1(x)either increases steadily or decreases steadily. But, since N +1is odd, PN+1(x) is negative for large negative values of x andpositive for large positive values. (Indeed, PN+1(x) is obviouslypositive and increasing for positive x.) Hence, the graphy = PN+1 (x) crosses the z-axis once, and once only, and theequation i# + 1 (x) = 0 has precisely one root.

(ii) Suppose that N is odd. By the assumption, the equation

PN(x) = 0

has exactly one real root, say £, which cannot be zero.Consider the equation

3m (*)-<>.where, as before, P^+1 (x) = PN (x),

so that

The curve y = PN+X{%) therefore has a turning value at x — £; and,as this is the only turning value, while PN+i {x) is large and positive(N+l being even) for large positive or negative x, the turningpoint is actually a minimum and gives the least value attainedby PN+1(x). Moreover,

gN+l


since PN(£) is zero. Hence, as N +1 is even, and £ is not zero,PN+I(£) is positive. Since the least value of PN+I(X) is positive,PN+I(X)

c a n never be zero.The result is therefore true for N +1 whether N is even or odd.But it is clearly true for w = 1. Hence it is true for n = 2,3,4,5,...,

and so generally.

EXAMPLES x

1. Prove that, if f(x) ==x2(l — x)2, then the roots of the equationf"(x) = 0 are distinct, and lie between 0 and 1.

Prove the corresponding result for the equation g'"(x) = 0,where g(x) = xz(l — x)s.

2. Indicate in a single diagram the relative positions of the rootsof the equations

, x2 ^ , x2 x* „

12.* Mean value theorems for two functions.(i) CAUCHY'S MEAN VALUE THEOREM. TO prove that, if the

functions f(x), g(x) have differential coefficients which do not vanishsimultaneously in the interval a, b, and if g(a) is not equal to g{b),then there is a number £ between a, b for which

f(b)-f(a) /'(g)g(b)-g(a) g'W

Introduce a function F(x) denned by the relation

F(z)=f(x)+Ag(x)

where A, B are constants chosen so that

F(b)=

These two equations can be solved for A, B since g(a), g(b) arenot equal.

By Rolle's theorem, there is a number £ between a, b such thatF(£) = 0. That is,

• This paragraph may be postponed, if desired.

MEAN VALUE THEOREMS FOR TWO FUNCTIONS 67

Moreover, g\t;) cannot be zero; if it were, this equation wouldmake /'(£) zero also, contrary to the first hypothesis. Hence wemay divide by </'(£), giving

But, by direct subtraction,

F(b)-F(a) = {f(b)-f(a)} + A{g(b)-g(a)},

so that, since F(b) = F(a) = 0 and g(b) —g(a) is not zero,

f(b)-f(a)= A

9(t>)-9(a)

HenceH e n c e

(ii) To prove that, if the functions f(x),g(x) have second differ-ential coefficients such that f"{x)9g"(x) do not vanish simultaneouslyin the interval a — h,a + h, and if g(a + h) + g(a — h) — 2g(a) is notzero, then there is a number £ between a — h,a + h for which

f(a + h)+f(a-h)-2f(a)=f"(£)g(a + h) + g(a-h)-2g(a) g"{^

Introduce a function F(x) defined by the relation

F(x) =f(x) + Ag{x) + Bx + C,

where A9B,C are constants chosen so that

F(a + h)=f(a + h) + Ag(a + h) + B(a + h) + C = 0,

F(a-h)=f(a-h) + Ag(a-h) + B(a-h) + C = 0,

F(a) =f{a) +Ag(a) +Ba +C = 0.

By Rolle's theorem, since F(a) = F(a + h) = 0, there is a number| x between a,a + h such that Ff(^±) = 0; and, since

F(a-h) = F{a) = 0,

there is a number £2 between a — h,a such that F'(£2) = 0.Now apply Rolle's theorem to the function F'{x), which

vanishes when x = £l9 £2« There is a number £ between fjl9 £2, andtherefore between a — h9a + h, such that F"(t;) = 0. That is


But /"(|),^"(£) are not both zero, by hypothesis, and so #"cannot vanish. Hence

nfl_ A

Moreover, from the three equations for A,B,C, we have

{f(a + h) +f(a - *) - 2f(a)} + A{g(a + A) + £(a - A) - 20(a)} = 0,

so that, since g(a + h) + gr(a — A) — 2gr(a) is not zero,

a - h) - 2g(a)

.no

13. Application to certain limits. To prove that, iff(x),g(x)are (continuous) functions such that f (a) = 0(a) = 0, andiff(a),gf(a)both exist (g'(a) =|= 0),

i i m M _x->ag(x) g(a)

Since f(a) = g(a) = 0, we have the relation

; — a

Now let x->a. The right-hand side tends to the ratio (which, byhypothesis, exists) of the two differential coefficients at x = a,

EXTENSION.* More generally, if f(x),g(x) are (continuous) func-tions such that f(a) = g(a) = 0, and if f'(x)jg'(x) tends to the limit Ias x tends to a, then ft \

lim M = I.x-*ag(x)

To prove this, we use Cauchy's mean value theorem (p. 66),t h a t / ( » ) / ( « )

=g{x)-g(a) g'(£)

* This extension may be postponed, if desired.

APPLICATION TO CERTAIN LIMITS 69

for a number £ between x, a. As x tends to a, the number £ musttend to a also, and so

lim M - lim MrM11111 ; - — 11111 —- -

NOTE. This extension (which the reader who has not yetstudied Cauchy's mean value theorem may accept withoutproof) leads to a 'continuation' process for evaluating the limitf(x)lg(x), assuming existence and continuity where necessary:

(i) Iff (a) = g(a) = 0, but g'{a) 4=0, then

(ii) If also /'(a) = g'(a) = 0, but g"(a)*0, then

so that ^ C ^ ,

and so on.

ILLUSTRATION 7. To evaluate the limit

,. 1 — cos a;lim ^—.

If f(x) = 1 - cos x, g(x) = x\

then/(O) = flr(O) = O. Also

/'(a:) = sina;, ^'(x) = 2a;,

so that/'(O) = g'(0) = 0. We therefore proceed to the next stage

/"(*) = coss, g"{x) = 2,

so that /"(0) = 1, ^'(0) = 2.

TT .. 1 — cos a; 1Hence lim — = - .


EXAMPLES XI

Use the method of § 13 to evaluate the following limits:

•. i- l~x* « , . (I-*)3

1. km - x . 2. km ^- f.

sin x3. lim —. 4. lim

x^2 x — 2 x->o x„ ,. x — sin# „ _. oi"«'6. lim 5—• 6. km

x->Q •0 a;—>• 0 *&

REVISION EXAMPLES I

'Alternative Ordinary' Level

1. Differentiate the following functions with respect to x:

(i) cos# + #sina;, (ii) (3a-1) (#-3), (iii) 1/(1 + x2).

2. Differentiate with respect to x:x— 1

(i) x2j(l-x), (ii) sin2a; cos2a, (iii) ^ ^ j -

3. Differentiate with respect to x:

,., (z2-l)(o;2-2) ....(i) ^ ^ *, (n) tan2a;.

4. Prove from first principles that the differential coefficient ofl/x2 with respect to x is — 2jxz.

Differentiate with respect to x:

5. Differentiate with respect to x:

(i) x*(l + x)2, (ii) sin2 2x, (iii) V(l-

6. Differentiate the following with respect to x:

(i) x2(3-2x), (ii) sin 2x cos a, (iii)

7. Find the equation of the tangent to y = x2 at the point (1,1)and of the tangent to y = \xz at the point (2, f).

Show that these tangents are parallel and that the distancebetween them is

EEVISION EXAMPLES I 71

8. The curve y = x(x - 1) (x - 2)

cuts the z-axis at the points 0(0,0),-4(1, 0),JB(2,0). Find theequations of the tangents at 0,A,B.

Find the coordinates of the point at which the tangent at Acuts the tangent at B.

9. Find the ^-coordinates of the points on the curve

at which the gradient is zero, and test whether y has a maximumor minimum value at each of the points that you have found.

Also find the ^-coordinate of the point at which the tangent isparallel to the tangent at the point (3,4).

Draw a rough sketch of the curve.

10. Find the equation of the tangent to the curve

at the point (— 1, | ) .Find the coordinates of another point on the curve where the

tangent is parallel to that at the point ( —1,£).

11. Find the coordinates of the points of intersection of theline 3y = x with the curve

y = x(l-x2).

If these points are in order P,O,Q, prove that the tangents tothe curve at P,Q are parallel, and that the tangent at 0 is per-pendicular to them.

12. Find the equation of the tangent to the curve

y = xz - 9x2 + 20# - 8at the point (1,4).

At what points of the curve is the tangent parallel to the line= 3?

13. The point (h,k) lies on the curve y = 2x2 +18. Find thegradient at this point and the equation of the tangent there.

Hence find the equations of the two tangents to the curvewhich pass through the origin.

72 REVISION EXAMPLES I

14. Prove that the equation of the tangent to the curve

y = 2xz + 2x2 -8x + 7

at the point (1,3) is 2x- y + 1 = 0.Find also the abscissae (^-coordinates) of the points on the

curve at which the tangents are perpendicular to the first tangent.

15. A curve whose equation is

I2y = axs + bx2 + ex + d

has the following properties:(i) it passes through the origin, and the tangent there makes

an angle of 45° with the axis OX;(ii) it is parallel to the axis OX when x = 1 and when x = 2;

determine a, 6, c, d and sketch the curve.

16. Find the maximum and minimum values of the expression

distinguishing maxima from minima.

17. Find the maximum and minimum values of the function

carefully distinguishing between them.Use these results to sketch the graph of the function, and find

the equation of the tangent at the point (0,2) on the graph.

18. Find the points on the curve

y = 2xz- 3x2- \2x +20

at which y has a maximum or minimum value.Use your results to make a rough sketch of the curve.

19. Find the points (x,y) on the curve whose equation is

at which y is a maximum or minimum.Use your results to draw a rough sketch of this curve.

20. Find the gradient of the curve

y = 2xz-5x2-4:x+12

KEVISION EXAMPLES I 73

at the point (2,0) and determine whether y has a maximum or aminimum value at this point. Illustrate your answer by a roughsketch showing the slope of the curve in the neighbourhood ofthe point.

21. Find the coordinates of the points on the curve

y = xz

at which the tangents are parallel to the #-axis.Show that the slopes of the tangents at points whose abscissae

(^-coordinates) are less than 1 are all positive, and that at pointswhose abscissae lie between 1 and 3 the slopes are negative.

Sketch the curve, and state what you infer as to the nature ofthe roots of the equation

x3-.6x2 + QX-2 = 0.

22. A particle is moving along a straight line in such a mannerthat its distance from a point 0 on the line at time t sec. is givenby x = pt2 + qt3, where p and q are constants. Find the velocityand acceleration of the particle at time t in terms of p, q, t.

Find also the values of p and q for which the maximum velocityis 48 ft./sec, this velocity being attained when t = 4.

23. OX and OY are two perpendicular straight roads, and Ais a fixed point on OX, distant a ft. from 0. A motor-cyclist P istravelling along OF at a constant speed v ft./sec, and at time tthe angle OAP is 0 radians. Prove that the rate of increase ofthe distance of P from A is vain 8 ft./sec. and that the rate ofincrease of the angle 0 is (v/a) cos2 0 radians/sec.

24. A particle moves along the #-axis in such a way that itsdistance x ft. from the origin after t sec. is given by the formulax = 21t — 2t2. What are its velocity and acceleration after 6f sec?

How long does it take for the velocity to be reduced from15 ft. per sec. to 9 ft. per sec, and how far does the particletravel meanwhile?

25. A point moves along a straight line so that, at the end oft sec, its distance from a fixed point on the line is t3 — 2t2 + t ft.Find the velocity and acceleration at the end of 3 sec.

26. A particle moves along a straight line so that its distance attime t from a given point 0 of the line is x, where x = t sin t + cos t.

74 BEVISION EXAMPLES I

Find its velocity and acceleration at time t.Prove that the particle is at rest at the times

t = = 0 , "jgTT, i f " } ~2î 2 ^ > • • • •

27. The angle C of the triangle ABC is always right. If thesum of CA, CB is 6 in., find the maximum area of the triangle.

If, on the other hand, the hypotenuse AB is kept equal to 4 in.and the sides CA, CB allowed to vary, find the maximum area ofthe triangle.

28. A man wishes to fence in a rectangular closure of area128 sq. ft. One side of the enclosure is formed by part of a brickwall, already in position. What is the least possible length offencing required for the other three sides? (Prove that yourresult gives a minimum.)

29. B is a point a miles due north of A, while C is 3a miles dueeast of B; P is a variable point on BC at a distance x miles from B.A man walks straight from A to P at 4 miles per hour and thenstraight from P to C at 5 miles per hour. Prove that the timefor the whole journey is, in hours,

Find what the value of x must be for the time taken on thewhole journey to be a minimum, and also find this minimum timein hours.

30. A statue 12 ft. high stands on a pillar 14 ft. high. A man,whose eye is 5 ft. above the ground, stands at a distance x ft.from the statue. Prove that the angle 6 which the statue subtendsat his eye is given by the equation

tan 0 s=

Find the value of x for which 0 is as great as possible, givingyour answer correct to one place of decimals.

31. A rectangular tank, open at the top, on a base x ft. byy ft. and of height x ft., is to be constructed of iron sheeting(whose thickness may be neglected) of total area 1350 sq. ft., sothat the volume of water which it can contain is a maximum.Find this maximum volume.

EEVISION EXAMPLES I 75

32. Post Office Regulations restrict parcels to a maximumlength of 3 ft. 6 in. and a maximum girth of 6 ft. Find themaximum permissible volume of a rectangular parcel.

Find also the length of the longest thin rod which can be packedinside a parcel of maximum permissible volume, giving youranswer in feet to three significant figures.

33. An open tank is to be constructed with a horizontal squarebase and four vertical rectangular sides. It is to have a capacityof 32 cu. ft. Find the least area of sheet metal of which it canbe made.

34. A piece of wire of length I is cut into two portions, thelength of one being x. Each portion is then bent to form theperimeter of a rectangle whose length is twice its breadth. Findan expression for the sum of the areas of these rectangles.

For what value of x is this area a minimum?

35. Prove that, if the sum of the radii of two circles remainsconstant, the sum of the areas of the circles is least when thecircles are equal.

36. A farmer has a certain length of fencing and uses it all tofence in two square sheep-folds. Prove that the sum of the areasof the two folds is least when their sides are equal.

37. A piece of wire of length I is cut into parts of lengths xand I — x. The former is bent into the shape of a square, and thelatter into a rectangle of which the base is double the height.Find an expression for the sum of the areas of these two figures.

Prove that the only value of x for which this sum is a maximumor a minimum is x = T?1; and find which it is.

38. The vertex and circumference of the base of a right circularcone lie on the surface of a sphere of radius R. The centre of thesphere lies inside the cone. It is a known result that, if the heightof the cone is x, its volume is \TTX2(2R — X). Prove that, if R isunaltered while x is increased by a small quantity h, the volumeof the cone is increased by \TTX{4^R — 3#) h approximately.

39. A cubical block of metal is heated and expands slightly.If its volume increases by h per cent, show that the length ofeach edge increases by \h per cent approximately.

By what approximate percentage is the surface area increased?

76 REVISION EXAMPLES I

40. A body consists of a cylinder of variable radius x in. andfixed length 10 in. closed at each end by a hemispherical cap ofradius x in. The volume of the body increases at the rate of144 cu. in. per sec. At what rate per sec. is x increasing whenz = 3 ?

41. It is known that, if I in. is the length of a pendulum andt sec. the time of one complete swing of the pendulum, I is pro-portional to t2. If the length of the pendulum is increased byh per cent, where h is small, show that the time of the swing isincreased by \h per cent approximately.

A clock loses 30 sec. per day of 24 hr. Should the pendulum belengthened or shortened to make the clock keep correct time, andby what percentage?

42. The height of a closed cylinder is 3 in., and remains constant.The radius of its base is 2-5 in. and it is increasing at the rate of0-01 in. per sec. At what rates are (i) the volume, (ii) the totalsurface of the cylinder increasing?

43. A sphere is expanding so that its surface is increasing atthe rate of 0-01 sq. in. per sec. Taking IT — -r, find the approxi-mate rates of increase of (i) its radius, and (ii) its volume at theinstant when its radius is 5 in.

44. The volume of a spherical lump of ice t hours after it hasbegun to melt is V cu. in., its surface is S sq. in. and its radius

i s r in . I f ^ - = - 3 £ , find ^ .at at

What is the rate of decrease of S in square inches per hourwhen r = 1 ?

45. A solid rectangular block of metal expands by being heatedthrough a certain range of temperature, the percentage increasein its length, breadth and thickness being the same. It is foundthat the percentage increase in its volume is h, where h is small.Find the approximate percentage increase in the length of theblock.

CHAPTER IV

THE IDEA OF INTEGRATION

1. The area 'under' a curve. The diagram (Fig. 31) repre-sents the graph of the function

for values of x between a, b. We assume for the moment that thecurve rises steadily from its value at a to its value at 6, and lies

t / !

77

Mo Mx M2 M3 M 4 M5 Mb M1 Mn

BFig. 31.

entirely above the #-axis* The ordinates AF,BO are drawn forx = a,b respectively.

Our problem is to find an expression for the area of the figureABGF.

The basic definition of an area to which we appeal is that, if arectangle is drawn with sides of lengths p, q, then its area is pq.

* An example will be given later (p. 84), when the reader has acquiredmore experience, to illustrate the treatment when some of the curve FOlies above, and some below, the ce-axis.

78 THE IDEA OF INTEGRATION

We therefore seek to attach rectangles as closely as may be tothe figure ABGF.

Divide the segment AB of the #-axis into n parts, not neces-sarily equal in length,

For convenience of notation, call A,B the points MQ)Mn; thenthe intervals are

MOMV MXM2, M2MS, ...,Mn_zMn_» Mn_xMn.

Through each point M0,Ml9M2,... draw the ordinate, meeting thecurve in points Po, Pv P2,...; let the lengths M0P0,M1P1,M2P2,...be denoted by the letters y^y^y^,.... (The diagram (Fig. 31)illustrates the case n = 8.)

Complete the two sets of rectangles

and

shown in the diagram. We propose to regard it as obviousintuitively that the area ABGF lies between the sum of theareas of the first set of rectangles and the sum of the areas ofthe second set.

Now write

± = 8xQ)M1M2 = hx^M^Mz = Sx2,M331^

Then the two sums of areas of rectangles are

y0 8x0 + y18x± + y2 8x2 + ys 8x3 + . . .

and yx 8x0 + y2 8x± + y3 8x2 + y± 8xz + . . . ,

or, in more concise notation,

n- l

o

n - l

0

THE AREA 'UNDER' A CURVE 79

Thus, if A is the required area,

n—1 n—L

We may now suppose that the number n of intervals is takenlarger and larger, while the length of each interval becomessmaller and smaller. Then the area of a typical rectangle of thefirst set, say MiMi+1Ni+1Pi becomes progressively nearer andnearer to the area of the corresponding rectangle MiMi+1Pi+1Li

of the second set. In the limit, as the number of intervals increasesindefinitely, their size decreasing indefinitely, the two sums

n-l n-l

approach equality, the sum of the areas of the small rectangles* covering' the arc FG shrinking to zero.

This common limiting value is known as the area of ABGF.

ILLUSTRATION 1. To find the area 'under' the curve

y = \^

between the ordinates x = l9x = 2.

Fig. 32.

Divide the interval between the two points ^4(1,0),B(2,0) inton equal parts (Fig. 32). The points of division occur where

n n n


In the notation of the preceding paragraph,

Sxo= Sxt= Sx2 = . . . = - ,

Hence

Now it is proved in text-books on algebra that

SIo

so that

14n2-9n+l

_ 7 _ 3 112 8n 24n2#

THE AREA 'UNDER' A CURVE 81

If n increases indefinitely, the number of intervals becomeslarger and larger while their magnitude becomes smaller and,smaller. In the limit, as n-+oo,

n-i 7

In the same way,n - l

o o *

It can be proved similarly thatn- l

i±l\%l~n ) n

Thus the two sums have the same limit, which is the area ofthe figure ABGF. Hence the area is y^ square units.

EXAMPLES In - l 7

1. Prove that 2 Vi+i &xi ~ To •o 12

2. Use the method of Illustration 1 (p. 79) to prove that the areas'under' the curves y = x and y = x3 between x = 1 and x = 2 aref and \ - respectively. [You are given that £ i 3 = JJfc2(i+ I)2.]

o

2. The integral. The ideas which we used in § 1 can be putin a somewhat more general form. Suppose that/(#) is a functionof x defined between the values x = a,b. As before, we divide theinterval into n parts, not necessarily equal, at points where xassumes in turn the values

a, Xl9 X2, #3, . . . , %n-2> xn-l> *•

For convenience of notation, we also write x0 = a, xn = 6.Consider a typical interval (xi9 xi+1)9 whose length we denote by

the symbol Sxi9 so that

Throughout that interval, f(x) assumes a succession of values, andthese will (for ordinary functions) have a greatest value and aleast value. We denote the least value of f(x) in the interval(xi9 xi+1) by the symbol mi9 and the greatest value by Mi%


Now form the two sumsn- l

0

These correspond to the sums

n - l n - l

respectively in § 1 (p. 78); but here we are not assuming thatthe least and greatest values mt, Mi in the interval (xi9 xi+1) occurat its end points.

Suppose that the number of intervals is taken larger and larger,while the magnitude of each interval becomes smaller and smaller.The two sums sn, Sn tend to limits which we call s, S respectively.We assume without proof the theorem that the values s, S areindependent of the way in which the subdivisions proceed totheir limit.

DEFINITION. When the two limiting sums s,S are equal, thefunction f(x) is said to be INTEGRABLE between a, b, and their commonvalue is called the integral of f(x) between the limits a,b.

Thus Illustration 1 (p. 79) tells us that the function \x2 isintegrable between the limits 1,2, and that the value of theintegral of \x2 between the limits 1,2 is yj.

COROLLARY. If we replace mi or Mi in the definitionswhere ^ is any point of the interval 8xi9 then

o

lies, by definition of mi9 Mi9 between the two sumsn- l n - l

2 mt 8xi9 2 ^h &xi*o o

n - lHence the sum 2 / (^) Sa

o

also has as its limit the integral off(x) between the limits a,b.

NOTATION 83

3. Notation. We have seen that the integral of a functionf(x) between the limits a, b is the common limiting value assumedby the sums

n - l

0

or, by the preceding corollary,

To imply that we are to sum over an 'infinite' number ofintervals, we replace the symbol of summation 2 by a symbol of

integration ; we also replace the particular xi by the current

variable x, and, instead of the notation Sxt for the very smallincrement at the point xi9 we write the symbol dx. The result isthe notation

/Ja

f(x)dx

to denote the integral of f(x) between the limits a, 6. For example(P- 8D.

COROLLARY (i). It is an immediate consequence of the definitionof an integral as summation over sub-intervals that

Ff{x)dx+ \f(x)dx= \f(x)dx.Ja Jc Ja

COROLLARY (ii). When a = 6, it is clear that%a

f(x)dx = 0.Ja

COROLLARY (iii). We can use Corollaries (i), (ii) to give us aninterpretation for the integral

Ff(x)dx (a<b)9Jb


which has not yet been defined. We adopt the meaning, consistentwith (i), that

[af{x)dx+ [hf{x)dx= [hf{x)dxJb Ja Jb

= 0 (by(ii)).

ra rbThat is to say, f{x)dx = - f(x)dx.

Jb JaThus the value of an integral is changed in sign by interchange of

the limits.

4. The sign of the area. The value of an integral, defined asthe limit of the sum n l

may easily be negative in cases where f(x) is not restricted to apositive value.

Suppose, for example, that the func-tion f(x), whose graph

is illustrated in the diagram (Fig. 33),is positive in the interval (a,c) andnegative in (c, 6), being zero at c wherethe curve crosses the x-axis. If A, B arethe values (essentially positive) of the two areas indicated, then

Fig. 33.

as before; but B

1 = [Cf{x)dxJa

= - f*f{x)dx,JC

since the area is the limiting value of the areas of a number ofrectangles of typical length S^ and 'height' - / ( ^ ) , this being thenumerical value of the negative number f(xi). Hence

\bf(x)dx= [°f(x)dx+ \bf{Ja Ja Jc

= A-B,

and this may be positive, negative, or zero.

THE SIGN OF THE AREA 85

It follows that it is always wise to sketch a diagram when cal-culating the area 'under9 a curve.

Further, and more detailed, treatment of areas contained byclosed curves will be given later (Vol. II, p. 128).

5. Definite and indefinite integrals. Consider the integral

[bf{x)dx.Ja

Its value depends on each of the limits a, b, so that it is, in fact,a certain function of a, 6.

Suppose that the lower limit a has an assigned value, but thatthe upper limit 6 is subject to variation. In order to imply thisability to vary, we change the name from 6 to x. The value ofthe integral is then a function of x, and we may write

rx

Ja)dx.

Having done this, it is customary to drop the bar from thesymbol x, thus denoting the upper limit by the letter x itself, andto write

Cx

u(x)= f(x)dx.

The symbol x is now doing double duty, as the variable in thefunction integrated, and as the upper limit of integration. Thisdouble use sometimes troubles beginners, and it should be watchedcarefully. The point is sometimes met by changing the name ofthe variable in the function integrated to the letter t. We thenhave

u(x)^ (Xf(t)dt.Ja

[It may be useful to point out explicitly that the value of anintegral is not affected by changing the name of the variableintegrated, the limits being unaltered. Thus

{*f(t)dt= (Xf{u)du.]Ja Ja


We now notice that a change in the lower limit a merely affectsthe value of u(x) to the extent of an additive constant. For letav a2 be selected as two values of the lower limit. By Corollary (i),p. 83, we have the relation

\Xf(x)dx= (af(x)dx+ (*f{x)dx9Jax Jax Ja%

and the expression f(x)dx is a constant, unaffected by a changeJax

in the value of the upper limit x in the other integrals.Finally, it is customary in many problems to leave the lower

limit unspecified. We then obtain an integral which might bewritten

/ •

f(x)dx,

but which is in practice written in the simple form

(x)dx

with limits omitted. This is a function of x, determined to withinan additive constant whose magnitude is arbitrary because of theunspecified lower limit.

f6The integral f(x)dx

Ja

with given limits is called a definite integral, and the integral

with unspecified limits is called an indefinite integral.

6. The evaluation of an integral. The process of calculatingthe value of an integral by subdividing an interval, summing, andproceeding to the limit is usually troublesome, and simpler alterna-tives must be found. In practice, it is often easier to find theindefinite integral first and then the definite. The fundamentallink is found in the following theorem:

THE EVALUATION OF AN INTEGRAL 87

/ / g(x) is a function of x defined as the integral of a continuousfunction f(x) in the form

then the differential coefficient g'(x) exists, and

In other words, g(x) is a function whose differential coefficientis/(z).

In order to prove this theorem, we go straight to the definitionof a differential coefficient, and consider

)-g(x).

If the lower limit of integration is a, then

rx+h rx(x + h)-g(x) = f(x)dx- f(x)dx

Ja Ja

Ix+hf(x)dx. (Corollary (i), p. 83)

X

Now suppose that x0 is a value of x at which f(x) takes its leastvalue in the interval (x, x 4- h), and that xx is a value of x atwhich f(x) takes its greatest value. Then, by the very definition

J ^x+hf(x)dx lies between hf(x0) and

X

hf(xx)—since h is the length of the interval of integration. Thus

Jmx+h

f(x)dxx

lies between hfixo)> hf(xi)>

g(z + h)-g{x)and so ^

lies between f(xo)> f(xi)*

Now take h to be progressively smaller and smaller. The valuesxOyx1 ultimately coincide with x itself, so that, since f(x) iscontinuous,


lies between two values which in the limit are each f(x). Hence

exists, and its value isf(x).

Thus 9f(x)=f(x).

We therefore have the rule:In order to integrate the function f(x), we must find a function

g(x) of which it is the differential coefficient. (Note the implicationthat g(x) is continuous; compare the footnote on p. 61.)

Note. If G is any constant,

so that g(x) is undetermined to the extent of an additive constant(compare p. 86).

Finally, we prove that the integral

g(x)= [f(x)dx

as evaluated by the rule $'(%) = f(x)

is unique apart from the additive constant.

Suppose that, on the contrary, h(x) is another function sucht h a t h'(x)=f(x).

Write u(x) = g(x) - h(x),

where u(x) is continuous since g(x), h(x) are.

Then u'{x) = g'{x)-h\x)

= 0.

Hence (p. 62) the function u(x) is constant.Another way of stating this result is that the equation

dx JK '

leads to the result y = \f(x)dx + O

uniquely.

THE LINK WITH DIFFERENTIALS 89

7. The link with differentials. Suppose that y is a functionof x whose differential coefficient is the given function f(x), so that

— = fix)

The differentials dy,dx satisfy (p. 43) the relation

or dy =f(x)dx.

We have also just proved that

y = jf(x)dx.

Hence we have the suggestive remark that the relation

dy =f(x)dx

between the differentials leads to the integrated relation

y = jf(x)dx.

This is the justification for the common sequence of argument:

therefore dy = f(x) dx,

therefore y = \f(x)dx.

8. The evaluation of a definite integral. Once we haveobtained (by methods to be given later) the indefinite integral

by finding a function g(x) whose differential coefficient is f(x), wecan evaluate the definite integral

l


by a simple rule. For suppose that

(x) = [Xf(x)dx.Ja

Then

= f(x) ~f(x) [P> 8 7 a n (* definition of g(x)]

= 0,

so that (p. 62) F(x) — g(x) = constant.

But (p. 83) F(x) = 0 when x = a, and the value of the constantis therefore — g(a), so that

= g(x)-g(a).rb

Hence f(x) dx = g{b) - g(a),Ja

each side being equal to F(b).Thus we have the rule:

hi order to evaluate the definite integral

\){x)dx,Ja

find a function g(x) whose differential coefficient is f(x); then

NOTATION. It is often convenient to write

to denote g{b) — g(a).

9. Some simple standard forms. The evaluation of

\x)dx

by finding a function g(x) whose differential coefficient is f(x) isnaturally a matter of some difficulty. A start can, however, bemade by inverting the formulae for differentiation given on p. 37.

SOME SIMPLE STANDARD FORMS 91

This gives us the formulae:

xn+1 „n+ 1

= — cos x -f G

secztanarefa; = sec x + (?.

sec2 x dx = tanz + C

cosecxcotxdx = —cosecx-f (7.

cosec2 x dx = — cot z + C*

We also have (pp. 40, 41)

= tan"1 a; + (7./ •

C dxIn the case of -T— %\ > *n e arbitrary constant may be taken as

a multiple of \TT and so the ambiguity of sign may be allowed for.

Note. The following results are also easily proved. If A is aconstant, then

Uf(x)dx = AJf(x)dx;

and iff(x),g(x) are integrable functions of #, then

f f fJ l J J


ILLUSTRATION 2. To find the height at time t of a particle pro-jected vertically upwards under gravity with speed F.

The physical law is that a particle moving freely under gravityis subject to an acceleration vertically downwards of amountusually denoted by g. In ordinary foot-second units, the valueof g is approximately 32.

Let x be the height of the particle above the ground at time t.d2x

Then (p. 48) its acceleration (vertically upwards) is -7-5-, so that

d(x) I, dxor JLJ. = -g * —

atHence, integrating to find the function x,

Now we are given that the velocity x is equal to V when t iszero, so that C — V. Hence

dx

Integrate to find x\ then

But x = 0 when t = 0 if the origin for x is taken at the point ofprojection, and so A = 0. Hence

x= Vt-\gt\

EXAMPLES

1. Prove by subdivision of the interval and actual summation

(i) dx= 1, (ii) xdx = $, (iii)J l Jo J2

2. Find the indefinite integrals

SOME SIMPLE STANDARD FORMS 93

3. Find the indefinite integrals

cos2#d#, sin|#6fo, 2cos2|a:^#.

4. Write out a formal proof of the theorems

(i) [bAf(x)dx = A[hf{x)dx,Ja Ja

(ii) [b{f{x) + g{x)}dx= \f(x)dx + f g(x)dxJa Ja Ja

based on the definition of § 2 (p. 81).

5. Evaluate the definite integrals

I x2dx, I x3 dx, I 5x*dx, \ (xs + x)dx.Ji J-i Jo J2


cosxdx, sinxdx, sin2A#d#, r 5.Jo Jo Jo 2 J0l+^2


f1 f2 f3/ 1\2

(1 -a ; ) 2 ^ , (x-l)(2-x)dx, U2 + ~2 dx-Jo Ji Ji \ x J

8. Find the value of

fin ^ [Insin2xdx+ cos2 a; da?.

Jo Jo

CHAPTER V

DEVICES IN INTEGRATION

T H E DIRECT EVALUATION of /(o;)^ by finding a func-tion g(x) with differential coefficient f(x) is possible only in thesimplest cases. A number of methods are, however, available toextend our scope considerably.

1. Substitution, or change of variable; indefinite integrals.Consider an integral such as

T _ _

Experience (gained by solving many similar examples) enablesus to see that the 'essential variable' is not really x, but x5+l.We therefore begin by finding the effect of replacing a;5+l by anew variable t, so that

Then dt = 5x*dx,

so that / =

_ - 1

The substitution of t for the function x5 + 1 has therefore led tothe evaluation of the integral.

As another example, consider

/ = sin7 x cos x dx.

The 'essential variable' is sin x> so we write

t = sin a;,

giving dt = oosxdx.

SUBSTITUTION—INDEFINITE INTEGBALS 95

Hence / = ifdt = |Z8if

= I sin8 x.

The theoretical basis on which the justification for the methodof substitution rests is as follows:

Let / =

be a given integral, and suppose that a certain function u(x)appears likely as an 'essential variable' for the integration. Write

Then

But

so that

or

t = u(x).

2=/(*) (p. 87).

fx-Ttfx (P'27>>// x dI '/ xf(x) = Ttu(x),

dl f(x)dt u'ixY

But the relation t = u(x) enables us to express x in terms of t,and so f(x)j{u'(x)} may be expressed as a function of t, say F(t).Then ,T

so that (p. 89) / = 1^(0*.

Thus the effect of the substitution

t = u(x)

is (i) to replace u'(x)dx by efa; (ii) to replace f{x)J{u'(x)} by thecorresponding function F(t) of £; (iii) to replace integration withrespect to x by integration with respect to L

f(%)Note. The expression , may look complicated; but the

u {x)whole virtue of the method lies in the selection of the functionu{x) in such a way that f{x)j{u'{x)} falls naturally into the form

96 DEVICES IN INTEGRATION

F(t) as a function of t. There is no point in remembering thisformula; it is the method of substitution that must be known.

COROLLARY. TO prove that, if a, b are constants, then

f 1

J a +Write t = ax + b, so that dt = adx. Then

\f(ax + b)dx = \f{t)— = -J J a i

For example, sin 6xdx = —\ cos 5x.

After a little practice, the reader should find that he uses thiscorollary so automatically that he is unaware of any substitutionat all.

EXAMPLES I

Find the following integrals by performing the substitutionsmentally:

C • f f .1. sin 2xdx. 2. \cos3xdx. 3. si

J J Jr r r

4. \sec24:xdx. 5. isechxt&nixdx. 6. \(zJ J Jr r r

7. \(x+3)3dx. 8. (#+5)4cte. 9. I —

J J JO*io. f * i i . r da; - r dx

c c cJ J J

_ ^ _ . 17. — . 18. .a;+l)2 J(4o;-3)3 J(J^-l)4

19

22

r r r. \J(x+l)dx. 20. (cc-l)fdoj. 21. -

J J h

f, fe . 23. f ——. 24. fJJ(2x+l) J (2x —3)f Ji

SUBSTITUTION — INDEFINITE INTEGRALS 97

Find the following integrals:

xdx

28. I T - — - A ^ . 29. I cos5 a; sin a; da;. 30. I sin3 x dx.

(3a;5 +1) 2 '

. cos5 x sin x dx. 30. sir

C C r SPO2 or dor31. sin2 x cos3 x dx. 32. tan23a;da;. 33. _ _ ! i — .J J J(l+tana;)3

r r34. (1 + sec x) sec x tan a; da;. 35. sec4 x dx.

36. sec2 x tan2 a; da;.

iVote. It is sometimes more convenient to make the substitutionfrom x to t in the alternative form

x = v{t).

We then have

ff(x)dx=jf{v(t)}v'(t)dt,

an integral with respect to t. This is, of course, easier to statethan the earlier form (p. 95), but in practice the substitutiont = u(x) is probably the commoner.

ILLUSTRATION 1. To find

dx

Let a;=3tan0

(remembering that tan2 0+ 1 = sec2 0),

so that dx = 3 sec2 0d0.

3sec20d0 f3sec20d081 sec4 0

T , f 3sec20d0 r:T h 6 n / = J(9tan20 + 9)2 = J-

^ f2cos20d0 = -^ f(l + cos20)d0


sin 9 cos 9 1Now

x

Hence / « ^

EXAMPLES II

Find the following integrals:

2 f-^f?*-• JV(l>)'

3. j £ * . 4.5- \^r^)dx' 6-

[For examples 7-10, compare the device at the start of Illustration 4,p. 100.]

f ads

J v ( i ) 'f f

J * J2. Substitution, or change of variable; definite in-

tegrals. The introduction of limits for the integral presents noessentially new feature, except, of course, that the original limitsfor x must now be replaced by corresponding limits for the newvariable t. (However, we add the warning that care must betaken in more advanced examples.)

ILLUSTRATION 2. To evaluate

sin4 x cos x dx.o

Write t = sin a;;then dt = cosxdx.

Hence Mn^xcosxdx = \t4dtMn^xcosxdx = \t4

SUBSTITUTION—DEFINITE INTEGRALS 99

Now as x, starting from zero, increases to \n> the new variable t%

starting from zero, increases to 1. Hence

sin4 x cos x dx = ft dtr\7t ri

sin4 x cos x dx =

15*

ILLUSTRATION 3. To evaluate.

Let

then dt =

f dx 1Hence JNow as x, starting from 2, increases to 8, the new variable t,starting from 12, increases to 42. Hence

10 I/ 12

10 [422 122J

L [ _360 [72 22

360 [ 196J

11568'



dx

It is a common device in such integrals to write t2 for theexpression under the square root. Thus

x = t2-l,

so that dx = 2tdt.

Also, as x rises from 3 to 11, t rises from 2 to 2^3. Hence

i11 dx 2tdt

_dt

Now write t — 2 tan 0,

so that dt = 2 sec2 0 0*0.

As t rises from 2 to 2^/3, 0 rises from \n to £TT. Hence

/ - 2 i - 2aec:edd.= \ d0

ILLUSTRATION 5. To ^nd fAe area of a circle of radius a.Referred to rectangular Cartesian axes with the origin at the

centre, the equation of the circle (Fig. 34) is

so that — x2),y = ±the two values of y for a given valueof x corresponding to the parts of thecircle 'above' and 'below' the #-axis.

The area of the whole circle is doublethat of the upper semi-circle, and so wemay take

ra- 2

J-a

l-x2)dx.Fig. 34.

SUBSTITUTION — DEFINITE INTEGRALS 101

Make the substitution x = a sin 2,

so that dx = a cos t dt;

the range ( — a, a) of # corresponds to the range ( —JTT, n-) of t.Hence

.4 = 2 a cos t.a cos tdt.J

= 2J-\n

[Note. Since a cos t is positive in the range (— J77, |TT), we havetaken the positive square root throughout the range.]

Thus A = a2 2 cos2 ^ = a2 (1 + cos 2*)dtJ J *

= 77a2.

EXAMPLES IIIEvaluate the following definite integrals:

1.

3.

5.

7.

9.

(*\nsin2 0 cos OdO.

Jof1 dx

Jo(l+*2)2'f *Jofin

sec4 Odd.Jo

f1

JoILLUSTRATION 6.* To evaluate

f* sin

2.

4.

6.

8.

10.

0d9

f\nsin* 0d9.

Jof1

<l(l-x2)dx.Jofin

sec2 0 tan2 Odd.Jo

\ ncos5 OdO.Jkn

r4ji

It is implicit in such a question that the positive square root isto be taken.

* Illustration 6 may be postponed, if desired.


Let t = + <](! - 2a cos 0 + a2),

so that t2 = 1 - 2a cos 0 + a2,

and 2£cft= 2asin Odd.

„ f sin0d0 1/7* 1 fHenCe JV(l~2acos0T^) = aJ T = aJ dt

Consider now the range of values of t as 0 runs from 0 to IT.When 0 = 0, we have the relation

^(1 —2a cos 0 + a2) = yj(l — 2a + a2) = ± (1 —a),

that sign being taken which makes the square root positive.When 0 = 77, we have

^(1 — 2a cos 0 + a2) = ^(1 + 2a + a2) = ± (1 +a),

that sign again being taken which makes the square root positive.Various cases must therefore be considered:

(i) When a lies between —1,1. The positive square roots are1 — a, 1 + a respectively, so that the value of the integral is

l+a

1-a

The value of t varies continuously from 1-a, through 1 (whencos 0 = |a), to 1 +a as 0 moves from 0 to TT.

(ii) When a is greater than 1. The positive square roots area — 1, a +1 respectively, so that the value of the integral is

P

The value of t rises continuously from a — I t oa + 1 as d movesfrom 0 to 7T.

SUBSTITUTION — DEFINITE INTEGRALS 103

(iii) When a (being negative) is less than — 1. The positivesquare roots are I—a and ~ ( l + a ) respectively. (For example,if a = — 4, the square roots are 1 — (— 4) and — (1 — 4), that is tosay, 5 and 3.) Hence the value of the integral is

{ ( l t t , ( l o ) } .

This value looks negative, but is actually positive, sin ce a is negative.The value of t falls continuously from 1— a to — 1— a as 6

moves from 0 to IT.SUMMARY. |The value of the integral is 2,2/a, — 2/a according as

— \<a<\,a>lia< — 1.EXAMPLES IV

Evaluate the integrals:sin Odd f» sin 6 d9 C» sin 0d9

J JC* sin Odd f» sin 6 d9 C

* Jo yl(5-±cosdy ' Jo V(lO + 6cos0)' " JoJo V(5O-14cos0)'

3. Integration by parts. A very important method of integ-ration follows as a direct result of integrating the formula for thedifferentiation of a product. If u(x), v(x) are given functions of x9t h e n d , , dv du

-1-(UV) = U^ + V-j-yax dx dxdu d . N dv

Hence, on integrating from an arbitrary lower limit a, and denotingthe variable of integration by the letter t (p. 85), we obtain theimportant formula

\ \ ° [X'{f)dt.[Xv(t)u\t)dt= \u{t)v(t)\°- [Xu{t)v'Ja L la Ja

We observe that the left-hand side is the integral of the product oftwo functions v> u', one of which can be recognized as the differentialcoefficient of a function u.

It is customary to quote this result in its 'indefinite integral* form

\v(x)u'(x)dx = u(x)v(x)— \u(x)v'(x)dx,

and we shall usually do so (Illustrations 7, 8); but note the warningwhich follows.


Warning. The omission of the term — u(a) v(a) may occasionallylead to curious results. For example, if v(x) = secx, u(x) = cosxy

we appear to have

sec x( — sin x) dx = cos x sec x — cos #(sec x tan x) dx,

or — tan#d# =1— tan#d#,

or 0=1 .

(The product cos x sec x is a constant rather than a genuinefunction of x, and would be cancelled in definite integration.)


\xsinxdx.

We have to integrate the product x sin x, where we recognizesin a: as the differential coefficient of the function — cos a;. Apply-ing the formula, we have

\xsinxdx = \x-T-( — cosx)dxJ J dxy

= #( — cosa;) — ( —cosa;).l.da;

\cosxdx

= — x cos x + sin x,

the desired value.It may be remarked that an alternative first step could be

/ •sin x -7- {\x2) dx

leading to sin x. \x* — \x2 -r- (sin x) dx

= \x% sin x — \ \x2 cos a; da;.

However, a glance at the integral

#2cos#<fo

will convince the reader that he has not made the problem anysimpler by this beginning.

INTEGRATION BY PARTS 105

The whole success of the method depends on an intelligentfirst step, and certainty of touch should be acquired by workingmany examples.


1= \sin-1xdx.

This example illustrates a case where the possibility of integ-ration by parts might pass unsuspected. We have

1= \8in-1x.l.dx= I sin-1 a;.-—J J dx dx

= [Zl^l = _ Uy = - t

Write J =

and put 1 — x2 = t/2,

so that — 2# da; = 2ydy.

Then

Hence / = x sin-1 x + J(l— x2).

EXAMPLES VFind the integrals:

f f1. #cosa;da;. 2. #2cosa;aa;.

3. \2xsec2xtanxdx. 4. Lrtan^xda;.

Evaluate:

5. a:cosa:ia;. 6. x2s\nxdx.Jo Jo

(l+#)2sin#da;, 8. (x — 2)2 cos xdx.o Jo


In conclusion, we remind the reader of the value of a goodtechnique in the normal processes of algebra and trigonometryFor example,

sin 3a; sin 5xdx = - {cos 2x — cos 8a;} dx

= - sin 2x — — sin 8a;.4 ID

Again, I sin4 x dx — - (2 sin2 xf dx

- I (1 - 2 cos 2a; + cos2 2x) dx

1 17 l

- J J 1 - cos x + -{i r/3 I \- - - 2 cos 2a; + - cos 4a; I4J \2 2 /- j-a; —sin2a; + -sin4a;>.4 12 8 1

EXAMPLES VIFind the integrals:

1. sin 6x cosxdx. 2. <

. si

vo&xdx.

3. sin2 4a; da;. 4. sin 4a; sin 2a: dx*

Evaluate:r\n fir

5. sin x cos 4a; dx. 6. sin3a;da;.Jo Jo

7. cos x cos |a; da;. 8. cos4 a; da;.Jo Jo

4.* Formulae of reduction.Consider the integral

1= &\nnxdx (n a positive integer).


FORMULAE OF REDUCTION 107

Writing this in the form

/ = sin*1"1 x. sin x dx,

and integrating by parts, we haver

I = — sin71"1 x cos x -f {n~\)&inn-2xco&x.GO&xdx

«= — sin71"1 x cos x + (n — 1) sin71""2 x( 1 — sin2 a:)

= — sin71"1 a; cos a; + (w — 1) sin71-2xdx — (n— 1) sinna;da;

as — sin71"1 a; cos a: + (n — 1) sinn-2a;da; — (n—1)1.

Hence nl = — sin71"1 a; cos a; + (n — 1) sinn~2 x dx,

_ sin71"1 a: cos x n—1 C . o 7or / = 1—— smn-2a;aa;.n n

In this way we have made the integral of sinna; depend on theintegral of sin71"2 a;, whose degree in sin a: is less than that of theoriginal. To emphasize this reduction of degree, we write

In= sin71 a; da;, /w_2= sinn~2a;efo;.

m _ sin71"1 x cos x n—lThus /„ = + _

It is implicit that n ^ 2.In the same way,

sin71"3 a; cos a; n — 3

_ sin71"5 a; cos a; n — 5

and so on.By repeated application of the formula, we reach a stage where

we must evaluate= — cosa:

when n is odd; or IQ= \dx = x,

when n is even.


A formula such as__ sin71™1 x cos x n—\

which enables us to reduce the index n by successive stages iscalled a FORMULA OF REDUCTION for the integral.


Write In= cos2nxdx.Jo

(This is a slight change of notation from the preceding, butperhaps a little more convenient.)

We have

Ir _ 21n— C U b

Jo\n [in

= [cos271"1 x sin x] — (2n — 1) cos271"2 x( — sin a:). sin x dxo Jo

= 0 + (2n - 1) I ""cos271-2 x( 1 - cos2 a:) da;,Jo

since cos271"-1 a; sin a; vanishes for x = 0, and, when n ^ 1, for a? = £TT.Hence

4 = (2TI

T 2 n - l

4 = ^^ -Applying this formula successively, we have

_ 2n~12n-3w - 2n 2n-2n~2

_ 2n-l2n-32n-5~~~2n~2n~^22n-4 "~3

2n-l2n-3 3~2^T27irr2""4 a

2n-12n-3 312 2 2 ' " 4 2 °'

FORMULA OF REDUCTION 109

where

Hence i, =2n(2n-2)...4:.2 2

EXAMPLES VII

Find the following integrals by the use of formulae of reduction:

1. I sin6 x dx. 2. cos7 x dx. 3. sin9 x dx.

Evaluate the following integrals by the use of formulae of

reduction:

cos8xdx. 5. Bin7xdx. 6. sin10xdx.o Jo Jo

7. f "w&xdx. 8. I ncos2n+1xdx. 9. | *Bin*n+1zdx.Jo Jo Jo


sin2m x cos2n+1 x dx.Jo

To imply dependence on both m and n, write

fliTThen Im n = sin2mxcos2na:cosa:dii;

Jo

= sin2wa;cos27la;d(sina;)Jo

s in 2 m + 1 x cos271 x\Jo

rL2m+1"

I

gm2w+i x u 2n cos211""1 x( — sin x) dx2m+l

0 + r f \ i1 Jo

110 DEVICES IN INTEGBATION

since sin2m+1a;cos27la: vanishes when x = 0, and, when n>0, forx = \TT. Hence

2n4 , n = o 7 — c o s 2 a?

"" 2 m + l

or (2m + 1 + 2TI) Jm> n

T - 2n T

° r m>n"~~ 2m + 2n+l m>n~

Reducing n successively, we have

_ _ 2n 2n-2m>n~ 2m + 2n+\2m + 2n^l m'n

2n 2n-2 2

finL,o= sir

JoAlso I,na= I sin2wla:cosa;da;

in ]

Hence lm%n

o2n(2n-2)...2

EXAMPLES VIII

Evaluate the following integrals by the use of formulae ofreduction:

Jmin [In

sin*xcos7xdx. 2. sin6 a; cos3 a; da;,o Jo

3. xm(l—x)ndx (by the substitution, x = sin28).Jo[in [in

4. sin2ma;cos2ma;da;. 5. sin2m+1a;cos2m+1a:da;.Jo Jo

6. f ^ l - « ) ! ( & . 7. r V ( l -Jo Jo

CHAPTER VI

APPLICATIONS OF INTEGRATION

I. Use in dynamics. Suppose that a particle P (Fig. 35) ismoving along a straight line so that at time t its distance from a

O *Fig. 35.

fixed point 0 of the line is z. We have seen that, if v9f are thevelocity and acceleration respectively, then (p. 48)

dx

dv _ d2xJ~~dt~~W

(i) If v is a known function of t, then x can be evaluated bymeans of the integral

x = \vdt\

and if / is a known function of t, then v can be evaluated bymeans of the integral

v = jfdt,

after which x can be determined by a further integration, as before.At each integration an arbitrary constant is introduced; its

value depends on given initial conditions for a particular problem.(ii) If v is a known function of x, then t can be evaluated by

means of the integralCdx

'-JT'If/ is a known function of x, we find v by means of the relation

f dv dv dx ^_ dv* ^ ~di ^ dx'~di "dx'Vy

112 APPLICATIONS OF INTEGRATION

SO that /

v2= Ufdx,Hence

so that v can be calculated.

ILLUSTRATION 1. Suppose that / is given as a function of t bymeans of the formula - 7 ±

f=kt cos t9

and that v = k, x = 0 when t = 0,

Then -^ = ktcost9

atr

so that

= kt sin £ — & sin t dt (by parts)

where, since v = k when £ = 0,

or

Hence v

dxthat is, "17 = kt sin t + k cos t9

Cut

so that # = k\ tsintdt + k I costfdfx = k itsintdt + k c

where D = 0 since a; = 0 when f = 0. Hence

x

ILLUSTRATION 2. Suppose that / is given as a function of x bymeans of the formula

/ = 2k2x(x2+l)

and that x = 0, v = k when £ = 0.

USE IN DYNAMICS 113

We have ^ ^ = 4P x(x2 + 1),ax

so that v2 = 42 (4a;3 + 4s) rte- " / <

where, from the initial conditions,

Hence v2 =

so that v = :

But v = k when x = 0, so the positive square root must be taken.Hence 7 / 9 1X

v = k(x2+l).dx

It follows that -=- = k(x2 +1),ctt

so that i

l x*

= tan"1 a; + D.

But a; = 0 when tf = 0, so that D = 0, and

Atf = tan"1 a;,

or x = tan if.. The formulae are relevant only up to time 7r[2k, when x

tends to infinity.

2. Area; Cartesian coordinates. We have already (pp. 77-85) discussed in some detail the meaning and evaluation of thearea 'under' the curve

For completeness, we repeat the formula

Cb

f(x)dxJa

for the area ABQP of the diagram (Fig. 36), reminding the readerto be careful about sign when the curve crosses the x-axis (p.84).


It may be useful to note also that, for the area bounded bya curve

two ordinates y = c, y = d, and the part of the t/-axis betweenthose ordinates (Fig. 37), the area is

f{y)dy.

o B *

Fig. 36. Fig. 37.

3. The area of a sector, in polar coordinates. The equationof a curve in polar coordinates is

where r is a single-valued function of 0. To find an expression forthe area of the sector OAB (Fig. 38), bounded by the radii OA,OBgiven by 0 •> a,j8 and the arc AB.

O

Fig. 38. Fig. 39.

METHOD I. Divide the arc AB into n parts at points

A = PQ, Ply P2, . . . , Pn-2> *n-V *n = B.

Write L PiOPi+1 = 80, (Fig. 39).

THE AREA OF A SECTOR 115Denote the least value of the function /(0) in 88i by the symbolmi and the greatest by Mit Then the area of a typical elementPiOPi+1 lies between that of a circular sector of angle 86i andradius miy and that of a circular sector of angle S^ and radius Mi%

Moreover, we proved (p. 100) that the area of a circle of radiusa is Tra2, so that, by proportion, the area of a sector of angle S0< is

or

Hence the area of the sector OAB lies between the two sums

n-l n-1

In the limit, as the number of intervals increases indefinitely,the size of each interval decreasing indefinitely, these two sums,for 'ordinary' functions /(0), approach (p. 82) the limit

so that we have the formula

Area OAB = \ [\f{8)fdd

-If r*dd.

METHOD II. The area of a sectorAOP (Fig. 40), where L xOP = 6, isa function of the angle 6, say A(8).Suppose that P' is a point on thecurve near to P, so that

LxOP' = 6+8d;

then the area of the sector AOP' is.4(0 + 80).

We make the postulate, in in-formal language, that, when P' isnear to P, the area of the sector Fig. 40.


POP' is not very different from that of the triangle POP'; inmore formal language, that

area of sector POP' __P'_^ p area of triangle POP'

-A(e) sector POP'

_ sector POP' APOP'A POP' # 88 '

where A POP' is the area of the triangle POP' so that

APOP' = \0P. OP' sin 80.

HenceA(0+89)-A(0) __ sector POP' sin80

S0 ~ APOP' ^ u r u r >

Let §0-^0, so that P'->P. Then

sector POP' , . , . , .= 1 (p°s t u l a t e );

lim

,. sinS0 , t rtrtVhm - — = 1 (p. 32).

Moreover, the limit of a product is the product of its limits,and so

A'(0) = \r\

so that A(d) =

between suitable limits.

Note. I t is usually advisable to make a rough sketch of thecurve before applying this formula.

THE AREA OF A SECTOR 117

ILLUSTRATION 3. To find the area of the curve (a limagon) forwhich r = 1 + f cos 0.

The curve (Fig. 41) may be sketchedby putting

0 = 0,^77, 177,J77,...,27T

in succession; note that

dr

Fig. 41.so that r decreases steadily between0,77, after which it increases again.The curve is symmetrical about the line 0 = 0.

The area is1 f2n

4 r2d0

1 f2^(

2Jo Idd

EXAMPLES T

Find the area of each of the following curves:

1. The circle r = 2 cos 0.

2. The cardioid r = 1 + cos 0.

3. The lima§on r = 3 - 2 cos 0.


4. Centre of gravity, or centroid. If a number of particlesPl9 P2, P3,..., of masses ml9 m2, ra3,..., are situated at points(#i>2/i)> ( 2*2/2)9 (^3^3) i n a plane (Fig. 42), their centre of gravity(more accurately, centre of mass) or centroid is defined to be thepoint G(£,7)), where

t _ m1x1 + m2

Writing M ==m1 + ra2 + m3+.. . for the total mass, we may expressthese relations in the form

M | = Em^fc, Jfi? = *Lmkyk,

where the summations extend over all the particles.

o

Fig. 42. Fig. 43.

Our purpose is to extend this definition to a lamina, such asthat shown in the diagram (Fig. 43), bounded by a closed curve.It is assumed that the lamina is made of uniform material. Forconvenience, we have placed it in the first quadrant, but thatrestriction is not necessary.

If the area is divided, in any way, into a large number ofelements, so that a typical element surrounds the point (xk,yk)and has area 8Ak, then, by obvious extension, the centre ofgravity is the point C?(£, 77), where

zy, £ J _!_/>• £ A _L w 9\ A _Lj . JU-t O / 1 1 "T~ JUa (J*LXc) ~X~ JVO O / I Q ~T~ • • •£ 1 1 ' & & ' o o 1

CENTRE OF GRAVITY, OR CENTROID 119

The sum 8At + 8A2 + 8A% +... is equal to A9 the area enclosed bythe curve. The remaining problem is to find an expression forthe sums v . A v

2uXk oAk9 z>yk oAk

as the number of elements of area is increased indefinitely whiletheir sizes decrease indefinitely.

We begin with the coordinate f. yAs a first step, consider the M

area 'under' the curve m]

O A / ^ \ B

Fig. 44.

wbetween the ordinates x =» a,x = 6, where f(x) is a single-valued function of x (Fig. 44).Divide the interval at the pointswith ^-coordinates

and draw the lines through these points parallel to the i/-axis.(Compare the introduction to area on p. 77.)

Confining our attention to a typical filament whose base, joiningthe points xi9 xi+1 is of length 8xi =• xi+1 — xi9 let us divide it intorectangles by lines parallel to Ox and take these rectangles as theelements 8 ^ of the definition. Denote the least and greatestvalues of f(x) in the interval by mi9 J^ respectively, and extendthe rectangles to the height J^. Then the contribution from thefilament to the sum 3Hxk 8Ak lies between xi(mi 8xt) and xi+1(Mi 8x4)9

so thatn _ 1

2 (xi<-0

In the limit, for ordinary curves, the two outer sums have thesame value, namely,

and so the formula for £ is

xf{x)dx9

xf(x)dx

Ja

b

f(x)dx


Proceeding now to the closed curve with which we began,suppose that it lies between ordinates v

ox = a, x

which it meets at P,Q respectively(Fig. 45). Suppose, too, that the 'upper'and lower' parts of the curve have theequations

o

Q

Fig. 45.

In practice, the two different forms for y may arise becausethe 'upper' and lower' curves are quite different; but they mayalso arise, perhaps more usually, because when the equation of thewhole curve is solved for y in terms of x, two functions areobtained distinguished from each other by the sign attached to asquare root. For example, if the bounding curve is the circle

y-2 = ± ,/{l-(^-.2)2} = ± J(4X

f(x) = 2 + V(4z - x2 - 3),

then

Thus

Then the contribution to *Lxk 8Ak from the area enclosed by thecurve is equivalent to that from the area under y = f(x) LESS

that from the area under y — g(x). Thus, if A is the area enclosedby the given curve,

rb rbA£ = xf(x)dx- xg(x)dx.

Ja JaThis formula may be expressed more concisely. Write

to denote the diflference between the two values of y corresponding

to x, so that \!f]=f(x)-g{x).

Then

rbx\

Ja[y]dx

In the same way, and with similar notation,

77 =

CENTRE OF GRAVITY, OR CENTROID 121

ALTERNATIVE EXPRESSION. Another formula for rj for a laminadefined by the area 'under9 a curve y =/ (#)•

Referring to the diagram (Fig. 44) for the area ABQP, considerthe contribution from the filament on 8xi towards the sum*Zyk8Ak required to calculate 7). If the points of division of thefilament are at heights

then a typical contribution is

where &/y

and y'j lies between yj and yi+1.Keeping 8xi constant, let the number of subdivisions of the

filament be increased indefinitely while their sizes decreaseindefinitely. Then the contribution from the filament lies between

rydySxit ydy8xi9Jo

or

Summing now for all 8xi and proceeding to the limit in the usualway, we obtain the integral

so that7b »

f(x)dxJa

or

Jaydx

For an oval curve not meeting the #-axis, the correspondingformula is

J[y]dx

where [y2] = {/(x)}* - {g (x)}«.


ILLUSTRATION 4. To find the centre of gravity (centroid) of thearea in the first quadrant bounded by the axes and the curve

y = 2 + x-x2.

The range for x is 0,2. The area is givenby the formula

ydx =Jo Jo

-x2)dx

Then

so that

10T"

M ydx= xydxJo Jo

-I83'

/3/ 3 ~ 6*

or *Fig. 46.

Also

so that

I P(4 + ix- 3x 2 - 2a;3 + x*)dx2Jo

• i t16

_ 16 / 1 0 _ 24

THE MOMENT OF INERTIA OF A LAMINA 123

5. The moment of inertia of a lamina.(i) THE MOMENT OF INERTIA ABOUT A LINE. If a number of

particles Pl9P2yPz,..., of masses mlim2,m3>..., lie in a plane (seeFig. 42, p. 118), their moment of inertia about a line I in theplane is defined to be the magnitude / given by the formula

where PuP&Pz, ••• a r e the perpendicular distances of PltP2iP^ ...from I.

Whenever possible, the line I is taken to be one of the axes ofcoordinates. Thus, if Iy is the moment of inertia about Oy, then

Iy = mxx\ + m2£§ + m3a;§ + ...,

where {%1,y1),(x2,y2),... are the coordinates of PX,P2,.. . ,Similarly, T

J9 Ix =The definition can be extended to a lamina bounded by a

closed curve (see Fig. 43, p. 118). The reasoning is exactlyanalogous to that just given (§ 4) in calculating centres of gravity,and need not be repeated. With the notation explained there,we have *&

Iy = %2

Ja

iX j%\x]dy.

(ii) THE MOMENT OF INERTIA ABOUT A POINT. The definitionof the moment of inertia of a plane system of particles PlyP2i...,of masses ml9m2,..., about a point 0 in the plane is very similar,

If the polar coordinates of P1 ?P2, . . . are (rl9 0^, (r2, 02),... when0 is pole, then T

IAlternatively, if the Cartesian coordinates of Pl9 P 2 , . . . are(xi>yi)> (#2>S/2)> •••> w i t n 0 as origin, then

/0 =* mx{x\ + y\) + m2(x\ + y\) +.. . .

It follows that Io = Iy + Ix9

so that the moment of inertia of a plane system of particles about apoint 0 of the plane is the sum of the moments of inertia of the


system about two perpendicular lines lying in the plane and passingthrough 0.

The extension to a plane lamina is immediate.

ILLUSTRATION 5. To find the moment of inertia about the y-axisof a uniform lamina of density p bounded by those parts of the x-axisand the curve y = cos x which lie between x = — \n and x = \n.(See Fig. 47.)

We have

= p\J-in[in

= p\x2 cos x dx

x2-r- (sinx)dxdx"

in r\np[x2sinx] —p\ sinx.2xdz

-in J-in

PIT2

2. cos xdx

-in

THE MOMENT OF INERTIA OF A LAMINA 125

ILLUSTRATION 6. To find the moment of inertia of a circle aboutits centre.

Suppose that the equation of thecircle (Fig. 48) is

Divide the area bounded by thecircle into concentric rings, of whicha typical one is enclosed by circlesof radii r, r + Sr. Then, by the defini-tion of moment of inertia about apoint, the moment for the ring is

Fig, 48.

where the number rx lies between r and r + 8r. Adding for all therings, and proceeding to the limit in which their number isincreased indefinitely,

Jo

- \Aa\where A is the area of the lamina.

COROLLARY. By symmetry, Iy = Ix. Hence

6. Volume of revolution. Let

be the equation of a given curve, where f(x) is a single-valuedfunction of x, and consider the usual area ABQP (Fig. 49) boundedby the curve, together with the lines x = a, x = b, y = 0.

Suppose that the curve does not cut the #-axis betweenx — a,x = b.

To find an expression for the volume generated by the completerevolution of this area about the axis Ox.

Suppose, as usual, that the interval a, b is divided into n partsat the points ,

and that

126 APPLICATIONS OP INTEGRATION

Suppose also that m^ Mi are the least and greatest values of thefunction f(x) in the interval 8xP Then the volume (Fig. 50) of thecorresponding element lies between that of a cylinder of height8xi and radius mi} and that of a cylinder of height 8x{ and radiusMP Hence the volume of revolution lies between the two sums

O

Fig. 49. Fig. 50.

If we now allow the number of subdivisions to increase indefin-itely, their lengths decreasing indefinitely, then, in ordinary cases,these two sums tend to the limit

fJa

orCb

7ry2dx,Ja

which is therefore the expression for the volume F.

Ch

Hence F = iry2dx.

ILLUSTRATION 7. To find the volumeof a sphere of radius a.

The sphere may be regarded asgenerated by the rotation of the'upper' semi-circle of the circle

about Ox (Fig. 51). Fig. 51.

Then

VOLUME OF REVOLUTION

V

127

[a= 7ry2dx

J-af° <jr(a2-x2)dx

J-a

[ 1 l a

4

7. The centre of gravity of a uniform solid of revolu-tion. We regard it as obvious that the centre of gravity of asolid of revolution, generated as in § 6, lies on the axis Ox.

To locate the centre of gravity, we use the same principle as in§ 4 (p. 118) for a lamina. In the present case, we see that, if $is the x-coordinate, then

where 5TJ is an element of volume generated by a small area of^-coordinate a:i. Dividing the area, and so the volume, into stripsas in § 6, and proceeding to the limit, we obtain the formula

where V ny2dx,J a

Cb

£7 = nxy2dx,

the volume of the solid.

ILLUSTRATION 8. To find the centre of gravity of a uniform hemi-sphere.

Regard the hemisphere as generatedby the rotation about the axis Ox (Fig.52) of that arc of the circle

which lies in the positive quadrant.Its volume, as above, is

- 7 r a s .3

O

Fig. 52.


2Hence = 7r xy2dx = TT\ x(a2 — x

Jo Jo

so that t-U

8. First theorem of Pappus. Suppose that a solid of revolu-tion is obtained by rotating the usual area ABQP about the #-axis(Fig. 53). We proved (p. 126) that, if PQ is the curve

the volume V so generated is given by the formula

7ry2dx.Ja

But we also proved (p. 121) that the^-coordinate t\ of the centre of gravityof the plane area is given by the formula

£ IV**

where A

Hence

Cb

= ydx is the area ABQP.Ja

V = 27rArj.

O

Fig. 53.

Since 27717 is the circumference of a circle of radius 77, we mayexpress this result as follows:

/ / a given area, lying on one side of a given line, is rotated aboutthat line as axis to form a solid of revolution, then the volume sogenerated is equal to the product of the area times the distancetravelled by its centre of gravity.

This rule enables us to calculate the volume when the centreof gravity is known; or, alternatively, to find the ^-coordinate ofthe centre of gravity of the area when the volume is known. Theresult can also be extended easily to prove that the volume ofthe solid, generated by the rotation of the area bounded by a

THE FIRST THEOREM OF PAPPUS 129

closed curve lying entirely 'above' the axis of rotation, is equalto the area times the distance travelled by its centre of gravity.

For example, the volume of the anchor-ring obtained by rotatingthe circle x2 + (y — b)2 = a2, where b>a, about the #-axis is(Tra2). 2TT6, or 2?72a26.

ILLUSTRATION 9. To find the centre of gravity of a semicircle ofradius a.

Rotate the semicircle about its bounding diameter. The solidgenerated is a sphere, whose volume is known to be f 7ra3.

The area of the semicircle is ^7ra2, and so the distance rotatedby the centre of gravity is

4 o / l 2 8na* = - a.

o

Hence 2 ^ = - a,o

4aor r] = —.

The centre of gravity therefore lies on the line of symmetry ofthe semicircle, at a distance

4a3TT

from its bounding diameter.

9. The moment of inertia of a uniform solid of revolu-tion about its axis. The definition given in § 5 (p. 123) for themoment of inertia of a system of particles P1?P2,..., of massesm1,m2,... about a line I holds equally well in space. If pvp2, ...are the distances of Pi,P2,... from Z, then

In order to effect the summation for a solid of revolution,when I is the axis, we refer to the diagram (Fig. 50) and notationof § 6. The volume is divided into a number of circular discs, ofwhich a typical one has radius between mi,Mi and breadth 8xt.The moment of inertia of this solid disc about its centre is there-fore between hxi times that of a circle of radius mi about itscentre and hxi times that of a circle of radius M^ Hence (p. 125)

n- l n - l

\TTM\ hx^


In the limit, these two sums are equal, and so

PJa

ILLUSTRATION 10. To find the moment of inertia of a solid sphereabout a diameter.

Suppose that the sphere is generated by rotating the circle

about the diameter Ox, taken to be the given diameter. Then

)*dx1 Ca

= -77 (a*-x* J-a

16a5l

- 5

where F = f7ra3 is the volume of the sphere.

10.* The area of a surface of revolution.LEMMA. The area of a right circular cone.Consider a right circular cone, of slant height Z, whose base is a

circle of radius r (Fig. 54a). If the cone is slit down a generatorand then opened up so as to lie in a plane, we obtain the sectorof a circle of radius I subtended by an arc whose length, equal


THE AREA OF A SURFACE OF REVOLUTION 131

to the circumference of the base of the cone, is 27rr (Fig. 546).Hence the area is the fraction (27rr/27rZ) of the area of a wholecircle of radius Z, so that

= irrl.

What we shall actually require is the area of a frustum of acone, as indicated in the diagram (Fig. 55). If rv lx and r2, Z2 are

Fig. 54a. Fig. 55.

the values of r, I for the two boundaries, then the area of thefrustum is , , T v

Now, by similarity, u

and so, on substituting for Z2, ll9 the area is

If we draw the 'half-way' circle through the middle of the frustum,its radius is £(r2 + fi) and perimeter ^(r^ + r-,). Hence the area ofthe frustum is

(perimeter of 'half-way' circle) x (slant height of the frustum).

We now proceed to our main task:

To find the area of a surface of revolution. Let f(x) be a positivesingle-valued function of # in a certain interval (a, 6), and rotate

132

the curveAPPLICATIONS OF INTEGBATION

through four right angles about the #-axis. We require anexpression for the area of the surface generated.

Suppose that the area traced out by the arc AP is 8 (Fig. 56);then S is a function of x which we denote by the symbol 8(x).

In the plane, let P = (x, y),Q = (x + h,y + Jc). Then

-81*) y

h

area traced by arc PQh

area traced by arc PQarea traced by chord PQ

area traced by chord PQX h

We assume as a matter of definition that

o M N

Fig. 56.

y area traced by arc PQ = ,fc->o area traced by chord PQ

Moreover, the area traced by the chord PQ is, in accordance withthe lemma,

Thus 8'{x) = lim

lim'

-8(x)

1 +

:ir .2y.l imA /( l + S

~A )dxso that,

between appropriate limits.We shall see in Volume II that the integral

^ \dx

THE AREA OF A SURFACE OF REVOLUTION 133

gives the length of the arc of the curve y =f(x). Denoting thislength by s, we may write the expression for 8 in the form

ILLUSTRATION 11. To find the area of that portion of a sphereof radius r which is cut off between two parallel planes at a distanceh apart.

The area is generated by rotatingthe arc AB of the circle

about the axis Ox (Fig. 57).Let the ^-coordinates of A,B be

a,b respectively, where

6 — a = h.

Differentiating the equation of thecircle with respect to x, we have Fig. 57.

so that

and

1 +

JHWH»/2»

taking positive values. Hence the area is

2nyi-\dx

fJc= 2>rrr(b-a)

= 2TTTA.

Note. This value depends only on r,h, and not on where theportion of the sphere is situated. It is actually equal to the areaof a circular cylinder of radius r and height h.

1 0


11. Approximate integration. The evaluation of a definiteintegral

(f(x)dxJa

may present considerable difficulty, but it is often easy to reacha good approximation to the result. In this and the followingparagraphs we give an account of some of the methods.

We proceed by regarding the integral as the area 'under' thecurve

/ (0between the limits x = a, x = b. In the diagram (Fig. 58), PA,QBare the ordinates x = a, x = b, of lengths ya, yb respectively, andthe curved hne joining A, Bis given by the equation y = f(x). Forease of exposition, we assume thatb > a and also that f(x) is positive inthe interval; the modifications forother cases can easily be obtained.

A crude approximation may befound by replacing the curve bythe straight line AB, thus replacingthe area under the curve by that ofthe trapezium PABQ, or

Fig. 58.

y

o

A

ya

pa I

B

yb

Q

To improve on this, we may divide the segment PQ into say,n equal parts at points given by x = xlyx2, ...,xn_1, and erectordinates to the curve, of lengths yv y2, ...,yn-x respectively. Thecurved segment AB may be replaced by the 'chain' of straightlines joining the 'tops' of these ordinates, and so we obtain ntrapezia, whose areas are respectively

b — a~2/T (

b — a b — a2n

b-a2n (

Adding, we obtain the approximation

b-a,2n • {(y +2/2+. - -

b-aOr {(sum of outside ordinates) + 2(sum of inside ordinates)}.

2n

APPROXIMATE INTEGRATION

For example, consider x*dx,Jo

where f(x) = #4,

6 = l,a = 0.

Divide the interval into ten equal parts at the points

xx = -1, x2 = -2, ..., x9= -9.

135

Then yx = -0001

y2 = -0016

yz = -0081

t/4 = -0256

2/5 = '0625

2/6 =

«/7 =

2/8 =

2/9 =

= 0, 2/6 = 1.

The approximate value of the integral is

^ { l + 2x 1= -20333.

= -1296

= -2401

= -4096

= -6561

•5333} = j& (4-0666)

The correct value is \\xh\ = J = -2.

12. Simpson's rule. An approximation which is often muchbetter than the 'trapezium' rule just given is obtained by replacingthe curve y = f(x) by a parabola, of the form

made to pass through the end points andone intermediate point of the curve.

[The 'trapezium' rule of § 11, is, of course,equivalent to replacing the curve y = f(x)by the straight line

y = A + Bx

ya

p

y

y

0

\

b-a- -

B

Q x

Fig. 59.

through the end points.]For ease of calculation, take fresh axes with the origin at the

middle point of PQ9 as in the diagram (Fig. 59). Let the ordinatesthrough P, Q, O be of lengths ya, yb, yx respectively. The parabolais to pass through the points (|(6~a),2/6), (0,^), (-!(&-a),2/a)-


Inserting the coordinates of the three 'guide' points in theequation y

of the parabola, and writing h = £(6 — a), we have

so that A = ylt

B = ~(yb-ya),

C=2fi(yb+ya-2y1).

Now the area under the parabola (and this area is not affectedby the simplified choice of axes) is

f* (A + Bx + Cx2)dxJ-h

-h

= J(6 — a) {(sum of outside ordinates)

+ 4 (middle ordinate)}.

For example, taking the integral

Joconsidered in § 11, we have

Va = °> Vb= !> Vi = ^ ,

so that the approximate value is

= -20833,

which is very good agreement for so simple a calculation.

S I M P S O N ' S RULE 137

Finally, we obtain SIMPSON'S RULE as follows:Divide the interval (a, 6) into 2n equal parts at the points

and let the corresponding ordinates be

Apply the preceding formula to the interval (a,x2), giving thecontribution

then to the interval (x2, #4), giving

then to the interval (#4, x6), giving

J3 = - ^ - (

and so on, up to

Adding, we obtain the approximation known as Simpson's rulefor 2n divisions:

—— {(sum of outside ordinates)

+ 2(sum of even ordinates)

+ 4(sum of odd ordinates)}.

Applying this rule to x*dx with ten divisions, we have theapproximation

= ^ ( 1 + 1-1328 + 3-8676) = ^(6-0004)

= -200013,

which is noticeably better than the corresponding 'trapezium'approximation (p. 135).


EXAMPLES IIObtain approximations to the following integrals by dividing

the interval of integration into ten equal parts and using (a) thetrapezium rule, (6) Simpson's rule.

1. Fx2dx. 2. [*x*dx. 3. [2{x+\ydx.JO J2 JO

4. F (a* + x)dx. 5. P (a* + x*)dx. 6. P^-x^x.

7. (x* + x)dx. 8. . 9. -9.Jo J o l + ^ Jil+a;2

13. Mean values. It is a matter of ordinary language that themean (or average) value of the seven numbers

1, 3, 8, 7, 11, 5, 4

1 + 3 + 8 + 7 + 11 + 5 + 4 39T - y

and that the mean value of the thirteen numbers

2, 2, 2, 3, 3, o, 3, o, 7, 9, 9, 9, 9,

where 2 occurs three times, 3 five times, 7 once and 9 four times,

3(2) + 5(3) + 7 + 4(9) _ 6 + 15 + 7 + 3618 3 + 5 + 1 + 4 "" 13

_ 64" 1 3 *

If the numbers yl9yz,..',yk appear nv n2,...,nh times respectively,the mean value is

n1 + n2+...+nk

If a plate is divided into k pieces, of uniform densities

and areas A l 9 A 2 , . . - , A k

respectively, the mean density is similarly

A1w1 + A2w2+...+Akwk

MEAN VALUES 139

If a man walks for \ hour at a speed of 4 m.p.h., for \ hour at5 m.p.h., rests for 20 mins., and then walks for l\ hours at3J m.p.h., his mean speed for the whole time is, in the same way,

t + 0 + ¥ si = 1022-A 2 ^ 31

m.p.h.

The last example serves to warn us of a danger: the distancetravelled is 2 miles at 4 m.p.h., l j miles at 5 m.p.h., rest for20 mins., and then 5£ miles at 3 | m.p.h. The mean speed for thewhole distance is

= 26168

= 3ff m.p.h.

In other words, the mean speed with respect to time taken isNOT the same as the mean speed with respect to distance covered.Where there is any possibility of doubt, the magnitude with respectto which a mean is calculated must be clearly stated.

These examples illustrate the evaluation of a mean for discretenumbers, and we assume that the ideas are familiar. Our problemis to extend the conception to the mean value of a function of acontinuous variable. In order to do this, we make a definition ofthe term 'mean', choosing that definition to fit in with the moreelementary considerations.

DEFINITION. The MEAN VALTJE of the function f(u)} defined for allvalues of u in the interval (a, 6), is the quotient

'6f(u)du

7=6du

Ja

r[[Compare


ILLUSTRATION 12. To find the mean density of a straight rod oflength 2a, given that the density at a distance kfrom the middle pointis p(a + k)2.

Take the middle point 0 of the rod as the origin for a coordinatex; then k = + x when x is positive and —x when x is negative.The interval of integration must therefore be divided into the twoparts (— a, 0), (0, a). Then the mean density is

Ya14 3

2a\3pa

= \Pa\

APPLICATION TO A VOLUME OF REVOLUTION.

To find the mean density of a solid of revolution whose axis isvertical, given that the density at height x above its base is

G(x)

(a function ofx only), the height of the solid being h.Suppose that the radius of the 'slice' (Fig. 60)

at height x is r, where

The volume between the base and that slice is(p. 126) _ p,

* Jo

so that (p. 87) -j- =Fig. 60.

The mean density with respect to volume is, by definition,

rJorJo

dv

MEAtf VALUES 141

where V is the total volume of the solid and where 0 is thefunction G(x) regarded as a function of v in virtue of the relationconnecting the two variables v, x.

Moreover, since — = 77T2, these integrals can be expressed in

terms of x in the form

Jo

rh »7rr2dx

Jowhere r = f(x)>

a given function of x.Hence the mean density can be calculated.

ILLUSTRATION 13. To find the mean density of a sphere of radius a,given that the density at a distance x from a given diametral planeis Ax2.

By the formula just obtained, with obvious modification, themean density is

[aAx27rr2dx

-aa

irr2dx

[a

J-

where, for a sphere, r2 + x2 = a2.

Now r irr2dx = | <n{a2-x2)dx = 7r|a23-

as is familiar.

Also I TTAX2 r2dx= I TTAX2 (a2 -x2)dxJ—a J—a

Hence the mean density is

\Aa\


APPLICATION TO A SURFACE OF REVOLUTION.* To find the meandensity of a surface of revolution whose axis is vertical, given that thedensity at height x above its base is

Q{x)

(a function of x only), the height of the surface being h.Suppose that the radius of the 'ring' at height x is r, where

r =/(*).

The area of surface between the base and that slice is (p. 132)

so

The mean density with respect to area is, by definition,

rA

GdSJo

J.AdS0

where A is the total area of the surface, and where G is thefunction G(x) regarded as a function of S in virtue of the relationconnecting the two variables 8, x.

Since -7- = &dx

(dr\2\

ME))'these integrals can be expressed in terms of x in the form

/ :

where r=f{x),

a given function of x.Hence the mean density can be calculated.

* This application may be postponed, if desired.

MEAN VALUES 143

ILLUSTRATION 14. A bowl of depth a is formed by rotating thatpart of the parabola y2 = 4ax

for which O^xâ about the x-axis. The density of the 'ring9 atposition x is a + x. To find the mean density of the bowl. (SeeFig. 61.)

Adapting the formula given above, themean density is

y2 = ±ax,Now

so that

and

Hence the mean density is

1+y2

(a + x)x

f % + x).

The numerator is

±7T*Ja I

The denominator is

a

3 (4A/2— 1)The mean density is thus - ) \ —~ a,

5 (^ A/J5 — 1 )

vFig. 61.

. V (^^ dx

.f | (a + x)* j- f 7ra3(4 ^2

a.f (a + z)*

- (a)*}

or, on multiplying numerator and denominator by 2 /2 + 1,

-).a = ^ ( 1 5 + 2y2)5(8-1) 35

144 APPLICATIONS OB1 INTEGRATION

EXAMPLES III1. ABC is an equilateral triangle of side 2a, and a point P is

taken on the side BC. Find the mean value of AP2 as P varieson BO.

2. Find the mean density of a rod AB of length 21, given thatthe density at a point P is AP2 + PB2.

3. Find the mean density of a thin hemispherical cap of radius a,given that the density at a distance x from the base is a -+- x.

4. Find the mean density of a thin bowl formed by rotatingthat part of the parabola y2 = lax for which 0 < x < a about thex-axis, given that the density of the ring at position x is (i) (a + x)2,

REVISION EXAMPLES II

'Alternative Ordinary9 Level

1. Integrate (2x — 3)2 and 2x* with respect to x.

2. Evaluate the integrals

x2dx, x*dx> I - T .J- i J- i Ji x2

How could you show graphically, without evaluation, that thevalue of the first integral is greater than the value of the second?

3. Find an expression which, when differentiated with respectto x, gives

Find the value of the integral

P (x+l)2dx.

4. Integrate with respect to x:

[inEvaluate sin 2x cos x dx*

siiJo

5. Integrate with respect to x:

(i) (z+1)3; (ii) cos2 3s.

Evaluate sin2 x cos x dxsirJo

BEVISION EXAMPLES II 145

6. Integrate the following with respect to x:

(i) sin2 x; (ii) —s—; (iii) sin 2 x cos x.x*

n a2 2xz\3—s + —ir\dx.

x2 a3/

8. A curve has a gradient which is given by -j- = 3a; —2; also(tX

this curve cuts the y-axis at the point where y = 4. Find theequation of the curve, and the equation of the normal at thepoint (0,4).

9. Find the equation of the curve which passes through thepoint (0,1) and whose gradient at the point (x, y) is given by

dy ,ax

Find the values of x for which y is a minimum and draw arough sketch of the curve.

x2

10. The gradient of a curve at the point (x} y) is 1 — — • Find theequation of the curve if it passes through the point (2,4).

Find the point of contact of the tangent which is parallel to thetangent at (2,4); also find the equations of both of these tangents.

11. Prove that the #-axis is a tangent to the curve y = (2x— I)2.Find the area bounded by this curve, the #-axis, and the line

x= 1.Find the area bounded by the curve, the cc-axis, and the tangent

at the point (1,1).

12. A curve passes through the point (2, J) and its gradient at

the point (x, y) is 1 —%• Kn ( i the equation of the curve.x

Where has y a minimum value on this curve?

13. Calculate the area bounded by the curve y = x2 + 2 and thelines y = x, x = 1 and x = 3.

Calculate the area in the first quadrant bounded by the curvey = x2 and the lines x = 0, y = 1 and y = 4.

146 REVISION EXAMPLES II

14. A particle starts with a velocity of 2 ft. per sec. and movesalong a straight line. Its acceleration in ft. per sec. per sec. aftert sec. is t + 3. Find its velocity at the end of 2 sec. and the distancetravelled in the next 2 sec.

15. A particle starts from rest and moves in a straight line. Itsvelocity in ft. per sec. is given to be 8/ — t2, where t is the time inseconds from the commencement of motion. How far will theparticle have moved in 3 seconds?

Find also its greatest distance from the starting point, and thevalue of t when this distance is reached.

16. The velocity of a train starting from rest is proportional tot2, where t is the time which has elapsed since it started. If thedistance it has covered at the end of 6 seconds is 18 ft., find thevelocity and the rate of acceleration at that instant.

17. A particle moves in a straight line with velocity It — t2 — 6ft. per sec. at the end of t seconds. What is its acceleration whent= 2 and when t = 4?

When t = 3 the particle is at A; when t = 5 the particle is at B.Find the length of AB.

For what values of t is the particle momentarily at rest?

18. A train starts from rest and its acceleration t sec. after thestart is £(20 — t) ft. per sec. per sec. What is its speed after 20 sec. ?

Acceleration ceases at this instant and the train proceeds at thisuniform speed. What is the total distance covered 30 sec. afterthe start from rest?

19. A body starts with velocity zero from a fixed point 0 andmoves in a straight line; its acceleration t sees, after it leaves 0 is3 — t ft. per sec. per sec. Find the velocity of the body 4 sec. afterleaving 0 and the distance travelled in the third second of themotion.

20. The velocity v of a particle moving in a straight line OA isgiven by the equation ,, 2.

CiyV j

dx '

where x is the distance from 0 and v = 3 when x = 2. Find vwhen x = 0; also find the greatest positive value which x attainsduring the motion.

REVISION EXAMPLES II 147

21. The velocity of a particle moving in a straight line isobserved to be 4£+4£2 — t3 cm. per sec. at the end of t sec. Findthe acceleration of the point after 4 sec, the distance of the pointafter 5 sec. from its position when t = 0, and the distance travelledin the fourth second.

22. A body starts from a point 0 and moves in a straight line.Its velocity at 0 is zero and its acceleration t sec. after leaving 0is 5 — f t ft. per sec. per sec. Find the greatest velocity attained bythe body on its outward journey, and its distance from 0 at theinstant when it begins to return towards 0.

23. A particle moving in a straight line has an acceleration of3 — t ft. per sec. per sec. at time t sec. When t = 1 the particle is atrest at a point A. Find for what value of t greater than 1 theparticle is again at rest and how far it is then distant from A.

24. The velocity v (in ft. per sec.) of a particle moving in astraight line is given by v = t2 — lt+ 10, where t is the number ofseconds which have elapsed since the particle passed through afixed point 0 on the line. Show that the particle is momentarilyat rest at each of two points A and B on the line and find thelength of AB.

25. The velocity v in feet per second of a point moving in astraight line is given by v = 3t2 + 2t + 1, where t is the time inseconds that has elapsed since a given instant. Find the accelera-tion when t = 2 and the distance covered between the times t = 2and t = 3.

26. A car starts from rest with a variable acceleration, itsacceleration after t seconds being (a — 3£) feet per second persecond. If the distance covered in the first 4 seconds is 88 feet,find the value of a.

27. From the point P(2,4) on the curve y = x2, PN is drawnperpendicular to the axis of x. Find the area bounded by PN, theaxis of x and the curve.

Find also the x and y coordinates of the centre of gravity of thisarea.

28. Find the area bounded by the curve

y=(x+l)(x-2)2

a n d t h e ic-axis f r o m x = — l t o # = 2 .

148 EBVISION EXAMPLES II

Find also the ^-coordinate of the centre of gravity of this area.

29. Calculate P x(x2-l)dx.

Find the area bounded by the curve y = x(x2 — 1) and the #-axis(i) between x = — 1 and x = 0 and (ii) between x = 0 and x = 1.Explain with the aid of a rough figure the connexion between yourresults and the value of the integral found in the first part.

30. Find the coordinates of the centre of gravity of the areawhich lies above the (positive) #-axis and below the curvey = x2(3-x).

31. Find the area bounded by the curve y = x2, the axis of xand the ordinates x = 1 and x = 2.

Find the x and y coordinates of the centre of gravity of this area.

32. Calculate the coordinates of the centre of gravity of the areaenclosed by the straight lines x = 0, y = 0 and the portion of thecurve y = 9 — x2 which lies in the first quadrant.

33. Find the ^-coordinate of the centre of gravity of the areabounded by the #-axis, the y-axis, and that portion of the curve

which lies in the first quadrant.

34. Find the area included between the axis of a; and the portionof the curve y = x2 — 9 below that axis.

Find also the coordinates of the centre of gravity of this area.

35. A flat thin plate of uniform density is bounded by thetwo curves y = x2, y = — x2 and the line x = 2. Find its area andthe coordinates of its centre of gravity.

36. Calculate the area above the #-axis bounded by the curvey = 2x(3 — x) and the #-axis.

Find both coordinates of the centre of gravity of this area.

37. A uniform lamina in the form of a quadrant of a circle ofradius a is bounded by radii OA and OB. Find the distance of thecentre of gravity of the lamina from OA and from OB.

By considering the rotation of the lamina about OA, prove thatthe volume of a hemisphere of radius a is \naz.

REVISION EXAMPLES II 149

38. The area enclosed by the parabola y2 = 4tx and the straightline x = 4 is rotated about the axis of x through two right angles.Find the volume of the solid so generated and the ^-coordinate ofits centre of gravity.

39. An area in the first quadrant is bounded by the ellipse4#2 + 9y2 = 36 and the axes of coordinates. This area is rotatedthrough four right angles about the #-axis. Find (i) the volumegenerated and (ii) the ^-coordinate of the centre of gravity of thisvolume.

40. The area bounded by the arc of the curve y=*x(3 — x)between the points when x = 0 and x = 2, the #-axis, and the linex = 2, is rotated about the #-axis. Find the volume of the solidof revolution so generated, and the ^-coordinate of its centre ofgravity.

41. The curve y2 = x2(2 — x) cuts the #-axis at the points given byx = 0 and x = 2. The area enclosed by the #-axis and the curvebetween these two points is rotated through four right anglesabout the axis of x so as to form a solid of revolution. Find thevolume of this solid and the ^-coordinate of its centre of gravity.

42. Solids of revolution are generated by rotating (i) about theic-axis the area bounded by the arc of the curve y = 2x2 between(0,0) and (2, 8), the line x = 2 and the #-axis; (ii) about the y-axisthe area bounded by the same arc, the line y = 8 and the y-axis.Calculate the volumes of the two solids so formed.

43. A cylindrical hole of radius 4 in. is cut from a sphere ofradius 5 in., the axis of the cylinder coinciding with a diameter ofthe sphere. Prove that the volume of the remaining portion ofthe sphere is 36TT.

44. Sketch the curve whose equation is y = xz and find the areabounded by the curve, the positive #-axis, and the straight linex = 2.

Find also the volume generated when this area is rotated aboutthe #-axis through four right angles.

45. A uniform solid right circular cone is of height h and theradius of its base is r. Find the volume by the methods of theintegral calculus.


Find also the distance of the centre of gravity of the cone fromits vertex.

46. The radius of a sphere is 5 in. Two parallel planes aredrawn at distances 2 and 3 in. respectively from the centre and1 in. apart. Use the calculus to determine the volume of the slice ofthe sphere between the two planes.

[Regard the sphere as formed by the rotation of the circlex2 + y2 = 25 about the #-axis.]

47. Sketch roughly the two curves

x2 + y2 = 25, x2 + 4y2 = 25.

A solid is formed by the revolution through four right anglesabout the o;-axis of the part of the area between the two curves inwhich y is positive. Find the volume of the solid.

48. Find the volume of the solid of revolution generated by therotation about the #-axis of the area bounded by the curve y2 = 2xand the line y = \x.

49. The ellipse ^5 + TI = l

a2 b2

is revolved about the #-axis. Find the volume of the solid soformed and the as-coordinate of the centre of gravity of its right-hand half.

50. The area between the circle x2 + y2 = 16 and the ellipse9x2 + I6y2 = 144 is rotated about the #-axis. Calculate the volumeof the solid of revolution so formed.

51. A solid of revolution is formed by rotating the portion ofthe curve y = a sin a; between x = 0, x = \TT about the #-axis.Find the volume of the solid and the distance of its centre of massfrom the origin.

52. Find the area included between the curve ythe axis of x, and the ordinates at x = l,x = 3. Show also thatthe area is exactly equal to that of a rectangle on the same base(of two units) whose height is £(2/i + 4t/2 + y3), where ylty2>yz arethe ordinates at # = !,# = 2, # = 3 respectively.

BEVISION EXAMPLES II

53. Apply Simpson's rule to calculate approximately

ri/O

making use of the table:

151

rJo

X

V(2+since)

0

1-414

\n

1-544 1-645

1*1-710

in

1-732

54. Prove that

ydx - a) {yb + ya

where y is a polynomial in x of the t ><2 degree, ya, yb are the valuesof y corresponding to the end points, and yv y2 are the values of ycorresponding to the points of trisection of the interval a, b.

Hence obtain an approximate value for

Jo55. Use Simpson's rule, taking five ordinates [four divisions], to

find an approximation to two decimal places to the value of theintegral

56. A river is 80 feet wide. The depth d in feet at a distancex feet from one bank is given by the following table:

a; 0 10 20 30 40 50 60 70 80d 0 4 7 9 12 15 14 8 3

Find approximately the area of the cross-section.

57. Establish Simpson's rule that, if

thenJo

)dx

Prove also that, if

f{x) ss A + Bx + Cx2 + Dx* + Ex\

then the error in still using that rule is yô -®-


58. The speed v in miles per hour of a train starting from rest isobserved at intervals of one minute:

J 0 1 2 3 4 5 6 7 8v 0 7 13 18 22 25 28 33 27

Estimate the distance covered in the eight minutes.

59. Show that constants a, 6,c exist such that, if f(x) is anypolynomial function of degree five (or less), then

J : f(x) dx = h{af(0) + b[f(h) + / ( - h)] + c[f(2h) + / ( --2h

Show also that, with these values of a, b, c but with f(x) = xe, the

expression on the right is equal to - f(x) dx.Vj-2h

60. Evaluate sin1 xdxsirJo

approximately by Simpson's rule, using five ordinates (i.e. atintervals of

61. Evaluate \~ 0sin20d00sirJo

to three decimal places both by an exact method and by Simpson'srule using five ordinates.

62. If y = a + bx + ex2 + dxz, prove that

J *3

owhere t/0, yv y2, yz on the values of y at x = 0, h, 2h9 3h respectively.

Hence approximate to the value of

o

63. Using Simpson's rule with seven ordinates, calculate thearea under the curve

between the ordinates x — 0, x = 6.

ANSWERS TO EXAMPLES

CHAPTER I

Examples I:1. (i) x*y-y* = 4, (ii) z* + y*-Jy = 2.2. a2 =

Examples II:1. 1; 1 and — 1; none; 1 and 3. 2. WTT; £(2w-f l)ir.

3. 0; 1 and - 1 .

Examples IV:1. h5. | .

Examples V l

1. 1.4. 1 and 2.

7. |(2W+1)T7.

10. i(4tt+l)7T.

Examples VI:

2.

6.

1. 3 at each value2. -2 ,0 ,2 ,4 .6. l<z<2.8. 2x-y-\ =

9. 4 z - 2 / - l =

Examples VII:9. 4, 20, 12.

- 1 .

4.

2.

5.

8.

11.

of #.

t# + */-f

1 and

nn.

1.

None.

-4 = 0,

3.

7.

_

y0, 2x-y

I.— 2

1. 3.

6.

9.

12.

= 0, 6x-y-= 0.

4. 0.8. 0.

3 and|(2TI +

3 and(2n+l

9 = 0.

- 3 .1)77.

— 3

154 ANSWERS TO EXAMPLES

Examples VIII:

1. Tangent 30x-y-45 == 0, normal x + 30y- 1353 = 0.Tangent 2Qx + y + 20 = 0, normal #-20?/+ 402 = 0.Tangent y = 0, normal x = 0.

2. Tangent 12a;-?/- 16 = 0, normal x+ 12?/-98 = 0.Tangent 3# — y 4- 2 = 0, normal x + 3y + 4 = 0.

Tangent y — 0, normal a: = 0.

3. (i) Tangent 12#-?/-24 = 0, normal x +12?/-2 = 0.(ii) Tangent ?/ + 8 = 0, normal x = 0.

CHAPTER II

-5 ) " 85)"8.

Exam/pies I:

1. 4a;3.

3.

5. 3x2(x+l)2(2x+l).

7. -or-2.

9.11.13.15.17.19.21.23. -1^(20*4-l)(4a;--1)"4.

Examples II:

1. 2cos2#. 2. —3 sin 3x.

3. 25cos5#. 4. sin 2x + 2# cos 2#.5. 2xQO&x — x2sinx. 6. 3cosZx — 9#sin3a;.7. 2(#+l)sin7#+7(a;+l)2cos7#. 8. 3 cos (3a:+ 5).9. 6(2a:+l)2. 10. -4(o; + 2)-5.

11. cosec x — x cosec a; cot x. 12. 1 + sin x + a: cos x.

2.4.6.8. -4a;-5.

10. -12. - 8 (14. far*.16. (2s+ 3)-*.18. -ix-K

20. (3s + 2) (2x +22.24.


13. sin 2a;. 14. - s i n 2a;,

15. 3 sin2 x cos x. 16. — 3 sin x cos2 x.

17. sin2 a; + a; sin 2a;. 18. 2x cos2 x — x2 sin 2x.

19. 2x cos2 2x -2x2 sin 4a;. 20. sin2 x +(1 + #) sin 2a;.

21. - s in (z - j7 r ) . 22. cos 2a;.

23. Jar*. 24. -Jar* .

25. §ar*sin2a; + a;f sin 2a;. 26. Jcosa;(sina;)-*.

27. (1 + Ja;cota;)(sina;)*. 28. 2a;(sin4a;)*(H-a;cot4a;).

29. — coseca;cota;. 30. sec a; tan a;.

31. sec2 a;. 32. — cosec2#.

33. -^-sin2a;0. 34. -^-

35. sina;° + ^-cosa;0 . 36.180^ c o s a ; . 36. ^ r180 loO

Examples III:

1. 5a;4, 20a;3, 60a;2.

2. 3a;2, 6a;, 6.

3. 1, 0, 0.

4. sinx + xcosx, 2cosx — xsinx, —3sinx — xcosx.

5. 2a; cos a; —a;2 sin a;, 2 cos x — 4a; sin x — x2 cos x}

— 6 sin x — 6a; cos x + x2 sin #.

6. sin2 x + a; sin 2a;, 2 sin 2a; + 2a; cos 2a;, 6 cos 2a; — 4a; sin 2a;.

7. 2 cos 2a;, - 4 sin 2a;, - 8 cos 2a;.

8 . - 4 sin 4a;, — 16 cos 4a;, 64 sin. 4a;.

9. sin 2a;, 2 cos 2a;, —4 sin 2a;.

10. - a r 2 , 2a;-3, - 6 a r 4 .

11. -2 (2a ; -3 ) - 2 , 8(2a;-3)-3, ~48(2a;-3)-4 .

12. -3a;-4, 12a;~5, -60a;-6.

Examples IV:

1. 2 sec 2a; tan 2a;. 2. 4 sec2 2a; tan 2x.

3. 4 sec2 2a; tan 2x. 4. 9 sec2 3a; tan2 3a;.

156 ANSWEKS TO EXAMPLES

5. cosec a; — a; cosec s c o t s . 6. 2x cot 2x — 2x2 cosec2 2x.

7. sec x tan2 a; + sec3 x. 8. Jsina;(cosa;)-*.

9 . - 6 cosec3 2x cot 2a;.

10. mxm-1 t an n a; + nxm tan71"1 x sec2 x.

11. — ma:"771"1 secw x + ma;~m secm a; tan x.

12. £

Examples V:

1. (i) 2,\

- 2

2

2. tanâ;* x

a;

6. 2a; cos*1 a;2 —

1

2a;3

8.

10.- 1

(ii) 2,

14. cosecâ; —-

16.

3.2x

5. 3a;2 sin-12a;+ -

2 t an- 1 a;

9.

11.2 sin-1 a:

1O , , x« 3a;(cos-1a;)2

12. (cos-1 a;)3 ~ ^ - . 13. 2a; sec-1 a;+ -

2a;3

15. 2a;(sin-1a;)2 +2a;2 sin"1 x

Examples VI:

1. 6 per cent.

3. 0-2.

2. 2 per cent.

4. 0-00023.


CHAPTER III

Examples I:

1. X = — 7TSin7Tf, X = — 7T2COS7r£.

2. x = Jhr cos |TT£, x = — JTT2 sin |7r£.

3. £ = 5-64J, a; = - 6 4 .

4. a; = 64*, a; = 64.

5. x = sin TT£ + JTT£ cos \nt, x — TT cos TT£ — JTT2 £ sin 17rt.

6. x = 2 ^ c o s ^ — 2 s i n | 7 7 ^ ,

^ = 2 cos TT£ — 2?r sin ^TTI — |T72 i2 cos \nt.

7. i = 2^-96, o;= 2.

8 . ^ = sin2 \nt + JTT^ sin nt, x = TT sin TT* + |TT2 t cos TT£.

Examples Hi

2. 1. 3. - 4 .

Examples III:

1. 0. 2. 1.

3 . - 1 and 1. 4. - 1 , 0, and 1.5. nTT + TT, all integral n. 6. 7i7r + \TT, all integral n.

7. nir, all integral w. 8. 2mr, all integral /i.

Examples IV:

4. (i) a;>i. (ii) All x,

(iii) |a?|>l. (iv) x>0.

(v) — 1 < x < 0 and # > 1. (vi) x < 1 and a; > 4.

Examples VI:

1. Minimum.2. Minimum.3. Minimum at x = 1, maximum at x = — 1.

4. Minimum at x = — 1 and 1, maximum at a; = 0.

5. Minimum at x = 2ri7r — JTT, maximum at a; = 271TT4- |7T.

6. Minimum at # = TITT — JTT, maximum at # = UTT + £77.


7. Minimum at x = (2^+1) TT, maximum at x = 2rnr.

8. Minimum at x = 2(2w + 1)TT, maximum at a; = 4TI7T.

10. (i) Maximum at x = — 2, inflexion at a; = 0, minimum at# = 2.

(ii) Maximum at x = — 1, inflexion at a; = — J, minimumat x = 0.

11. Maximum at x = 2mr -f-~> 2mr +—, 2WTT + — .4 4 4

Minimum at x = 2wn + —-, 272JT + — , 2w7r-h — .

4 4 4

12. (i) Maximum at # = 1, minimum at x = 2.

(ii) (a) z < l andz>2 . (6) l<ic<2.

Examples IX:

2. (i) | = 1 . (ii) | = 2. (iii) | = 1 .

Examples XI

1. f. 2. 0. 3. it. 4. 1. 5. \. 6. 1.

REVISION EXAMPLES I

— 2 r1. (i) zcosz. (ii) 6x-10. (iii)

(L+X )

43. (i) 2a; - - ^ . (ii) 2 tan a; sec2 a;.

4. (i) &*-* (") C°SX • («i) - 7 - ~ —•v ^ JL T~ b i l l .& y \\U/ —X I

— 2r5. (i) a;2(l+ x) (3 + 5x). (ii) 2sin 4a;. (iii) -j—

6. (i) 6a;(l—a;), (ii) 2 cos 2a; cos a; — sin 2a; sin x. (iii) —.

7. 2a ; - i / - l = 0, 6x-3y-8 = 0.

8. 0, 2a:-i/ = 0; ^4, x + y- 1 = 0; 5 , 2a; - t / -4 = 0. Intersection


9. Maximum at x = 0, minimum at x = 2. Parallel tangent at

10. ±x- = 0. Point (§, ) .

12. 5x-y-l = 0, (2,4), (4 , -8) .

13. Gradient 4&, tangent y — k = 4Ji(x — h). Tangents through

origin y = ± 12a;.

14. # = - and - - .

15. a = 2, 6 = - 9 , c = 12, d = 0.16. Maximum value 9 at x = — 1; minimum value — 7 at x = 1

and at x = — 3.17. Maximum value 4 at # = — 2; minimum value 0 at x = 2.

Tangent 3# + 2i/ - 4 = 0.

18. (-1,27) and (2,0).

19. (1,6) and (3,2).

20. Gradient zero, minimum.

21. (1,2) and (3, —2). Three positive roots, one between 0 and 1,

one between 1 and 3, and one greater than 3.

22. Velocity = 2pt + 3qt29 acceleration = 2p + 6qt, p = 12, g = - 1.

24. Velocity = 0, acceleration = — 4. 1 | seconds, 18 feet.

25. Velocity 16 ft./sec. Acceleration 14 ft./sec.2

26. Velocity = t cos t. Acceleration = cos t — tsin t.

27. f sq. in., 4 sq. in.

28. 32 ft.

4a 3a29. x = —. Time — hours.

o 4

30. x = 13-7.

31. 4500 cu. ft.

32. Volume 7£ cu. ft. Length 4-09 ft.

33. 48 sq. ft.34. Area = ^ (I2 - 2lx + 2x2). x = - .

2


37. Area = ^x2 + 3 - (I — a;)2. Minimum.

39. f A per cent.

40. — cu. in. per sec.

41. Shortened, -7-2- per cent.

42. (i) 0*15 77 cu. in. per sec. (ii) 0-16 77 sq. in. per sec.

43. (i) 0-00008 in./sec. (ii) 0-025 cu. in./sec.

44. — = — 3 in. per hour. 24?7 sq. in. per hour.(tt

45. \h per cent.

CHAPTER IV

Examples II:

3. A + 1 sin 2x, A —2 cos \x,

5. 8 | , 0, 1, 66.

213527 ' *' *' "405~*

8. 4TT.

CHAPTER V

Examples I:

1. C — Jcos2#. 2. (7+Jsin3a?.

3 f7 — 9 PDQ •l-/>* ^1-

7. C + |(a; + 3)4. 8.

11. C-i(a;-l)-2. 12. O-i

13. C + 4(2z+l)3. 14.


15. C+f(|z + 7)5. 16.17. C-£(4a;-3)-2. 18.19. C + §(z+l)*. 20.21. C + 2(#+l)*. 22.23. C-(2s-3)-*. 24.25. C-Kxt+l)-1. 26.27- 0—Aâ^+l)-1. 28.

O l t vy ~|~ "R" o l l l *t/ """" •g' S i l l •*/• tJ4,

33. (7 —^(1+tanx)~2. 34.

35. (7 + tanx + Uan3a;. 36.

Examples II:

2. (7 + 1 sin-1 a; - \x{ 1 - a;2)*

3. C + a; —tan-1*.4. 2*5.

6.

7. O -8. 0 - | ( 1 8 +a;) (9-a;)*.9. C — »

10.

Examples III:

1. i 2. f 3. | + 1. 4. y. 5. g + .

6-*- '•*• 8 - 1 5 - l 6 0 V 3 ' 9 ' ^ 1 0 ' -

Examples IV:

1. 1. 2. | . 3. f.


Examples V:

1.

3.

5.

7.

C + a; sin a; -f cos #.

(7 +a; sec2 a; —tana;.

^+2.-2.

Examples VI:

1.

3.

5.

c — A c o s 6#~~1 c o s ^(7 + \x — -fa sin 8a:.

__i_ # 6. - .

Examples VII:

1.

2.

3.

A

7.

q

2.

4.

6.

8.

2.

4.

7.

C + x2 sin x + 2x cos a; — 2 sin a;.

C + \x* tan-1 x-\x + \ tan"13.

77 77

6-&r + J,A

(7 + lx + Jsin2a;.C + £ sin 2a; — 2 s i n 6a;.

«• 8- r.-

^ 1 . 5 5 . 5 . 5(7 — - sm5 a: cos a; — — sm3 a: cos a ; - - s m 3 ; cos x-\-—zX.

D 24 ID ID

n 1 . , 6 .\J T^ ~ Ol l l JL> \j\Ji5 Jb ^ ~~ o l l l Jb \j

7 35

C — - sin8 a; cos a; — — sin6 x cosy DO

315 S m "

35TT 16

256* ' 35

(2w- l ) (2n -3 ) . . . 3 . 1 77

(2TI+1) (2^-1) . . .5 .3"

Examples VIII:

1.

A

6.

16 o 21155' "" 63#

(2m- l ) 2 (2m-3) 2 . . . 3 2 772 2 m ( 2 m ) ! * 2 '

o 12 .10 2'19 .17 7*

ccc

, 8 . 2 16 .# + ^^ sin x cos2 a; + —: sin x.

35 35

128>sa; — —— cos a;.

315

6 63n6 ' 512*

Q

(m + n + l ) ! #

(m!)2

' 2(2m+l)!#

7 7 7 r

2 1 0 '


CHAPTER VI

Examples I:

1. 7T. 2. frr. 3. UTT.

Examples II:

1. (a) 0-335, (6) 0-333. 2. (a) 60-1, (6) 60-0.

3. (a) 8-68, (6) 8-67. 4. (a) 0, (b) 0.

5. (a) 0-680, (6) 0-667. 6. (a) -0-248, (6) -0-250.

7. (a) 648, (6) 638. 8. (a) 0-694, (6) 0-693.

9. (a) 0-322, (6) 0-322.

Examples III:

1. ±g-a\ 2. |/2. 3. fa.

4- C) ^2(31 + 6V2). (ii) ~ (325 + 6p).

REVISION EXAMPLES II

2. f, f,

4. (i) iz3-!a;4 + ia;5. (ii) - l S i n 2 ^ (iii) f.

5. (i) i(x+l)*. (ii) ix + sin6x. (iii)

6. (i) Ix-~lsm2z. (ii) -xz. (iii) -fcos3:r.

7. 10a.

8. y = f#2-2#-f4, 2y-x-8 = 0,

9. y s= x4 —|a;2+ 1, minima at a; = + J.

10. 2, = * - - + -^-, (-2,1), *+y-6-0at (2,4), x + y - \ - 0

at (~2,|).

11. i,A- 12. y = a? + i + l , ( l , 3 ) .


13. 8|, 4 | . 14. 10 ft. per sec, 3l£ ft.

15. 27 ft., 851 ft. after 8 sec.

16. 9 ft. per sec, 3 ft. per sec per sec.

17. 3 ft. per sec per sec when t = 2, —1 ft. per sec per sec

when t = 4. AB = 11 | ft. At rest when J = 1,6.

18. 50 ft. per sec, 1166| ft. 19. 4 ft. per sec, 4£ ft.

20. v = ± 5, max. x = f.

21. — 12 cm. per sec. per sec, 60^ cm., 19^ cm.

22. 31-25 ft. per sec, 520f ft. 23. t = 5; 5£ ft.

24. *= 2,5; 4Jft.

25. 14 ft. per sec. per sec, 25 ft. 26. 15.

28. 6f, h

30. (I ft).

32. (f, V ) .

34. 36, (0, - ^ ) .

36. 9, (|, f ) .

27.

29.

31.

33.

35.

37.

39.

41.

44.

46.

48.

i(i)

i¥

(i)

\«

4,

-%-•

(if).1, (ii) h

(II, I§)•

.(1,0).

87T, (ii) |

38. 32TT, f.

40. ^77, f i

42. (i) i|-87T, (ii) 16TT.

45. j7rr2A, p .

47. 125TT.

49. %7rab\ fa.

50. H^TT. 51. ^ - , ~ + - .4 4 77

52. * p . 53. 2-55.

54. 0-292. 55. 0-84.

56. 710 sq. ft. 58. 2-7 miles.

59. a = ft, b = ff, c = ft- 60. 0-385.

61. Exact method 0-867. Simpson's rule 0-865.

62. 50-3. 63. 342.

INDEXAcceleration, 48, 111Approximate integration, 134Area

of circle, 100polar coordinates, 114sign of, 84of surface of revolution, 130'under' a curve (Cartesian co-

ordinates), 77, 113

Centroid (centre of gravity), 118of solid of revolution, 127

Circle, area of, 100Concavity, 54Continuity, 12

denned, 13Continuous variable, 3Curve, to sketch, 59

Decreasing function, 52Derivative, 20Differential coefficient, 19

of cos x, 34of determinant, 45of 'function of function', 27of higher order, 35of product, 26of sinx, 33standard forms, 36of sum, 26of xn, 30

Differentials, 42, 89Discrete variable, 3Dynamical illustrations, 48, 111

Formulae of reduction, 106Function, 1

decreasing and increasing, 52

Gradient, 17Graph, 5

Increasing function, 52Inflexion, 55Integral, 81

approximate value, 134defined, 82definite and indefinite, 85evaluation, 86, 89formulae of reduction, 106by parts, 103standard forms, 90substitution, 94, 98

Inverse circular functions, 38

Limit, 5denned, 7, 8off{x)jg(x) whenf (a) =g(a) =0, 68of sin xIx, 32theorems on, 25

Maxima and minimato determine, 51to discriminate, 56introductory illustration, 49

Mean values, 138defined, 139

Mean value theorem, 61Cauchy's, 66

Moment of inertia, 123of solid of revolution, 129

Normal, 23

Pappus, 128Parameter, 2Parts, integration by, 103

Rate of change, 14Real roots of f(x) = 0, f{x) = 0, 63Reduction, formulae of, 106Rolle, 60

Simple harmonic motion, 49Simpson's rule, 135Sketching of simple curves, 59Solid of revolution

centroid (centre of gravity), 127mean density, 140moment of inertia, 129volume, 125

Substitutiondefinite integrals, 98indefinite integrals, 94

Surface of revolutionarea, 130mean density, 142

Tangent, 23

Variables, 1discrete and continuous, 3

Velocity, 48, 111Volume of revolution, 125

An Analytical Calculus Volume I … · AN ANALYTICAL CALCULUS VOLUME I. AN ANALYTICAL CALCULUS FOR SCHOOL AND UNIVERSITY BY E. A. MAXWELL Fellow of Queens' College, Cambridge VOLUME

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