Coroneos Level 1 HSC

( ). ,,

I '

CERTIFICATE

COURSE

•

I '

CERTIFICATE

COURSE

•

'I•

for transmission by post as a book

ALL RIGHTS RESERVED

Set Up By

BIRRONG SECRETARIAL SERVICE

P R E F A C E

This volume is the second of two which together cover the Level I syllabus in Mathematics for the Higher School Certificate. Since it is intended to be used in conjunction with "A Higher School Certificate Course in Mathematics" Form 6 Level 2F by J, Coroneos, no part of the 2F syllabus is included.

As in Volume I, a large number of carefully graded ex­ ercises has been provided. In using these, however, teachers should exercise discretion, because of the considerable vari­ ation between students in speed of working and the amount of practice needed to consolidate learning. In addition, the time available for the course will vary from school to schooL As a guide some examples have been denoted by t for ;upplementary work and

*

for the more challenging problem.

While all answers have been carefully checked, the authors would appreciate notification of any errors which have remained undetected.

Sydney

CONTENTS.

CHAPTER l. CALCULUS

Rolle's Theorem - Mean Value Theorem of the Differential Calculus The Definite Integral as the Limit of a Sum

-�: o

::::.::,'::,::::: it

: I:;·:1.::;,:-::}V: l

::x� e

:, e

m0::

and Even Functions Lengths of Arcs •••••••••••••• page 1

CHAPTER 2. INTEGRATION

Table of Standard Integrals Change of Variable (Easier) Use of Partial Fractions to Integrate Rational Fractions Trigonometric Integrals

T

_{ypes ax}f f(x) dx f f(x) dx - Integration by Parts -2_+bx+c'

lax2_+bx+c

Change of Variable (Harder) - Reduction Formulae Im-proper Integrals - Methods of Integration (Surnmary) •• page 48

CHAPTER 3. GROUPS, ISOMORPHISM, COMPLEX NUMBERS

Groups Isomorphism - Groups of Order 6 Com-plex Numbers . . . , . . . , . . . , . . . , page 104

CHAPTER 4. APPLICATION OF MATRICES

TO GEOMETRY AND PROBABILITY

Conic Sections Standard Equations (Terms Used etc) -Change of Coordinate System (Translation, Reflection, Rotation, Affine Change) - Reduction of Quadratic Forms by Change of Origin - Case 1: ab-h2 _{# 0 (The Ellipse} and Hyperbola); Case 2: ab-h2 _{= 0 (The Parabola)} Degenerate Conics - 2 x 2 Matrices and Quadratic Forms -Reduction of Quadratic Forms by Rotation of Axes Reduction of Equations of Conics Similar Matrices

Newton's First Two Laws of Motion Work done = Change in Kinetic Energy Potential Energy Conservation of Energy Dynamical Problems involving Variable Acceleration Resisted Motion under Gravity ,,,,,,page 192

CHAPTER 6. SEQUENCES AND SERIES

Infinite Sequences - Meaning of a Divergent Sequences Finding Sequences Some Special Limits Difference Method Series The Comparison Test - The Series

Limit - Convergent and Limits of Convergent Infinite Series of Positive Terms

-E n-p

=l Absolute

and Conditional Convergence - un as a function of x .• page 223

CHAPTER 7. SOME SPECIAL SERIES

Series for log(l+x) Use in Calculating Logarithms Series for tan-lx Use in Calculating TI Series for eX _{Calculation of e Series for sin x and cos x} -Taylor's Series for Polynomials .•••.•••.••••••••••.•. page 286

CHAPTER I.

CALCULUS.

PRELIMINARY NOTES

The Level 1 Course, as set down in the Syllabus, is a self-contained course in itself. In theory, a level 1 student should cover this syllabus in a continuous manner, from the first to the last topic, However, in practice, few separate le'1el 1 groups exist; in the main, level 1 students take 9 periods together with a level 2F class, and two separate periods on level 1 work.

Of the topics in the level 1 Syllabus, the only ones which are common to the level 2F Syllabus, and to which new material has been added, are Calculus and Dynamics (Topics 3, 5 of the level 1 Syllabus). In these sections, it will be assumed that level 1 students have covered the appropriate level 2F work, and this work will not be re-treated here, Because of this fact, the treatments of both the Calculus and Dynamics may seem somewhat disjointed, as only the level 1 new items will be developed in this text.

1.

ROLLE'S THEOREM, AND THE MEAN VALUE THEOREM

OF THE DIFFERENTIAL CALCULUS.

A.

( i)

ROLLE IS THEOREM

�� rex.1

STATEMENT OF THEOREM

It f(x) is aontinuous tor the interval as x s b, and t1_{(x) exists tor the interval as x .s: b,} and it t(a) = t(b) = O, then there is at least one value of x, say c, where a< c < b, such that f 1 (c) = 0,

The result is obvious from the sketches above, In each figure (i), (ii), the curve y = f(x) crosses the x axis at x = a and x = b, i.e. f(a) = f(b) = O. Since the curve is continuous, it must pass from a to b without break. In each case, the curve has at least one stationary point in the in­ terval a< x < b. That is, there is at least one value of x in the interval a< x <,b, at which the tangent is parallel

.2Y

to the x axis, i.e. dx = O. If we denote this value of x by c or �,then f' (c) = 0 = f' (0. (Fig (iii) shows that if f(x) is not continuous for a .'.: x .'.: b, and although f(a) = f(b) O, then f'(x) is not necessarily zero in the interval a< x < b� However, the facts that f(a) = f(b) = O, and f(x) is continuous for the closed interval a� x � b, are not enough to guarantee that f'(c) = O, where a� c � b.

Thus, in figures (iv) (iv), (v), f' (x) does not exist at x = c in the interval a < x < b; i.e. f(x) is not dif­ ferentiable at x = c. It is obvious that there need not be a

point in the interval (a,b) for which f'(x) O.

Further, in the theorem, (vi) although f(x) is continuous

for the aZosed interval [a,b], f'(x) need exist only for the open interval (a,b).

Thus, in fig (vi) for the semi-circle y = �;

X

f' (a), f' (-a) are undefined -a. 0 o.

(the tangent to the curve is

ROLLE

Is

THEOREM

EXAMPLE 1. Verify Rolle's Theorem for f(x) = (x+1)(x-1)(x-2), where

(i) a= -1, b = 1 (ii) a= 1, b = 2 (iii) a= -1, b=2 Theoretically Now f(x) = (x+l) (x-1) (x-2) is a polynomial

in x, and is thus continuous for all values of x; (not just for a .:5 x .:5 b),

Also f'(x) = --2.[x3_-2x2_{-x+2] 3x}2_{-4x-1, which} dx

exists for all values of x, since it is also a poly­ nomial in x (again, not just for a < x < b),

Now f' ( x) O when x = 4 ± /i'6+I2 = .l( 2+v7) or_{6 3} t<2-v7), and f(x) 0 when x = -1,1,2.

(i) When a = -11 b = 1, we know that f(-1) = f(l) = o; f(x) is continuous for -1 .:5 x .:5 1, and f'(x) exists for -1 < x < 1. Hence, the requirements of Rolle's Theorem are satisfied, and thus there exists c, where -1 < c < 1, such that f' ( c) = O. Here c = t<2-v7),

(ii) When a= 1, b = 2, we know f(l) = f(2) = o, f(x) is continuous for 1 .:5 x .:5 2, and f' (x) exists for 1 < x < 2, Hence, there exists s such that f'(s) = O, where l<s<2, Here s = .l(z+/7),₃

(iii) vlhen a= -1, b = 2, the criteria of Rolle's Theorem are satisfied, and there exists c where -1 < c < 2, such that f'(c) = O. In actual fact, two values of c, namely t(2-v7), t(2+v7) are possible,

Geometrically� The re­ sults above are obvious from the sketch of

y=(x+l)(x-l)(x-2),

In each of the in- ---h-�-.-++,,,.----+-'-..,_��-x tervals (-1,1),

(1,2) there is a point where the tangent to the curve is parallel to the x axis; whilst for the in­

terval (-1,2) there are two such points.

EXAMPLE 2 Discuss the application

. . x2_-5x+4

funat�ons (1) f(x)

=

_{X-u 7}

of Rolle's Theorem to the (ii) f(x) = (x-�ix-4) (iii) f(x)

Method

---(i) Here f(x) = (x-l)(x-4) and thus x-3

However, f(x) is dis- _L:\ continuous at x

=

3,

which is in the in-terval [1,4],

Hence, Rolle's Theorem does not apply, and

1-(x-2/h

f (1) f(4)

o.

there does not exist ----1----"''---...J_z. �--4�---"'-c, where 1 < c < 4,

I such that f'(c) =

o.

�

The sketch of _ (x-1) (x-4) Y - _x-3 this clearly.

shows

I

I

·

(x-1) (x-4)

(ii) Here f(x) = _x+3 , and thus f(l) = f(4) = O, whilst f(x) is discontinuous at x

=

-3, which, however, is not in the interval [1,4]. This means that Rolle1_{s Theorem} may apply here, but we still have to test the existence of f'(x) in the interval (1,4).

f, ( ) x2+6x-19 1 1 . d h f, ( ) Now x =

(x+3) 2 , on ca cu at ion, an t us x exists for the interval (1,4), [Actually, f(x) is differentiable everywhere except at x

=

-3.]

Hence, all the conditions for isfied, and thus there exists at which f'(!;)

=

O.

-x. _-3

I

I

I

_,

Rolle's Theorem are sat­ !;, in the interval (1,4),

(iii)

ROLLE Is THEOREM

The actual value of

s may be found by solving the equa­

tion x2_{+6x-19 = 0, This gives x = -3±2./'f, whence the} appropriate value of

s =

-3+2/7. These results are obvious from the sketch.

[Note the other value where f' (x) =

o,

bears no relation to Rolle's Theorem. It, of course, signifies that there is another stationary point on the curve, in addition to the one mentioned at s,J

Here f(x} = 1-(x-2)2/3 _2/

Now

f(x) = O where (x-2) 3 = 1, i.e. where (x-2)2 = 13_, and this gives x-2 = ±1, i.e. x 3 or 1. Also, f(x) is continuous for all values of x, and certainly for 1 .$ X .$ 3.

-2

By calculation f'(x) = _{3(x-2) /.J, which exists for all} values of x except for x = 2. That is, f(x) is differ­ entiable everywhere except at x = 2, which is in the interval (1,3),

Hence, Rolle's Theorem cannot be applied here. The sketch of2

y = 1-(x-2) h makes these points clearly.

[Note that 2 f"(x)

9 (x-2) Y3 which is positive for all x except 2. Thus the curve y = f(x) is always concave up, except at x = 2.J

EXERCISES SET lA

1. Verify Rolle's Theorem for each of the following functions, for the interval O .$ x .$ 1, by actual calcu­ lation, and also by drawing a graph.

2.

(a) f(x) x(l-x) (b) f(x) = x2_(1-x) (c) f(x) = x(l-x)(l+x)

Verify Rolle's Theorem, by actual by a sketch, for the function a = 1!. b = 311._{2' 2}

5

3. Discuss the application of Rolle's Theorem to the func-tions

(a) f(x)

=

x_x-22-4x (c) f(x)

=

4-(8-x)�3

(b) f (x) = x2_x+2 -4x (d) y = tan x

4. The curve (y+1)3

=

_x2 _{passes through the points (1,0)} and (-1,0). Does Rolle's Theorem justify the conclusion that� vanishes for some value of x in the interval -1 .5. X ,:. 1?

Give reasons for your answer, and sketch the curve.

12_,

THE MEAN VALUE THEOREM OF THE DIFFERENTIAL CALCULUS

If f(x) is continuous for a s; x s; b, and f'(x) exists for a < x < b, then there is at least one value of x, say c, where a < c < b, such that f(b) - f(a) (b-a) f'(c).

The meaning of this result can be easily seen by study­ ing the diagrams below,

T

(ii) B

A

R

In the sketches, y = f(x) is continuous for the closed interval [a,b], and f'(x) exists for the open interval (a,b) {and could exist for [a,b)}. It is obvious that there exist point(s) Pon the curve, at which the tangent is parallel to the join AB.

Now the gradient of the join AB gradient of the tangent at P (where x

f(b)-f(a) b-a c) is f' (c), since these lines are parallel to each

f(b)-f(a) b-a

MEAN VALUE THEOREM

f1 ( _{c) , which may be rewritten as}

f(b)-f(a) = _{(b-a) f' (c), where a < c < b.}

Notes. 1.

2.

3.

In fig (i), there is only one such value of c, whereas there are four in fig (iii). The theorem says at least one value of c for the interval (a,b).

t, .. ,,1 Figures (iii), (iv) �'11

1 show cases where this theorem cannot j be applied.

In fig (iii), f'(x) j does not exist at Al

x = _{d, and in r---.----1} fig (iv), f(x) is

not continuous at x = d. (In both O, cases a< d < b).

The Mean Value Theorem can be readily proved by the use of Rolle' s Theorem,

Thus, in the figure, y = f(x) obeys the conditions f(x) is continuous for [a,b], and A f'(x) exists for (a,b). The equation of the chord AB is Y-f(a) f(b)-f(a) ( ) _b-a x-a '

b

Qv)J

I

I

I ��fC1e)

I \

A

I I

I

I hT Now if S is the point {x,f(x)} u C. :x \) on the curve y =_{tf(x), then}

1,,

1<RS = >�TS-*TR = (f_'. (�) - [f(a) ··)+ f(b),_{,· b-a} f�a) (x-a)] F(x) say. ,_/

Since *RS is continuous for [a,b], then F(x) is contin­ uous for [a,b].

Further, F(a) = F(b) -o'

Also F'(x) = f'(x)

-= O. {Check this by substitution}, since *TS= *TR at endpoints A,B. [f(b)-f(a) l] '•.• .. ,, .. ,''' (L),

b-a and this exists for a< x < b. Hence, F(x) obeys the conditions,

a L x Lb, Thus, by Rolle's Theorem, there exists c, where a -t. c L. b, such that F'(c) = O.

Now F' (c) = f' (c) - [f(b)-f(a)] from eqn. (L)_{L b-a}

J,

and ,', f'(c) = f(b)-f(a) _b-a , since F'(c) = .

o.

4. The Hean Value Theorem can be expressed in different ways.

( o:) If a = x, b = x+h, then the _Pf) result becomes 1f'1lx. f(x+h)-f(x)=[(x+h)-x]. f' (c), where x < c < x+h

i.e. f(x+h) f(x) + hf' (c) (S) Since x < c < x+h,

c = x

+

eh,

where 0<8<1 Hence, the result in ( o:) be- 11'1.J

·'!'''\} comes

f(x+h) = f(x) + hf1_(x+eh), where O < 9 < 1.

c..

EXAMPLE 1. Discuss the application of the Mean Value Theorem to the following functions, for the intervals stated. (i) _{f(x) = x}3_{; a= -2, b=2}

(ii) f(x) = x2/3_{; a= -27, b = 27} (iii) _{f(x) 1 a= -1, b = 5=}

;;

(iv) _f(x) = _{lx-1; a} = _{1, b} = 10

F!here possible, find c so that f(b)-f(a) a < C < b.

Solution

(b-a) f I (c),

(i) Now f(x) = _x3 _{is continuous for all x, ·and} contin-uous for -2 � x � 2, Also f'(x) = 3x2 _{exists for all x,} and .·. exists for -2 < x < 2. Thus, the conditions of the Jvlean Value Theorem are satisfied, and there exists c, where -2 < c < 2, such that

f(2) - f(-2) = [2 - (-2)) f'(c).

MEAN VALUE THEOREM

Thus _{8 - (-8)}

=

_4,3c2_{, whence c}

=

_{± �}

Both values of c lie in the interval -2 < c < 2, they are both illustrations of the

Mean Value Theorem, j This result is easily seen

from a sketch.

There are two positions in�-t-�+-!!!f-:�,;Z:,;�--'-� the interval -2 < x < 2,

at which the tangent is parallel to the join AB.

(ii)

Here f(x)

c

x

7/3 is contin­ uous for all x, and ,•, for the interval -27 � x � 27,

2 However, f'(x) = �/ and_{3x 3}

does not exist for x

=

O, which is in the interval -27 < X < 27,

Thus, the Mean Value Theorem cannot be applied for (-27,27). This is ob­ vious from the sketch, since there is no point in the interval -27 < x < 27, for which the tangent is -21 parallel to AB,

(iii)

If f(x)

=

i,

discontinuous which lies in -1 < X < 5,

then f (x) is at X = 0, the interval

Hence, Theorem for the

the Mean Value cannot be applied interval -l<x<S, This can be seen from the sketch.

0 ₂₁

(iv)

For f(x)

=

lx-1,

the domain of x is x ?: 1, and f (x) is continuous for 1 � x � 10.

Also f 1

(x) = 2

;!_

1, which exists for l < x < 10 (but not for x = 1).

Thus, there exists c, where l < c < lo, such that

f(lO) Now f(lO) =

- f(l)

=

_{(10-1) • 21!-1} /9 = 3, f(l) o.

•• •. • • • • • • • • • • (L)

Thus the result (L) becomes

9 1 3 - O = 2 · v'c-1'

hence lc-1 =

f,

and thus c = 1

! ,

which lies in the required interval. This can be seen from the sketch. At P, where c - 3�, the tangent is parallel to the join AB. {Note that at A, the tangent is vertical, f I (1)

�

A

0 C:::.6fal. )OX

is undefined.}

EXAMPLE 2. If f' (x) = O at aU points in the intewal a S x Sb, show that f(x) is constant in the interval. Solution Since f'(x) exists for

a� x Sb, then f(x) must be continuous for this interval. {The converse of course is

not tr.ue,

above.} see f(x) =

/is

Hence, a value c exists, for a< c < b, such that

f(b)-f(a) _{f I ( C),} b-a

�::.JCx-)

a. But, f'(c) = O,

and .·. f(b)-f(a)

=

o,

hence f(b) = f(a).

I

I

I

I

I

C b )(

This is true for aU points c in the interval a< c < b, and thus f(x) is constant for all x in the interval. [This is a familiar result, but the actual proof of it depends on the Mean Value Theorem.]

EXAMPLE 3. Show for the function f(x) = sin-1x in the inter-sin-1h 1

val (0, h), where O<h<l, that h = �, where O<c<h. _h y 1-(Y-Hence, deduce that h < sin-1h <

-;rr::::,;z•

1. I

MEAN VALUE THEOREM

and thus for O < x < h,

where h < 1. -- --1T/z ':l 1

Further, f'(x) = �· and since the domain of

1 _{..:-:;!.1 __} -==,...,,.i,.:;;;:�=-!ll-l-lf=i'l is -1 < X < 1,

then f'(x) exists for 0 < x < h, where h < 1. Hence, by the Mean Value Theorem, there exists c, where O<c<h,

such that f(bl-f(a) = _{f'(c), where b} -a

i.e. such that f(h�=�(O) = f'(c) sin-1h - sin-10 1

-oT/z __ _

h, a 0

i.e. such that _h = _�,

sin-1h 1

II

so _h = _�

i.e.

Now

since c > 0

and

li=c"2'" < /1-oz '

1 > 1

.. � 71-W

Also

li=c"2'"

>

/i=hT,

since c < h d • 1 < 1

an··��

Thus since O < c < h 1 < 1 _�� < 1 sin-1h 1

Hence since _h = �·where O < c < h

d 1 1 1

an _{< � < /l-h2'} sin-1h 1 1 < h <

7l=li2"

i.e. _{h < sin-lh < �' (h is positive).}

Using the find C in (i) f (x) (ii) f (x) (iii) f(x) (iv) f (x)

EXERCISES SET lB

result f(b)-f(a) = (b-a) f' (c), where a<c< b each of the following cases.

3x2_{+4x-3; a} = 1, b = 3 x3_{; a = O, b = 6}

7T

sin x; a = O, b - 2 sin x; a = O, b = s;.

2.

3,

4,

Verify the Mean Value Theorem for each of the functions below on the specified intervals.

j (i) f(x) = h-2x; [-1,3] (ii) f(x) (3x-4)3_; 1

[1,2] If f(x) = _{x + 1 show that f(b)-f(a)}

x' b-a 1 - -;;-2, where

a < C < b,

Find c if a \, b 8.

(ii)

(iii)

_1/ If a < O <b and f(x) = x 3_, no c that satisfies f(b)-f(a) where a < c < b.

show that there is (b-a) f'(c),

Illustrate with a sketch of the graph,

If a < 0 <b and f(x)

=

xV3_{, show that there is a} value of c that satisfies f(b)-f(a) = (b-a)f'(c), where a < c < b, even though the function fails to have a derivative at x =

o.

Illustrate with a sketch, _{X -8 3}

If f(x) = _{�-2-, x 1 2, show that there is a value}

x-of c to satisfy the equation

f(b)-f(a) = (b-a) f'(c),

where a = _{-2, b} = _{3 and a < c < b, even though} f(x) is not continuous at x

=

2,

Illustrate by means of a sketch,

{Note the lack of a converse to the Mean Value Theorem,

Thus the fact that c exists such that f(b)-f(a) _{= f'(c) for a < c < b,}

b-a

does not mean that f(x) is continuous in [a, b] or that f'(x) exists in (a, b).

These points are illustrated in parts (ii),(iii) above.}

5, The Mean Value Theorem may be written in the form f(a+h) "' f(a)+hf' (0 .. .. .. .. .. .. .. .. .. . . . .. (1) where� lies between a and a+h.

Take f(x) = x5_+x2_{, a} = 2, � = a = 2.

Put h = O•l, 0•5, l•O respectively in (1), and so find f(2•1), f(2•5), f(3).

Compare your answers with those given by tables.

6, The method illustrated in question 5, gives estimates in cases of small increases or decreases.

7.

8.

9.

f·)

J

MEAN VALUE THEOREM

(i) For example, to find 6/65 approximately, we could 1�

take f(x) = x 6, a.= 64, h = 1 and since

s

lies in the interval 64 <

s

< 64+1, then we could take

s

= 64 as an approximation.

Hence, from the equation (1) in question 5, .we have (64+1) % 'f 641/6 + 1 • �

6.64 6 Thus show 6/65 'f 2•00521.

(ii) Similarly obtain estimates for (a) /g:"I° (b) 3/f:g If f'(x) > 0 at all points in the interval a� x � b, prove that f(x) is strictly increasing in this interval.

tan-1b - tan-1a 1

Show that _{b-a = l+c2• where O < a < c < b.}

b-a ₁ 1 b-a

Hence prove that l+b2 < tan- b - tan- a< l+a2·

Further, by substituting appropriate values for a and b, show that

f

_{+ 2; < tan-1 (%) <}

f

+

i,

Apply the Mean f(x) = log X in hence show that Use this result

Value Theorem to the interval (a,b),

a b

1 - -< log(-) _{b a} < - -b _a 1 to prove that 6 < log

the where 1.

1,2 Use the Mean Value Theorem to prove that (i) _{l+h < log(l+h) < h,} h if 0 < h < 1 (ii) l�-h2 > sin-1h > h, if O < h < 1.

function a> o, and

1 < -5·

11. Show that the Mean Value Theorem can be expressed in the form f(a+h) = f(a) + hf'(a+6h), where O < e < 1. If f(x) = /l+x, write down f'(x), and show that

/l+h = 1 + zl�+eh' where O < e < 1.

h � h

If h > O, deduce that 1 + � < vl+h < 1 + 2. *12. Find e in terms of x, h from the formula

f(x+h) = f(x) + hf' (x+eh), where o < e < 1 where (i) f(x) = l. (ii) f(_X x) = ex

*13. State the Mean Value Theorem for the differential

calculus.

Use it to prove that if, for a< x < b, f(x) and g(x)

are continuous and differentiable functions, and if

f' (x) g1 _{(x) for a < x < b, then f(x)-g(x) = C for}

a < x < b, where C is a constant,

THE DEFINITE INTEGRAL DEFINED AS THE LIMIT OF A SUM

This item has already been treated in the level 2F

Syllabus, but will be dealt with again in more detail ,

EXAMPLE 1. If f(x) is a continuous

prove that j_ n;l f(!:.J < _j

1

f(x) d:c

increasing function,

1 n r

< -

l:

f(-)

n _r=O n

o n r=l n

Hence show that

�� n;l � f(�} = ( l

f(x)

r=_O

Jo

O

'fn 24,, =,h

Solution

�. n 1 d:c = t/l,m

l: -

_{f r!:.J}

n-m r=l n n

/

/

)'..

The interval [O,l] is divided into n equal parts, each

of width h = 6x = 1..

n

Let the sum of inner rectangles be s,

the sum of outer rectangles be S, and the area between the

curve y = f(x), the x axis and the ordinates x=O, x=l be A,

then from the sketch s < A< S.

Now s

= 1.

,

f(O) + 1.

,

f(l.) + 1.

,

f(l) + ... + 1. f(n_-l)

n n n n n n n

= 1.[f(Q)+f(l.)+f(l) _{n n n n} + ... + f(n-1)]_n

n-1

=

1.

l:

f(.!.), n r=O n

1 DEFINITE INTEGRALS AND SUMS

Also A =

t

f(x) dx

Further

s

1 _n

=l

n n-1

Thus l _E

n r=O

f(l_{) +} l

n n f(l)n n

E f (.!:) r=l n

I:

f(.E)_n < f (x) dx

+ l _f(1)

n n

< l n _E

n r=_l

+ ...

f (.!:)_n

+l _{. f (!!.)}

n n

When the number of divisions becomes greater and great-er, the inequality above becomes closer and closer to

equality. 1

lim n-1 1 r

l

Thus, n-¥,, E0 n f(n) = f(x)

EXAMPLE 2.

(i) (ii) Solution Lim (i) _n-¥,o Our Now

l. n 1

r= ₀ dx = n-¥,, r=l n n �m E - f(:£_)

Use lim n-¥,, lim n-¥,, n-1

the result above to evaluate

n-1 _1_

E

r=O lii"2-r2

{(bx)_· 5 _{+ l_(2bx)}5 _{+ 1_(3bx)}5

2 3

where n.bx = 3.

l

+ . • . + l(nbx) 5},_n

J,

f(x)

E f (.!:) =_n dx, by example 1. r=O

aim is to set up lim n-1 1 } { E n-¥,o r=O

lri'1=?2

n-1

lim E 1 _{n-+oo r}

lri'2-=r2

is this form.

=_{O n-1}

1 lim {

n-¥,o E n�/ 2_r=O n }

n-1 lim { E l

n-¥,o r=O n

1. n-1 1 _{1m E _} n-¥,o _r=O n

l

J

-b

dx

0 l

[sin-1x] ₀ 15

1 h-(r/n)2}

f (*), where

f(x) = h_=:x2

(ii) The expression

�,!:

{(6x)5_+\(26x)5_{+ f(36x)}5 _{+ ••• + �(n6x)}5_}, lim { � l.(r6x)5_}

n-+«> r=l r

lim{ � l.(r . l.)s} since 6x = _l n-+«> _r=l r n ' n

1. n 1 = 1m{ i:: _ n-+<x> r=l n

where f(x)

[tt

0

48•6, on calculation.

EXAMPLE 3.

By considEring the value of /

2

0

lx(2-x) dfXJ as the

Zimit of a sum, prove that

Zim l.

1

n-+oo n x=l

/r(n-r)

n2

=

-Solution

Geometrically f:/x(2-x) dx represents the area between the curve y = /x(2-x), the x axis and the ord­ inates x = O, x ₌ 2.

Now y = lx(2-x) is a

semicircle,

radius 1 unit. centre (1, 0) and

{This can be seen since y2 _{= x(2-x), i.e. x}2_+y2_{-2x =}

o,

i.e. (x2_-2x+l)+y2 _{= 1, i.e. (x-1)}2_+y2 _{= 1.}

The equation (x-1)2_+y2

(1,0) and radius 1 unit. 1 represents a

circle,

centre

However,

y2 _{= 1-(x-1)}2_{, i.e. y = ±/1-(x-1)}2_,

i.e. y = ±!x(2-x) and in this form, the equation rep­ resents the semi-circles

�\

y = +/x(2-x) and y = -/x(2-x)

0 ).

�

o�__/)

2

Thus the value of 1 lx(2-x) dx is the �

� measure of the area of the semi-circle

o

1

16

• 1).;//

DEFINITE INTEGRALS AND SUMS

1 stated above, _{This area = 211.1}2

f:/x(2-x) dx •

f•

Hence,

Divide the interval y

[0,2] into n equal

parts. Each part ' f 'd h 2 is o wi t - u_n nits.

=

l

•

[f(l)+f(!!.)+f(.2.)+ ••• +f(2n)]

n-- n rt n n

11

=

2

_{sq units.}

=

l �

f(2r) = �

l.

/

2r(i- 2r), since f(x) = lx(2-x) n

r=l n r=_{l n} n n n

2 n _!!. lr(n-r)

E

=

E

r=_l n /2r( 2n-2r)

n n _{r=l n} n

n

lr(n-r)

4

.

E

n2

r=_l

limit as n + co

Hence, taking the

1' n

•• • 4 • im E lr(n-r) 11 . lim � n

-w> r=l n2 =2;1.e. n-w> r=l,.,

EXERCISES SET lC

If f(x) is a continuous function, �rove that

lim l[f(l)+f(l)+f(l)+ ••• +f(!!)] I f(x) dx.

n+oo n n n n n

0 Hence, evaluate the following

lim 1 . 11 . 211 311 n11

(a) -[sin - + sin - + sin - + ••• + sin -] n

-w> n n n n n

( c)

�!:

gh

[

If

+

fi

+ .fS + • .. + in]

n

(d) lim{ _n-w>

rt2+i2"

n + n _n2+22 + n_n2+32 • •.

+

+ n _n2_+n2 }

2.

4.

(i)

(ii)

(iii)

If 6X = xm - xm-l

and x = mh for m

n

+

_{n } , if p > -1} p

h =

!!.

_n' 1,2, ••• ,n lim

then h->O E (2mh+3)h

m=l is equal to an expression involving a definite integral. Find the value of

;: i :h i : t :: r

::�w that lim � h(l+rh) 3 =

f 3

x3dx,

n-+oo _r=l

1 and hence evaluate this limit.

n _h

ht(2!�)2 1

If Sn =

r�l (2+rh)2 + (2+2h)2 + ••••

..• +

_{(Z!nh)2} where h} = .!n'

find the value of�!::'. Sn•

By considering the value off: il-x2 dx as the limit of

a sum, show that

1· 1 n-1 im _E n-+oo � r=O

Prove, by considering the area under the curve

1 lim n n1 1

(i !

/

y = _{-;2 from x} = 1 to _X = 2, _that

E�--n-+oo r=l(�+E) - 2 lA

(ii) y

(iii) y

1 = 0 1,

= �from x to X = n-1 1

that lim E ₌

n-+oo _r=O 2

il+x from x O to x = 1,

that lim � /n+r

n-+oo r=l n

%

l(u'z - 1) 3

2

lim 1 1 1 1

J dx 5. Prove that _{n-+oo (n+l} + _n

+2 + n+3 + • • • + 2n) = �

. 1 Prove also that for all positive integral values of n

11 I 1 1 1 1

\ 1 - zJ+_{(3 - 4 \}+ _"' + _{2n-1 - 2n}

6.

DEFINITE INTEGRALS AND SUMS

1 1 1 1 \.. l 1

[l + 2 + 3 + 4 + ••• + 2n] - 2

[z

+ 4 +

1 1 1

= n+ 1 + · n+2 + ' ' ' + 2n

H ence, e uce t at t e d d h h i f. n inite series -2+3-4+ ....l 1 1 1 has a finite sum. What is this sum?

(i)

*(ii)

Using the result log x =

[

x

_dt

e _{1 t'}

show that logexn _{= n logex'} Also show,

log 4 > 1. _e with the aid of

X > 0,

a diagram, that

Deduce that logex increases indefinitely as x in­ creases,

If p and n are that lo (n+p)

ge n-1

positive integers, n > 1, show >1+ _1_ +...l....+_{n n+l n+2}

1 + +l

Hence deduce t.hat

limcl + _1_ + L + ... + ...1.)

n� n n+l n+2 nq

any given positive integer,

+ - > log ( !!:!:P.!.:!:.)_{n+p n}

logeq, where q is

Defining log x =

I

x

du, prove that

e _{1 u}

-l+-1+_1+ _{2 3 4 '' '} +-1<1 _{n oge}n < + 2 + 3 +l 1 1 n-1 1

and hence deduce that O < E - - log n < 1, r=_{l r} e

1

+ n-1'

8. Find approximations to the area of the region bounded by the arc of the curve y = sin x0 _{from x} = Oto x= 10, the x axis and the ordinate x = 10; (use 1 unit = 1°), (i) from the sum of the areas of the five rectangles,

(ii)

each of width 2 and of heights sin 2r0 _for r = 0,1,2,3,4,

from the sum of the areas each of width 1 and of s = 0,1,2, ... ,9.

of the ten rectangles, heights sin s0 _for

(iii) from the sum of the areas of the ten trapezia each of width 1 and of heights sin s0 _{on one}

side and sin(s+l)0 on the other, s=0,1,2,3, ••• ,9,

By dividing the interval O 5 x :S 10 into n equal parts,

lim n 10 °

show that n+w { E • s in(__r _)}

r=l n n J::in x

0 _dx,

Assuming that n = 3·142, cos 10°

approximate value of this limit. •9848,determine the

THEOREMS ON THE DEFINITE INTEGRAL

(;)

i:

f(x) dx +

J:

f(x) dx •

l

e

f(xl dx This result is obvious from the sketch, ,ince S1 • J:f(x) dx; S2 • I:f(x) dx; Si+ S2

s.

b C

rf(x) dx. a

( r i) l f m, M are the sma I I est and greatest va I ues of the continuous function Hx)

then m(b-a)

.s

I' Hxl a

In the sketch, y

=

f(x) is a function which is con­ tinuous for all x in the in­ terval a S x 5 b.

It is o bvious that

in the interval a 5 X !:. b 1

dx

:s

M(b-a)

i.)

>-°'� �

-o;-.----,..b

<

r

f(x) dx < M(b-a) , , , .. , .. (n) a

i.e. m(b-a)

When f(x) is constant for the interval

� /�� ,/

LJ0� f0(>--)� a� x .Sb, then m M and thus

� �

m(b-a)

=

/bf(x) dx = M(b-a) .•.•.••.••• (8) �

THEOREMS ON DEFINITE INTEGRALS

Thus, from (a), (S); m(b-a) :5 t f(x) dx .S M(b-a) a

{Note: The equalities occur only if f(x) is constant.}

Solution Now f1 _{(x) -(3 + 2 cos x)-2 • - 2 sin x} 2 sin x

(3 + 2 cos x)2

7T < < 1[ 2 . 2 For the interval 3 - x - 2, sm x > 0 and (3 + 2 cos x) > O

Hence f'(x) > O, and thus f(x) is steadily increasing over the interval given.

Thus the least value of f(x) occurs x =

f,

_{which gives f(x) = -3-+-\---,-TI/3} = -· 1

cos

4'

greatest value of f(x) occurs f(x) _{3 + 2 cos 7!/2 = 3'}1 1

at

1

The sketch of y = ---_{3 + 2} COS X

for ..'!!. _{� x �..'!!.is shown.}

3 2

By the theorem above, or ob­ viously by consideration of areas again,

-2!. 24

J TI/2'

dx < _{-'-3-+- 2}-=

c=-o-s-x 7T

I

3

< j 7T

/\ 7T

I

3

dx 7T

+ 2 COS X < 18

7T x=-2'

at the end whilst which

(iii) THE MEAN VALUE THEOREM OF THE INTEGRAL CALCULUS

point the gives

An extension of the theorem above is now illustrated. In the sketch, y = f(x) is an integrable function,

It is easily seen that there exists a number c, where a< c < b, such that the area of the rectangle with dimensions (b-a) and f(c) is equal to the area under the curve y = f (x) from

x = a to x = b.

i.e. area of rectangle PQTV area bounded by y

=

f(x), the x axis and the ordin­

ates x

=

a, x

=

b. ..._..._ __ h;-;;--.----JU.v.!.4,z

i.e. (b-a) f(c) "

r

f(x) dx, where a < C < b. a

--���

STATEMENT OF THE THEOREM

If f(x) is continuous for as x � b, then there

a value of x, say c, where

r

f(x) dx (b-a) f(c).

a

EXAMPLE Find Um _x-+3

3 {-1-f aos2

x-3

X Solution ₃

Now lim {-1- j cos2 Tit dt}

x+3 x-3 4

a < C <

Tit

dt}

4

b, such that. is

'!im 1 2 TIC

3 {--3 , (3-x) cos -4}, where x < c < 3, by

x+ :x:- the Mean

Value Theorem of the Integral calculus. [Note y cos2 _1I!,4 is continuous and differentiable for all values of t,]

lim _{2 TIC}}

x+3 {- cos �, where x < c

2 3TI

cos �· since as x + 3, c + 3 [See sketch.] 1 2 1

-(- 72) = - 2'

22

THEOREMS ON DEFINITE INTEGRALS

(iv) _{If g1Cx), f(x), g2(x) are continuous functions}

over the interval a � X.:, b,

91(x) .:: f(x) � 92<x), for a .:,

theo

I'

91 (xl dx � a

This result is obvious from the diagram, where g1(x) � f(x).:: g2(x) for all x in the domain a.:, x.:, b. The area "under" the curve y

=

g1(x) is less than the area "under" y

=

f(x) which is less than the ' area "under" y = g2 (x),

r f(x) dx

::

a

Hence

I'

g1 (x) dx , a

I.'

f (x) dx <

I«

f'.,cx) dx

a a

If g1(x) = f(x) = g2(x),

and if X .:, b

f

\

2cx1

a

then fb&1(x) dx a f

b

f(x)

a a

dx • f'.,(x) dx

a

Thus

f

_{'.1 (x)} a

dx.

EXAMPLE 1. Show that I

rr

I

\

a:

x dx > _loge

f

TT/ it

Solution For the interval {.:, x.:, {, tan x is an increas-ing. function, _{and t'an { .:: tan x � tan ]:}

i. e. 1 � tan x .:: /3.

Hence, on dividing by x, (x > O in the domain stated) . 1. < tan x < T3

• • X - X - X

Thus, for { � x.::

f,

By the J 7f / 3

tan x theorem above, _{-x- dx >}

J

TT/3

� dx ... (L)

TT/ 4

{Note that since tan__J£ is not

1

1 X

equal to - for _X all x in the domain, then the area

TT/4

g.::: �x:

�7

'under' the tan x curve is _X

- ___..

/J

I

greater than the correspond-ing area 'under' the l. curve.}

1r/3 X 0

J

7f / 3

Thus

. tan

x x dx > [log x]_{e 7r/ 4} , from (L) 7f/ 4

log .'.!!. - _log .'.!!.

3 4

log(3 • �) 4 loge (3)

EXAMPLE 2 _{Prove that j}l x4_{l+x2 dx} (1-x)4 = _{7 - TT} 22

I

I

Show that �1�4(1-x�4dx <

j

1

x4(l-x)4 dx <

j

0 1

x4(J-x)4

2 _1+x2 dx,

0 ₂₂ 0

1 < 22 _ _ 1_ and henae deduce that 7 - 630 < TT 7 1260 Solution Now ( 1-x) 4

Thus x4_(1-x)4

x4_(1-x)4

and •. • _l+x2 1

Henoo

I

1-4x+6x2_-4x3_+x4_,

by the Binomial Theorem x4_-4x5_+6x6_-4x7_+x8 _{•••••••••••••• (A)} x8_-4x7_+6x6_-4x5_+x4

X +1

x6_-4x5_+sx4_-4x2_{+4 - �}

X +1'

by long division. [Check this.] -- fE7 7 4x_{[ - 6 + 5 -3 + 4x- 4 tan}6 Sxs 4x3 -1x J l

0

1 2 4 _.'.!!.

=

_{j - 3 + 1 - 3 + 4 - 4 . 4}

22

THEOREMS ON DEFINITE INTEGRALS

Further, when o S x � 1, • •• 0 S x2_{S 1}

and thus 1 S l+x2 _{$ 2}

Taking reciprocals, > 1 >

l

_(B) 1 - l+x2 - 2 • • • • • • · ·' ·' • • • • • �fultiplying by x4_(1-x)4 _{which is fositive or zero, for all x,} we obtain from (B) x4_(1-x)4 > x (l-x)4 � x4(1-x)4

and thus ,J:x4(

;�x;4dx'

I:

x'i!:��:::

'I:

x4:1-x)4dx ,,,(C) Now, from (A),

1

x4(1-x) 4dx • �5

-t

x6 +

f

x7

-1

x6 +

l]:

1

=

630, on calculation. Thus .l2 _630 1_ < - -22 7 7f < 630, from (C)1

i.e. _{- 1260} 1 > 1r - � 22 > - _{630, on multiplying by -1}1 22 1

Henae, 7 - 630 < rr 22 <----_{7 1260} 1

EXERCISES SET 10

1, Let m, M be the smallest and greatest values of the integrable function f(x) in a S x Sb, Using a suitable diagram, show that m(b-a) i J:f(t) dt i M(b-a),

2.

3.

If x > O, deduce that -2L. S log(l+x) 2 x._l+x

Show that f(x)

=

_{1 � sin x} steadily decreases as x in-creases from O to rr/2. Write down the greatest and least values of f(x) in this interval.

Deduce that, if OS X _{- 2'} < .2!.

dx _{< .2!.} 1 + sin x 2 J

rr/2 then

f

<

0

x2_-x Find the greatest and least values of e

-k x2_-x domain O $ x $ 2.

f2

Hence show that 2e 4 _<

0 e dx < 2e2

25

4, State the Mean Value Theorem of the Integral Calculus,

Use this theorem to determine the following limits.

(i)

;:_�fx�Z f !si_{X X} n 1T� dt}

(ii) lim [-x- I 1 f(t) dt]_{x+x1 x-x1} X

5. Use the Mean Value Theorem of the Integral Calculus to

prove that

fx+ h

(i) ��

l

t

u+��2+11 = ll+x 2 _{- x}

(ii) J:x2dx .'., for ,ome a (OS a S 1), 1

and hence show that a= 73•

6. Draw a sketch showing that, if f(x) > g(x) > 0 for a 5 x Sb, then Jbf(x) dx > J:g(x) dx.

Prove that f

1r/

Zsi; x dx < loge3.

7,

8.

9,

10.

1r/6

From a diagram, show that Q < X <

i'

_r1r/6 Hence prove that

Jo

x2sin

::.: ',:,: jf ,

J:'��:i�

s

x t

::

sin x < x < tan x, if

X dx < ,,:;, <

r:'t=

X dXo

greater, sin x or lsin x?

1T

lies between 1 and 2.

Prove that 3 < /9 + 16 sin2_{x < 5, and hence deduce that}

3; <

J:

1' 1, + 16 sin2x dx < ';.

If n > 2 and O < Hence

/

�prove that dx 1T

� < -- < 6' _{O /1-x}n

x < 1, show that O l < _1_

<-1-/1-xn _!i-x2

26

11.

12.

THEOREMS ON DEFINITE INTEGRALS

Evaluate flu

X e -x2 dx.

Show that for u > 1,

fu

e-x2 dx < �(e-1 - e-u2}

hence that 1

0 < I l

°' -xe 2 dx < ₂1 _e,

Show that 1

0 I: x3 _(1-x) 3_{dx < /}

and deduce that

Jo

1553 <--₂₂₄₀

x3_(1-x) 3

f l

3 3

l+x dx < 0

x (1-x) dx,

THEOREM _{d� {}

f:

_{f(tl dt) f(x)}

We assume that f(t) is an integrable function of t, over the domain a .:S t .:S x, where "a" is constant.

Now f:f(t) dt represents the area bounded by the curve

y = f(t), the t axis, and the ordinates t = a, t = x, Since

this area varies with x, th� f:f(t) dt is a function of x,

say A(x).

{For

simplicity;

in the diagram, we take f(t) > O for the interval a .:St� x, but the proof is still valid if f(t) .:SO in [a,x].}

II

Now d! (I: f(t) dt) - d! {A(x)) -:;a�.__..._.J__.-4_-L.,.._...:J.,�

'f��(t)

_{lim rA(x+h)-A(x)} b d f' 'ti} h-+o

l

h , y e 1n1 on ,

J�

h

f(t) dt -

J:

f(t) dt

lim

h-+o _h

lim 1 jx+h = h-+0 h

X

f(t) dt

Now, by the Mean Value Theorem of the Integral Calculus,

fx+h

x f(t) dt = [(x+h)-x],f(I:;), where x <I:;< x+h.

•• • •• • •. (P)

Hence (P) becomes

.� {f(t) dt} _{h+O {h ' h f(I:;)}, where x <I:;< x+h}lim 1 lim {f(s)}

h+O

lim {f(I:;)}. l:;+x

f(x)

See figure above.

EXAMPLE _1." If y = r/l sin 7 t dt, find the value of

'ff;

when

X

= 2..

X

dx a - I:f(x) dx

sin7_{x, since f(t) = sin}7_{t and}

a =-

1T

3 Thus \•,hen x = .'.!!. _<!Y_ = - sin 7 .'.!!:. = - (1) 7 = -1.

· ' 2' dx 2

EXAMPLE 2. Find for what value of x, the d.efinite integral [",;:;;;du has its greatest value, and find this value.

Solution

THEOREMS ON DEFINITE INTEGRALS

j

x 1-u Let f(x) =

0

h+u du

:. f' (x) " d: f

r,t:::

du] • 1

t;'.

by the theo<em above Now stationary values of f(x) occur when f'(x) = O,

. 1-x

i.e. when ll+x = O, i.e. when x = 1.

h '( ) - 1-.Ll=.tl_ >

Furt er, _{f 1-e - 7f+(l-e) O, assuming e > O}

and t _ 1-(1+£)

f (l+e) - ll+(l+e) < O

Hence, the stationary value at x = 1 is a maximum. 1f

[Note from 1-u Y = ll+u'

the sketch of the curve crosses the u axis where u = 1, and for u >l, y < O. Hence for O <xi 1, the value of

f;;i:�

du is

�:,:::: x•�:, 1:•r:�� n

::

is decreasing numerically.

l

The actual greatest value of ;

;i:�

du occurs when x = 1.] To dete,,,,/,ne this grea

r '

\_:•lue, we have to evaluate

Now I:

1i��

du •

r

,;,: u

�u. Check this step I

{

1

{2(l+u)-� - (l+u)�} du

[20(l;u)� _ (l+u)

%]

1

'2

%

4(12-1) -1(212-1) 0 2(4/z-s)

NOTE ON THE THEOREM

It is the use of the theorem above which enables us to

show that

J:

f(x) dx • F(b) - F(a), where F(x) is a primitive

function of f(x).

Since F(x) is any primitive function of f(x), then d

dx [F(x)] = f(x).

Now

a!

_{ix f(t) dt} • f(x), by the theorem above,

and hence d! [F(x) - J:f(t) dt] = O

, ', F(x) -

J'

f(t) dt • c, where C is a constant.

Now when :iC =a a, • ', F(a) -

J:

f(t) dt = C

i.e. C = F(a)

Hence F(x) -

J:

f(t) dt = F(a), and

thus

I.'

f ( t) dt • F(x) - F(a). I' f(t) �t • F(b) - F(a),

Howeve:

f:

f(t) dt =

I:

f(x) dx = � f(z) dz etc. and .', f:f(x) dx = F(b) - F(a)

Thus (i) r (x2_{-3x+7) dx}

[1 ·

53 _-

f .

₅2 _{+ 7 • 5] -}

[f .

_z3 _-

f .

₂2 _{+ 7 • 2]}

since here F(x) = 1 x3 _{- l x}2 _{+ 7x}

3 2

F(b) F(a)

F(5) =

t'

53 _-

t

F(2) l 2 _{= 3 .} 3 _{- 2}

2

30

52 _{+ 7 5}

THEOREMS ON DEFINITE INTEGRALS

(ii) The area bounded by the curve y = x3_{, the x axis and the}

ri

d

::::·:

[�4]:

XS:,

2

Un::,,

noting F(x) = \x4 is a

prim-itive function of x3• _{...;O�c:::::::..-�====��}

Hence r x'dx • \i

15 =4

z4 - !;r. 14

[F(b)=F(2); F(a)=F(l)]

3

Thus the area is 3

4

sq. units.

EXERCISES SET lE

1. _{(a) Evaluate f:t,dt, and verify that d: {f:t}3 _{dt) x'}

(b) If f(x) = jx (sin t + et) dt, verify by direct in -11/ 3

2.

3.

4.

s.

tegration and then differentiation, that f' (x) sin x + ex,

Write down f'(x) if

(a) f(x) f >1-t2 _{dt (b) f(x)}

C'

_{log tan t dt}

f x

lx

-7e u (c) f(x)

5t

2sin(l+_t2_{) dt (d) f(x)} sec u du

(i) If g(x) = J: 11 - sin t dt, find g' (�)

(11) If f(x) •

f:

,-t',o, t dt, find f' (D)

(iii) If y "

r

,x' dx, find ¥x when X • 2

f 71

/6

..

h

71

If y =

x tan x dx, find the value of dx7 when x =

4.

If U = x2 , rf(t) dt, prove that X, :� - 2u = X3,f(x),

6. If A(x) is the area under the curve y = f(t) from t = a d2_w

to t = x, and w = x, A(x), then dx2 equals

(i) A(x) + xA'(x) (iii) 2f(x) + xf'(x) (v) xf"(x) + f(x), Select the correct answer.

(ii) f' (x)

(iv) xf"(x) + f' (x)

7, Two variables x and y are related by the equation

X = J

OY

1i!!

t

z

dx _<:I_2_y

Find dy and dx2' [Give both answers in terms of y,]

8. Find the derivative, with respect to x, of r(l-t)et dt,

and hence find when this definite integral has its

greatest value.

Proof

Noting that _{d: (}t et - et)

=

t et, determine the great

-est value off: (1-t)et dt.

THEOREM, f: f( x) dX f: f( a-x) dx

In l:f(x) dx, Now when x =

o,

u

dx d Also - = -(a-u)_{du du}

put x = a-u

a and when X = a, u

-1.

o.

Thus f:f(x) dx • f:f(x) dx • du_du = I: f(a-u) • -1, du

f ;f(a-u) du

t

f(a-x) dx, reverting to the variable x.

THEOREMS ON DEFINITE INTEGRALS

{Note: The use of this theorem enables us to determine the value of certain definite integrals, even though we are actually unable to obtain the primitive functions. See ex­ amples 2, 3 below.}

EXAMPLE 1. Evaluate

I:

x2_{• lt-x dx}

Solution

J:,

2_{.11-x dx •}

I:

_(1-x)2

11-(1-J

:) dx, {since f(x) dx =

I:

f(a-x)

I:

(1-x)2_,x'i _dx 0

f 0 1

x\l-2x+x2_{) dx}

f:

(x\ - 2x 3Jz + x 5/z) dx

=l_i+l

_{3 5 7} _li

105 on evaluating.

EXAMPLE 2. _{Show that ls�n x + lcos x = 4} !

1112 _{/sin x}

dx 11

f 11/2

�

0

ff

ifsin(.1!. _{- x) dx}

S ovu �on ow 1 t, N tsin x dx . = -,====2=----;=====-v'sm x + v'cos x _{O /sin(.'.!!. - x) + /cos (.1!.- x)}

Thus

o 2 2

f 11/2

�

d

0 � +

rsti"'i"

X

v'sin x + rctis x

dx}

f

11/2 �

dx

i _o

{f

1r/2 �

dx

>;i l

O

rsfri""i'

+ v' COS X + J

11/2 _{� dx}

1

O rsrn4t + �S X

.k: � +

rcos-x

d )

11/2

2

7s':Fn""i + � X

0

33

EXAMPLE 3,

Solutim

J:

f

x sin x

Eva Zuate

1 + _{COS X} 2 dx

X sin X :X _ /

11

(11-x) sin(11-x) dx 1 + cos2_x

-0 1 + cos2(11-x)

Thus

2

Hence f

11

x sin x dx _ 1

+

cos2_x

-0

f 11

x sin x dx 1 + cos2_x

0

-1 11

11 [-tan (cos x)] ₀

EXERCISES SET lF

Check this carefully!

1. _{Prove that J:f(x) dx • J:f(a-x) dx.}

2.

3.

Hence evaluate _{1 2}

(i) J:x(l-x)10_{ax (ii) f}

0x(l-x) o/, dx (iii) f0

x,l2-x dx

Show that !11

0

12

cos x - sin x d _ f1112 cos x - sin 1

+

sin 2x x - - 1

+

sin 2x

f 11 0/ 2

0

and hence that cos x - sin x dx_{1 + sin 2x}

o.

x, using the theorem above,

Prove that the value of each of the definite integrals

r

sin

J:

COS X dx

(i) cos _{cos x} X -

+

_sin X _X dx (ii) ₃

+

_sin2_x

0

('

cot x

(iii) tan x -_tan2_x

+

_cot2_x dx

is zero, 0

4. Show that definite

5. Prove that of m.

ODD AND EVEN FUNCTIONS

J:��n2_{e dO • J:':}

J

o::: dB, and hence that :ach integral is\

0 (sin

2_{e+ cos}2_{8)d8 =}

4

JTI/2 , � sin dx - ..'!!:. for all real values 0 sinmx + cos� - 4'

6. Show that sin(3_{; - x) = - cos x, and find a similar}

re-3Tr sult for cos(2 -Hence evaluate

x) •

J::

sin x + cos x 3 + sin 2x dx.

7, Prove that

J:x

_{cos 2x dx = i}rr(rr-x) cos 2x dx, and hence evaluate

J:x

cos 2x dx,

8. Show that

)Tr/:

!x. = _jn/_l

2

+

dx sin x cos x

0 _{0 X}

9.

10.

Noting that cos x = 2 cos2

2 - 1, evaluate this inte-gral.

Show that Jn/4 cosec 2x - 1 dx = _frr

o/4 1 - cos 2x d cosec 2x + 1 1 + cos 2x x'

0

j""

and hence evaluate this integra:. O I 2Tfx cos x dx - . Prove that _{1 + . z sin x} = rr[tan 1(sin

0

tan2_{x dx,}

2_n x)] O

0

ODD AND EVEN FUNCTIONS

DEFINITION f(x) Is said to be an ( i) even function, if f(-x) = f(x) (ii) odd function, if f(-x) = -f(x).

Examples of

(i) even functions are f(x) = x4; -x2

since when f(x)

=

e say, -(-x)2 _-x2 f(-x)

=

e

=

e

f(x)

=

cos x; f(x)

= f(x)

-x2 e

(ii) odd functions are f(x) = x3; f(x) = sin x; f(x) since when f(x) = sin x say,

f(-x) = sin(-x) = - sin x = -f(x).

Note: Many functions are neither even nor odd.

Examples are f(x)

=

ex; f(x)

=

l-3x; f(x)

=

x3+2x-7; since when f(x) x3+2x-7, say

f(-x) (-x)3+2(-x)-7

=

-x3-2_x-7

=

-(_x3+2_x+7)

1' f(x) or -f(x)

PROPERTIES OF ODD AND EVEN FUNCTIONS

(i)

EVEN FUNCTIONS

(ii)

1

The sketches of 3 even functions are shown above. In each case, note that

f(-a)

=

f(a). [This is the definition of an even

f a

f(x) dx

=

2 faf(x) dx. [;::: ti

:::�lt can be easily -a o seen by consideration of areas.] (iii) the y axis is an axis of symmetry of the curve y = f (x).

(ii)

/fa)

ODD AND EVEN FUNCTIONS

The sketches of 3 odd functions are shown above. In each case, note that

(i) f(-a) = -f(a). [This is the definition of an odd function.]

(ii)' f(O) = o; i.e. the curve y f(x) passes through

origin.

{Proof: In f (-a) -f(a), put a = O

f(-0) -f(O), i.e. f (0) -f(O). Hence 2f(O) O, whence f(O) o.} (iii)

J::(

x) dx = O. [Consider areas in the sketches.]

(iii) COMBINATIONS OF ODD AND EVEN FUNCTIONS

The most important results here are:

the

the product of (a)two odd or two

function. even functions is an even

(b)an odd and an

function.· even function is an odd Proof that the product of two odd functions is an even function.

Let f(x), g(x) be odd functions, i.e. f(-x) = -f(x) and g(-x) = -g(x). We wish to prove that h(x) = f(x) g(x) is an even function; i.e. we have to show that h(-x) = h(x).

(a)

Now h(-x) f(-x) g(-x) -f(x) .-g(x),

since f(x), g(x) are odd functions, f(x) g(x)

h(x).

Hence h(x) is an even function.

Proofs of the other results are very similar.

Thus, x3_{cos x}

cos x is even.

Hence

j

!_/, �

c:: d

x f

�

cti

:� ·

-rr/2

since x3 _{is odd and}

(b) Further, x sin-1x is an even function, since both x and sin-1x are odd functions.

I

_{Hence J�}

EXAMPLE. Use the properties of odd and even functions to evaluate j\1+x3_{) 3 dx.}

-1

Solution Now if f(x) = (l+x3₎ 3

then f(-x) = [1+(-x)�]3 _{= (1-x}3₎3

Thus f(x) is neither an odd nor an even function. In thi

l

case we expand (l+x3₎3_,

Hence 1(1+x3₎3_{dx =}

I

1_(1+3x3_+3x6_+x9_{) dx}

-l

= f1: dx + 3 J1x3_{dx + 3 J} 1

x6_{dx +j} 1

x9_dx

,-'j:,

dx + 0-: 3. 2

J:

x6:: + O

-I

[Note that x3 _,x9 _{are odd functions whilst 1 = x}o_{, x}6 _are

�::

n

j

(

�:::�::�

J

.

2[x]: + _6��

1

• 2�

-1 0

{Note carefully the difference between

evaluating Jaf(x) dx

-a

and finding the area under the curve y

ordinates x = -a, x = a. f(x) between

re;le,

-x2

-x

o,

_{since x is odd function.}

x e dx = e an

-2

tr

_11:::11:.e -x'I..

Howeveri, the area

bounded by the

-2

curve _-x2 X.

y = X e

'

the x axis, and the ordinates X •

-r,

X" :

[ -x2_f

-,

is 2

0 x e -x dx 2 -fe 0 (1 e-4) sq. units, on calculation}

ODD AND EVEN FUNCTIONS

EXERCISES SET l G

1.) Determine which of the following functions are even, and which are odd.

(a) f (x) x2sin x (b) f (x) = x3-x

(c) f (x) Hex + e-x) (d) f(x) = _lxl

(e) f (x) sin(5x2) (f) f(x) = x3tan-1_x

(g) f(x) Cos3x (h) f (x) = cos-1x (i) f(x) = a_a2 2 _{+ x}- x2 ₂

2. If f(x), F(x) are even functions and g (x) , G(x) are

functions, show that

( i) f (x) F(x) is even

(iii) f(g(x)) is even

(ii) f(x) g(x) is odd

(iv) g( G(x)) is odd,

{Note if, for example, f(x) = co

j x, g(x) = x

3 _then

f(g(x)) = cos x3 _{whilst g(f(x) = cos3x.}}

odd

3. Use the properties of odd and even functions to evalu­

ate the following.

(a) /TT/Z sin x dx

-TT/2 f

TT/2

(b) COS X dx

-TT/2

(d) [:'.; _:�n ₃

� �� •. (,) .f:'.;ts'x dx

(g)

f

1 x2 tan x dx (h)

f

TT sin x cos x dx

\

1

-1 -TT

(j) fTT/\ sec4_{x dx (k) j\osx(e}x_-e-x_)dx

-TT/2 -4

fl x7dx f

TT

��n3_{x dx}

(m)\ a2+_x2 (n) 1 + cos_2x

4.

-1 -TT/2

/TT/4

(c) tan2x dx

-TT/4

(f) /TT/Zsin9x dx

-TT/2

• (i)/JTT�fn3x cos5_{x dx}

-TT/2

(l) f ::3_,la2_{=;z2 dx}

f 1 3 -x2

(o) · x e dx

-1

Evaluate ₃

ta). f x5(1-_x2)

-3 dx f x

4_(1-x6) d_x

. (b)

-1

5

(c) f x2(1-x7) d_x

-s

(d)J::(a2_-x2_{) dx (e) r(a}2_{-x2) dx (f) r (x+x}7₎ 5 _dx

-a -1

39

(g)

11

I (l+x5)2dx (h) f (1 - 2

-1

f 11/2

-11

(i) (1 + sin 2x)3

-11 /2

dx.

ex_-1

(i) _{Show that ex+l is an odd function, and hence}

fl ex_-1

5.

evaluate �� dx

(ii) Prove that

/a dx-l:� l

Ja �x -�. x2+a2 - x2+a2 - 2a

(iii) Show that -a

J'

i!x�)'

_{Odx • 2 r 1 dx • 2,}

-1 0

ARCS OF CURVES

This section is concerned with calcu­ lating the length of a continuous and differentiable curve y = f(x).

Consider the arc AB, where A, B have abscissae a, b respectively.

Take any set of points Po,P1,P2,••••• ••• , Pn-l• Pn on the arc AB, and consider the sum of the lengths

••• , Pn_1B [Note Po= A, Pn - B.]

If there are many points

Pi

(i=0,2,3, ••• ,n) on the arc AB, then the length of the arc AB is approximately the sum of the lengths AP1, P1P2, P2_{P3, •••, Pn_1B.}

We

define

the length of the arc AB as the limit of tl;, sum of the lengths of the chords, as we make the number.of points greater and greater.

Take a typical chord PiPi+l'

where

Pi,Pi+l

have coord­ inates (x,y) and (x+ox, y+oy) respectively.

LENGTH OF ARCS

Now l*P_iP_i+