integral calculus

Integration

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Integral calculus Integration is the inverse process of differentiation.

Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation.Let us consider the following examples:We know that,

23

3

cossin

xx

dx

d

edx

de

xdx

xd

xx

We observe that, the function cos x is the derived function of sinx. We say that sinx is an anti derivative (or an integral) of cos x. Similarly , in (2) and (3), and ex are the anti derivatives of x2 and ex

--------------- (1)

--------------- (2)

--------------- (3)

3

3x

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we note that for any real number C, treated as constant function, its derivative is zero and hence, we can write (1), (2) and (3) as follows :

23

3 ,cos)(sin xC

x

dx

deCe

dx

dxCx

dx

d xx

Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually , there exist infinitely many anti derivatives of each of these functions which can be obtained by choosing C arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary constant. We introduce a new symbol, namely , which will represent the entire class of anti derivatives read as the integral of f with respect to x

dxxf )(

An antiderivative of f `(x) = f(x)The indefinite integral: `( ) ( )f x dx f x c


You need to remember all the integral identities from higher.

1

11

cos( ) sin( )

1sin( ) cos( )

n naax dx x c

n

ax b dx ax b ca

ax b dx ax b ca

A definite integral is where limits are given.

This gives the area under the curve of f `(x) between these limits.


Standard formsFrom the differentiation exercise we know:

2

( )

1(ln )

(tan ) sec

x xde e

dxd

xdx xd

x xdx

This gives us three new anti derivatives.

2

1ln

sec tan

x xe dx e c

dx x cx

x dx x c

Note: when 0 but when 0x x x x x x

Example1 23 3

0 11. Find x xe dx e dx

1 2 23 3 3

0 1 0

x x xe dx e dx e dx 2

3

0

13

xe 61 1

3 3e


52. Find dx

x5 1

5dx dxx x

5lnx c 5lnx c

23. Find tan x dx(We need to use a little trig here and our knowledge of integrals.)

From a few pages ago we know 2sec tanxdx x c

2 2sin cos 1x x 2 2

2 2 2

sin cos 1cos cos cos

x xx x x

2 2tan 1 secx x

2 2tan sec 1x dx x dx

2 2tan sec 1 tanxdx x dx x x c © iTutor. 2000-2013. All Rights Reserved

Integration by SubstitutionWhen differentiating a composite function we made use of the chain rule.

3(2 3)y x Let 2 3u x 3y u

23dy

udu

and 2dudx

dy dy dudx du dx

23 2u 26u 26(2 3)x

When integrating, we must reduce the function to a standard form – one for which we know the antiderivative.

This can be awkward, but under certain conditions, we can use the chain rule in reverse.


If we wish to perform we can proceed as follows.( ( )). `( )g f x f x dxLet ( )u f x then `( )du f x dx

The integral then becomes which it is hoped will be astandard form.

( )g u du2 31. Find ( 3)x x dx

2Let 3u x 2du xdx

2 3 31( 3)

2x x dx u du 41

8u c Substituting back gives,

2 41( 3)

8x c


2 3 52. Find 3x ( 4)x dx3Let 4u x 2then 3du x dx

2 3 5 53x ( 4)x dx u du 616u c

3 6( 4)6

xc

Putting the value of u

33. Find 8cosx sin x dxLet sinu x then cosdu xdx

3 38cosx sin 8xdx u du 42u c

42sin x c © iTutor. 2000-2013. All Rights Reserved

For many questions the choice of substitution will not always be obvious.

You may even be given the substitution and in thatcase you must use it.

4ln1. Find . Let ln

xdx u x

x

1du dx

x

4ln4

xdx udu

x 22u c 22(ln )x c

2 12. Find (1 ) . Let sinx dx u x cosdx udu

2 2(1 ) 1 sin cosx dx u udu 2cos u du

sinx u Substituting gives,

We can not integrate this yet. Let us use trig.

2 1cos (1 cos2 )

2u u


1 1cos2

2 2u du

1sin2

2 4u

u c

1.2sin cos

2 4u

u u c 1sin cos

2u u u c

1 21sin 1

2x x x c

Now for some trig play…..

sin , but what does cos equal?x u u

1sin cos

2u u u c

2 2sin cos 1u u 2 2cos 1 sinu

2cos 1 sinu u

2 1(1 ) sin cos

2x dx u u u c


2

23. Find . Let x 2sin

4

xdx

x

2cosdx d

2 2

2 2

4sin .2cos

4 4 4sin

xdx d

x

2

2

4sin .2cos

2 1 sind

2

2

2sin .2cos

cosd

24sin d

2 1 1sin cos2

2 2

2 2cos2 d 2 sin2 c

We now need to substitute theta in terms of x.

2 2sin cos c

2sinx

sin2x

1sin

2x

2

22 2sin cos

4

xdx c

x

2 24sinx 24 4cos

2 24cos 4 x 22cos 4 x

21cos 4

2x

1 212sin 2. . 4

2 2 2x x

x c

1 22sin 42 2x x

x c


Now for some not very obvious substitutions……………….

51. Find sin x dx5 4sin sin sinx x x 2 2sin (sin )x x 2 2sin (1 cos )x x

5 2 2sin sin (1 cos )x dx x x dx Let cosu x sindu x dx

5 2 2sin (1 )x dx u du 2 4(1 2 )u u du 2 41 2u u du 3 52 1

3 5u u u c

3 52 1cos cos cos

3 5x x x c


12. Find

1dx

x Let 1u x

121

2du x dx

122dx x du 2 1dx u du

1 2( 1)

1

udx du

ux

22 duu

2 2lnu u c

2 2 2ln 1x x c


Substitution and definite integrals Assuming the function is continuous over the interval, then exchanging the limits for x by the corresponding limits for u will save you having to substitute back after the integration process.

22 3

1

1. Evaluate (2x+4)(x 4 )x dx2Let 4u x x 2 4du x dx When 2, 12; 1, 5x u x u

2 122 3 3

1 5(2x+4)(x 4 )x dx u du

124

5

14u

5027.75

Special (common) formsSome substitutions are so common that they can be treated as standards and, when their form is established, their integrals can be written down without further ado.

1( ) ( )f ax b dx F ax b c

a

`( )ln ( )

( )f x

dx f x cf x

21`( ) ( ) ( ( ))

2f x f x dx f x c


Area under a curve

a b

y = f(x)

( )b

aA f x dx a b

y = f(x)

( )b

aA f x dx

Area between the curve and y - axis

b

a

y = f(x)

( )b

aA f y dy


1. Calculate the area shown in the diagram below.

5

2

y = x2 + 1

2 1y x 2 1x y

1x y

15 2

2( 1)A y dy

532

2

2( 1)

3y

14 units squared.

3


Volumes of revolutionVolumes of revolution are formed when a curve is rotated about the x or y axis.

2b

aV y dx 2d

cV x dy


1. Find the volume of revolution obtained between x = 1 and x = 2 when the curve y = x2 + 2 is rotated about

(i) the x – axis (ii) the y – axis.

2 2

1( )i V y dx

2 4 2

1( 4 1)x x dx

25 3

1

45 3x x

x

32 32 1 12 1

5 3 5 3

3263 units

15

( ) when 1, 3 and when 2, 6ii x y x y

6

3( 2)y dy

62

3

22y

y

918 12 6

2

315 units

2

6 2

3V x dy


The End

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