Class: B.Sc CS.

Subject: Discrete Mathematics

Unit-2

RAI UNIVERSITY, AHMEDABAD

UNIT- 2: Definite Integration and it’s Application

Introduction: In Mathematics, whenever possible, we think of an Opposite or

inverse Process of a given Process. For example, Subtraction is inverse process of addition and division is inverse process of multiplication. Similarly, integration is inverse process of differentiation.

Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function like this:

What is the area under y = f(x)?

There are two types of integration:1. Indefinite Integration.2. Definite Integration.

Definition : Indefinite Integration If f (x) is the derivative of a differentiable function F (x) with respect to x in the intervalI⊂R .

i.e. if ddxF ( x )= f ( x ) , x∈ I

then F (x) is called integral or primitive of f (x) with respect tox .

Symbolically it is written as

∫ f ( x )dx=F (x )

Read as “F ( x ) is the integral of f ( x ) with respect to x .

Here the sign '∫ ' denotes the Process of integration and it is called

sign of integration.

The function f ( x ) whose integration is to be obtained is called the integrand.

The Process of finding integrals of given functions is called definite integration.

A function whose integration can be obtained is called an integrable function.

Notation:

After the Integral Symbol we put the function we want to find the integral of (called the Integrand), and then finish with dx go in the xdirection (and approach zero in width).

And here is how we write the answer:

We wrote the answer as x2 but why + C? It is the "Constant of Integration". It is there because of all the functions

whose derivative is 2 x :

So when we reverse the operation (to find the integral) we only know 2x, but there could have been a constant of any value.

So we wrap up the idea by just writing + C at the end.

Standard Formulae Of Integration:

Sr. No. Function Integration1 ∫ xndx xn+1

n+1+C

2 ∫1dx x+C

3 ∫ 1xdx ln|x|+C

4 ∫ cosxdx sin x+C

5 ∫ sinxdx −cosx+C

6 ∫ sec2 x dx tan x+C

7 ∫ cosec2 x dx −cot x+C

8 ∫ sec x tan x dx sec x+C

9 ∫ cosecx cot xdx −cosec x+C

10 ∫ exdx ex+C

11 ∫ ax dx ax

ln a+C

12 ∫ dx

x2−a21

2alog|x−ax+a |+C

13 ∫ dx

a2−x21

2alog|a+ xa−x|+C

14 ∫ dx

a2+x2

1a

tan−1 xa+c

15 ∫ dx

√ x2+a2log|x+√ x2+a2|+C

16 ∫ dx

√ x2−a2log|x+√ x2−a2|+C

17 ∫ dx

√a2−x2sin−1 x

a+C

18 ∫ dx

|x|√x2−a2

1asec−1 x

a+C

19 ∫√x2+a2dx x2

√x2+a2+ a2

2ln|x+√x2+a2|+k

20 ∫√x2−a2dx x2

√x2−a2+ a2

2ln|x+√ x2−a2|+k

21 ∫√a2−x2dx x2

√a2−x2+ a2

2sin−1 x

a+k

Working Rule Of Integration:

Example based on the above Rules:

Example-1. Solve the integral ∫ x5dx .

Solution: ∫ x5dx= x

5+1

5+1+k

¿=1

6x6+k¿

Example-2. Solve the integral ∫ x32 dx .

Solution: ∫ x32 dx=

x32+1

32+1

+k

¿

¿ =x

52

52

+k=25x

52 +k

Example-3. Solve the integral ∫(4 cosx−3 ex+2

√1−x2 )dx.

Solution: ∫(4 cosx−3 ex+2

√1−x2 )dx

¿4∫ cos x dx−3∫ ex dx+2∫ dx

√1−x2

=4 sinx−3ex+2 sin−1 x+k Methods of integration

1. Integration by substitution—

If integrand is not in the standard form and can be transformed to integral form by a suitable substitution then such process of integration is called as integration by substitution.

Take these steps to evaluate the integral ∫ f {g ( x ) }g ' ( x )dx , where f and g 'are

contineous function—

Step 1: Substitute z=g(x ) and dz=g ' ( x )dxto obtain the integral ∫ f ( z )dz .

Step 2: Integrate with respect to .

Step 3: Replace z by g(x ) in the result.

Example—Solve the following integral∫ 1

cos22θdθ .

Solution: ∫ 1

cos22θdθ

=∫ sec2 2θdθ

¿∫ sec2 z .12dz [ z=2θ

⟹dz=2dθ

⟹dθ=12dz]

¿12∫ sec

2 z . dz

¿12

tan z+k

¿12

tan 2θ+k

Example— Evaluate ∫ (2 x+3 )7dx .

Solution— ∫ (2 x+3 )7dx

=∫ z7 12dz Let 2 x+3=z

= 12∫ z

7dz 2dx=dz

= z8

16+k dx=1

2dz

= (2x+3 )8

16+k

Example— Integrate ∫ ex

e2x+9dx .

Solution— ∫ ex

e2x+9dx

=∫ z

z2+9

1zdz Let ex=z

= ∫ 1

z2+9dz ex dx=dz

= ∫ 1

z2+ (3 )2dz dx= 1

exdz

= 13ta n−1( z3 )+k dx=1

zdz

= 13ta n−1( ex3 )+k

2. Integration by method of parts—

If u and v are functions of x then the integral of product of these functions is given by—

∫u v dx=¿ (v∫udx )−∫ dvdx (∫udx )dx¿

This rule is called as ‘integration by method of parts’.

The choice of first and second integral is given by—

I— Inverse

L—Logarithmic

A—Algebraic

T—Trigonometric

E—Exponential

The term appearing first in this series have to take first integral (u).

Example—Integrate ∫ x cos x dx .

Solution: ∫ x cos x dx

=( x∫cos x dx )−∫ dxdx (∫cos x dx )dx

¿ xsin x−∫ sin xdx ¿ x sin x+cos x+k

Integration of terms involving multiple trigonometric functions—

1. sinmx cos nx=12

.2 sinmx cosnx=12

[sin (m+n ) x+cos (m−n ) x ]

2. sinmx sinnx=12

[cos (m−n ) x−cos (m+n ) x ]

3. cosmx cosnx=12

[cos (m−n ) x−cos (m+n ) x ]

3. sin2 x=12¿

4. cos2 x=¿ 12¿¿

5. sin3 x=14

¿

6. cos3 x= 14

¿

Example-1. Evaluate ∫sin 3 x cos2 xdx .

Solution: ∫sin 3 x cos2 xdx

¿12∫ 2 sin3 x cos2 xdx

¿12∫ (sin 5 x+sin x )dx

¿12 (−1

5cos 5x−cos x)+k

¿−110

¿

Example-2. Integrate ∫sin3 x dx .

Solution: ∫sin3 x dx

¿14∫ (3 sin x−sin 3x )dx

¿14 (−3cos x+ 1

3cos3 x)+k

¿−34

cos x+ 112

cos3 x+k

2. Definite integration

Definition— If f (x) is continuous in the interval [a ,b] and F (x) is an antiderivative of f (x) , then

∫a

b

f ( x )dx=F (b )−F (a)

Elementary properties of definite integral—

1. ∫a

b

f ( x )dx=−¿∫b

a

f ( x )dx ¿

2. ∫a

b

f ( x )dx=∫a

b

f ( y )dy=∫a

b

f ( z )dz

3. ∫c

c

f ( x )dx=0

4. ∫a

b

f ( x )dx=∫a

α

f ( x )dx+∫α

b

f (x )dx ,a<α<b

5. ∫a

b

f (a+b−x )dx=∫a

b

f ( x )dx

6. ∫0

a

f (a− x )dx=∫0

a

f ( x )dx

7. ∫−a

a

f (x )dx={2∫0a

f ( x )dx if f is even function

0 if f is odd function

8. ∫0

2a

f ( x )dx=¿ {2∫0a

f ( x )dx if f (2a−x )=f (x )

0 if f (2a−x )=−f (x)

¿

5.3 Standard properties of definite integral—

1. If f (x)≥0 is defined and continuous on[a ,b], then ∫a

b

f ( x )dx ≥0

2. If f (x)≤0 is defined and continuous on[a ,b], then ∫a

b

f ( x )dx ≤0

3. If f (x)≤g (x) is defined and continuous on[a ,b],

then ∫a

b

f ( x )dx ≤∫a

b

g ( x )dx

4. If  f (x) and g(x ) be two functions defined and continuous on [a, b] then—

(∫a

b

f (x ) g ( x )dx )2

≤∫a

b

( f ( x ) )2dx∫a

b

(g ( x ) )2dx

5. The mean value theorem—

Statement— If f (x) is continuous on [a ,b], then at least one value of x=c in (a ,b)

such that—

f ( c )= 1b−a∫a

b

f ( x )dx

Where f ( c ) is called as average or mean value of f ( x ) on [a ,b ] .

Example-1 Integrate ∫1

2

x3dx .

Solution:∫1

2

x3dx

¿|x4

4 |1

2

=24

4−14

4 =

154

Example-2 Integrate ∫0

π2

sin x dx

Solution:∫0

π2

sin x dx

= |−cos x|0

π2

= (−cosπ2 )— cos0

¿0−(−1 )=1

Example-3 Integrate ∫1

e

logx dx

Solution: ∫1

e

logx dx

= [ logx∫ dx−∫ d ( logx )dx

(∫dx )dx ]1

e

= [ xlogx−∫ 1xx dx]

1

e

= [xlogx−∫dx ]1e

= [ xlogx−x ]1e

= (e loge−e )−( log1−1 )

= (e−e )+1=1

Example-4 Evaluate ∫0

π2

sin 3 x cos2 xdx

Solution:∫0

π2

sin 3 x cos2 xdx

¿ 12∫

0

π2

2sin3 x cos2 xdx

¿ 12∫

0

π2

(sin 5 x+sin x )dx

¿12 (−1

5cos 5x−cos x)

0

π2

¿12 [(−1

5cos5

π2−cos

π2 )−(−1

5cos 0−cos 0)]

= 12 [ (0−0 )−(−1

5−1)]

= 35

Example based on the Partial Fraction:

Example-1. Evaluate ∫ (2 x−3)(x−2)(x+3)

dx

Solution: Observe that the factors in the denominator are( x−2 )and(x+3) so, we can write

(2x−3)(x−2)(x+3) =

A(x−2)

+ B(x+3) , where A and B are numbers.

Now, multiplying both sides by the common denominator (x−2)(x+3)

2 x−3=A (x−2)+B (x+3)

2 x−3= (A+B ) x−(2 A−3 B)

By comparing L.H.S & R.H.S— A+B=2 …eqn (1) 2 A−3 B=3 …eqn (2) Now, solving eqn (1) & eqn (2) –

A=95 and B=

15

Hence, putting these results together we have

(2x−3)

(x−2)(x+3) =95

(x−2)+

15

(x+3)

∫4

5 (2 x−3)(x−2)(x+3)

dx=95∫

4

51x−2

dx+ 15∫

4

51x+3

dx

∫ (2 x−3)(x−2)(x+3)

dx=95

log ( x−2 )45+ 1

5log (x+3 )4

5

∫ (2x−3 )( x – 2 ) (x+3 )

dx=¿ 95

[ log (5−2 )−log ( 4−2 ) ]¿

+15

[ log (5+3 )−log (4+3 ) ].

∫ (2x−3 )( x – 2 ) (x+3 )

dx=¿ 95 [ log( 3

2 )]+ 15 [ log( 8

7 )]¿ Example -1. Evaluate ∫

0

π2

secxsecx+cosecx

dx

Solution:

I = ∫0

π2

secxsecx+cosecx

dx

I=¿ ∫0

π2 sec( π2 −x)sec ( π2 −x )+cosec( π2 −x)

dx ¿]

I=∫0

π2

cosecxsecx+cosecx

dx

Adding (1) and (2) we get

2I=∫0

π2secx+cosecxsecx+cosecx

dx

2 I=∫0

π2

1dx

2 I=[x ]0π2

2 I=π2

i .e . I= π4

Example: Evaluate ∫0

πxsinx

1+sinxdx

Solution: I=∫0

πxsinx

1+sinxdx

I=∫0

π (π−x )sin (π−x)1+sin(π−x)

dx

I=∫0

π (π−x )sinx1+sinx

dx

I=∫0

ππsinx

1+sinxdx−∫

0

πxsinx

1+sinxdx

I=π∫0

πsinx

1+sinxdx−I

∴2 I=π∫0

πsinx

1+sinxdx

2 I=π∫0

π

[ sinx1+sinx

¿×1−sinx1+sinx

]dx¿

2 I=π∫0

πsinx (1−sinx)

1−sin2 xdx

2 I=π∫0

πsi nx−sin2 x

cos2 xdx

2 I=π∫0

π

[ sinxcos2 x

¿−sin2 xcos2 x

]dx ¿

2 I=π∫0

π

[ 1cosx

.¿sinxcosx

−( sinxcosx )2

]dx ¿

2 I=π∫0

π

[ secxtanx−tan2 x ] dx

2 I=π∫0

π

¿¿

2 I=π∫0

π

¿¿

2 I=π [secx−tanx+x ]0π

2 I=π [ ( sec π−tan π+π )−( sec 0−tan 0+0 ) ] 2 I=π [ (−1−0+π )− (1−0+0 )]

∴2 I=π (π−2)

∴ I= π2(π−2)

Application of Definite Integration:

We shall now discuss some geometrical applications of definite integrals

which are useful in engineering problems. We shall discuss the use of

definite integrals for finding area and volume.

(A) Area:

The Area bounded by a curve y=f ( x ) , the X-axis and the ordinates x=a

and x=b is given by the definite integral A=∫a

b

f ( x )dx

The area is obtained by dividing the given figure in small parallel strips

with equal bases and taking the sum of the areas of the strips when the

number of strips is very large.

If a function y=f (x ) is continuous on the interval [a, b] then the area

bounded by the curve y=f ( x ) , the X-axis and the lines x=a and x=b is

given by A=|∫a

b

f ( x )dx|

If a function x=g( y) is continuous on the interval [c, d], then the area

bounded by the curve x=g ( y ) , the Y-axis and the lines y=c and y=d is

given by A=|∫c

d

g ( y )dy|Area between two curves:

(1). If two curves y=f 1(x ) and y=f 2(x ) intersect only in two points x=a and

x=b then the area bounded by the two curves is given by

A=¿

(2). Similarly, if two curves x=g1( y ) and x=g2( y ) intersect in the points y=c

and y=d , then the area bounded by the two curves is given by

A=¿

Example-1. Find the area of the Standard circle x2+ y2=a2 ,(a>0).

Solution: The equation of the circle is x2+ y2=a2

∴ The centre of the circle is O(0,0) and its radius is a .

The circle is Symmetric with respect to both the axes in (a ,0) ,

(0 , a) ,(−a ,0)∧(0 ,−a).

As the circle is symmetric with respect to the Origin, its total area is four

times its area in the first quadrant. For this the limits are x=0and x=a .

Now y2=a2−x2 ∴ y=√a2−x2

If A1 is the area of region in the first quadrant, then

A1=∫0

a

ydx=∫0

a

√a2−x2dx

A1=[ x2 √a2−x2+a2

2sin−1 x

a ]0a

.

A1=[ a2 √a2−a2+a2

2sin−1 a

a ]0a

−[0+0 ].

A1=a2

2sin−1 1.

A1=a2

2.π2=π a

2

4.

∴ Total area of the circle = A=4 A1=4 ( π a2

4 )=π a2

Example-2:  Determine Area of the region enclosed by y=x2 and y=√x.

Solution: First of all, just what we mean by “area enclosed by”. 

This means that the region we’re interested in must have one of the two curves on every boundary of the region. 

So, here is a graph of the two functions with the enclosed region shaded.

The limits of integration for this will be the intersection points of the two curves.  In this case it’s pretty easy to see that they will intersect at x=0 and x=1  so these are the limits of integration.

A=∫a

b

[ f 1 ( x )− f 2 ( x ) ]dx

A=∫0

1

(√ x¿−x2)dx¿

A=∫0

1

√x dx−∫0

1

x2dx

A=23

[ x 32 ]0

1

−13

[ x3 ]01

A=23

[1−0 ]−13[1−0]

A=23−1

3

A=13

(B). Volume of a Solid revolution:

The Solids generated by revolving a plane area about a fixed line not intersecting the plane area are called solids of revolution.

Let y=f (x ) be a continuous function on [a, b] and let f (x)>0. The volume of the solid generated by the revolution of the area bounded by the curve y=f ( x ) , the X-axis and the lines x=a and x=b about the X-axisIs given by

V=π∫a

b

y2dx=π∫a

b

[ f ( x ) ]2dx

Let x=g( y) be a continuous function on [a, b] and let g( y )>0. The volume of the solid generated by the revolution of the area bounded by the curve x=g ( y ) , the Y-axis and the lines y=c and y=d about the Y-axisIs given by

V=π∫c

d

x2dy=π∫c

d

[ g ( y ) ]2dy

Volume between two curves:(1). If two different curves y=f 1(x ) and y=f 2(x ) intersect in two distinct points whose x−¿coordinates are a and b, then the volume of the solid of revolution Obtained by revolving the area bounded by the two curves about the X-axis is given by

V=π∫a

b

[ {f 1 ( x ) }2−{ f 2 ( x ) }2 ]dx(2). Similarly, for two different curves x=g1( y ) and x=g2 ( y ) , the volume of the solid generated is given by

V=π∫c

d

[ {g1 ( y ) }2−{g2 ( y ) }2 ]dy

Example-1. Find the volume of the sphere having radius a .

Solution: The sphere is symmetrical about X-,Y- and Z- axes. Also it is Symmetrical with respect to the origin. If we rotate the semi-circle given by

the equation x2+ y2=a2 about the X-axis, we get the sphere with radius a . The limits are – a to a .

Now x2+ y2=a2⟺ y2=a2−x2

The required volume is

V=π∫−a

a

y2dx

V=π∫−a

a

(a2−x2 )dx

V=2π∫0

a

(a2−x2 )dx

V=2π [a2 x−x3

3 ]0

a

V=2π [a3−a3

3 ] V=2π [ 2a3

3 ] V= 4

3π a3

∴ Volume of the sphere =43π a3

Example-2. Find the volume of the solid generated by revolving the region bounded by the graph of y=x , y=0 , x=0 andx=2. At the solid

Solution: we shall now use definite integrals to find the volume defined above. If we let f (x)=x according to 1 above, the volume is given by the definite integral.

Volume = π∫a

b

y2dx

V=∫0

2

π x2dx

V=π∫0

2

x2dx

V=π [ x3

3 ]0

2

V=π [ 83−0 ]

V=8 π3

Example-3.Obtain the volumes of the solids of revolution obtained by

revolving the ellipse x2

a2 + y2

b2 =1 about (i) the X-axis (ii) the Y-axis

Solution: The given ellipse intersects the X-axis in the points is (a, 0) and (-a, 0) and the Y-axis in the points is (0, b) and (0, -b).

The centre of the ellipse is O(0,0).

Now x2

a2 + y2

b2 =1

y2=b2(1− x2

a2 ) and x2=a2(1− y2

b2 ) Rotation about the X-axis: The volume of the solid generated by revolving

the given ellipse about the X-axis is given by

V=π∫−a

a

y2dx

V=2π∫0

a

y2dx (∴ y2 is even function)

V=2π∫0

a

b2(1− x2

a2 )dx V=2π b2∫

0

a

(1− x2

a2 )dx V=2π b2

a2 ∫0

a

(a2−x2)dx

V=2π b2

a2 [a2 x−x3

3 ]0

a

V=2π b2

a2 [a3−a3

3 ] V=2π b2

a2 .2a3

3

V= 43πab2

Rotation about the Y-axis: The volume of the solid generated by revolving the given ellipse about the Y-axis is given by

V '=π∫−b

b

x2dy

V '=2 π∫0

ba2

b2(b2− y2 )dy

V '=2π a2

b2 [b2 y−y3

3 ]0

b

V '=2π a2

b2 [b3−b3

3 ] V '=4

3π a2b

Reference Book and Website Name:1. Advance Mathematics (Group-2) – Atul Prakashan2. Engineering Mathematics – N.P.Bali and Dr. Manish Goyal3. www.mathisfun.com 4. www.mathworld.com 5. www.analyze.math.com 6. https://images.search.yahoo.com

EXERCISE-2

Q-1. Integrate the following function w.r.t. x:

1. ∫ (1+logx )2

xdx

2. ∫ 6 x+2

3 x2+2 x+5dx

3. ∫ xn logxdx

4. ∫ xta n−1 xdx

5. ∫ x+3( x−1 )(x−2)

dx

Q-2. Evaluate the following:

1. ∫0

π3

tanxdx

2. ∫a

blogxxdx

3. ∫3

8xdx

( x−1 )(x−2)

4. ∫0

5 √5−x√5−x+√ x

5. ∫0

π2

sin 5 xcos3 xdx

Q-3. Find the Area enclosed by the parabola x2=4 y and the line y=x .

Q-4. Find the Area enclosed by the parabolas y2=4ax and x2=4 ay .

Q-5. Find the volume of the regular cone having radius of base r and height h.

Q-6. A line x=a intersects the parabola y2=4ax . If the area enclosed by the

parabola , the line x=a and the X-axis is revolved about the X-axis, find the volume

of the solid of revolution.