5.

3 Definite Integrals
Definite Integrals

If f is continuous at every point of the interval [ a , b ] and if F is any

antiderivative of f on [ a , b ] , then


a
f ( x) dx  F (b)  F (a)
 Properties of Definite Integrals

If f and g are integrable on closed interval [a,b] and

c is any constant, then
b b b
1.   f x   g x dx   f x dx   g x dx
a a a

2. b b
 cf x dx  c  f x dx
a a
 b c b
 f x dx   f x dx   f x dx
3.
a a c
for a  c  b

a
 f x dx  0
4

a
b a
5.
 f x dx   f x dx
a b
Example 1

b

2
x dx
a
Solution
b
b 2  x3 

a
x dx   
 3  a
3 3
b a
 
3 3
 Example 2

5 2
Given ∫ 2 f(x)dx = 4, find ∫ 5 f(x)dx
Solution

2
∫ 5 f(x)dx = -4

b a
∫ a f(x)dx = - ∫ b f(x)dx
 Example 3

2
Find ∫ 2x dx
0
Solution
2
∫ 0 2 dx 2
∫ 0 2 dx
2
= [ 2x ] 0 =2(2+0)
= (2)(2) – (2)(0) =4
=4

b
∫ a c dx = c ( b + a ),
where c is any constant.
 Example 4

2
Find ∫ 2 7 dx
Solution
2
∫ 2 7 dx
2
= [ 7x ] 2
= 7(2) – 7(2)
=0
a
∫ a f(x)dx = 0
 Example 5

  1 dx
2
x
1
Solution
2
x3 
   x
 3 1

8  1 
   2     1
3  3 
10

3
 Example 6

Evaluate 3

( x  1 ) dx
3

1
Solution

3
 x4 
   x

 4 
 1
81 1
 (  3)  (  1) = 24
4 4
 Example 7
1

Evaluate 2x
e dx
0
Solution

1
e 
 
2x
1 2
=    e 1
 2 0 2
 Example 8
1

  1 dx
2 3
Evaluate 3 x x
1
Solution
Let u = x3 + 1 , du = 3x2 dx
When x = -1 , u = ( -1 ) 3 + 1 = 0
When x = 1 , u = ( 1 )3 + 1 = 2

1 2
 3 x 2
x 3
 1 dx  u du
1 0
2
2 3 
 u 2

3 0
2  32 3  4 2
 2  0 2

3  
3
 Exercise
1

3x
Evaluate xe dx
0

Answer : 3
2e 1

9 9
i. Area (between the curve and the x-axis)

Formula :
f(x)
b
1. A   y dx
a
a b
b c
2. A   f ( x)dx   f ( x)dx
a b

a b c
 Example 1
Find the area of the region between the curve of y = 8x – 2x 2, x = 2,
x = 3 and the x-axis.
Solution

y = 8x – 2x 2, x = 2, x = 3

So, the area of the region is

3
2 8x  2 x
2
dx

 4x  2 2
3
3 3
x 2 
22

3
 Example 2

Find the area of the region between the curve of y = x 2 – 4x, the x-axis
and the lines x = 0 and x = 4

Solution

y = x 2 – 4x
So, the area of the region is

4 2
 0
x  4 x dx
2 4
 x 4x 
3
  
3 2 0
64
 unit 2
3
ii. Area (Between the curve and the y-axis)

• Formula :

d
d
A x dy
c c
 Example 4

What is the area between the curve of y = (x – 2) 2 and

the y-axis from y = 0 to y = 4 ?

Solution
Area
4
1
= (y
0
2
 2) dy

4
2 3 
=  y 2
 2 y
3 0
 2 32 
=  3 4  24  0 = 40
unit 2
  3
iii. Area (Between two curves)

y2

y1

a b

b
A   y 2 - y1 dx
a
 Example 5

Find the area enclosed by the curve y = (x – 3) 2 and the

line x + y = 9
Solution 5

    2
(9 x ) ( x 3) dx
0
5
   x 2  5 x dx
0
5
 x 5x 
3 2
   
 3 2 
0
125 125 125
   units 2
3 2 6
 Example 6

Find the area bounded by the curve of x = y2 ,

and the curve x = 8 – y2

x = 8 – y2 x = y2
Solution

The intersect points are,

y  8 y
2 2

2y2  8  0

( y  2)( y  2)  0

y  2 or y  2

Thus the limits of the integration is

[-2, 2]
The area bounded are,
2

 [(8  y )  y ] dy
2 2

2
2
  (8  2 y 2 ) dy
2


 8y  y 2
3

3 2
2

 16   16  64
 16      16    units 2
 3  3 3
 Exercise
1. Find the area enclosed by the lines 13
x = 1, x = 2, y = 3x and the curve y = x 2 answer : unit 2
6

2. Find the area enclosed by the curves 1

y = 3x – x 2 and y = 2x 2 answer : unit 2
2

3. The area enclosed by the curve x = y 2 and

the line y = x – 2 is rotated about the
y-axis.
72
answer   unit 3
5
Volume Of Solids of Revolution
Definition :

(a) Volume of a solid of revolution (Revolve about the x-axis)

The volume of the solid generated by revolving the region about the
x- axis between the graph of the continuous function y = f (x) and the
x - axis from x = a to x = b is

b
Volume    2
( radius function ) dx
a y
b y =f(x)
V    2
( y ) dx
a
b a b x
V     f ( x) 
2
dx
a
Definition :
(b) Volume of a solid of revolution (Revolve about the y-axis)

The volume of the solid generated by revolving the region about the
y axis between the graph of the continuous function x = g (y ) and
the y - axis from y = c to y = d is

b
Volume    2
( radius function ) dy
a
b y x =g(y)
V    2 d
( x ) dy
a
c
b
V     g ( y )  2
dy x

a
 Example 1

Find the volume of the solid of revolution formed by revolving the graph
of f(x) = 4 – x 2 about the x -axis in the first quadrant.
Solution
2
V    ( f ( x)) 2 dx
0

 4  x 
2
 2 2
dx
0
2
  
0
( 16 - 8x 2  x 4 ) dx
2
 8x x  3 5
  16x -  
 3 5 0
256 
 unit3
15
 Example 2

The curve y  x , the line x = 2 , and the x – axis form the sides of a
bounded region R . Find the volume of the solid generated by
revolving R about the x- axis .
Solution
2
V  
0
y 2 dx
2
  
0
( x ) 2 dx
2
x  2
  
 2 0
4 
    0
2 
 2  unit3
 Example 3
Find the volume of the solid obtained by rotating the region bounded
by y = x 3 , y = 8 and x = 0 around the y – axis .
Solution

y
y=8
8
x=0
y = x 3 or x = 3 y

0 x 0 x
x3= y
x 3 y  g (y)  3 y
b
V    g ( y)  2 dy
a
8
  ( 3
y )2 dy
0


8 2
  y 3
dy
0

 3y 3  5 8

   
 5 
0

96 
 unit3
5
More general solids of revolution
Volume of solids generated by revolving the region
between two curves
a) Revolve about x –axis
Suppose that y 1 = f (x ) and y 2 = g (x ) are nonnegative continuous
function such that , f (x ) ≥ g (x ) for a ≤ x ≤ b and let R be the
region enclosed between the graphs of these functions and the lines
x = a and x = b .
The volume of the solid generated by revolving R about the x- axis is

b
V    ( y 1  y 2 ) dx
2 2

a
    f (x)   g ( x)  dx
b
V  2 2

a
 y
y = f (x)
x=b
R
x =a y=g(x)

a b
 Example 4
Find the volume of the solid generated by revolving the region bounded
by the line y = x + 2 and the parabola y = x2.
Solution

x + 2 = x2 y
x2 – x – 2 = 0
(x – 2)(x + 1) = 0

x = 2 , x = -1

-1 2 x
 
2
V  ( y1  y 2 ) dx
2
π b
a

2
 π  [(x2) 2
 (x 2 ) 2 ] dx
-1
2
 (x  2 ) 3
x  5
 π   
 3 5 1

 64 32  1 1 
      
3 5 3 5 
72
 units3
5
b) Revolve about y - axis
Suppose that x 1 = f (y ) and x 2 = g ( y) are nonnegative continuous
function such that , f (y ) ≥ g ( y) for c ≤ y ≤ d and let R be
the region enclosed between the graphs of these functions and the
lines y = c and y = d.

The volume of the solid generated by revolving R about the y-axis is

  dy
d
V    x1  x 2
2 2

    f  y   g  y  dy
d
V 2 2

c
 Example 5
Find the volume of the solid generated when the region bounded by
y 2 = 4x and y = 2x - 4 is revolved about the y - axis.
Solution
y2 = 4x , y = 2x - 4
y2 y4 y
x x
4 2
y2 y  4 4

4 2
y2 – 2y – 8 = 0 x
-2
(y – 4)(y + 2) = 0

y = 4 , y = -2
 4  y  4  2   2 
V  
2

 2 
     dy
 4  
1 1 4
4
   
 y 4 2
 y  dy
2 4 16 
 1 1 5
4
    
y4 
3
y 
 12 80   2
  128 64   2 2 
        
 3 5   3 5 
144
 units3
5
 Example 6

Find the volume of the solid generated by revolving the region enclosed
by y2 = x and y = - x + 2 about the indicated line :
(a) y – axis (b) x = -2
Solution
(a) y2 = x ; y = -x + 2 , y-axis

y y2 = 2 - y
y2 + y – 2 = 0
(y + 2)(y – 1) = 0
1
y = -2 , y = 1
x

-2
The volume of the solid generated

 
1
V    (2  y ) 2  ( y 2 ) 2 dy
2

   4  4 y  y 
1
2
 ( y ) dy
4

2
1
 y y  3 5
  4 y  2 y 
2
 
 3 5  2

72
 units 3

5
(b) y2 = x ; y = -x + 2 , x = -2

R(y) = (2 – y) + 2 = 4 - y
1 r (y) = y2 – (-2) = y2 + 2

r(y) x

-2
R(y)

 
1
The volume of the solid genetared    12  8 y  3 y 2  y 4 dy
 
1
2
V    (4  y )  ( y  2) dy
2 2 2
 y 5

1

2   12 y  4 y  y  
2 3

 5  2
 
1
   16  8 y  y 2  ( y 4  4 y 2  4) dy 162
2
 unit 3
5
 Example 7

Find the volume of the solid generated by revolving the region enclosed
by y = x2 + 1 and y = x + 3 about the indicated line :
(a) x – axis (b) y = 7
Solution
(a) y = x2 + 1 ; y = x + 3 , x - axis
x2 + 1 = x + 3
y
x2 - x – 2 = 0
(x -2)(x + 1) = 0

x=2 ; y=5
x = -1 ; y = 2

-1 2 x
The volume of the solid genetared

 
2
   ( x  3) 2  ( x 2  1) 2 dx
1

2
   ( x 4  x 2  6 x  8)dx
1

2
 x x 5 3

     3x  8 x 
2

 5 3  1

67
 units 3
5
(b) y = x2 + 1 ; y = x + 3 , y=7
y
y=7

R(y)
r(y)
R(y)= 7 – (x2 + 1) = 6 – x2
r(y) = 7 – (x + 3) = 4 - x

-1 2
 
2

The volume of the solid genetared    20  8 x  13x 2  x 4 dx

1

 
2
   (6  x )  (4  x) dx
2
 x 
2 2 2 3 5
13x
  20 x  4 x 
2
 
1
 3 5  1

 
2
  36  12 x 2  x 4  (16  8 x  x 2 ) dx  198 3
5
units
1
 Exercise
1. Prove that the area enclosed by the curve
1
y
( x  1)(3  x)
1
the line x = 2 and the x- and y-axes is ln 3 unit2
2

2. Show that the area enclosed by the line y = 4 – 2x and

the curve y = 4 – x2 is 4 unit2
3
3. The two curves y = 12x – x2 and y = x2 + 16 intersect at
the points A and B.
(a) Find the coordinates of A and B
(b) Sketch the two curves on the same set of axes.
(c) Calculate the area enclosed by the two curves.

4. Sketch the curve y 

12 for values between 0 and 4.
2 x
Hence, deduce that
4
12
8 dx  24
0
2 x
5. The two curves y2 = 4x – 8 and y2 = 2x intersect at the
points P and Q.

(a) Find the coordinates of P and Q.

(b) Sketch the curves on the same axes.

(c) Calculate the volume of the solid generated

when the area bounded by the curves is rotated
throught 3600 about the y-axis.