BAM FORM 5 – INTEGRATION

DEFINITE INTEGRAL

The definite integral is given by

Where    a = is the lower limit

             b = is the upper limit

Note:

The arbitrary constant is not shown in the definite integral

Examples

           ^1.       

       Solution

         

        

     =  –

    =  –

    = 31  – 10

    = 21     

           
2.     

          

= 8 – 1

= 6

 

EXERCISE

Find the value of

1.     

2.     

3.     

4.     

5.     

Solution

1.          

 –

(27 – 18 +15) – (8 – 8 + 10)

(27 – 3) – (8 + 2)

= 14

2.          

(   – 2) – (-  –  )

(   – 2) – (

 + 4

= 5

3.       

(  – (

(8 – 4 + 6) – (1 – 1 + 3)

(8 + 2) – (1+2)

= 7

4.     

        Solution

(2(2) +) – (2(-3) + 7  )

(4 + 14) – (-6 +)

18 – (25)

= -7

INTEGRATION BY SUBSTITUTION METHOD

Integrate the following with respect to x

1. (3x-8)⁶

2. 4

3. x

Solution

           1.      dx

Let u = 3x – 8

      du=3dx

             du =

             

            =  du

            =  x  + c

            =   + c

    2.     dx

Let u = 1-x

      du=-dx

       dx

      Let u = 1 – x

        dx = (- du )

      = -4  du

      =  + c

     =-4 x   + c

    =   + c

    = –   + c

   3

Let u=  + 1

      du= 2xdx

      dx =

       =

       =  du

      =  x   + c

    =   + c

     =   + c

Exercise

Determine the integral of each of the following

         1.       dx

         2.       dx

         3.      dx

APPLICATION OF INTEGRATION

To determine the area under the curve

Given A is the area bounded by the curve y=f(x) the x -axis and the line x=0 and x=b where b> a

 

The area under that curve is given by the define definite integral of f(x) from a to b

          = f (b) – f (a)

Examples

1. Find the area under the curve f(x) =x²+1 from x=0 to x=2

2. Find the area under the curve f(x) = from x=1 to x=2

3. Find the area bounded by the function f(x) =x ²-3, x=0, x=5 and the x- axis

Solution

1.      f(x) =  + 1

      y intercept=1

A = dx

    =

    = ( ) – ( + 0)

    = ( + 2) – (0)

    = 4 or  sq. units

2.      f(x)=

A =

   =

=

=  =

=   sq.units

3.      f(x) = x²-3

Where:  y intercept =-3

             X intercept = and x=

A = A₁+A₂

= ₊ 

=  +

=  –   +

=  +

= 2 +

= 6 +  + 6 

= 12 + 80 sq.units

EXERCISE

1.      Find the area between y = 7-x²  and the x- axis from x= -1 to x=2

2.      Find the area between the graph of y=x² x – 2 and the x- axis from x= -2 to x=3

Solution

^1.        y =7-x²

Where y- intercept =7

             X- Intercept =

A = A₁+A₂

=  +

=  +

= (⁺6.67) + (11.3 – 0)

= ⁺ 6.67 + 11.3

=17.97 sq units

Volume of the Solids of Revolution

The volume,V of the solid of revolution is obtained by revolving the shaded portion under the curve, y= f(x) from x= a to x =b about the x -axis is given by

similarly, when the region is rotated about y- axis from y =a to y=b we shall have obtained the volume, V by
.
Example 1

Find the volume of revolution by the curve y=x² from x=0 to x=2 given that the rotation is done about the the x- axis
                                                                                                           
                                                                                                              
                                        

but y=x²

=
=

 .
Exercise

  1. Find the volume obtained when each of the regions is rotated about the x – axis.

  a) Under  y= x³, from x =0 to x=1
  b) Under y²= 4-x, from x=0 to x=2
  c)Under y= x², from x=1 to x=2
  d)Under y= √x, from x=1 to x=4

2. Find the volume obtained when each of the region is rotated about the y-axis.
a) Under y= x², and the y-axis from x=0 to x=2
b) Under y= x³, and the y-axis from y=1 to y=8
c) Under y= √x, and the y-axis from y=1 to y=2

.