Application of integration

Chapter Six
Applications of integration

6.1 Area between two curves
The area A of the region bounded by the curves y  f (x ) and
y  g (x ) , and the lines x  a and x  b , where f and g are
continuous and f (x )  g (x ) for all x [a,b ] , is
b
A   [f (x )  g (x )]dx
a

Example 1. Find the area of the region bounded above by y  x 2  1 ,
bounded below by y  x , bounded on the sides by x  0 and x  1 .
Solution. The area is given by
1
1 1  x3 x 2  1 1 5
A   [( x  1)  x]dx   ( x  x  1)dx     x     1 
2 2
0 0
3 2 0 3 2 6




Example 2. Find the area of the region bounded above by y  e x , bounded
below by y  x , bounded on the sides by x  0 and x  1 .
Solution. The area is given by
1
1  x x 2   1 12   0 02  1 3
A   [e  x]dx  e     e     e    e   1  e 
x
0
 2 0  2  2 2 2




1

, Example 3. Find the area of the region enclosed by the y  x 2 and
y  2x  x 2 .
Solution. We find the point of the intersection of the two curves of y  x 2
and y  2x  x 2 . Then x 2  2x  x 2 and 2x 2  2x  0 . Thus
2x (x  1)  0 and x  0 or x  1 . The point of intersection are (0,0) and
(1,1) . The area is given by
1
1 1  2 2 x3  2 1
A   [(2 x  x )  x ]dx   (2 x  2 x )dx   x 
2 2 2
 1 
0 0
 3 0 3 3




1. If we are asked to find the area between the curves y  f (x ) and
y  g (x ) where, f (x )  g (x ) for some values of x but
g (x )  f (x ) for the others values of x , then we split the region S
into several regions S 1 , S 2 , with areas A1 , A 2 , . Then the area is
A  A1  A 2  . Since
f (x )  g (x ); f (x )  g (x )
f (x )  g (x )  
 g (x )  f (x ); g (x )  f (x )



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