So, the required area is
= 
2
2
40
a b
a
x dx
a
−
∫
= 
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a


−
+




 (Why )
= 
2
1
4
0
sin
1
0
2
2
b
a
a
a
−




× +
−










= 
2
4
2 2
b a
ab
a
π =π
Fig 8 6
Fig 8 7
APPLICATION OF INTEGRALS         363
Alternatively, considering horizontal strips as
shown in the Fig 8