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If f and g are continuous functions on `[ 0, pi]` satisfying `f(x) +f(pi-x) =1=g (x)+g(pi-x)` then `int_(0)^(pi) [f(x)+g(x)]`dx is equal to

A

`pi`

B

`2pi`

C

`(pi)/(2)`

D

`(3pi)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`I=overset(pi)underset(0)int {f(x)+g(x)}`
`rArr I=overset(pi)underset(0)int {f(pi-x)+g(pi-x)}dx`
`rArr I=overset(pi)underset(0)int {1-f(x)+1-g(x)}dx [ :. f(pi-x)+f(x)=1 and g(pi-x)+G(x)=1]`
`rArr I=overset(pi)underset(0)int2-{f(x)+g(x)]dx`
`rArr I= 2 pi- I rArr 2I = 2pi rArr I=pi`
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