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Iff(pi)=2int0^pi(f(x)+f^(x))sinxdx=5,t h...

`Iff(pi)=2int_0^pi(f(x)+f^(x))sinxdx=5,t h e nf(0)` is equal to (it is given that `f(x)` is continuous in `[0,pi])dot` 7 (b) 3 (c) 5 (d) 1

A

7

B

3

C

5

D

1

Text Solution

Verified by Experts

The correct Answer is:
B
`int_(0)^(pi) [f(x)+f''(x)]sinx dx`
`=int_(0)^(pi)f(x)sinx dx+int_(0)^(pi) f''(x)sin x dx`
`=(f(x)(-cosx))_(0)^(pi)+int_(0)^(pi)f'(x)cosxdx`
`+sinxf'(x)|_(0)^(pi)-int_(0)^(pi) cos xf'(x)dx`
`=f(pi)+f(0)=5` (given)
`:. f(0)=5-f(pi)=5-2=3`
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CENGAGE-DEFINITE INTEGRATION -Exercise (Single)
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