Question

If ƒ : R®R, g : R®R are continuous functions then the value of the integral is -

• A. 1
• B. 0
• C. –1
• D. p

Answer 1

Answer: (B)

0

Answer 2

Let F(x) = [ƒ(x) + ƒ(–x)) (g(x) – g(–x)) so

F(–x) = (ƒ(–x) + ƒ(x)) (g(–x) – g(x)) = – F(x)

Hence $\int _ { - \pi / 2 } ^ { \pi / 2 } f ( \mathrm { x } ) \mathrm { dx }$ = 0. (using Property 11).