Q. If $f(x)\ and\ g(x),\ x\epsilon R$ are continuous functions, then value of integral $\int_{\frac{- \pi}{2}}^{\frac{\pi}{2}}\left\lbrack \left\{ f(x) + f( - x) \right\}\left\{ g(x) - g( - x) \right\} \right\rbrack\ dx$ is

A) $\pi$
B) $\frac{\pi}{2}$
C) 1
D) 0
Explanation

(d)
Let $I = \int_{\frac{- \pi}{2}}^{\frac{\pi}{2}}{\left\{ f(x) + f( - x) \right\}\left\{ g(x) - g( - x) \right\}}\ dx$
Again, let
$h(x) = \left\{ f(x) + f( - x) \right\}\{ g(x) - g( - x)\}$
$\Rightarrow h( - x) = \left\{ f( - x) + f(x) \right\}\{ g( - x) - g(x)\}$
$\Rightarrow h( - x) = - h(x)$
Hence, $h(x)$ is an odd function
$\therefore\ I = 0$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

Want more practice?
Practice Integrals MCQs