Let f : R $\to$ R and g : R $\to$ R be continuous functions, then the value of the integral $\int_{- \frac{π}{2}}^{\frac{π}{2}} [f (x) + f (- x)] [g (x) - g (- x)] d x =$ [IIT 1990; DCE 2000; MP PET 2001]

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Let f:R→R and g:R→R be continuous functions. Then, the value of the integral ∫π−π (f(x)+f(−x)](g(x)−g(−x)] dx is

The correct option is A \NLet h(x)=(f(x)+f(−x)](g(x)−g(−x)]∴h(−x)=(f(−x)+f(x)](g(−x)−g(x)]⇒h(−x)=−h(x)Hence, h(x) is an odd function.∴∫π−π (f(x)+f(−x)](g(x)−g(−x)] dx =0

Let $f : R \to R a n d g : R \to R$ be continuous functions. Then, the value of the integral
$\int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx$ is

The correct option is A \N
$Let h (x) = (f (x) + f (- x)] (g (x) - g (- x)] ∴ h (- x) = (f (- x) + f (x)] (g (- x) - g (x)] \Rightarrow h (- x) = - h (x) Hence, h (x) is an odd function . ∴ \int_{- π}^{π} (f (x) + f (- x)] (g (x) - g (- x)] dx = 0$