Let $$f:R \to R$$ and $$g:R \to R$$ be continuous functions. Then the value of the integral $$\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} \left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx$$ is-
A.
$$\pi $$
B.
$$1$$
C.
$$-1$$
D.
$$0$$
Answer :
$$0$$
Solution :
We have,
$$\eqalign{
& I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} \left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx \cr
& {\text{Let }}F\left( x \right) = \left( {f\left( x \right) + f\left( { - x} \right)} \right)\left( {g\left( x \right) - g\left( { - x} \right)} \right) \cr
& {\text{then }}F\left( { - x} \right) = \left( {f\left( { - x} \right) + f\left( x \right)} \right)\left( {g\left( { - x} \right) - g\left( x \right)} \right) \cr
& = - \left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right] \cr
& = - F\left( x \right) \cr} $$
$$\therefore F\left( x \right)$$ is an odd function,
$$\therefore $$ We get $$I=0$$
Releted MCQ Question on Calculus >> Definite Integration
Releted Question 1
The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-