Q. If $f(x) = f(a - x)and\ g(x) + g(a - x) = 2,$then the value of $\int_{0}^{a}{f(x)g(x)dx\ }is$
A) $\int_{0}^{a}{f(x)dx\ }$
B) $\int_{0}^{a}{g(x)dx\ }$
C) $\int_{0}^{a}{\lbrack g(x) - f(x)dx\ }$
D) $\int_{0}^{a}{\lbrack g(x) + f(x)dx\ }$
Explanation
(a)
Let $I = \int_{0}^{a}{f(x)g(x)dx}$
$= \int_{0}^{a}{f(a - x)g(a - x)dx}$
$\Rightarrow \ I = \int_{0}^{a}{f(x)\left\lbrack 2 - g(x) \right\rbrack dx}$
[$\because f(x) = f(a - x),g(x) + g(a - x) = 2\rbrack$
$= \int_{0}^{a}{2f(x)dx - \int_{0}^{a}{f(x)g(x)dx}}$
$\Rightarrow \ I = \int_{0}^{a}{2f(x)dx - I}$
$\Rightarrow \ I = \int_{0}^{a}{f(x)dx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.