Question:
Let $f$ and $g$ be continuous functions on $[0$, a such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$,
then $\int_{0}^{a} f(x) g(x) d x$ is equal to :-
Correct Option: , 2
Solution:
$I=\int_{0}^{a} f(x) g(x) d x$
$I=\int_{0}^{a} f(a-x) g(a-x) d x$
$I=\int_{0}^{a} f(x)(4-g(x) d x$
$I=4 \int_{0}^{a} f(x) d x-I$
$\Rightarrow I=2 \int_{0}^{a} f(x) d x$