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: , 3
Solution:
$f(x)=f(a-x)$
$g(x)+g(a-x)=4$
Let, the integral,
$I=\int_{0}^{a} f(x) g(x) d x$
$=\int_{0}^{a} f(a-x) \cdot g(a-x) d x$
$\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$\Rightarrow \quad I=\int_{0}^{a} f(x)[4-g(x)] d x$
$\Rightarrow \quad I=\int_{0}^{a} 4 f(x) d x-\int_{0}^{a} f(x) \cdot g(x) d x$
$\Rightarrow \quad I=\int_{0}^{a} 4 f(x) d x-I$
$\Rightarrow \quad 2 I=\int_{0}^{a} 4 f(x) d x$
$\Rightarrow \quad I=2 \int_{0}^{a} f(x) d x$