Let f and g be continuous functions on

$f(x)=f(a-x)$

$g(x)+g(a-x)=4$

Let, the integral,

$I=\int_{0}^{a} f(x) g(x) d x$

$=\int_{0}^{a} f(a-x) \cdot g(a-x) d x$

$\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$

$\Rightarrow \quad I=\int_{0}^{a} f(x)[4-g(x)] d x$

$\Rightarrow \quad I=\int_{0}^{a} 4 f(x) d x-\int_{0}^{a} f(x) \cdot g(x) d x$

$\Rightarrow \quad I=\int_{0}^{a} 4 f(x) d x-I$

$\Rightarrow \quad 2 I=\int_{0}^{a} 4 f(x) d x$

$\Rightarrow \quad I=2 \int_{0}^{a} f(x) d x$