Show that $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,if $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

(N/A) Let $I = \int_{0}^{a} f(x) g(x) \, dx$ ..... $(1)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we have:
$I = \int_{0}^{a} f(a-x) g(a-x) \, dx$
Given $f(x) = f(a-x)$,this becomes:
$I = \int_{0}^{a} f(x) g(a-x) \, dx$ ..... $(2)$
Adding $(1)$ and $(2)$,we obtain:
$2I = \int_{0}^{a} \{f(x) g(x) + f(x) g(a-x)\} \, dx$
$2I = \int_{0}^{a} f(x) \{g(x) + g(a-x)\} \, dx$
Given $g(x) + g(a-x) = 4$,we substitute this into the integral:
$2I = \int_{0}^{a} f(x) \times 4 \, dx$
$2I = 4 \int_{0}^{a} f(x) \, dx$
Dividing by $2$,we get:
$I = 2 \int_{0}^{a} f(x) \, dx$