If $f(x)$ and $g(x)$ are continuous functions in $[0, a]$ such that $f(x) = f(a-x)$ and $g(x) + g(a - x) = a$ then $\int\limits_0^af(x)g(x)dx =$

$\frac{a}{2}\int\limits_0^af(x)dx$

The correct answer is Option (3) → $\frac{a}{2}\int\limits_0^af(x)dx$

Given: $f(x) = f(a-x)$ and $g(x) + g(a-x) = a$

Consider $I = \int_0^a f(x) g(x) \, dx$

Substitute $x \to a-x$: $dx = -dx'$, limits: $x=0 \to a$, $x=a \to 0$

$I = \int_0^a f(a-x) g(a-x) \, dx = \int_0^a f(x) g(a-x) \, dx$ (using $f(a-x)=f(x)$)

Add the two expressions:

$2I = \int_0^a f(x) [g(x) + g(a-x)] \, dx = \int_0^a f(x) \cdot a \, dx = a \int_0^a f(x) \, dx$

Thus: $I = \frac{a}{2} \int_0^a f(x) \, dx$

Answer: $\displaystyle \int_0^a f(x) g(x) \, dx = \frac{a}{2} \int_0^a f(x) \, dx$