If $f(a-x) = f(x)$, then $\int\limits_0^axf(x)dx$ is equal to

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

The correct answer is Option (1) → $\frac{a}{2}\int\limits_0^af(x)dx$

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

Let $I=\int_{0}^{a}x f(x)\,dx$.

Substitute $x=a-t \Rightarrow dx=-dt$.

Then, $I=\int_{a}^{0}(a-t)f(a-t)(-dt)=\int_{0}^{a}(a-t)f(a-t)\,dt$.

Since $f(a-t)=f(t)$,

$I=\int_{0}^{a}(a-t)f(t)\,dt=a\int_{0}^{a}f(t)\,dt-\int_{0}^{a}t f(t)\,dt$.

$I=a\int_{0}^{a}f(x)\,dx-I$.

$2I=a\int_{0}^{a}f(x)\,dx$.

$I=\frac{a}{2}\int_{0}^{a}f(x)\,dx$