If $f(a-x) = f(x)$, then $\int\limits_0^axf(x)dx$ is equal to
Answer & explanation
Correct answer: option 1
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 $f(a-t)=f(t)$,
$I=\int_{0}^{a}(a-t)f(t)\,dt=a\int_{0}^{a}f(t)\,dt-\int_{0}^{a}t f(t)\,dt$.
$I=a\int_{0}^{a}f(x)\,dx-I$.
$2I=a\int_{0}^{a}f(x)\,dx$.
$I=\frac{a}{2}\int_{0}^{a}f(x)\,dx$