Question

Prove that ${\int }_0^{2a}f\left(x\right)dx={\int }_0^a\left[f\right(x)+f(2a-x\left)\right]dx$

Answer 1

L.H.S. = ${\int }_0^{2a}f\left(x\right)dx$

By property, the integral limits can be split from $0$ to $a$ and $a$ to $2a$.

$\therefore$ L.H.S. $={\int }_0^{a}f\left(x\right)dx+{\int }_a^{2a}f\left(x\right)dx$

In the second integral, substitute $x=2a-t$, then $dx=-dt$

L.H.S. $={\int }_0^{a}f\left(x\right)dx+{\int }_a^{0}f(2a-t)(-dt)={\int }_0^{a}f\left(x\right)dx-{\int }_a^{0}f(2a-t)dt$

Interchanging the limits of the second integral and changing the variable $t$ to $x$, we get

L.H.S. $={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-x)dx$

$={\int }_0^a\left[f\right(x)+f(2a-x\left)\right]dx$

$=$ R.H.S.

Hence proved.