Question

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

Answer 1

R.H.S. $={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-x)dx$
$=I_1+I_2$
For $I_2={\int }_0^{a}f(2a-x)dx$
Put $2a-x=t$
$\therefore dx=-dt$
When $x=0$, then $t=2a$
When $x=a$, then $t=a$
${\int }_0^{a}f(2a-x)dx={\int }_{2a}^{a}f\left(t\right)(-dt)$
$=-{\int }_{2a}^{a}f\left(t\right)dt$ .... $(\because {\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x\left)dx\right)$
$\therefore {\int }_0^{a}f(2a-x)dx={\int }_a^{2a}f\left(x\right)dx$
$\therefore I_1+I_2={\int }_0^{a}f\left(x\right)dx+{\int }_a^{2a}f\left(x\right)dx$
$={\int }_0^{2a}f\left(x\right)dx=$ L.H.S.
$\therefore {\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-x)dx$