Question

If f (x) is a continuous function defined on [0, 2a]. Then, prove that
$\underset{0}{\overset{2a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}\left\{f\left(x\right)+f\left(2a-x\right)\right\}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{2a}f\left(x\right)dx \\ \mathrm{By}\mathrm{Additive}\mathrm{property} \\ I={\int }_0^{a}f\left(x\right)dx+{\int }_a^{2a}f\left(x\right)dx \\ \mathrm{Consider}\mathrm{the}\mathrm{integral}{\int }_a^{2a}f\left(x\right)dx \\ \mathrm{Let}x=2a-t,\mathrm{then}dx=-dt \\ \mathrm{When}x=a.t=a,x=2a,t=0 \\ \mathrm{Hence},{\int }_a^{2a}f\left(x\right)dx=-{\int }_a^{0}f\left(2a-t\right)dt \\ ={\int }_0^{a}f\left(2a-t\right)dt={\int }_0^{a}f\left(2a-x\right)dx \\ \mathrm{Therefore} \\ I={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(2a-x\right)dx \\ ={\int }_0^a\left\{f\left(x\right)+f\left(2a-x\right)\right\}dx \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$