Question

If f is an integrable function such that f(2a − x) = f(x), then prove that
$\underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

Answer 1

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

Hence Proved