Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{-a}^{a}f\left(x\right)dx \\ \mathrm{By}\mathrm{Additive}\mathrm{property} \\ I={\int }_{-\mathrm{a}}^0\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x} \\ \mathrm{Let}x=-t,\mathrm{then}dx=-dt, \\ \mathrm{When}x=-a,t=a,x=0,t=0 \\ \mathrm{Hence}{\int }_{-\mathrm{a}}^{0}f\left(x\right)\mathit{d}x=-{\int }_{\mathrm{a}}^0\mathit{\text{f}}\left(-t\right)\mathit{d}t \\ ={\int }_0^{\mathrm{a}}\mathrm{f}\left(-\mathrm{t}\right)\mathit{d}t={\int }_0^{\mathrm{a}}\mathrm{f}\left(-\mathrm{x}\right)\mathit{d}x\left(\mathrm{Changing}\mathrm{the}\mathrm{variable}\right) \\ \mathrm{Therefore}, \\ I={\int }_0^{\mathrm{a}}\mathrm{f}\left(-\mathrm{x}\right)d\mathrm{x}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x} \\ ={\int }_0^{\mathrm{a}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(-\mathrm{x}\right)\right\}\mathrm{dx} \\ \mathrm{Hence},\mathrm{proved}.\mathrm{} \end{gathered}$$