Question

If f(x) is a continuous function defined on [−a, a], then prove that 

$$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}\{f\left(x\right)+f(-x)\}dx$$

Answer 1

$$LetI={\int }_{-a}^{a}f\left(x\right)dx$$
$$\text{By Additive property}$$
$$I={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$$
$$Letx=-t,thendx=-dt,$$
$$Whenx=-a,t=a,x=0,t=0$$
$$Hence{\int }_{-a}^{0}f\left(x\right)dx=-{\int }_a^{0}f(-t)dt$$
$$={\int }_0^{a}f(-t)dt={\int }_0^{a}f(-x)dx.......................\left(\text{Changing the variable}\right)$$
Therefore,
$$I={\int }_0^{a}f(-x)dx+{\int }_0^{a}f\left(x\right)dx$$
$$={\int }_0^a\{f\left(x\right)+f(-x)\}dx$$
$$\text{Hence, proved}.$$