Question

Show that:
${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$, if $f\left(x\right)$ is an even function. $I=0$, if $f\left(x\right)$ is an odd function.

Answer 1

Let $I={\int }_{-a}^{a}f\left(x\right)dx={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

Now, if $f\left(x\right)$ is even, $f\left(x\right)=f(-x)$

$\therefore I={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$\therefore I={\int }_{-(-a)}^{-0}f(-x)(-dx)+{\int }_0^{a}f\left(x\right)dx$

(substituting $x$ by $-x$)

$\therefore I={\int }_0^{a}f(-x)dx+{\int }_0^{a}f\left(x\right)dx$

but $f\left(x\right)=f(-x)$

$\therefore I={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$\therefore I=2{\int }_0^{a}f\left(x\right)dx$ if $f\left(x\right)$ is even.

Now, if $f\left(x\right)$ is odd, $f\left(x\right)=-f(-x)$

$\therefore I={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$\therefore I={\int }_{-(-a)}^{-0}f(-x)(-dx)+{\int }_0^{a}f\left(x\right)dx$ (substituting $x$ by $-x$)

$\therefore I={\int }_0^{a}f(-x)dx+{\int }_0^{a}f\left(x\right)dx$

but $f\left(x\right)=-f(-x)$

$\therefore I=-{\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$\therefore I=0$ if $f\left(x\right)$ is odd.

Hence, proved.