Question

If the function f (x), be an even function of x, then prove that :
${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$.

Answer 1

we have,

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

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

Let $z=-x\Rightarrow dz=-dx$

when $x=-a,z=a$ and $x=0,z=0$ and $x=-z$

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

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

Since, ${\int }_a^{b}f\left(z\right)dz={\int }_a^{b}f\left(x\right)dx$

$I=2{\int }_0^{a}f\left(x\right)dx$