Question

If $f(a-x)=-f\left(x\right)$, then ${\int }_0^{a}f\left(x\right)dx=0$.

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

Put $x=-t$

$\Rightarrow dx=-dt$

When $x=-a,t=a$ and when $x=0,t=0$

$I={\int }_a^{0}f(-t)(-dt)$

$=-{\int }_a^{0}f(-t)dt$

$={\int }_0^{a}f(-t)dt...........$ since ${\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx$

$={\int }_0^{a}f(-x)dt..........$ since ${\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(t\right)dt$

Thus,

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

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

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

If $f\left(x\right)$ is an odd function, then $f(-x)=-f\left(x\right)$

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

Hence proved.

Thus, the given statement is true.