Question

If f is an integrable function, show that
(i) $\underset{-a}{\overset{a}{\int }}f\left(x^2\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x^2\right)dx$

(ii) $\underset{-a}{\overset{a}{\int }}xf\left(x^2\right)dx=0$

Answer 1

(i)
$$\begin{gathered} I={\int }_{-a}^{a}f\left(x^2\right)dx \\ \mathrm{Here}g\left(x\right)=f\left(x^2\right) \\ \Rightarrow g\left(-x\right)=f{\left(-x\right)}^2=f\left(x^2\right)=g\left(x\right)i.e,g\left(x\right)\mathrm{is}\mathrm{even} \\ \mathrm{Therefore} \\ I=2{\int }_0^{a}f\left(x^2\right)dx\left[\mathrm{Using}{\int }_{-a}^{a}g\left(x\right)dx=2{\int }_0^{a}g\left(x\right)dx\mathrm{when}g\left(x\right)\mathrm{is}\mathrm{even}\right] \end{gathered}$$


(ii)
$$\begin{gathered} I={\int }_{-a}^{a}xf\left(x^2\right)dx \\ \mathrm{Let}g\left(x\right)=xf\left(x^2\right) \\ \Rightarrow g\left(-x\right)=\left(-x\right)f{\left(-x\right)}^2=-\left(x\right)f\left(x^2\right)=-g\left(x\right)i.e,g\left(x\right)\mathrm{is}\mathrm{odd} \\ \mathrm{Therefore} \\ I=0\left[\mathrm{Using}{\int }_{-a}^{a}g\left(x\right)dx=0\mathrm{when}g\left(x\right)\mathrm{is}\mathrm{odd}\right] \end{gathered}$$