Question

The value of the integral $\underset{-1}{\overset{1}{\int }}x\left|x\right|dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{-1}{\overset{1}{\int }}x\left|x\right|dx \\ \mathrm{Let}f\left(x\right)=x\left|x\right| \\ f\left(-\mathrm{x}\right)=\left(-x\right)\left|-x\right| \\ =-x\left|x\right| \\ =-f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=0 \end{gathered}$$​

​Hence, the value of the integral $\underset{-1}{\overset{1}{\int }}x\left|x\right|dx$ is 0.