Question

$\underset{-1}{\overset{1}{\int }}x\left|x\right|dx.$

Answer 1

$$\begin{gathered} \left|x\right|=\left\{\begin{array}{l}-x,-1<x<0\\ x,0<x<1\end{array}\right. \\ \therefore x\left|x\right|=\left\{\begin{array}{l}-x^2,-1<x<0\\ x^2,0<x<1\end{array}\right. \\ \mathrm{Now},{\int }_{-1}^{1}x\left|x\right|dx \\ ={\int }_{-1}^0-x^{2}dx+{\int }_0^1{x}^{2}dx \\ =-{\int }_{-1}^0{x}^{2}dx+{\int }_0^1{x}^{2}dx \\ =-{\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1 \\ =-\left(0+\frac{1}{3}\right)+\left(\frac{1}{3}-0\right) \\ =0-\frac{1}{3}+\frac{1}{3}-0 \\ =0 \end{gathered}$$