$\text{ Evaluate }{\int }_{-1}^2 \left|x^3-x\right|dx$

$$\begin{gathered} f\left(x\right)=x^3-x=x\left(x^2-1\right)=x(x-1)(x+1) \\ \text{ Sign scheme of }f\left(x\right)\text{ is as shown in the following figure. } \end{gathered}$$

$\text{ From the sign scheme, we have }$

$\begin{array}{c}{\int }_{-1}^2 \left|x^3-x\right|dx={\int }_{-1}^0 \left(x^3-x\right)dx+{\int }_0^1 -\left(x^3-x\right)dx+{\int }_1^2 \left(x^3-x\right)dx\\ ={\left[\frac{x^4}{4}-\frac{x^2}{2}\right]}_{-1}^0+{\left[\frac{x^2}{2}-\frac{x^4}{4}\right]}_0^1+{\left[\frac{x^4}{4}-\frac{x^2}{2}\right]}_1^2\\ =-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\\ =-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+2-\frac{1}{4}+\frac{1}{2}=\frac{11}{4}\end{array}$