\(\int_{-1}^{2} \left|x^{3}-x\right|dx\) is equal to

\(\frac{11}{4}\)

$|x^3-x|=\left\{\begin{matrix}x^3-x,&-1≤x<0\\x-x^3,&0≤x≤1\\x^3-x,&1<x≤2\end{matrix}\right.$

\(\int\limits_{-1}^{2} \left|x^{3}-x\right|dx\)

$=\int\limits_{-1}^{0}x^3-xdx+\int\limits_{0}^{1}x-x^3dx+\int\limits_{1}^{2}x^{3}-xdx$

$=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0}+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_{0}^{1}+\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{1}^{2}$

$=-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{16}{4}-\frac{4}{2}-\frac{1}{4}+\frac{1}{2}$

$=\frac{11}{4}$