Evaluate: ∫_-1^2|x^3-x| d x

Here, \(\displaystyle x^{3}-x \geq 0\) on [-$\displaystyle 1$, $\displaystyle 0$], \(\displaystyle x^{3}-x \leq 0\) on [$\displaystyle 0,1$] and \(\displaystyle x^{3}-x \geq 0\) on \(\displaystyle (1,2)\) So, \[\begin{aligned} I & =\int_{-1}^{0}\left(x^{3}-x\right) d x+\int_{0}^{1}\left(x-x^{3}\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x \\ & =\frac{x^{4}}{4}-\left.\frac{x^{2}}{2}\right|_{-1} ^{0}+\left|\frac{x^{2}}{2}-\frac{x^{4}}{4}\right|_{0}^{1}+\left|\frac{x^{4}}{4}-\frac{x^{2}}{2}\right|_{1}^{2} \\ & =\frac{1}{4}+\frac{1}{4}+\frac{9}{4} \\ & =\frac{11}{4} \end{aligned} \]