Evaluate $\int_{-1}^{2}\left|x^{3}-x\right| d x$

(D) We note that $x^{3}-x \geq 0$ on $[-1,0]$,$x^{3}-x \leq 0$ on $[0,1]$,and $x^{3}-x \geq 0$ on $[1,2]$.
Using the property of definite integrals,we split the interval:
$\int_{-1}^{2} |x^{3}-x| dx = \int_{-1}^{0} (x^{3}-x) dx + \int_{0}^{1} -(x^{3}-x) dx + \int_{1}^{2} (x^{3}-x) dx$
$= \int_{-1}^{0} (x^{3}-x) dx + \int_{0}^{1} (x-x^{3}) dx + \int_{1}^{2} (x^{3}-x) dx$
$= \left[ \frac{x^{4}}{4} - \frac{x^{2}}{2} \right]_{-1}^{0} + \left[ \frac{x^{2}}{2} - \frac{x^{4}}{4} \right]_{0}^{1} + \left[ \frac{x^{4}}{4} - \frac{x^{2}}{2} \right]_{1}^{2}$
$= [0 - (\frac{1}{4} - \frac{1}{2})] + [(\frac{1}{2} - \frac{1}{4}) - 0] + [(4 - 2) - (\frac{1}{4} - \frac{1}{2})]$
$= -(-\frac{1}{4}) + (\frac{1}{4}) + (2 - (-\frac{1}{4}))$
$= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{3}{4} + 2 = \frac{11}{4}$