Evaluate $\int\limits_{-1}^{2} |x^3 - x| \, dx$

$\frac{11}{4}$

The correct answer is Option (2) → $\frac{11}{4}$

We note that $x^3 - x \geq 0$ on $[-1, 0]$ and $x^3 - x \leq 0$ on $[0, 1]$ and that $x^3 - x \geq 0$ on $[1, 2]$. So by $P_2$ we write

$\int\limits_{-1}^{2} |x^3 - x| \, dx = \int\limits_{-1}^{0} (x^3 - x) \, dx + \int\limits_{0}^{1} -(x^3 - x) \, dx + \int\limits_{1}^{2} (x^3 - x) \, dx$

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

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

$= -\left( \frac{1}{4} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + (4 - 2) - \left( \frac{1}{4} - \frac{1}{2} \right)$

$= -\frac{1}{4} + \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + 2 - \frac{1}{4} + \frac{1}{2} = \frac{3}{2} - \frac{3}{4} + 2 = \frac{11}{4}$