Evaluate: ∫₋₁² |x³ - x| dx

$x^3 - x = x(x-1)(x+1)$, which is $\ge 0$ on $[-1, 0]$, $\le 0$ on $[0, 1]$ and $\ge 0$ on $[1, 2]$. $I = \int_{-1}^0 (x^3 - x)\, dx - \int_0^1 (x^3 - x)\, dx + \int_1^2 (x^3 - x)\, dx$ $\int (x^3 - x)\, dx = \frac{x^4}{4} - \frac{x^2}{2}$ $I = \left[0 - \left(\frac14 - \frac12\right)\right] - \left[\left(\frac14 - \frac12\right) - 0\right] + \left[(4 - 2) - \left(\frac14 - \frac12\right)\right]$ $= \frac14 + \frac14 + \frac94$ Hence $I = \frac{11}{4}$.