Explanation

(b)
$\int_{- 1}^{2}{|x|^{3}dx = \int_{- 1}^{0}{- x^{3}dx + \int_{0}^{2}{x^{3}dx}}}$
$= - \left\lbrack \frac{x^{4}}{4} \right\rbrack_{- 1}^{0} + \left\lbrack \frac{x^{4}}{4} \right\rbrack_{0}^{2}$
$= - \frac{1}{4}\lbrack 0 - 1\rbrack + \frac{1}{4}\lbrack 2^{4}\rbrack$
$= \frac{1}{4} + 4 = \frac{17}{4}$