Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 73

$$\frac{1}{2}$$

Given $$\int_{-1}^{1}|x^3| d x $$ Since $$x^3 > 0\ \ \text{ for }\ \ x>0 \\ x^3< 0\ \ \text{ for } \ \ x<0$$ Then \begin{align*} \int_{-1}^{1}\left|x^{3}\right| d x&=\int_{-1}^{0}-x^{3} d x+\int_{0}^{1} x^{3} d x\\ &= \int_{0}^{-1}x^{3} d x+\int_{0}^{1} x^{3} d x\\ &=\frac{1}{4}x^{4} \bigg|_{0}^{-1}+ \frac{1}{4}x^{4}\bigg|_{0}^{1}\\ &=\frac{1}{4}+\frac{1}{4}\\ &=\frac{1}{2} \end{align*}