Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 49

$$\frac{46}{3}$$

\begin{align*} \int_{-3}^{3} | x^2-4|dx& = \int_{-3}^{2}( x^2-4)dx-\int_{2}^{3}( x^2-4)dx\\ &= \frac{x^{3}}{3}-\left.4 x\right|_{-3} ^{2}+4 x-\left.\frac{x^{3}}{3}\right|_{2} ^{3} \\ &= \frac{46}{3} \end{align*}