Answer
$\frac{22}{3}$
Work Step by Step
We have
$$
\int_{0}^{3}|x^2-1| d x= \int_{0}^{1}|x^2-1| d x+\int_{1}^{3}|x^2-1| d x\\
= -\int_{0}^{1}x^2-1\ d x+\int_{1}^{3}x^2-1\ d x\\{4}x^4|_{0}^{3}\\
=-(\frac{1}{3}x^3 +x)_{0}^{1}+(\frac{1}{3}x^3 +x)_{1}^{3}\\
=-\frac{1}{3}-1+\frac{27}{3}+1-\frac{1}{3}-1=\frac{22}{3}.
$$
