Answer
\[\frac{23}{3} \]
Work Step by Step
Let \[I=\int_{0}^{3}|x^2-4|dx\]
\[|x^2-4|=\left\{\begin{array}{cc} -(x^2-4)\;\;\;\;
: 0\leq x<2\\
x^2-4\;\;\; :2\leq 3\end{array}\right.\]
\[\Rightarrow I=-\int_{0}^{2}(x^2-4)dx+\int_{2}^{3}(x^2-4)dx\]
\[I=-\left[\frac{x^3}{3}-4x\right]_{0}^{2}+\left[\frac{x^3}{3}-4x\right]_{2}^{3}\]
\[I=-\left[\frac{8}{3}-8\right]+\left[9-12-\left(\frac{8}{3}-8\right)\right]\]
\[I=-\frac{2(8)}{3}+2(8)-3\]
\[\Rightarrow I=13-\frac{16}{3}=\frac{23}{3}\]
Hence $I=\frac{23}{3}$.
