Answer
$\int_{0}^{3} \vert x^2-4 \vert~dx = \frac{23}{3}$
Work Step by Step
Note that the function $~~x^2-4~~$ is negative on the interval $0 \leq x \lt 2$
We can evaluate the integral:
$\int_{0}^{3} \vert x^2-4 \vert~dx$
$= \int_{0}^{2} (4-x^2)~dx+\int_{2}^{3} (x^2-4)~dx$
$= (4x-\frac{x^3}{3})~\vert_{0}^{2}+ (\frac{x^3}{3}-4x)~\vert_{2}^{3}$
$= [4(2)-\frac{(2)^3}{3}] - [4(0)-\frac{(0)^3}{3}]+ [\frac{(3)^3}{3}-4(3)] - [\frac{(2)^3}{3}-4(2)]$
$= (\frac{16}{3})+ (-3) - (-\frac{16}{3})$
$= \frac{16}{3}-\frac{9}{3} +\frac{16}{3}$
$= \frac{23}{3}$
