Answer
$\displaystyle \frac{64}{3}$
Work Step by Step
The indefinite integral is over the interval [0,4],
where f(x) has a zero, x=3.
$f(x)=\left\{\begin{array}{lll}
x^{2}-9, & when & 4 \geq x \geq 3\\
-(x^{2}-9) & when & 0 \leq x < 3
\end{array}\right.$
Apply Th.4.6 from section 4-3, Additive Interval Property
$\displaystyle \int_{0}^{4}f(x)dx=\int_{0}^{3}f(x)dx+\int_{3}^{4}f(x)dx$
$\displaystyle \int_{0}^{4}|x^{2}-9|dx=\int_{0}^{3}(9-x^{2})dx+\int_{3}^{4}(x^{2}-9)dx$
Use the table "Basic Integration Rules", p.246
$=[9x-\displaystyle \frac{x^{3}}{3}]_{0}^{3}+[\frac{x^{3}}{3}-9x]_{3}^{4}$
$=(27-9)+(\displaystyle \frac{64}{3}-36)-(9-27)$
$=\displaystyle \frac{64}{3}$
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