Answer
$\displaystyle \frac{19}{4}$
Work Step by Step
The area between two graphs over [a,b], where one graph is above the other, is given with
$\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$.
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For $x\in[-2,0]$ and $x\in[2,3],\ \ $
$y_{above}=\displaystyle \frac{x^{3}}{3}-x,\ \ y_{below}=\frac{x}{3}$
$A_{1}=\displaystyle \int_{-2}^{0}[(\frac{x^{3}}{3}-x)-(\frac{x}{3})]dx=\frac{1}{3}\int_{-2}^{0}(x^{3}-4x)= \frac{1}{3}\left[\frac{x^{4}}{4}-\frac{4x^{2}}{2} \right]_{-2}^{0}$
$= \displaystyle \frac{1}{3}\left[\frac{0-16}{4}-2(0-4) \right] = \frac{4}{3}$
$A_{3}=\displaystyle \int_{2}^{3}[(\frac{x^{3}}{3}-x)-(\frac{x}{3})]dx=\frac{1}{3}\int_{2}^{3}(x^{3}-4x)= \frac{1}{3}\left[\frac{x^{4}}{4}-\frac{4x^{2}}{2} \right]_{2}^{3}$
$= \displaystyle \frac{1}{3}\left[\frac{81-16}{4}-2(9-4) \right] = \frac{25}{12}$
In the middle interval, $x\in[0,2],$
$y_{above}=\displaystyle \frac{x}{3},\ \ y_{below}= \frac{x^{3}}{3}-x$
$A_{2}=-\displaystyle \frac{1}{3}\int_{0}^{2}(x^{3}-4x)=- \frac{1}{3}\left[\frac{x^{4}}{4}-\frac{4x^{2}}{2} \right]_{0}^{2}$
$= - \displaystyle \frac{1}{3}\left[\frac{16-0}{4}-2(4-0) \right]= \frac{4}{3}$
Total area = $A_{1}+A_{2}+A_{3}=\displaystyle \frac{4}{3}+ \frac{25}{12}+\frac{4}{3}=\frac{19}{4}$
