Answer
$16$
Work Step by Step
The area between two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
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For $x\in[-2,0],\ \ y_{above}=2x^{3}-x^{2}-5x,\ \ y_{below}=-x^{2}+3x$
$A_{1}=\displaystyle \int_{-2}^{0}[(2x^{3}-x^{2}-5x)-(-x^{2}+3x)]dx$
$=\displaystyle \int_{-2}^{0}(2x^{3}-8x)dx$
$= \displaystyle \left[ 2\cdot\frac{x^{4}}{4}-4x^{2} \right]_{-2}^{0}$
$=\displaystyle \frac{0-16}{2}-4(0-4)$
$=8$
For $x\in[0,2],\ \ y_{above}=-x^{2}+3x,\ \ y_{below}=2x^{3}-x^{2}-5x$
$A_{2}=\displaystyle \int_{0}^{2}[(-x^{2}+3x)-(2x^{3}-x^{2}-5x)]dx$
$=-\displaystyle \int_{0}^{2}(2x^{3}-8x)dx$
$= -\displaystyle \left[ 2\cdot\frac{x^{4}}{4}-4x^{2} \right]_{0}^{2}$
$=-[\displaystyle \frac{16-0}{2}-4(4-0)]$
$=8$
Total area = $A_{1}+A_{2}= 8+8=16$
