Answer
$\displaystyle \frac{38}{3}$
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 $x\in[-3,-2],\ \ y_{above}=x^{2}-4, \ \ y_{below}=-x^{2}-2x$
$A_{1}=\displaystyle \int_{-3}^{-2}[(x^{2}-4)-(-x^{2}-2x)]dx$
$=\displaystyle \int_{-3}^{-2}[2x^{2}+2x-4 ]dx$
$= \displaystyle \left[ 2\cdot\frac{x^{3}}{3}+x^{2}-4x \right]_{-3}^{-2}$
$=2\displaystyle \cdot\frac{-8+27}{3}+(4-9)-4(-2+3)$
$=\displaystyle \frac{11}{3}$
For $x\in[-2,1],\ \ y_{above}=-x^{2}-2x,\ \ y_{below}=x^{2}-4$
$A_{2}=\displaystyle \int_{-2}^{1}[(-x^{2}-2x)-(x^{2}-4)]dx$
$=-\displaystyle \int_{-2}^{1}[2x^{2}+2x-4 ]dx$
$=-\displaystyle \left[ 2\cdot\frac{x^{3}}{3}+x^{2}-4x \right]_{-2}^{1}$
$=-(2\displaystyle \cdot\frac{1+8}{3}+(1-4)-4(1+2)$
$=9$
Total area = $A_{1}+A_{2}=\displaystyle \frac{11}{3}+9=\frac{38}{3}$
