Answer
$\displaystyle \frac{49}{6}$
Work Step by Step
The area between two graphs over [a,b], where one graph is above the other, is given with
$\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
---
For $x\in[-2,-1]$ and $x\in[2,3],\ \ $
$y_{above}=-x+2,\ \ y_{below}=4-x^{2}$
$A_{1}=\displaystyle \int_{-2}^{-1}[(-x+2)-(4-x^{2})]dx=\int_{-2}^{-1}[x^{2}-x-2]dx= \left[\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x \right]_{-2}^{-1}$
$= \displaystyle \frac{-1+8}{3}-\frac{1-4}{2}-2(-1+2) =\frac{11}{6}$
$A_{3}=\displaystyle \int_{2}^{3}[(-x+2)-(4-x^{2})]dx=\int_{2}^{3}[x^{2}-x-2]dx= \left[\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x \right]_{2}^{3}$
$= \displaystyle \left[ \frac{27-8}{3}-\frac{9-4}{2}-2(3-2) \right]=\frac{11}{6}$
In the middle interval, $x\in[-1,2],$
$y_{above}=4-x^{2},\ \ y_{below}= -x+2$
$A_{2}=\displaystyle \int_{-1}^{2}[(4-x^{2})-(-x+2)]dx=-\int_{-1}^{2}[x^{2}-x-2]dx= -\left[\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x \right]_{-1}^{2}$
$=- \displaystyle \left[ \frac{8+1}{3}-\frac{2+1}{2}-2(2+1) \right]=\frac{9}{2}$
Total area = $A_{1}+A_{2}+A_{3}=\displaystyle \frac{11}{6}+ \frac{9}{2}+\frac{11}{6}=\frac{49}{6}$
