# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 59

The area of the shaded region is $38/3$.

#### Work Step by Step

1) Calculate points of intersection between 2 curves: From the graph, the curves intersect at one point $(1,-3)$. There is one more point of intersection whose $x=(-3,0)$ which we would find out here. $$x^2-4=-x^2-2x$$ $$2x^2+2x-4=0$$ $$x^2+x-2=0$$ $$(x-1)(x+2)=0$$ $$x=1\hspace{1cm}\text{or}\hspace{1cm}x=-2$$ - For $x=-2$: $y=(-2)^2-4=0$ So the second point of intersection is $(-2,0)$. 2) Calculate the area The given graph indicates that there are two parts to the shaded region whose area needs to be calculated. In the first part, from $x=-3$ to $x=-2$ (intersection point), the region is bounded above by $y=x^2-4$ and below by $y=-x^2-2x$. The area of this part would be $$A=\int^{-2}_{-3}[(x^2-4)-(-x^2-2x)]dx=\int^{-2}_{-3}(2x^2+2x-4)dx$$ In the second part, from $x=-2$ to $x=1$, the region is bounded below by $y=x^2-4$ and above by $y=-x^2-2x$. The area of this part would be $$B=\int^{1}_{-2}[(-x^2-2x)-(x^2-4)]dx=\int^{1}_{-2}(-2x^2-2x+4)dx$$ Therefore, the total area of the shaded region is $$T=A+B=\int^{-2}_{-3}(2x^2+2x-4)dx+\int^{1}_{-2}(-2x^2-2x+4)dx$$ $$A=\Big(\frac{2x^3}{3}+x^2-4x\Big)\Big]^{-2}_{-3}+\Big(\frac{-2x^3}{3}-x^2+4x\Big)\Big]^{1}_{-2}$$ $$A=\Big(\frac{2}{3}\times(-8)+4+8\Big)-\Big(\frac{2}{3}\times(-27)+9+12\Big)+\Big(-\frac{2}{3}-1+4\Big)-\Big(-\frac{2}{3}\times(-8)-4-8\Big)$$ $$A=\Big(-\frac{16}{3}+12\Big)-\Big(-18+21\Big)+\Big(-\frac{2}{3}+3\Big)-\Big(\frac{16}{3}-12\Big)$$ $$A=\frac{20}{3}-3+\frac{7}{3}+\frac{20}{3}$$ $$A=\frac{47}{3}-3=\frac{38}{3}$$

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