## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 60

#### Answer

The area of the shaded region is $16$.

#### Work Step by Step

The given graph indicates that there are two parts to the shaded region whose area needs to be calculated. In the first part, from $x=-2$ to $x=0$, the region is bounded above by $y=2x^3-x^2-5x$ and below by $y=-x^2+3x$. The area of this part would be $$A=\int^{0}_{-2}[(2x^3-x^2-5x)-(-x^2+3x)]dx=\int^{0}_{-2}(2x^3-8x)dx$$ In the second part, from $x=0$ to $x=2$, the region is bounded below by $y=2x^3-x^2-5x$ and above by $y=-x^2+3x$. The area of this part would be $$B=\int^{2}_{0}[(-x^2+3x)-(2x^3-x^2-5x)]dx=\int^2_0(-2x^3+8x)dx$$ Therefore, the total area of the shaded region is $$T=A+B=\int^{0}_{-2}(2x^3-8x)dx+\int^2_0(-2x^3+8x)dx$$ $$T=\Big(\frac{x^4}{2}-4x^2\Big)\Big]^0_{-2}+\Big(-\frac{x^4}{2}+4x^2\Big)\Big]^2_0$$ $$T=0-\Big(\frac{16}{2}-4\times4\Big)+\Big(-\frac{16}{2}+16\Big)-0$$ $$A=-(8-16)+(-8+16)$$ $$A=8+8=16$$

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