## 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: 55

#### Answer

The area of the shaded region is $4/3$.

#### Work Step by Step

In this exercise, since the given curves are described by functions of $y$, we use the formula $$A=\int^d_c[f(y)-g(y)]dy$$ to calculate the area, in which $f$ denotes the right-hand curve and $g$ the left-hand curve. Here, the shaded region is bounded on the right by the curve $x=12y^2-12y^3$ and on the left by the curve $x=2y^2-2y$ and runs from $y=0$ to $y=1$. Therefore, the area of the shaded region is $$A=\int^{1}_{0}\Big((12y^2-12y^3)-(2y^2-2y)\Big)dy$$ $$A=\int^{1}_{0}(-12y^3+10y^2+2y)dy$$ $$A=\Big(\frac{-12y^4}{4}+\frac{10y^3}{3}+\frac{2y^2}{2}\Big)\Big]^{1}_{0}$$ $$A=-3(1^4)+\frac{10}{3}(1^3)+1^2$$ $$A=-2+\frac{10}{3}=\frac{4}{3}$$

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