## 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: 54

#### Answer

The area of the shaded region is $1/12$.

#### 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 $x=y^2$ and on the left by the curve $x=y^3$ and runs from $y=0$ to $y=1$. Therefore, according to the definition, the area of the shaded region is $$A=\int^{1}_{0}(y^2-y^3)dy$$ $$A=\Big(\frac{y^3}{3}-\frac{y^4}{4}\Big)\Big]^{1}_{0}$$ $$A=\frac{1}{3}(1-0)-\frac{1}{4}(1-0)$$ $$A=\frac{1}{12}$$

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