University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 74


The area of the region is $9/2$.

Work Step by Step

$x=y^2$ and $x=y+2$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ here. 2) Find the limits of integration: To find the limits of integration, we find intersection points with respect to $y$. $$y^2=y+2$$ $$y^2-y-2=0$$ $$(y+1)(y-2)=0$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=2$$ So the upper limit is $2$ and the lower limit is $-1$. 3) Find the area: The region from $y=-1$ to $y=2$ is bounded on the right by $x=y+2$ and on the left by $x=y^2$. Therefore, the area of the region is $$A=\int^2_{-1}(y+2-y^2)dy$$ $$A=\frac{y^2}{2}+2y-\frac{y^3}{3}\Big]^2_{-1}$$ $$A=\Big(2+4-\frac{8}{3}\Big)-\Big(\frac{1}{2}-2+\frac{1}{3}\Big)$$ $$A=\frac{10}{3}-(-\frac{7}{6})=\frac{9}{2}$$
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