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 338: 66


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

Work Step by Step

$y=x^2-2x$ and $y=x$ 1) Draft the graph The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area. 2) Find the limits of integration: We can find the limits of integration by finding points of intersection between the curve and the line. $$x^2-2x=x$$ $$x^2-3x=0$$ $$x(x-3)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=3$$ So, the upper limit is $3$ and the lower one is $0$. 3) Find the area: Looking at the draft, we see that the region from $x=0$ to $x=3$ is bounded above by $y=x$ and below by $y=x^2-2x$. So, according to definition, the area of the region is $$A=\int^3_{0}[x-(x^2-2x)]dx=\int^3_{0}(-x^2+3x)dx$$ $$A=-\frac{x^3}{3}+\frac{3x^2}{2}\Big]^3_{0}$$ $$A=-\frac{27}{3}+\frac{27}{2}-0$$ $$A=\frac{9}{2}$$
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