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: 64


The area of the region is $32/3$.

Work Step by Step

$y=2x-x^2$ and $y=-3$ 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. $$2x-x^2=-3$$ $$x^2-2x-3=0$$ $$x=-1\hspace{1cm}\text{or}\hspace{1cm}x=3$$ So, the upper limit is $3$ and the lower one is $-1$. 3) Find the area: Looking at the draft, we see that the region is bounded above by $y=2x-x^2$ and below by $y=-3$. So, according to definition, the area of the region is $$A=\int^3_{-1}[2x-x^2-(-3)]dx=\int^3_{-1}[2x-x^2+3]dx$$ $$A=\Big(x^2-\frac{x^3}{3}+3x\Big)\Big]^3_{-1}$$ $$A=(9-9+9)-(1-\frac{(-1)}{3}-3)$$ $$A=9-(-\frac{5}{3})=\frac{32}{3}$$
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