Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 9

Answer

$A = \int_1^2 {\left( {3x - {x^2} - 2} \right)} dx$

Work Step by Step

$$\eqalign{ & {\text{Let }}y = 2 - x{\text{ and }}y = 2x - {x^2} \cr & {\text{Find the intersection points between the curves}} \cr & y = y \cr & 2 - x = 2x - {x^2} \cr & {x^2} - x - 2x + 2 = 0 \cr & {x^2} - 3x + 2 = 0 \cr & {\text{Factoring}} \cr & \left( {x - 2} \right)\left( {x - 1} \right) = 0 \cr & {x_1} = 1{\text{ and }}{x_2} = 2 \cr & {\text{We obtain the interval }}\left[ {1,2} \right] \cr & {\text{The enclosed area is shown in the graph below}}{\text{.}} \cr & 2x - {x^2} \geqslant 2 - x{\text{ on the interval }}\left[ {1,2} \right] \cr & {\text{Therefore}} \cr & {\text{The area is given by}} \cr & A = \int_1^2 {\left[ {\left( {2x - {x^2}} \right) - \left( {2 - x} \right)} \right]} dx \cr & A = \int_1^2 {\left( {2x - {x^2} - 2 + x} \right)} dx \cr & A = \int_1^2 {\left( {3x - {x^2} - 2} \right)} dx \cr} $$
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