Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Review - Exercises: 1

Answer

$ \frac{8}{3} $

Work Step by Step

Step 1 of 2: Consider the function $ y=x^{2},y=4x-x^{2}. $ The region enclosed by the curves $ y=x^{2},y=4x-x^{2} $ is shown below: Step 2 of 2: From the above graph the point of intersection of the curves $ y=x^{2},y=4x-x^{2} $ are $ \left( 0,0 \right) and $ $ \left( 2,4 \right) $ . Consider $ y=4x-x^{2}=f \left( x \right) $ $ y=x^{2}=g \left( x \right) $ The area between the curves $ y=f \left( X \right) ,y=g \left( x \right)$ and between x=a and x=b is $ A= \int _{0}^{2} \vert f \left( x \right) -g \left( x \right) \vert dx $ . Here the area of the region bounded by the curves $ y=x^{2},y=4x-x^{2} $ and between $ x=0 $ and $ x=2 $ is $ A= \int _{0}^{2} \vert 4x-x^{2}-x^{2} \vert dx $ $ = \vert 2x^{2}-\frac{2x^{3}}{3} \vert _{0}^{2} $ $ = \vert \left( 2 \left( 2 \right) ^{2}-\frac{2 \left( 2 \right) ^{3}}{3} \right) - \left( 2 \left( 0 \right) ^{2}-\frac{2 \left( 0 \right) ^{3}}{3} \right) \vert $ $ =\frac{8}{3} $ Hence the required area of the given region is $ \frac{8}{3} $.
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