Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 2

Answer

$ \frac{55}{12} $

Work Step by Step

Step 1 of 3: Area: Find the area bounded by the curves $ =\sqrt[]{x} $ , $ y=-\sqrt[3]{x} $ ,$ y=x-2 $ Step 2 of 3: Find the point of intersections of two curves. $ \sqrt[]{x}=x-2 \rightarrow x=4 $ $ -\sqrt[3]{x}=x-2 \rightarrow x=1 $ Thus the curves intersect at $ =1,x=4 \rightarrow y=-1,y=2 $ . Step 3 of 3: Area of the shaded region is $ \int _{-1}^{0} \left( y+2+y^{3} \right) dy+ \int _{0}^{2} \left( y+2-y^{2} \right) dy $ . $ \int _{-1}^{0} \left( y+2+y^{3} \right) dy+ \int _{0}^{2} \left( y+2-y^{2} \right) dy= \left[ \frac{y^{2}}{2}+2y+\frac{y^{4}}{4} \right] _{-1}^{0}+ \left[ \frac{y^{2}}{2}+2y-\frac{y^{3}}{3} \right] _{0}^{2} $ $ =-\frac{1}{2}+2-\frac{1}{4}+\frac{2^{2}}{2}+4-\frac{2^{3}}{3} $ $ =\frac{55}{12} $ Hence, the required area is $ \frac{55}{12} $
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