Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 31

Answer

$\frac{6}{5}$
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Work Step by Step

Step 1. Graph the function as shown in the figure; we can identify two enclosed regions and their areas can be found by integrating with respect to $y$ in the interval of $[-1,0]$ and $[0,1]$. Step 2. The total area of the enclosed regions is given by $A=\int_{-1}^0(y^{2/3}-y)dy+\int_{0}^1(y^{2/3}-y)dy=(\frac{3}{5}y^{5/3}-\frac{1}{2}y^2)|_{-1}^1=(\frac{3}{5}(1)^{5/3}-\frac{1}{2}(1)^2)-(\frac{3}{5}(-1)^{5/3}-\frac{1}{2}(-1)^2)=\frac{3}{5}+\frac{3}{5}=\frac{6}{5}$
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