Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.1 Double Integrals over Rectangles - 15.1 Exercises - Page 1040: 25

Answer

$$\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v = \frac{31}{30} $$

Work Step by Step

Given $$\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v$$So we have \begin{aligned} I&=\int_{0}^{1} \int_{0}^{1} v\left(u+v^{2}\right)^{4} d u d v\\ &=\int_{0}^{1}\left[\frac{v\left(u+v^{2}\right)^{5}}{5}\right]_{0}^{1} d v \\ &= \int_{0}^{1} \frac{v\left(1+v^{2}\right)^{5}}{5}\ dv-\int_{0}^{1} \frac{v\left(0+v^{2}\right)^{5}}{5} d v \\ &= \int_{0}^{1} \frac{v\left(1+v^2\right)^5}{5} \ dv- \int_{0}^{1} \frac{ v^{11}}{5} d v \\ &= \frac{1}{10} \int_{0}^{1}2v\left(1+v^2\right)^5 \ dv- \int_{0}^{1} \frac{ v^{11}}{5} d v \\ &= \frac{1}{60} \left(1+v^2\right)^6|_{0}^{1}- \frac{ v^{12}}{60} |_{0}^{1}\\ &= \frac{1}{60} \left(1+1\right)^6-\frac{1}{60} \left(1 \right)^6- \frac{ 1^{12}}{60}-0\\ &= \frac{2^6}{60} -\frac{1}{60} - \frac{ 1}{60}\\ &= \frac{2^6-1-1}{60} \\ &= \frac{2^6-2}{60} \\ &= \frac{31}{30} \\ \end{aligned}
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