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: 26


$$\int_{0}^{1} \int_{0}^{1} \sqrt{s+t} d s d t=\frac{8}{15} \cdot(2 \sqrt{2}-1)$$

Work Step by Step

Given $$\int_{0}^{1} \int_{0}^{1} \sqrt{s+t} d s d t$$So we have \begin{aligned} I&=\int_{0}^{1} \int_{0}^{1} \left(s+t\right)^{\frac{1}{2}} d s d t\\ &=\frac{2}{3}\int_{0}^{1} \left(s+t\right)^{\frac{3}{2}}| _{0}^{1}\ \ d t\\ &=\frac{2}{3}\int_{0}^{1} \left(1+t\right)^{\frac{3}{2}} \ d t- \frac{2}{3}\int_{0}^{1} \left(t\right)^{\frac{3}{2}} \ \ d t\\\\ &=\frac{4}{15} \left(1+t\right)^{\frac{5}{2}} |_{0}^{1}- \frac{4}{15} \left(t\right)^{\frac{5}{2}} |_{0}^{1}\\\\ &=\frac{4}{15} \left(1+1\right)^{\frac{5}{2}} - \frac{4}{15} \left(1\right)^{\frac{5}{2}}- \frac{4}{15} \left(1\right)^{\frac{5}{2}}+0\\\\ &=\frac{4}{15} \left((\sqrt 2)^{5} -2\right)\\ &=\frac{4}{15} \left(4\sqrt 2 -2\right)\\ &=\frac{8}{15} \left(2\sqrt 2 -1\right)\\ \end{aligned}
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