Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 847: 31


$$6 \ln 6-2 \ln 2-5 \ln 5 $$

Work Step by Step

\begin{align*} \int_{1}^{2} \int_{0}^{4} \frac{d y d x}{x+y}&=\int_{1}^{2}\left(\int_{0}^{4} \frac{d y}{x+y}\right) d x\\ &=\left.\int_{1}^{2} \ln (x+y)\right|_{y=0} ^{4} d x\\ &=\int_{1}^{2}(\ln (x+4)-\ln x) d x \end{align*} Use \begin{aligned} \int_{1}^{2} \int_{0}^{4} \frac{d y d x}{x+y} &=(x+4)(\ln (x+4)-1)-\left.x(\ln x-1)\right|_{1} ^{2} \\ &=6(\ln 6-1)-2(\ln 2-1)-(5(\ln 5-1)-(\ln 1-1)) \\ &=6 \ln 6-2 \ln 2-5 \ln 5 \end{aligned}
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