Calculus: Early Transcendentals 8th Edition

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

Chapter 15 - Section 15.1 - Double Integrals over Rectangles - 15.1 Exercise - Page 1000: 34

Answer

$$\ \iint \limits_{R} \frac{1}{1+x+y} d A = 6\ln 6-5 \ln 5 - 4 \ln 4 +3\ln 3 $$

Work Step by Step

Given$$\ \iint \limits_{R} \frac{1}{1+x+y} d A, \quad R=[1,3] \times[1,2]$$ So, we get \begin{aligned} I&= \int_{1}^{2} \int_{1}^{3} \frac{1}{1+x+y} dx \ dy \\ &=\int_{1}^{2} \left[ \ln(1+x+y) \right]_{1}^{3} dy \\ &=\int_{1}^{2} \left[ \ln(1+3+y) \right] dy - \int_{1}^{2} \left[ \ln(1+1+y) \right] dy \\ &=\int_{1}^{2} \left[ \ln(4+y) - \ln(2+y) \right] dy \\ \end{aligned} So by the partition technique let $$u= \ln(4+y) - \ln(2+y)\\ \Rightarrow du=\left( \frac{1}{4+y}- \frac{1}{2+y}\right)dy$$ $$dv= dy \Rightarrow v=y $$ So, we get \begin{aligned} I&=uv|_{1}^{2}- \int_{1}^{2} vdu\\ &= y\ln(4+y)|_{1}^{2} - y \ln(2+y)|_{1}^{2}- \int_{1}^{2} \left( \frac{y}{4+y}- \frac{y}{2+y}\right)dy\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( \frac{4+y-4}{4+y}- \frac{2+y-2}{2+y}\right)dy\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( 1-\frac{ 4}{4+y}- 1+\frac{ 2}{2+y}\right)dy\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( -\frac{ 4}{4+y}+\frac{ 2}{2+y}\right)dy\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \left( -4 \ln (4+y) +2\ln (2+y)\right)_{1}^{2}\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \left( -4 \ln 6 +2 \ln 4+4\ln 5- 2 \ln3\right)\\ &= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3+4 \ln 6 -2 \ln 4-4\ln 5+ 2 \ln3 \\ &= 6\ln 6-5 \ln 5 - 4 \ln 4 +3\ln 3 \\ \end{aligned}
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