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: 32

Answer

$$ \iint \limits_{R} \frac{x}{1+x y} d A= 2 \ln 2-1$$

Work Step by Step

Given$$\ \iint \limits_{R} \frac{x}{1+x y} d A, \quad R=[0,1] \times[0,1]$$ So, we get \begin{aligned} I&= \int_{0}^{1} \int_{0}^{1} \frac{x}{1+x y} dy \ dx \\ &=\int_{0}^{1} \left[\ln (1+x y)\right]_{0}^{1} dx \\ &=\int_{0}^{1} \left[\ln (1+x )\right] dx -\int_{0}^{1} \left[\ln (1)\right] dx \\ &=\int_{0}^{1} \left[\ln (1+x )\right] dx-0\\ &=\int_{0}^{1} \left[\ln (1+x )\right] dx\\ \end{aligned} So by partition technique Let $$u=\ln (1+x ) \Rightarrow du= \frac{1}{1+x }dx$$ $$dv= dx \Rightarrow v=x $$ So, we get \begin{aligned} I&=\int_{0}^{1} \left[\ln (1+x )\right] dx\\ &=uv|_{0}^{1}- \int_{0}^{1} vdu\\ &=x \ln(1+x)|_{0}^{1}- \int _{0}^{1}\frac{x}{1+x }dx\\ &= \ln(2)-0 - \int _{0}^{1}\frac{1+x-1}{1+x }dx\\ &= \ln(2) - \int _{0}^{1}\left[1-\frac{ 1}{1+x } \right]dx\\ &= \ln(2) - \left[x-\ln (1+x ) \right]_{0}^{1} \\ &= \ln(2) - \left[1-\ln (1+1 ) -0+\ln1\right] \\ &= \ln(2) - 1+\ln (2 ) \\ &= 2 \ln(2) - 1 \\ \end{aligned}
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