Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 875: 33


$\frac{3}{2}ln 3-1$

Work Step by Step

$\int^{2}_{0} \int^{1}_{0} \frac{x}{1+xy} dx dy$ =$\int^{1}_{0} \int^{2}_{0} \frac{x}{1+xy} $ =$\int^{1}_{0}(ln(1+xy)]^{y=2}_{y=0})dx$ =$\int^{1}_{0} ln(1+2x)dx$ =$\frac{(1+2x)}{2}[ln(1+2x)-1]^1_0$ =$\frac{3}{2}ln3-1$
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