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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 11



Work Step by Step

$\int^{1}_{0} \int^{2x}_{ x/2} 1dy dx+\int^{2}_{0} \int^{3-x}_{x/2} $ =$\int^{1}_{0} [y]^{2x}_{x/2}dx+\int^{2}_{0} [y]^{3-x}_{x/2}$ =$\int^{1}_{0} (\frac{3}{2}x)dx+\int^{2}_{0} (3-\frac{3}{2}x)dx$ =$[\frac{3}{4}x^2]_0^1+[3x-\frac{3}{4}x^2]^2_1$ =$\frac{3}{2}$
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