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: 2

Answer

4

Work Step by Step

$\int^{2}_{0} \int^{4}_{2x}dydx$ =$\int^{2}_{0}(4-2x)dx$ =$[4x-x^2]^2_0$ =4
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