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 999: 9

Answer

$24\sqrt 2$

Work Step by Step

Plug in the limits for x and y into the integral: $\int\int\sqrt 2 dxdy$ With y limits from -1 to 5, and x limits from 2 to 6 Integrate the integral with respect to x first: $\int[\sqrt 2 x]dy$ with x limits from 2 to 6 $=\int4\sqrt 2 dy$ Integral with respect to y: $4\sqrt 2 y$ with limits -1 to 5 $=24\sqrt 2$
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