University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 2

Answer

$4$

Work Step by Step

Our aim is to integrate the integral as follows: $ Area =\iint_D dA $ $\int^{2}_{0} \int^{4}_{2x}\space dy \space dx =\int^{2}_{0}(4-2x) dx $ or, $=[4x-x^2]^2_0$ or, $=(4)(2)-(2)^2$ or, $=4$
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