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 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 50

Answer

$\dfrac{e^{8}-1}{4}$

Work Step by Step

Consider: $I= \int_0^{4} \int_0^{\sqrt {4-y}} \dfrac{x e^{2y}}{4-y} dx dy$ or, $=\int_0^{4} [ \dfrac{x^2 e^{2y}}{4-y} ]_0^{\sqrt {4-y}}dx dy$ or, $= \int_0^{4} \dfrac{(\sqrt {4-y})^2 e^{2y} }{2 (4-y)} dy$ or, $=\int_{0}^{4} \dfrac{e^{2y}}{2} dy$ or, $[\dfrac{e^{2y}}{4}]_0^4=\dfrac{e^{8}-1}{4}$
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