Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - 15.7 Exercises - Page 1049: 5

Answer

$\dfrac{5}{3}$

Work Step by Step

Here, we have $ \int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz=- \int_{1}^2 \int_{0}^{2z} [xe^{-y}]_{0}^{\ln x} dx dz$ $\implies \int_1^2[-x+\dfrac{x^2}{2}]_0^{2z} dz= -\int^{1}_{2}[2z-2z^2] dz$ $\implies - [z^2-\dfrac{2}{3}z^3]_{1}^2=-[4-\dfrac{16}{3}-1+\dfrac{2}{3}]$ Thus, we have $\int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz=\dfrac{5}{3}$
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