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 882: 27



Work Step by Step

After integrating the inner $dv$ integral, we get: $\iint_{R} f(u,v) dA= \int_{0}^{1} [\dfrac{v^2}{2}-vu^{1/2}]_0^{1-u} du$ or, $= \int_{0}^{1 } [1/2) (1-u)^2-(1-u) \sqrt u ] du$ or, $= \int_{0}^{1 } \dfrac{1-4x^3-3x+6x^2}{3} dx$ or, $= \dfrac{1}{2}+\dfrac{1}{6}-\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{2}{5}$ Thus, we have $I=\dfrac{-1}{10}$
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