Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 25


$\dfrac{8}{35} $

Work Step by Step

$$I=\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{x+1} (2x-y-z) \ dz \ dy \ dx \\=\int_{0}^{1} \int_{0}^{x^2} [\dfrac{3x^2}{2}-\dfrac{3y^2}{2}] \ dy \ dx\\=\int_{0}^{1} \dfrac{3x^4}{2} \ dx -\int_{0}^{1} \dfrac{x^6}{2} \ dx \\=(\dfrac{3}{2}) \times \int_{0}^{1} x^4 \ dx-(1/2) \times [\dfrac{x^5}{5}]_0^1 \\=\dfrac{3}{2} \times [\dfrac{x^5}{5}]_0^1 -\dfrac{1}{2} [\dfrac{x^5}{5}]_0^1 \\=\dfrac{3}{10}-\dfrac{1}{14} \\=\dfrac{8}{35} $$
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