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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 18

Answer

$\frac{32}{3}$

Work Step by Step

$\int_{0}^{2} \int^{0}_{x^2-4}dydx+\int_{0}^{{\sqrt{x}}}dydx$ =$\int_{0}^{2}(4-x^2)dx+\int_{0}^{4}x^{1/2}dx$ =$[4x-\frac{x^3}{3}]^2_0+[\frac{2}{3}x^{3/2}]^4_0=(8-\frac{8}{3})+\frac{16}{3}=\frac{32}{3}$
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