Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 980: 8

Answer

\[\int_{-3}^{4}{\int_{2{{x}^{2}}}^{2x+24}f(x,y){dy}dx}\]

Work Step by Step

\[\begin{align} & y=2x+24,\text{ }y=2{{x}^{2}} \\ & \text{Let }y=y \\ & 2x+24=2{{x}^{2}} \\ & 2{{x}^{2}}-2x-24=0 \\ & {{x}^{2}}-x-12=0 \\ & x=-3,\text{ }x=4 \\ & \text{From the graph, the region }R\text{ is} \\ & R=\left\{ \left( x,y \right):2{{x}^{2}}\le y\le 2x+24,-3\le x\le 4 \right\} \\ & \text{Then,} \\ & \int_{-3}^{4}{\int_{2{{x}^{2}}}^{2x+24}f(x,y){dy}dx} \\ \end{align}\]
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