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 981: 37

Answer

\[\int_{0}^{1}{\int_{y}^{2-y}{f\left( x,y \right)dx}dy}\]

Work Step by Step

\[\begin{align} & y=-x+2\Rightarrow x=-y+2 \\ & \text{From the graph, the region }R\text{ is} \\ & R=\left\{ \left( x,y \right):y\le x\le -y+2,\text{ 0}\le y\le 1\text{ } \right\} \\ & \text{Then,} \\ & \int_{0}^{1}{\int_{y}^{2-y}{f\left( x,y \right)dx}dy} \\ \end{align}\]
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