Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.1 Double Integrals over Rectangles - 15.1 Exercises - Page 1040: 22


Given $$\int_{0}^{1} \int_{0}^{2} y e^{x-y} d x d y=\left(e^{2}-1\right)(1-\frac{2}{e}) $$

Work Step by Step

Given $$\int_{0}^{1} \int_{0}^{2} y e^{x-y} d x d y$$So we have \begin{aligned}I&=\int_{0}^{1} \int_{0}^{2} y e^{x-y} d x d y \\ &=\int_{0}^{1}\left[y e^{x-y}\right]_{0}^{2} d y\\ &=\int_{0}^{1} y e^{2-y}-y e^{0-y} d y\\ &=\int_{0}^{1} y\left(e^{2-y}-e^{-y}\right) d y\\ &=\int_{0}^{1} y\left(e^{2} \cdot e^{-y}-e^{-y}\right) d y\\ &=\left(e^{2}-1\right) \int_{0}^{1} y e^{-y} d y \end{aligned} So by partition technique Let $$u=y \Rightarrow du= dy$$ $$dv=e^{-y }dy \Rightarrow v= -e^{-y }$$ So, we get \begin{aligned} I&=\left(e^{2}-1\right)\left[uv|_{0}^{1}- \int_{0}^{1} vdu\right]\\ &=\left(e^{2}-1\right)\left[y e^y|_{0}^{1}- \int e^{-y } d y \right ] \\ &=\left(e^{2}-1\right)\left[ -y e^{-y }|_{0}^{1}+ e^{-y }|_{0}^{1}\right ]\\ &=\left(e^{2}-1\right)\left[ -e^{-1 }-0-(e^{-1 }-1) \right ] \\ &=\left(e^{2}-1\right)(1-\frac{2}{e})\end{aligned}
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