## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 37

#### Answer

$${e^2} - 3$$

#### Work Step by Step

\eqalign{ & \iint\limits_R {x{e^{xy}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 2,\,\,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr & or \cr & \iint\limits_R {x{e^{xy}}}dydx \cr & \cr & {\text{replacing the limits for the region }}R \cr & = \int_0^2 {\int_0^1 {x{e^{xy}}dy} dx} \cr & = \int_0^2 {\left[ {\int_0^1 {x{e^{xy}}dy} } \right]dx} \cr & {\text{solve the inner integral, treating }}x{\text{ as a constant}} \cr & = \int_0^1 {x{e^{xy}}dy} \cr & {\text{integrate by using the rule }}\int {{e^u}} du = {e^u} + C \cr & = \left[ {{e^{xy}}} \right]_0^1 \cr & {\text{evaluating the limits in the variable }}y \cr & = {e^{x\left( 1 \right)}} - {e^{x\left( 0 \right)}} \cr & {\text{simplifying}} \cr & = {e^x} - 1 \cr & \cr & \int_0^2 {\left[ {\int_0^1 {x{e^{xy}}dy} } \right]dx} = \int_0^2 {\left( {{e^x} - 1} \right)dx} \cr & {\text{integrating}} \cr & = \left[ {{e^x} - x} \right]_0^2 \cr & {\text{evaluate}} \cr & = \left( {{e^2} - 2} \right) - \left( {{e^0} - 0} \right) \cr & {\text{simplifying}} \cr & = {e^2} - 2 - 1 \cr & = {e^2} - 3 \cr}

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