## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 46

#### Answer

$$\frac{{{e^5}}}{5} - \frac{{{e^3}}}{3} + \frac{2}{{15}}$$

#### Work Step by Step

\eqalign{ & \int_0^1 {\int_{2x}^{4x} {{e^{x + y}}} dydx} \cr & = \int_0^1 {\left[ {\int_{2x}^{4x} {{e^{x + y}}} dy} \right]dx} \cr & {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr & = \int_{2x}^{4x} {{e^{x + y}}} dy \cr & {\text{integrate}} \cr & = \left( {{e^{x + y}}} \right)_{2x}^{4x} \cr & {\text{evaluating the limits in the variable }}y \cr & = {e^{x + 4x}} - {e^{x + 2x}} \cr & = {e^{5x}} - {e^{3x}} \cr & \cr & \int_0^1 {\left[ {\int_{2x}^{4x} {{e^{x + y}}} dy} \right]dx} = \int_0^1 {\left( {{e^{5x}} - {e^{3x}}} \right)dx} \cr & {\text{integrating by using }}\int {{e^{ax}}} dx = \frac{1}{a}{e^{ax}} \cr & = \left( {\frac{{{e^{5x}}}}{5} - \frac{{{e^{3x}}}}{3}} \right)_0^1 \cr & {\text{evaluate}} \cr & = \left( {\frac{{{e^{5\left( 1 \right)}}}}{5} - \frac{{{e^{3\left( 1 \right)}}}}{3}} \right) - \left( {\frac{{{e^{5\left( 0 \right)}}}}{5} - \frac{{{e^{3\left( 0 \right)}}}}{3}} \right) \cr & {\text{simplifying}} \cr & = \frac{{{e^5}}}{5} - \frac{{{e^3}}}{3} - \frac{1}{5} + \frac{1}{3} \cr & = \frac{{{e^5}}}{5} - \frac{{{e^3}}}{3} + \frac{2}{{15}} \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.