## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 513: 17

#### Answer

$${\left( {\ln 3} \right)^2}$$

#### Work Step by Step

\eqalign{ & \int_1^3 {\int_1^3 {\frac{1}{{xy}}} } dydx \cr & = \int_{16}^{25} {\left[ {\int_1^3 {\frac{1}{{xy}}} dy} \right]} dx \cr & {\text{solve the inner integral}} \cr & \int_1^3 {\frac{1}{{xy}}} dy = \frac{1}{x}\int_1^3 {\frac{1}{y}} dy \cr & = \frac{1}{x}\left[ {\ln \left| y \right|} \right]_1^3 \cr & = \frac{1}{x}\left( {\ln 3 - \ln 1} \right) \cr & = \frac{{\ln 3}}{x} \cr & {\text{then}} \cr & \int_1^3 {\int_1^3 {\frac{1}{{xy}}} } dydx = \int_1^3 {\frac{{\ln 3}}{x}} dx \cr & {\text{integrating}} \cr & = \ln 3\left[ {\ln \left| x \right|} \right]_1^3 \cr & = \ln 3\left( {\ln 3 - \ln 1} \right) \cr & = {\left( {\ln 3} \right)^2} \cr}

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