## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 37

#### Answer

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

#### Work Step by Step

\eqalign{ & \int_{ - 3}^{ - 2} {\frac{{dx}}{x}} \cr & {\text{integrate using the rule }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C{\text{; then}}{\text{,}} \cr & \int_{ - 3}^{ - 2} {\frac{{dx}}{x}} = \left( {\ln \left| x \right|} \right)_{ - 3}^{ - 2} \cr & {\text{use the fundamental theorem of calculus }} \cr & \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\left( {{\text{see page 281}}} \right) \cr & {\text{then}}{\text{,}} \cr & = \ln \left| { - 2} \right| - \ln \left| { - 3} \right| \cr & {\text{simplifying}} \cr & = \ln 2 - \ln 3 \cr & {\text{use the quotient property for logarithms}} \cr & = \ln \left( {\frac{2}{3}} \right) \cr}

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