## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 38

#### Answer

$$\ln \left( {\frac{2}{5}} \right)$$

#### Work Step by Step

\eqalign{ & \int_{ - 1}^0 {\frac{{3dx}}{{3x - 2}}} \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = 3x - 2,{\text{ so that }}du = 3dx \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ }}u = 3\left( 0 \right) - 2 = - 2 \cr & \,\,\,\,\,\,{\text{If }}x = - 1,{\text{ }}u = 3\left( { - 1} \right) - 2 = - 5 \cr & {\text{write the integral in terms of }}u \cr & \int_{ - 5}^{ - 2} {\frac{{du}}{u}} \cr & {\text{integrate}} \cr & = \left( {\ln \left| u \right|} \right)_{ - 5}^{ - 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 & = \ln \left| { - 2} \right| - \ln \left| { - 5} \right| \cr & {\text{simplifying}} \cr & = \ln 2 - \ln 5 \cr & {\text{use logarithmic property}} \cr & = \ln \left( {\frac{2}{5}} \right) \cr}

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