Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 63

Answer

$$\frac{{{3^{1 + \ln 2}} - 1}}{{\ln 3}}$$

Work Step by Step

$$\eqalign{ & \int_1^{2e} {\frac{{{3^{\ln x}}}}{x}} dx \cr & {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr & {\text{express the limits in terms of }}u \cr & x = 2e{\text{ implies }}u = \ln \left( {2e} \right) = 1 + \ln 2 \cr & x = 1{\text{ implies }}u = \ln \left( 1 \right) = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_1^{2e} {\frac{{{3^{\ln x}}}}{x}} dx = \int_0^{1 + \ln 2} {{3^u}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{{{3^u}}}{{\ln 3}}} \right)} \right|_0^{1 + \ln 2} \cr & {\text{evaluate limits}} \cr & = \frac{{{3^{1 + \ln 2}}}}{{\ln 3}} - \frac{{{3^0}}}{{\ln 3}} \cr & {\text{simplify}} \cr & = \frac{{{3^{1 + \ln 2}} - 1}}{{\ln 3}} \cr} $$
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