Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 70

Answer

$$2$$

Work Step by Step

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