Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - Review Exercises: 57

Answer

$$\ln 4$$

Work Step by Step

$$\eqalign{ & \int_{{e^2}}^{{e^8}} {\frac{{dx}}{{x\ln x}}} \cr & {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr & {\text{express the limits in terms of }}u \cr & x = {e^2}{\text{ implies }}u = \ln {e^2} = 2 \cr & x = {e^8}{\text{ implies }}u = \ln {e^8} = 8 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{{e^2}}^{{e^8}} {\frac{{dx}}{{x\ln x}}} = \int_2^8 {\frac{1}{u}} du \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\ln \left| u \right|} \right)} \right|_2^8 \cr & {\text{use the fundamental theorem}} \cr & = \ln \left| 8 \right| - \ln \left| 2 \right| \cr & {\text{properties of logarithms}} \cr & = \ln \left( {\frac{8}{2}} \right) \cr & = \ln 4 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.