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: 65

Answer

$$\frac{{32}}{3}$$

Work Step by Step

$$\eqalign{ & \int_1^{{e^2}} {\frac{{{{\left( {\ln x} \right)}^5}}}{x}dx} \cr & {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr & {\text{express the limits in terms of }}u \cr & x = {e^2}{\text{ implies }}u = \ln \left( {{e^2}} \right) = 2 \cr & x = 1{\text{ implies }}u = \ln \left( 1 \right) = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_1^{{e^2}} {\frac{{{{\left( {\ln x} \right)}^5}}}{x}dx} = \int_0^2 {{u^5}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{{{u^6}}}{6}} \right)} \right|_0^2 \cr & {\text{evaluate limits}} \cr & = \frac{{{2^6}}}{6} - \frac{{{0^6}}}{6} \cr & {\text{simplify}} \cr & = \frac{{64}}{6} \cr & = \frac{{32}}{3} \cr} $$
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.