Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 88

Answer

$$2\ln 2 - \frac{3}{2}$$

Work Step by Step

$$\eqalign{ & \int_1^2 {\left( {\frac{2}{s} - \frac{4}{{{s^3}}}} \right)ds} \cr & {\text{write }}\frac{4}{{{s^3}}}{\text{with negative exponent}} \cr & = \int_1^2 {\left( {\frac{2}{s} - 4{s^{ - 3}}} \right)ds} \cr & {\text{integrate using the logarithmic rule and the power rule for integration}} \cr & = \left. {\left( {2\ln \left| s \right| - 4\left( {\frac{{{s^{ - 2}}}}{{ - 2}}} \right)} \right)} \right|_1^2 \cr & = \left. {\left( {2\ln \left| s \right| + \frac{2}{{{s^2}}}} \right)} \right|_1^2 \cr & {\text{using the fundamental theorem}} \cr & = \left( {2\ln \left| 2 \right| + \frac{2}{{{{\left( 2 \right)}^2}}}} \right) - \left( {2\ln \left| 1 \right| + \frac{2}{{{{\left( 1 \right)}^2}}}} \right) \cr & = 2\ln 2 + \frac{1}{2} - 2 \cr & = 2\ln 2 - \frac{3}{2} \cr} $$
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.