Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises - Page 446: 67



Work Step by Step

Evaluate the integral $\int_1^2\frac{dt}{8-3t}$. $\int_1^2\frac{dt}{8-3t}=ln[\frac{8-3t}{-3}]_1^2$ $=-\frac{1}{3}[ln2-ln5]$ Thus, $=\frac{1}{3}[ln5-ln2]$ Hence, $\int_1^2\frac{dt}{8-3t}=\frac{1}{3}[ln(\frac{5}{2})]$
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