Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - 8.3 Continuous Money Flow - 8.3 Exercises - Page 447: 1


a.$ \approx 6883.4$; b. $\approx 15319.26$

Work Step by Step

a. $P=\int ^{10}_{0} 1000e^{-0.08t}=1000\frac{e^{-0.08t}}{-0.08}|^{10}_{0}=-12500(e^{-0.8}-e^{0})=6883.39 \approx 6883.4$ b. $A=e^{(0.08)10}\int^{10}_{0}1000e^{-0.08t}dt = e^{0.8}.1000\int^{10}_{0}e^{-0.08t}dt=e^{0.8}.1000.\frac{1}{-0.08}(e^{-0.08t}|^{10}_{0}) \approx 15319.26$
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.