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: 9


a. 29.88; b. $66.5

Work Step by Step

a. $P=\int ^{10}_{0} 25t.e^{-0.08t}=25\int ^{10}_{0} t.e^{-0.08t}=25 (-12.5te^{-0.08t}-156.25e^{-0.08t})|^{10}_{0} \approx 29.88$ b. $A=e^{(0.08)10}\int^{10}_{0}25te^{-0.08t}dt = e^{0.8}.25\int^{10}_{0}te^{-0.08t}dt=e^{0.8}.25.(-12.5te^{-0.08t}-156.25e^{-0.08t}|^{10}_{0}) \approx 66.5$
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