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: 3

Answer

a. 3441.7; b. $7659.63

Work Step by Step

a. $P=\int ^{10}_{0} 500e^{-0.08t}=500\frac{e^{-0.08t}}{-0.08}|^{10}_{0}=-6250(e^{-0.8}-1)=3441.7 $ b. $A=e^{(0.08)10}\int^{10}_{0}500e^{-0.08t}dt = e^{0.8}.500\int^{10}_{0}e^{-0.08t}dt=e^{0.8}.500.\frac{1}{-0.08}(e^{-0.08t}|^{10}_{0}) \approx 7659.63$
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