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 448: 16

Answer

a. 45231.8; b. 41469.1; c. 38121.7;

Work Step by Step

a. $P=\int ^{6}_{0} 8000e^{-0.02t}dt=8000(\frac{e^{-0.02t}}{-0.02})|^{6}_{0} \approx 45231.8$ b. a. $P=\int ^{6}_{0} 8000e^{-0.05t}dt=8000(\frac{e^{-0.05t}}{-0.05})|^{6}_{0} \approx 41469.1$ c. b. a. $P=\int ^{6}_{0} 8000e^{-0.08t}dt=8000(\frac{e^{-0.08t}}{-0.08})|^{6}_{0} \approx 38121.7$
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