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 - Chapter Review - Review Exercises - Page 456: 46

Answer

$$ \left( f(x)=5000, \quad r =8, \quad 9 \% \right) $$ If $f(t)=5000$ is the rate of continuous money flow at an interest rate $r=0.09$ for $T=8$ years, then the present value is $$ \begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\ &=\int_{0}^{8} 5000 e^{-0.09 t} d t \\ & \approx \$ 28,513.76 \end{aligned} $$

Work Step by Step

$$ \left( f(x)=5000, \quad r =8, \quad 9 \% \right) $$ If $f(t)$ is the rate of continuous money flow at an interest rate $r$ for $T$ years, then the present value is $$ P=\int_{0}^{T} f(x) e^{-rt}dt. $$ If $f(t)=5000$ is the rate of continuous money flow at an interest rate $r=0.09$ for $T=8$ years, then the present value is $$ \begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\ &=\int_{0}^{8} 5000 e^{-0.09 t} d t \\ &=\left.\frac{5000}{-0.09} e^{-0.09 x}\right|_{0} ^{8} \\ &=\frac{-5000}{0.09} e^{-0.72}+\frac{5000}{0.09}\\ & \approx \$ 28,513.76 \end{aligned} $$
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