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: 47

Answer

$$ \left( f(x)=25,000, \quad r =12, \quad 10 \% \right) $$ If $f(t)=25,000$ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=12$ years, then the present value is $$ \begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\ &=\int_{0}^{12} 25000 e^{-0.1 t} d t \\ & \approx \$ 174701.45 \end{aligned} $$

Work Step by Step

$$ \left( f(x)=25,000, \quad r =12, \quad 10 \% \right) $$ 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 $f(t)=25,000$ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=12$ years, then the present value is $$ \begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\ &=\int_{0}^{12} 25000 e^{-0.1 t} d t \\ &=\left.\frac{25,000}{-0.1} e^{-0.1 t}\right|_{0} ^{12} \\ &=250000(-e^{-01.2}+1) \\ &=250000(-0.3012+1) \\ & \approx \$ 174701.45 \end{aligned} $$
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.