Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.5 Compound Interest and Present Value - Exercises - Page 356: 20



Work Step by Step

Since \begin{align*} \begin{aligned} \left(-\frac{e^{-r t}(1+r t)}{r^{2}}+C\right)^{\prime} &=\frac{-1}{r^{2}}\left(-r \cdot e^{-r t}(1+r t)+e^{-r t}(0+r)\right)\\ &=\frac{-1}{r^{2}}\left(-r e^{-r t}-r^{2} t e^{-r t}+r e^{-r t}\right) \\ &=\frac{-1}{r^{2}}\left(-r^{2} t e^{-r t}\right) \\ &=t e^{-r t} \end{aligned} \end{align*} Then $$\int t e^{-r t} d t=-\frac{e^{-r t}(1+r t)}{r^{2}}+C$$ Since for $$ R(t)=5000+1000 t, \ \ \ \ \ T=5,\ \ \ \ r=0.05$$ Then \begin{aligned} P V &=\int_{0}^{T} R(t) e^{-r t} d t \\ &=\int_{0}^{5}(5000+1000 t) e^{-0.05 t} d t \\ &=5000 \int_{0}^{5} e^{-0.05 t} d t+1000 \int_{0}^{5} t e^{-0.05 t} d t\ \ \text{integrate by parts }\\ &= \left.\frac{5000}{-0.05} e^{-0.05 t}\right|_{0} ^{5}+ \left.1000\left(-\frac{e^{-0.05 t}(1+0.05 t)}{0.05^{2}}\right)\right|_{0} ^{5}\\ &= 22119.92+10599.61\\ &=\$32719.53 \end{aligned}
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