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Question

Find the future value of an income stream of $2000 per year, deposited into an account paying 2% interest per year, compounded continuously, over a 15-year period.

Solution

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Answered 4 months ago
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The deposit of P(t)ΔtP(t) \Delta t has a time period of (Mt)(M-t) years to earn interest and

Future Value of Money deposited inn interval t to t+Δt[P(t)Δt]er(Mt)t \text { to } t+\Delta t \approx[P(t) \Delta t] e^{r(M-t)}

Summing all over subintervals

 Total future value P(t)Δter(ut) dollars. \text { Total future value } \approx \sum P(t) \Delta t e^{r(u-t)} \text { dollars. }

As the length of the subdivisions tend towards zero

 Future value =0MP(t)e(u1)dt dollars \text { Future value }=\int_{0}^{M} P(t) e^{(u-1)} d t \quad \text { dollars }

In other words

 Future Value =ent. Present Value \text { Future Value }=e^{n t} . \text { Present Value }

Now calculating future value

0101000e0.08(100)dt dollars =1000e0.8010e0.08tdt dollars =1000e0.8[e0.88t0.08]010 dollars =1000e1030.08[e0.81] dollars =12500(2.225541)(10.449329)=$15,319.26\begin{align*} \int_{0}^{10} 1000 e^{0.08(10-0)} d t \text { dollars }&=1000 e^{0.8} \int_{0}^{10} e^{-0.08 t} d t \text { dollars }\\ &=1000 e^{0.8}\left[\frac{-e^{-0.88 t}}{0.08}\right]_{0}^{10} \text { dollars }\\ &=-\frac{1000 e^{103}}{0.08}\left[e^{-0.8}-1\right] \text { dollars }\\ &=12500(2.225541)(1-0.449329)\\ &=\$ 15,319.26 \end{align*}

The future value of income stream is =$15,319.26=\$ 15,319.26

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