Section 9.6: Applications of definite integrals
Income streams

If at time $t$ the rate of income generated by some enterprise is given by the value of the function $f(t)$, then the total income generated between time $a$ and time $b$ is $\int_a^b f(t) dt$.

Valuation of future payments

What value should one assign to an expected future payment?

Review of continuously compounded interest

Recall that if one has a principal of $P$ which earns continuously compounded interest at a rate of $r$, then after $t$ years, the investment would be worth $P e^{rt}$.

Present value

The present value of a payment of $$A$ made $t$ years in the future is the amount $P$ for which with a principal of $$P$ dollars invested for $t$ years with continuously compounded interest of rate $r$ one would earn $$A$.

From the formula for continuously compounded interest, we conclude that $P e^{rt} = A$ so that $P = A e^{-rt}$.

Valuation of an income stream

Suppose that some enterprise produces income at a steady rate of $$A$ per year. Of course, this income stream over the next $T$ years will produce $$AT$, but how much is it worth in present dollars?

A solution

We may approximate the continous income stream as one that is paid in discrete increments.

Suppose that between now and $T$ years from now $N$ payments are made at uniform intervals. Then, the length of time between each payment is $\Delta = \frac{T}{N}$ years.

The $i^\text{th}$ payment of $A \Delta$ is made at time $i \Delta$. As such, if we assume an interest rate of $r$, it has a present value of $(A \Delta) e^{i \Delta}$.

So, the sum of the present values is $\sum_{i=1}^N A e^{-i \Delta r} \Delta$.

Solution, continued

The expression

$$\sum_{i=1}^N A e^{-i \Delta r} \Delta$$

is the right-hand approximation to

$$\int_0^T A e^{-rt} dt$$

Example

Assuming an interest rate of $5$%, compute the present value of a constant income stream of $$100,000$ per year for $10$ years.

Solution

$$\begin{eqnarray*} \int_0^{10} 100,000 e^{-(0.05)t} dt& = & -2,000,000 e^{-(0.05)... ...000,000 e^{-\frac{1}{2}} + 2,000,000 \\ & \approx & 786,938.68 \end{eqnarray*}$$

Valuation of fluctuating income streams

If the income stream varies as a function of time, so that at time $t$, $A(t)$ is the rate at which the payments are made, and the interest rate also varies (possibly) as a function of time, given by the function $r(t)$, then the present value of the income stream over the next $T$ years is

$$\int_0^T A(t) e^{-tr(t)} dt$$

Example

Suppose the interest rate is constantly $5$% and the income stream is given by the function $A(t) = 1000 + 50t$. What is the present value of this income stream over the next $10$ years?

Solution

$$\begin{eqnarray*} \int_0^{10} (1000 + 50t) e^{-t(0.05)}dt & = & 1000 \int_0^{10} e^{\frac{-t}{20}} dt + 50 \int_{0}^{10} t e^{\frac{-t}{20}} dt \end{eqnarray*}$$

We compute that

$$\begin{eqnarray*} 1000 \int_0^{10} e^{\frac{-t}{20}} dt & = & -20,000 e^{-\frac{... ... = & -20,000 e^{\frac{-1}{2}} + 20,000 \\ & \approx & 7,869.38 \end{eqnarray*}$$

Integrating by parts, with $u = t$ and $dv = e^{\frac{-t}{20}} dt$, so that $du = dt$ and $v = -20 e^{\frac{-t}{20}}$.

So,

$$\begin{eqnarray*} \int t e^{\frac{-t}{20}} dt & =& -20 t e^{\frac{-t}{20}} + 20 ... ...dt \\ &=& -20 t e^{\frac{-t}{20}} - 400 e^{\frac{-t}{20}} + C \end{eqnarray*}$$

Thus, $50 \int_{0}^{10} t e^{\frac{-t}{20}} = -30,000 e^{\frac{-1}{2}} + 20,000 \approx 1804.08$.

So, the total present value is $\approx \$ 1804.08 + \$ 7869.38 = \$ 9673.46$.