Application of Integration by Parts

Income Streams

Definition and Notation

An income stream refers to income that is generated continuously and transferred into an account that earns interest at a fixed rate. Interest is assumed to be compounded continuously.

• \(R(t)\) = rate at which income is generated (in dollars per year)

• \(T\) = length of time (in years) of the income stream

• \(r\) = annual interest rate (compounded continuously)

Future Value

The future value of an income stream is the total of all of the money transferred plus all of the interest earned and is given by

\[A = e^{rT}\int_0^T R(t) e^{-rt} ~dt.\]

Present Value

The present value of an income stream is the principal investment, \(PV\), that yields the same accumulated value as the income stream when \(PV\) is invested for a period of \(T\) years at the same interest rate and is given by

\[PV = \int_0^T R(t) e^{-rt} ~dt.\]

Example 1

Compute future and present value of an investment

Suppose an investment is expected to generate income at the rate of

\[R(t) = 5 + 3t\]

thousands of dollars per year for the next 10 years. Find the present and future values from this investment if the prevailing interest rate is 2% per year compounded continuously.

Step 1:   Write the present and future values of the income stream as definite integrals.

\[PV = \int^{10}_0 \left( 5 + 3t \right)e^{-0.02t} ~dt ~~~~~~~ A = e^{0.2}\int^{10}_0 \left( 5 + 3t \right)e^{-0.02t} ~dt\]

Step 2:   Compute \(\displaystyle \int \left( 5 + 3t \right)e^{-0.02t}~dt\) using integration by parts.

Pick \(u\) and \(dv\) and compute \(du\) and \(v\). (Recall \(\int e^{ax} ~dx = \frac{1}{a}e^{ax}+C\).)

\[\begin{align*} u &= 5 + 3t & dv &= e^{-0.02t}~dt \\ du &= \frac{d}{dt}(5 + 3t) ~dt = 3~dt & v &= \int e^{-0.02t}~dt = -\frac{1}{0.02}e^{-0.02t} = -50e^{-0.02t} \end{align*}\]

\[\begin{align*} \int \left( 5 + 3t \right)e^{-0.02t}~dt &= uv-\int v~du\\ &= (5+3t)(-50e^{-0.02t}) - \int (-50 e^{-0.02t})3~dt\\ &= -50(5+3t)e^{-0.02t} + 150\int e^{-0.02t}~dt && \hbox{simplify}\\ &= -50(5+3t)e^{-0.02t} + 150(-50e^{-0.02t}) + C\\ &= -50(155 + 3t)e^{-0.02t}+C \end{align*}\]

Step 3:   Compute the present value using the answer to Step 2.

\[\begin{align*} PV &= \int^{10}_0 \left( 5 + 3t \right)e^{-0.02t}dt\\ &= -50(155 + 3t)e^{-0.02t}\Big|^{10}_{0} && \hbox{using Step 2}\\ &= -50(185) e ^{-0.2} - (-50(155)e^{0})\\ &= -9250e^{-0.2} + 7750 &&\hbox{since $e^0 = 1$}\\ &\approx 176.74053 && \hbox{in thousands of dollars} \end{align*}\]

Therefore, the present value of this income stream is approximately $176,740.53.

Step 4:   Compute the future value using the answer to Step 3.

\[\begin{align*} A &= e^{0.2}\int^{10}_0 \left( 5 + 3t \right)e^{-0.02t}dt\\ &= e^{0.2}(-9250e^{-0.2} + 7750) && \hbox{using Step 3}\\ &= -9250e^{0.2-0.2} + 7750e^{0.2}\\ &= -9250e^{0} + 7750e^{0.2}\\ &= -9250 + 7750e^{0.2} &&\hbox{since $e^0 = 1$}\\ &\approx 215.87138 && \hbox{in thousands of dollars} \end{align*}\]

Therefore, the future value of this income stream is approximately $215,871.38.