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# Reading: Continuous Income Stream

In precalculus, you learned about compound interest in that really simple situation where you made a single deposit into an interest-bearing account and let it sit undisturbed, earning interest, for some period of time.

## Compound Interest Formulas

Let P = the principal (initial investment), r = the annual interest rate expressed as a decimal, and let A(t) be the amount in the account at the end of t years.
• Compounding n times per year: $A(t) = P(1 + \frac{r}{n})^{nt}$
• Compounded continuously: $A(t) = Pe^{rt}$
If you’re using this formula to find what an account will be worth in the future, t > 0 and A(t) is called the future value. If you’re using the formula to find what you need to deposit today to have a certain value P sometime in the future, t < 0 and A(t) is called the present value. You may also have learned somewhat more complicated annuity formulas to deal with slightly more complicated situations—where you make equal deposits equally spaced in time. But real life is not usually so neat. Calculus allows us to handle situations where “deposits” are flowing continuously into an account that earns interest. As long as we can model the flow of income with a function, we can use a definite integral to calculate the present and future value of a continuous income stream. The idea is that each little bit of income in the future needs to be multiplied by the exponential function to bring it back to the present, and then we’ll add them all up (a definite integral).

## Continuous Income Stream

Suppose money can earn interest at an annual interest rate of r, compounded continuously. Let F(t) be a continuous income function (in dollars per year), that applies between year 0 and year T. Then the present value of that income stream is given by $PV = \int_{0}^{T}F(t)e^{-rt}dt$. The future value can be computed by the ordinary compound interest formula $FV = PVe^{rt}$. This is a useful way to compare two investments—find the present value of each to see which is worth more today.

### Example

You have an opportunity to buy a business that will earn $75,000 per year continuously over the next eight years. Money can earn 2.8% per year, compounded continuously. Is this business worth its purchase price of$630,000?

First, please note that we still have to make some simplifying assumptions. We have to assume that the interest rates are going to remain constant for that entire eight years. We also have to assume that the $75,000 per year is coming in continuously, like a faucet dripping dollars into the business. Neither of these assumptions might be accurate. But moving on: The present value of the$630,000 is . . . $630,000. This is one investment, where we put our$630,000 in the bank and let it sit there. To find the present value of the business, we think of it as an income stream. The function F(t) in this case is $75,000 dollars per year, r = .028, and T = 8: [latex-display] PV = \int_{0}^{8}7500e^{-.028t}dt \cong 672,511.66 [/latex-display] The present value of the business is about$672,500, which is more than the $630,000 asking price—this is a good deal. I used technology to compute the value of this definite integral. For many of the integrals in this section, you won’t be able to use antiderivatives. But technology will work quickly, and it will give you an answer that’s good enough. ### Example A company is considering purchasing a new machine for its production floor. The machine costs$65,000. The company estimates that the additional income from the machine will be a constant $7000 for the first year, then will increase by$800 each year after that. In order to buy the machine, the company needs to be convinced that it will pay for itself by the end of 8 years with this additional income. Money can earn 1.7% per year, compounded continuously. Should the company buy the machine?

#### Solution

Assumptions, assumptions. We’ll assume that the income will come in continuously over the 8 years. We’ll also assume that interest rates will remain constant over that 8-year time period. We’re interested in the present value of the machine, which we will compare to its $65,000 price tag. Let t be the time, in years, since the purchase of the machine. The income from the machine is different depending on the time: From t = 0 to t = 1 (the first year), the income is constant$7000 per year. From t = 1 to t = 8, the income is increasing by \$800 each year; the income flow function F(t) will be $F(t) = 7000 + 800(t-1)=6200+800t$. To find the present value, we’ll have to divide the integral into the two pieces, one for each of the functions: $PV = \int_{0}^{1}7000e^{-0.017t}dt + \int_{1}^{8}(6200+800t)e^{0.017t}dt \cong 70166$ (Again, I used technology to evaluate these integrals. This is an example where you can’t use antiderivatives.) The present value is greater than the cost of the machine, so the company should buy the machine.