We are continuing our study of the applications of integration. Today, we'll be looking at future
value and present value. Our goals are to solve continuous money flow problems,
find present value and future value, and find the accumulated present value of an investment.
Earlier in the semester and in your previous algebra class, you considered a one-time deposit
into an account that paid continuous interest for a set time. You may recall this given formula.
Well, now we want to consider a situation where we have a continuous money flow rather than a
one-time deposit over a period of time earning continuous interest. We will call this the future
value of a continuous money flow. If the yearly flow of money into an investment is given by
some function r of t, then the future value of the continuous money flow at interest rate k
compounded continuously over t years is given by this definite integral. Now remember, r of
t is the money flow per year. Occasionally, r of t will be referred to as p sub zero or p naught.
Find the future value of a continuous money flow of one thousand dollars per year
being invested at six percent continuous interest for eight years.
The future value will be equal to the integral from zero to eight because it's eight years.
r of t is the amount put into the account per year, one thousand dollars
times e raised to the power of k times t, which is zero point zero six t dt.
Now in order to evaluate this integral, you may recall that you would have to do a u substitution.
The value of u would be zero point zero six t, so d u would be equal to zero point zero six dt.
Now the way I like to show this is I would divide both sides of this by zero point zero six,
therefore I can replace dt with du divided by zero point zero six.
Let's go ahead and show the substitution. The next step I would factor out the one thousand
and I can rewrite this as e to the power of u. Now when I replace dt, I replace it with du
divided by zero point zero six, so I'd put the du here and I would factor out the
zero point zero six there. Now we can go ahead and integrate.
Of course, the antiderivative of e to the u is e to the u, but u is technically zero point zero
six t, and we need to evaluate this at eight and zero. Let's go ahead and show another step.
So first, we replace t with eight, then replace t with zero.
Remember, e to the zero would be one. I've already calculated this to save some time. We will check
it on the calculator in a moment. Ten thousand two hundred sixty-seven dollars and ninety-one cents.
Let's check it now. Remember to evaluate a definite integral. If you hit math, option nine,
that brings up the function integration command and we can type in our integrand.
Now I'm going to use x instead of t just for convenience.
And that verifies our answer is correct. Okay, let's go ahead and take a look at another example.
What should the continuous cash flow be per year, sometimes referred to as p sub zero or r of t,
so that the continuous money flow over ten years at an interest rate of eight percent will be ten
thousand dollars? So in this problem, they're telling us we want ten thousand dollars in the
future. How much should we put into the account per year? So we're actually trying to find r of t.
Okay, we need to evaluate this definite integral which again is going to require
a u substitution, very similar to the last example.
So let's go ahead and show our substitution.
Again, we can factor out the r of t.
Now remember that when we replace d of t,
we're going to get a d u divided by zero point zero eight, which I can factor out.
Okay, to clean this up, what I might do now is multiply both sides by
zero point zero eight. That would give us eight hundred.
Is equal to r of t times the integral of e to the u d u.
Now we'll integrate. Eight hundred would equal r of t
times the antiderivative of e to the u, which is e to the u,
which is actually e to the power of zero point zero eight t evaluated at ten and zero.
Okay, now I've already evaluated this to save a little bit of time. Again,
we get eight hundred is equal to r of t times one point two two five five.
And if we divide both sides by that value,
we would have our annual money flow and r of t
is equal to six hundred fifty-two dollars and eighty-five cents, which means if we deposit
six hundred fifty-two dollars and eighty-five cents into an account every year over ten years
and it earns eight percent continuous interest, we will end up with ten thousand dollars.