Present and Future Value Example 5

Uploaded by TheIntegralCALC on 23.08.2010

Hi everyone. Welcome back to WeÕre going to be doing another present value
problem today. This one is going to be using the formula that IÕve written here on the
board because the problem asks us to suppose that money is deposited steadily into an account
at a constant rate of $15,000 per year for five years.
Find the present value of this income stream if the account pays 7.5 percent interest compounded
continuously. So the problem gives us several clues to know that weÕre using this formula.
The obvious giveaway is that weÕre depositing income into the account steadily at a constant
rate of $15,000 per year and since we have an income stream as opposed to just depositing
money once and then not ever touching it again, we need to use this S of t here which represents
the stream and that automatically comes with the integral here and then of course weÕre
depositing this money steadily and it asks us directly to find the present value.
So we know itÕs the present value formula as opposed to the future value formula with
the similar income stream e to the negative r or itÕs r times capital T minus lowercase
t. So the problem clearly indicates that weÕre supposed to use this formula and letÕs go
ahead and start plugging in values and then all we need to do is obviously simplify and
So present value because it asks us to find the present value. WeÕre depositing money
into the account for five years so weÕre plugging in 5 for capital T because thatÕs
the time that this is all going on. WeÕre depositing $15,000 per year so $15,000 is
our income stream. We obviously leave e alone. Our rate is 7.5 percent interest. So we have
a negative sign here that comes along with this formula.
So negative 0.075 and then t is what we leave alone because we have d lowercase t here so
we know that t is the variable that weÕre going to need to solve for. So weÕve plugged
in all of the information that was given to us in the problem and now we need to take
the integral and solve.
So first of all, to take the integral when we have e to some coefficient multiplied by
our variable, we simply need to take the integral. Divide this coefficient here by the coefficient
in the exponent and then leave this as is.
So our integral is going to look like this. 15,000 over negative 0.075 e and then we leave
this, negative 0.075t and weÕre going to be evaluating that on the range zero to 5.
So we go ahead and put zero to 5 and now what we need to do is plug in 5 and zero here and
So when weÕre evaluating a definite integral like this, weÕre plugging in these two values,
we always plug in the top number first. So, present value equals 15,000 over negative
0.075 e to the negative 0.075 and then plug in 5 here for t. Then from that, we subtract
what we get when we plug in the bottom number.
So 15,000 over negative 0.075 e to the negative 0.075 times zero. So this is how we evaluate
the definite integral. Plug in the top number and subtract. Plug in the bottom number. So
once weÕve done that, we just simplify and I guess one of the easiest ways to do that,
I will just go ahead and simplify a couple of things and then you can plug it to your
First of all, we have minus and negative here so we can take away that negative and make
this a plus and then we have negative 0.075 times zero. So this here will be zero and
anything raised to the zero power is 1. So this is all going to go away here because
thatÕs the same thing as saying that this 15,000 over 0.075 is multiplied by 1 which
is redundant. So we can drop it and so weÕve simplified somewhat and from this point, I
would just use your calculator to solve and the answer that we get is that the present
value is $62,542.14 of a constant, continuous income stream, $15,000 per year deposited
for five years at an interest rate of 7.5 percent compounded continuously.
So thatÕs it. ThereÕs our answer. Hope that helps. I will see you guys next time. Bye.