Uploaded by TheIntegralCALC on 23.08.2010

Transcript:

Hi everyone. Welcome back to integralCALC.com. 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

solve.

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

solve.

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

calculator.

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.

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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

solve.

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

solve.

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

calculator.

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.

1