Problem 35
Question
It is estimated that \(t\) years from now, a certain investment will be generating income at the rate of \(f(t)=8,000+400 t\) dollars per year. If the income is generated in perpetuity and the prevailing annual interest rate remains fixed at \(5 \%\) compounded continuously, find the present value of the investment.
Step-by-Step Solution
Verified Answer
The present value of the investment is $320,000.
1Step 1: Identify the given rate function
Given the rate function for the income generated by the investment: \[ f(t) = 8,000 + 400t \]
2Step 2: Interest rate information
The prevailing annual interest rate is given as 5% compounded continuously. This means the interest rate can be represented as a continuous compounding rate, which is: \[ r = 0.05 \]
3Step 3: Present Value (PV) formula for continuous income
The present value function for a continuous income stream is given by: \[ PV = \int_{0}^{\infty} e^{-rt} f(t) \, dt \]
4Step 4: Substitute the given functions into the PV formula
Now, substitute the given values: \[ PV = \int_{0}^{\infty} e^{-0.05 t} (8,000 + 400 t) \, dt \]
5Step 5: Split the integral
The integral can be split into two simpler integrals: \[ PV = \int_{0}^{\infty} e^{-0.05 t} \, 8,000 \, dt + \int_{0}^{\infty} e^{-0.05 t} \, 400 t \, dt \]
6Step 6: Evaluate the first integral
Evaluate the first integral: \[ \int_{0}^{\infty} e^{-0.05 t} \, 8,000 \, dt \] Using the formula for the integral of the exponential function: \[ \int e^{at} \, dt = \frac{e^{at}}{a} \] we get: \[ \int_{0}^{\infty} e^{-0.05 t} \, 8,000 \, dt = \frac{8,000}{0.05} = 160,000 \]
7Step 7: Evaluate the second integral
Evaluate the second integral: \[ \int_{0}^{\infty} e^{-0.05 t} \, 400 t \, dt \] Using the formula for integration by parts: \[ \int u \, dv = uv - \int v \, du \] Let: \[ u = 400t \quad du = 400 \, dt \] \[ dv = e^{-0.05t} \, dt \quad v = \frac{e^{-0.05t}}{-0.05} \] Applying integration by parts, we get: \[ uv - \int v \, du = 400t \frac{e^{-0.05t}}{-0.05} \bigg|_{0}^{\infty} - \int_{0}^{\infty} \frac{400}{-0.05} e^{-0.05t} \, dt \] \[ = \frac{-400}{0.05} t e^{-0.05t} \bigg|_{0}^{\infty} + 8,000 \frac{1}{0.05} \] \[ = 0 + 160,000 \]
8Step 8: Add the results of the integrals
Sum the results of both integrals: \[ PV = 160,000 + 160,000 = 320,000 \]
Key Concepts
continuous compoundingexponential function integrationintegration by parts
continuous compounding
Continuous compounding is a method where the investment's interest is calculated and added back to the principal at an infinitely small interval.
Unlike regular compounding periods such as annually, quarterly, or monthly, continuous compounding uses the mathematical constant **e**.
This is particularly useful for modeling scenarios where growth needs to be as accurate as possible.
In our problem, the interest rate is compounded continuously, represented using the formula:
\[e^{rt}\]
Here, **r** is the interest rate and **t** is time.
For an interest rate of 5%, the continuous compounding factor becomes \[e^{-0.05t}\] as used in our present value calculation.
This ensures that every tiny fraction of time is considered for compounding interest.
Unlike regular compounding periods such as annually, quarterly, or monthly, continuous compounding uses the mathematical constant **e**.
This is particularly useful for modeling scenarios where growth needs to be as accurate as possible.
In our problem, the interest rate is compounded continuously, represented using the formula:
\[e^{rt}\]
Here, **r** is the interest rate and **t** is time.
For an interest rate of 5%, the continuous compounding factor becomes \[e^{-0.05t}\] as used in our present value calculation.
This ensures that every tiny fraction of time is considered for compounding interest.
exponential function integration
When working with continuous streams of income, integrating exponential functions becomes extremely useful.
The exponential function often appears in finance due to the nature of continuous compounding interest.
One key integral formula that was used in this problem is: \[ \int e^{at} \, dt = \frac{e^{at}}{a} \] This formula helps simplify the calculation of present values.
For example, in our exercise, to integrate \[\int_{0}^{\infty} e^{-0.05t} \, dt \] for the constant income portion, we employed this exponential function integral.
By substituting *a* with *-0.05*, we obtained: \[ \int_{0}^{\infty} e^{-0.05t} \, 8,000 \, dt = \frac{8,000}{0.05} = 160,000 \] Knowing how to integrate these functions allows us to find the present value of perpetually growing investments.
The exponential function often appears in finance due to the nature of continuous compounding interest.
One key integral formula that was used in this problem is: \[ \int e^{at} \, dt = \frac{e^{at}}{a} \] This formula helps simplify the calculation of present values.
For example, in our exercise, to integrate \[\int_{0}^{\infty} e^{-0.05t} \, dt \] for the constant income portion, we employed this exponential function integral.
By substituting *a* with *-0.05*, we obtained: \[ \int_{0}^{\infty} e^{-0.05t} \, 8,000 \, dt = \frac{8,000}{0.05} = 160,000 \] Knowing how to integrate these functions allows us to find the present value of perpetually growing investments.
integration by parts
Integration by parts is a technique used to integrate products of functions.
It's particularly useful when working with functions that are difficult to integrate directly.
The formula for integration by parts is: \[\int u \, dv = uv - \int v \, du \] To apply this, we need to identify parts of our integral as *u* and *dv*.
In our exercise, we integrated \[ \int_{0}^{\infty} e^{-0.05t} \, 400t \, dt \] by setting: *u = 400t* *dv = e^{-0.05t} \, dt* This gives: *du = 400 \, dt* and *v = \frac{e^{-0.05t}}{-0.05}*.
Substituting these into the integration by parts formula, we get:
\[ 400t \frac{e^{-0.05t}}{-0.05} \bigg|_{0}^{\infty} - \int_{0}^{\infty} \frac{400}{-0.05} e^{-0.05t} \, dt \]
Evaluating the definite integral results in simplifying complex financial calculations, as seen in the final step where both parts were added to yield the present value. By understanding and applying this method, you can handle intricate integrals involving products of polynomial and exponential functions.
It's particularly useful when working with functions that are difficult to integrate directly.
The formula for integration by parts is: \[\int u \, dv = uv - \int v \, du \] To apply this, we need to identify parts of our integral as *u* and *dv*.
In our exercise, we integrated \[ \int_{0}^{\infty} e^{-0.05t} \, 400t \, dt \] by setting: *u = 400t* *dv = e^{-0.05t} \, dt* This gives: *du = 400 \, dt* and *v = \frac{e^{-0.05t}}{-0.05}*.
Substituting these into the integration by parts formula, we get:
\[ 400t \frac{e^{-0.05t}}{-0.05} \bigg|_{0}^{\infty} - \int_{0}^{\infty} \frac{400}{-0.05} e^{-0.05t} \, dt \]
Evaluating the definite integral results in simplifying complex financial calculations, as seen in the final step where both parts were added to yield the present value. By understanding and applying this method, you can handle intricate integrals involving products of polynomial and exponential functions.
Other exercises in this chapter
Problem 33
Either evaluate the given improper integral or show that it diverges. $$ \int_{-\infty}^{\infty} x^{3} e^{-x^{2}} d x $$
View solution Problem 34
Either evaluate the given improper integral or show that it diverges. $$ \int_{-\infty}^{\infty}\left(e^{x}+e^{-x}\right) d x $$
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