Problem 18 - Calculus and its applications

Question

Find the accumulated present value of each continuous income stream at rate \(R(t),\) for the given time \(T\) and interest rate \(k\) compounded continuously. $$R(t)=\$ 6400 t, \quad T=20 \mathrm{yr}, \quad k=3.8 \%$$

Step-by-Step Solution

Answer

The accumulated present value is approximately $246,975.

Step 1: Understand the Formula

The present value of a continuous income stream that generates income at a rate \( R(t) \), over a period \( T \), with a continuous compound interest rate \( k \), is given by the formula \( PV = \int_{0}^{T} R(t) e^{-kt} \, dt \).

Step 2: Substitute Known Values

Substitute the given values into the formula. Here, \( R(t) = 6400t \), \( T = 20 \), and \( k = 0.038 \). The integral becomes \( \int_{0}^{20} 6400t e^{-0.038t} \, dt \).

Step 3: Integration by Parts

Use integration by parts to solve the integral \( \int u \, dv = uv - \int v \, du \). Let \( u = 6400t \) and \( dv = e^{-0.038t} \, dt \). Then, \( du = 6400 \, dt \) and \( v = -\frac{1}{0.038} e^{-0.038t} \).

Step 4: Apply Integration by Parts

Apply integration by parts: \( \int_{0}^{20} 6400t e^{-0.038t} \, dt = \left[ -6400t \frac{1}{0.038} e^{-0.038t} \right]_0^{20} + 6400 \int_{0}^{20} \frac{1}{0.038} e^{-0.038t} \, dt \).

Step 5: Evaluate the Definite Integral

Calculate the definite integral: \( \left[ -6400t \frac{1}{0.038} e^{-0.038t} \right]_0^{20} = -6400 \frac{1}{0.038} \left[ 20 e^{-0.038 \times 20} - 0 \right] \). Simplify this expression and compute the result.

Step 6: Compute the Remaining Integral

Calculate the remaining integral: \( 6400 \int_{0}^{20} \frac{1}{0.038} e^{-0.038t} \, dt = \left[ -\frac{6400}{(0.038)^2} e^{-0.038t} \right]_0^{20} \). Evaluate this expression by substituting the limits.

Step 7: Final Calculation

Combine both evaluated parts to get the present value. First, substitute the calculated values back into the expressions from steps 5 and 6, evaluate each, and sum them up to find the total accumulated present value.

Key Concepts

Integration by Parts

Integration by parts is a technique from calculus used to evaluate integrals where the integrand is the product of two functions. The classic formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]This method is particularly useful when you have an integrand that is difficult to solve directly. In this problem, we're looking at the integral \( \int_{0}^{20} 6400t e^{-0.038t} \, dt \), which can be tackled by integration by parts.
Let's break it down:

• **Choose ** \( u = 6400t \) and **thus** \( dv = e^{-0.038t} \, dt \).
• **Differentiate** to find \( du = 6400 \, dt \).
• **Integrate** \( dv \) to find \( v = -\frac{1}{0.038} e^{-0.038t} \).

Now substitute into the formula, using the limits of integration from 0 to 20. This helps to transform the complex integral into simpler parts:
\[ \int u \, dv = -\frac{6400}{0.038} t e^{-0.038t} \bigg|_0^{20} + \frac{6400}{0.038} \int_{0}^{20} e^{-0.038t} \, dt \]

Continuous Compounding

Continuous compounding refers to the concept where an investment's interest is calculated and added back to the principal balance continuously, at every conceivable moment of time. This results in the maximum possible return from interest, as even the smallest amounts of interest begin earning further interest immediately.
With continuous compounding, we use the formula \( A = Pe^{rt} \), where:

• \( A \) is the amount of money accumulated after a certain time.
• \( P \) is the principal amount.
• \( r \) is the annual interest rate.
• \( t \) is the time the money is invested for.

In present value problems like this one, continuous compounding affects how we discount future income streams to their present value.
The integration method accounts for the continuous compounding by using the exponential function \( e^{-kt} \) in the integral, which discounts the income stream \( R(t) \) at any given time \( t \). In our example, \( k \) is 3.8%, thus \( k = 0.038 \). Adjusting for continuous compounding ensures the present value reflects the most accurate worth of future payments as of today.

Definite Integral Evaluation

Evaluating a definite integral involves computing the integral of a function over a specific interval, producing a numerical value rather than a function. In this exercise, we begin with the expression:\[ \int_{0}^{20} 6400t e^{-0.038t} \, dt \]The procedure takes into account the limits of integration, transforming the integrand correctly via integration by parts and evaluating the limits to find a specific numerical value.
First, you use integration by parts to split the integral into more manageable components. Once the computation is set up, the evaluation involves substituting the upper and lower bounds into the antiderivative.

• **Integrate** within the bounds from 0 to 20 using the transformations provided by integration by parts.
• Carefully **substitute** these limits back into separate parts of the integral.

This integral evaluates to quantify the total accumulated present value of a continuous income stream over the time given. By processing through the upper bound (20 years in this case) and lower bound (at year 0), you can determine the amount that represents this income's present value, fully integrating both mathematical and economic concepts.