Problem 111 - Applied Calculus

Question

BUSINESS: Capital Value of an Asset The capital value of an asset (such as an oil well) that produces a continuous stream of income is the sum of the present value of all future earnings from the asset. Therefore, the capital value of an asset that produces income at the rate of \(r(t)\) dollars per year (at a continuous interest rate \(i\) ) is $$ \left(\begin{array}{c} \text { Capital } \\ \text { value } \end{array}\right)=\int_{0}^{T} r(t) e^{-i t} d t $$ where \(T\) is the expected life (in years) of the asset. Use the formula in the preceding instructions to find the capital value (at interest rate \(i=0.06\) ) of an oil well that produces income at the constant rate of \(r(t)=240,000\) dollars per year for 10 years.

Step-by-Step Solution

Answer

The capital value is $2,683,200.

Step 1: Identify the Given Values

We are given the income function of the asset as a constant rate, \( r(t) = 240,000 \) dollars per year. The interest rate is continuous at \( i = 0.06 \) and the expected life of the asset is \( T = 10 \) years.

Step 2: Set Up the Integral

The formula for the capital value of an asset in continuous terms is \( \int_{0}^{T} r(t) e^{-it} \, dt \). Substitute \( r(t) = 240,000 \), \( i = 0.06 \), and \( T = 10 \) into the formula, resulting in the integral \( \int_{0}^{10} 240,000 \, e^{-0.06t} \, dt \).

Step 3: Integrate the Function

The integral is of the form \( \int e^{-at} \, dt \), which integrates to \( -\frac{1}{a}e^{-at} + C \). Therefore, integrate \( \int_{0}^{10} 240,000 \, e^{-0.06t} \, dt = 240,000 \left[ -\frac{1}{0.06} e^{-0.06t} \right]_{0}^{10} \).

Step 4: Evaluate the Definite Integral

To evaluate the definite integral, substitute the upper and lower bounds into the antiderivative. This yields \( 240,000 \left(\left[-\frac{1}{0.06} e^{-0.06 \times 10}\right] - \left[-\frac{1}{0.06} e^{-0.06 \times 0}\right]\right) \).

Step 5: Simplify the Expression

Simplify the expression: \( 240,000 \left( -\frac{1}{0.06} e^{-0.6} + \frac{1}{0.06} e^{0} \right) = 240,000 \left( -\frac{1}{0.06} e^{-0.6} + \frac{1}{0.06} \right) \).

Step 6: Calculate the Numerical Value

Compute the numerical value: \(-\frac{1}{0.06} e^{-0.6} \approx -5.49 \) and \(\frac{1}{0.06} = 16.67 \). Then, \( 240,000 (16.67 - 5.49) \approx 240,000 \times 11.18 = 2,683,200\).

Step 7: Conclude the Calculation

The calculated capital value of the oil well after evaluating and simplifying the integral at an interest rate of 6% for 10 years is \( 2,683,200 \) dollars.

Key Concepts

Understanding Continuous Income Stream

A continuous income stream refers to a consistent flow of money generated over time by an asset. In this context, the asset, such as an oil well, produces earnings at a constant rate represented by the function \( r(t) \). A continuous income stream indicates that the asset generates income all year round at the same rate, making it predictable and stable. This makes it easier to calculate the present value of future earnings, as the constant rate simplifies mathematical modeling. In our exercise, the oil well has an income stream of \( r(t) = 240,000 \) dollars per year, showing that annually, without interruption, the income generated by the oil well remains constant.

Interest Rate and Its Role

The interest rate \( i \) is crucial in determining the present value of future income. This rate reflects the time value of money, highlighting that money available today is worth more than the same amount in the future due to its potential earning ability. With a continuous interest rate, money's growth potential is assessed constantly over time. In this scenario, an interest rate of \( i = 0.06 \) or 6% is applied. This percentage indicates how much each unit of money will grow per year due to investment returns or inflation, thus affecting how future earnings are discounted back to their present value. The lower the interest rate, the higher the present value of future income becomes, and vice versa.

Definite Integral in Calculating Present Value

The definite integral is a fundamental tool in calculus for calculating the area under a curve, which in financial terms can represent the total accumulated value. It is used here to sum up the present value of an income stream over a specific period. By integrating the function \( r(t) e^{-it} \) from 0 to \( T \), where \( T \) is the time duration of future income, we calculate the overall capital value of the asset. In our example, the integration takes the form \( \int_{0}^{10} 240,000 \, e^{-0.06t} \, dt \). This setup helps transform the continuous stream of future income into a present value by mathematically accounting for the decay of money value over time due to the interest rate.

Performing Capital Value Calculation

Capital value calculation involves determining the present value of a stream of future income an asset generates. This calculation uses the formula \( \int_{0}^{T} r(t) e^{-it} \, dt \), where each future income component is discounted back to its present value using an exponential decay factor. In this exercise, after computing the integral and simplifying the expression, the numerical solution of \( 240,000 (16.67 - 5.49) \) results in a capital value of \( 2,683,200 \) dollars. Evaluating this integral with respect to its bounds considers the entire time span the asset will generate income, providing a comprehensive present value that represents the total economic worth of future cash flows at a specified interest rate. This process ensures the investor understands the true value of the continued earnings expected from their asset.