Question

Present Value In Exercises 25 and 26, use a program similar to the Simpson's Rule program on page 906 with \(n=8\) to approximate the present value of the income \(c(t)\) over \(t_{1}\) years at the given annual interest rate \(r\). Then use the integration capabilities of a graphing utility to approximate the present value. Compare the results. (Present value is defined in Section 12.1.) $$ c(t)=200,000+15,000 \sqrt[3]{t}, r=10 \%, t_{1}=8 $$

Short answer

Approximating the integral using both Simpson’s rule and a graphing utility should give us a very close estimation of the net present value. Simpson’s rule should give us a bit more accurate result due to it using parabolic approximations. The precise values can only be obtained by performing the actual computations.

Step-by-step solution

Understand and set up the integral

Set up the integral for the present value \(P\) of the income stream \(c(t)\) over 8 years using the continuous compounding formula: \[P = \int_{0}^{8} c(t)e^{-rt} dt\] where \(c(t) = 200,000 + 15,000 t^{1/3}\) and \(r = 0.1\) (as 10% interest rate should be written in decimal form).

Simplify the function to integrate

Simplify the function to integrate by substituting \(c(t)\) and \(r\) into the continuous compounding formula. This leads to the integration function: \[P = \int_{0}^{8} (200,000 + 15,000 t^{1/3})e^{-0.1t} dt\]

Approximate using Simpson's Rule

Apply Simpson's rule to approximate the integral. The general form of Simpson's rule is: \[P \approx \frac{t_{1} - 0}{N} [ \frac{f(0) + 4f(\frac{t_{1}}{N}) + 2f(2\frac{t_{1}}{N}) + \cdots + 4f((N-1)\frac{t_{1}}{N}) + f(t_{1})}{3}]\] Set up and solve this with the given function to approximate the present value. Remember that in this case \(t_{1} = 8\) and \(N = 8\).

Approximate using a graphing utility

Next, use a graphing utility to approximate the definite integral. Observe the graph of the function over the interval [0, 8] and use the integration function of the graphing utility to obtain an estimate of the integral.

Compare the results

Finally, compare the results obtained from Simpson's method and the graphing utility approximation. These methods should both give similar results.

Key concepts

Simpson's Rule

When it comes to estimating the value of a definite integral, such as the present value of a cash flow over time, Simpson's Rule provides an efficient method for achieving an approximation. This technique utilizes parabolic arcs to approximate the area under a curve between two points, boosting accuracy over simpler methods like the trapezoidal rule.

Simpson's Rule is expressed mathematically as: \[P \approx \frac{b - a}{3N} \left( f(a) + 4\sum_{i=1}^{N/2} f(a + 2i\Delta x) + 2\sum_{i=1}^{N/2 - 1} f(a + (2i + 1)\Delta x) + f(b) \right)\]where \(a\) and \(b\) are the lower and upper limits of integration, \(N\) is an even number of intervals, \(\Delta x = \frac{b-a}{N}\), and \(f(x)\) is the function being integrated.

Applied to our exercise, we compute the present value by partitioning the interval from 0 to 8 (years) into eight segments and evaluating the income function at these strategically chosen points. The rule requires the end points, the mid points and the points at the halfway mark between each midpoint and the end. In doing so, we capture a more accurate picture of the function's behavior over the interval. The 'checkerboard' pattern of coefficients (1, 4, 2) is what ensures the higher degree of accuracy by giving more weight to certain intervals.

Continuous Compounding Formula

The continuous compounding formula is central to understanding present and future value calculations in finance. Continuous compounding assumes that interest is added to the principal not once a year or once a month, but an infinite number of times per instant. Mathematically, continuous compounding utilizes Euler's number \(e\), the base of the natural logarithms.

The formula for the continuous compounding is given by: \[A = Pe^{rt}\]where \(A\) is the amount of money accumulated after a certain time, including interest, \(P\) is the principal amount, \(r\) is the annual interest rate (expressed as a decimal), and \(t\) is the time in years. The continuous compounding formula can be used to calculate the present value by adjusting for the fact that money expected in the future is worth less than the same amount in hand today. This is done by modifying the formula to account for a negative growth rate, effectively moving the value from future dollars to today's dollars. The exponential function models the impact of compounding, while the negative exponent reflects the discounting effect when calculating present value. The function \(c(t)\) in our exercise is given an extra dimension of complexity, growing not at a constant rate but according to a specific function of time.

Definite Integral Approximation

Definite integrals represent the accumulation of quantities, such as the present value of an income stream over a fixed period. However, evaluating them analytically can sometimes be complex or impossible, especially when dealing with intricate functions or those without a primitive function. In such cases, numerical approximation methods like Simpson's Rule become invaluable.

The goal of a definite integral approximation is to find an estimate that is close to the actual area under the curve. For our present value calculation, the integral takes the form: \[P = \int_{a}^{b} c(t)e^{-rt} dt\]where the range from \(a\) to \(b\) is the time over which the income is considered, and the function \(c(t)e^{-rt}\) represents the changing income over time discounted by a continuous interest rate. By approximating the integral, we estimate the present value without having to determine an antiderivative of the function, which can be complex or unknown. Approximation methods like Simpson's Rule use points sampled from the function being integrated to construct a series of simple shapes (like rectangles or trapezoids in other methods) or segments of parabolas to estimate the area under the curve, offering a practical solution to an otherwise mathematically demanding problem.