Problem 102 - University Calculus: Early Transcendentals

Question

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100.200\), and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=\frac{1}{\sqrt{1-x^{2}}} \text { on }\left[0, \frac{1}{2}\right]$$

Step-by-Step Solution

Answer

The average value of the function with 1000 partitions helps solve \( f(x) = \text{Average} \) over the interval \([0, \frac{1}{2}]\).

Step 1: Plot the Function

We want to plot the function \( f(x) = \frac{1}{\sqrt{1-x^{2}}} \) over the interval \([0, \frac{1}{2}]\). Use a Computer Algebra System (CAS) or a graphing tool to visualize the curve of the function. The function starts at \( x=0 \) and ends at \( x=\frac{1}{2} \). Observe the behavior of the function within this interval.

Step 2: Partition the Interval

Divide the interval \([0, \frac{1}{2}]\) into subintervals with the lengths determined by \( n=100, 200, \) and \( 1000 \). For each value of \( n \), calculate the width of the subinterval \( \Delta x = \frac{1/2 - 0}{n} = \frac{1}{2n} \). Choose the midpoint \( x_i = \frac{i - 0.5}{n} \) for \( i = 1, 2, ..., n \).

Step 3: Evaluate the Function at Midpoints

For each \( n \), calculate \( f(x_i) = \frac{1}{\sqrt{1-(x_i)^{2}}} \) at the midpoint \( x_i \) of each subinterval. Store these values for further calculations.

Step 4: Compute the Average Value

For each partition size, compute the average value of the function values obtained in Step 3 using the formula \( \text{Average} = \frac{1}{n} \sum_{i=1}^{n} f(x_i) \). Especially note the average value for \( n=1000 \).

Step 5: Solve for f(x) Equals Average Value

Use the average value from \( n=1000 \) calculated in Step 4 and set \( f(x) = \text{Average} \). Solve the equation \( \frac{1}{\sqrt{1-x^2}} = \text{Average} \) for \( x \). Find the solution(s) for \( x \) over the interval \([0, \frac{1}{2}]\).

Key Concepts

Midpoint Rule

The Midpoint Rule is a popular numerical method for estimating the value of a definite integral. It works by partitioning the interval into smaller subintervals and then evaluating the function at the midpoint of each subinterval. This provides a more accurate approximation than simply evaluating at the endpoints.

To apply the Midpoint Rule, follow these key steps:

• First, determine the number of subintervals, denoted by \( n \).
• Calculate the width of each subinterval \( \Delta x \), as presented in the form \( \Delta x = \frac{b-a}{n} \), where \( a \) is the start point and \( b \) is the end point of the interval.
• Select the midpoint of each subinterval using \( x_i = a + (i - 0.5) \cdot \Delta x \). Here, \( i \) ranges from 1 to \( n \).
• Evaluate the function at each midpoint and sum up these values.
• The sum gives an estimate of the integral when multiplied by \( \Delta x \).

By incorporating these steps, you can compute an accurate approximation of the integral without requiring direct analytical integration.

Partitioning Intervals

Partitioning intervals is a fundamental component of numerical evaluation methods, such as the Midpoint Rule. By dividing the interval into subintervals, we can estimate the behavior of a function with more specificity.

Here's how partitioning works:

• The interval is divided into \( n \) equally spaced subintervals.
• The width \( \Delta x \) of each subinterval is determined by the formula \( \Delta x = \frac{b-a}{n} \).
• This method allows more opportunities to sample the function and observe changes within each fragment.
• Increased numbers of \( n \) lead to more refined calculations, capturing more intricacies of the function's behavior.

Adjusting the partition size means striking a balance between computational effort and precision of results. Using 100, 200, or even 1000 partitions, as seen in the context of the original exercise, demonstrates the increasing accuracy this method can provide.

Average Value of a Function

The average value of a function over a specific interval can provide valuable insights into the overall behavior of the function throughout that interval. In numerical methods, particularly with partitioning and the Midpoint Rule, calculating the average value forms an integral part of the analysis.

The formula for calculating the average value of a function \( f(x) \) over an interval \( [a, b] \) is:

• \( \text{Average} = \frac{1}{n} \sum_{i=1}^{n} f(x_i) \)

Here, \( x_i \) are the midpoints of the subintervals, and \( n \) is the number of subintervals. This approach gives a numeric approximation of how the average height of the function's graph behaves across the interval.A crucial part of the original exercise involved determining the average value specifically for 1000 subintervals. This step provides a deeper understanding of the function's behavior and is used to solve for specific values that equate to this average.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is an incredible tool for students and professionals alike when dealing with complex mathematical computations, such as plotting functions and evaluating them at various intervals. These systems can automate calculations that would otherwise be tedious and error-prone, especially when dealing with numerous iterations needed in partitioning subintervals.

Using CAS in exercises like the original problem allows us to:

• Effortlessly plot the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \) across its specified interval \([0, \frac{1}{2}]\).
• Handle multiple computations over large numbers of partitions (e.g., 100, 200, 1000) efficiently.
• Ensure precision and accuracy in evaluating function values at midpoints and calculating average values.

By utilizing CAS, you not only save time but also enhance the accuracy of your solutions, making it an essential asset in tackling diverse mathematical challenges.