Problem 23 - Thomas Calculus

Question

In Exercises \(23-26,\) use a CAS to perform the following steps. a. Plot the functions over the given interval. b. Subdivide 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. $$ f(x)=\sin x \quad \text { on } \quad[0, \pi] $$

Step-by-Step Solution

Answer

The average value of \(\sin x\) on \([0, \pi]\) for \(n=1000\) is approximately 0.6366. Solve \(\sin x = 0.6366\) to find \(x\) values in \([0, \pi]\).

Step 1: Plot the Function

First, we plot the function \(f(x) = \sin x\) over the interval \([0, \pi]\). You can use any software capable of plotting, such as a Computer Algebra System (CAS) or graphing calculator. The plot demonstrates the sine wave starting at 0, reaching its peak at \(x = \frac{\pi}{2}\), and returning to 0 at \(x = \pi\).

Step 2: Subdivide the Interval and Evaluate Function

Next, we subdivide the interval \([0, \pi]\) into 100, 200, and 1000 subintervals. For each subdivision, find the midpoint \(x_i\) of each subinterval, where \(x_i = a + (i-0.5)\frac{(b-a)}{n}\) with \(a=0\) and \(b=\pi\). Evaluate \(f(x_i) = \sin x_i\) for each midpoint.

Step 3: Compute the Average Value of Function

For each subdivision, compute the average value of \(f(x)\) using the midpoints. The average is calculated as \(\frac{1}{n} \sum_{i=1}^{n} \sin(x_i)\) for \(n = 100, 200, \) and \(1000\). Specifically for \(n=1000\), execute this computation to obtain the most precise average value.

Step 4: Solve the Equation with Average Value

With \(n=1000\), determine the average value \(A\) from step 3, then solve the equation \(\sin x = A\) for \(x\). Use a CAS to find the solutions in the interval \([0, \pi]\). This solution provides the points where the sine function equals the average value.

Key Concepts

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a powerful tool used in mathematical computations to perform symbolic manipulation and numerical calculations. These systems are designed to help with a variety of tasks, such as solving equations, plotting graphs, and performing calculus operations like integration and differentiation.

• Ease of Use: Most CAS software provides an intuitive user interface, allowing users to input complex functions and get results quickly.
• Versatility: They can handle a range of mathematical problems, from simple arithmetic to more complex calculus operations.
• Accuracy: CAS systems perform operations with high precision, crucial for intricate calculations.

For students, using a CAS can simplify the task of plotting functions, such as the sine function over a given interval, and solve equations efficiently. It's like having a math assistant that ensures precision and aids in visualizing mathematical concepts.

Subintervals

Subdividing an interval into smaller parts, known as subintervals, is a common practice in numerical methods. It helps in approximating functions and calculating integrals. In the exercise, the interval \([0, \pi]\) is divided into subintervals to aid in analyzing the sine function.
Subintervals are determined by splitting the range into equal lengths. For example, with 1000 subintervals, each subinterval length is \(\frac{\pi}{1000}\). Here's how it helps:

• Improved Accuracy: Smaller subintervals lead to more accurate representations of functions.
• Increased Detail: More subintervals provide detailed insight into the behavior of a function over an interval.

Using the midpoint of each subinterval is one method in numerical integration to evaluate the function accurately. This approach ensures that calculations reflect the average behavior of the function within each segment.

Average Value of Function

The average value of a function over a particular interval gives insight into the "central" or typical value of the function over that range. In numerical integration, especially with large numbers of subintervals, this average is crucial for understanding the function's general level.
Here's how to calculate it:

• Find the value of the function at each midpoint of the subintervals.
• Sum up all these midpoint values.
• Divide by the total number of intervals (\(n\)) to get the average value.

Mathematically, the average value is calculated as \(\frac{1}{n} \sum_{i=1}^{n} f(x_i)\), where \(x_i\) represents the midpoints. In the problem, this calculation with \(n = 1000\) subintervals provides a precise average, approaching the true average as \(n\) increases.

Midpoint Rule

The Midpoint Rule is a method used in numerical integration to approximate the area under the curve of a function. It involves evaluating the function at the midpoints of subintervals, providing a simple yet effective means of approximation.
Here’s how it works:

• Select Midpoints: For each subinterval of the range, determine the midpoint.
• Evaluate Function: Calculate the function's value at each midpoint.
• Approximate Integral: The sum of these function values multiplied by the subinterval width approximates the integral.

The formula is \(\int_{a}^{b} f(x) \, dx \approx \sum_{i=1}^{n} f(x_i) \cdot \Delta x\), where \(x_i\) are midpoints, and \(\Delta x\) is the subinterval width. It simplifies integration, especially when used in combination with a CAS, giving a visual grasp on function behavior across intervals.