Problem 91 - Thomas Calculus

Question

In Exercises \(89-92,\) 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. $$ f(x)=x \sin \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right] $$

Step-by-Step Solution

Answer

Plot, partition, evaluate midpoints, compute average, solve for \(x\).

Step 1: Plot the function

First, we need to plot the function \[ f(x) = x \sin \left(\frac{1}{x}\right) \]over the interval\[ \left[\frac{\pi}{4}, \pi\right] \]This plot will give us a visual idea of how the function behaves over the given interval. This can be done using a CAS (Computer Algebra System) like Desmos or GeoGebra by entering the function equation and defining the x-axis limits from \(\frac{\pi}{4}\) to \(\pi\).

Step 2: Partition the interval

Divide the interval \(\left[\frac{\pi}{4}, \pi\right]\) into \(n=100, 200,\) and \(1000\) equal subintervals. We find the length of each subinterval by calculating:\[ \Delta x = \frac{\pi - \frac{\pi}{4}}{n} \]This will give the width of each subinterval for different values of \(n\). Evaluate the function \[ f(x) = x \sin \left(\frac{1}{x}\right) \] at the midpoint of each subinterval.

Step 3: Evaluate midpoints

For each subinterval, calculate the midpoint using \[ x_i = \left(a + \frac{i-0.5}{n} \cdot (b-a)\right) \]where \(a = \frac{\pi}{4}\) and \(b = \pi\). Evaluate \[ f(x_i) = x_i \sin \left(\frac{1}{x_i}\right) \]for each midpoint \(x_i\).

Step 4: Compute the Average Value

To find the average value of the function from the calculated values at the midpoints, use:\[\text{Average value} = \frac{1}{n} \sum_{i=1}^{n} f(x_i)\]Compute this average for \(n=100\), \(n=200\), and \(n=1000\). Particularly focus on \(n=1000\) for subsequent steps.

Step 5: Solve for \(x\)

Using the average value calculated with \(n=1000\) subintervals, solve the equation\[ f(x) = \text{Average value} \]This involves solving \[ x \sin \left(\frac{1}{x}\right) = \text{Average value} \]for \(x\). Use a CAS to find the solution for \(x\) within the interval \(\left[\frac{\pi}{4}, \pi\right]\).

Key Concepts

Midpoint Rule

The Midpoint Rule is a numerical method used to approximate the value of an integral, which can be thought of as the area under a curve. In this context, instead of analyzing the entire curve, we take a piece-by-piece approach by dividing the interval into smaller, manageable subintervals. Each subinterval's value is approximated by evaluating the function at the midpoint and multiplying it by the interval's width.

To use the Midpoint Rule, you start by identifying the midpoints, which can be calculated using the formula:

• Find each midpoint: \( x_i = a + \frac{i-0.5}{n}(b-a) \)
• Here, \(a\) and \(b\) are the boundaries of the interval, and \(n\) is the number of subintervals.

Multiplying the function value at each midpoint by the width of the subinterval provides an estimate of the area (or the integral). This technique can be very efficient, especially when dealing with functions that are computationally challenging to integrate analytically.

This averaging method helps give us a more accurate estimate of the integral, especially when used with a large number of intervals, such as 1000, as in this exercise.

Average Value of a Function

Understanding the average value of a function over a particular interval helps in determining how the function behaves in that domain. The average value can be thought of as the constant value the function would have if you stretched the total output evenly across the interval length.

The formula for calculating the average value of a function on an interval \([a, b]\) is given by:

• \( \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

For practical purposes, especially when dealing with estimations like in the exercise, you might employ the midpoint evaluation. Here, after determining the function values at the midpoints of subintervals, the average can be directly calculated using:

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

This technique is particularly useful when opting for numerical methods, offering insights into the function's overall behavior without exact analytic integration.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a type of software designed to perform symbolic mathematical computations. They are incredibly helpful tools for students and professionals working on complex mathematical problems as they can plot functions, solve equations, and perform calculus-related tasks efficiently.

In this exercise, using a CAS—such as Desmos or GeoGebra—enables users to:

• Effortlessly plot the function \( f(x) = x \sin \left(\frac{1}{x}\right) \), observing its behavior over the interval \([\frac{\pi}{4}, \pi]\).
• Divide and evaluate the interval at various levels of subintervals, like 100, 200, and 1000, to facilitate deeper analysis.
• Handle complex calculations required to find function values at midpoints and average values across subintervals.

Utilizing a CAS helps achieve accuracy, especially for settings requiring heavy computation, and can streamline the learning and analysis process significantly.

Partition of Intervals

The partition of intervals refers to dividing a given interval into a number of smaller, equal sections to analyze the behavior of a function systematically. This process is fundamental in numerical methods, including the Midpoint Rule and Integral calculus approximations.

For any interval \([a, b]\), the interval is divided into \(n\) equal parts, and the width of each subinterval \( \Delta x \) is:

• \( \Delta x = \frac{b-a}{n} \)

This division allows for easier computation of function values at distinct points, such as midpoints, without evaluating over the entire range directly. In practice, partitioning into finer intervals improves the precision of numerical approximations and integrations.

Appropriate partitioning is crucial, especially when using numerical integration techniques. Smaller partitions often provide more accurate results, as seen with partition sizes like 1000 in the exercise, enhancing the reliability of such estimations.