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Chapter 5: Problem 94

If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises \(89-94 .\) Use \(n=4,10,20,\) and 50 subintervals of equal length in each case. $$ \int_{1}^{2} \frac{1}{x} d x \quad(\text { The integral's value is about } 0.693 .) $$

Short Answer

Expert verified
Riemann sums approximate 0.693 using 4, 10, 20, 50 subintervals, each result gets closer as \( n \) increases.

Step by step solution

01

Understanding the Problem

We are asked to approximate the integral \( \int_{1}^{2} \frac{1}{x} \, dx \) using Riemann sums. Riemann sums approximate the area under a curve by summing the areas of rectangles, with more rectangles providing a better approximation. We need to perform this process for \( n = 4, 10, 20, \) and \( 50 \) subintervals.
02

Determine Rectangle Width

Calculate the width \( \Delta x \) of each rectangle. For an interval \([a, b]\) divided into \( n \) subintervals, \( \Delta x = \frac{b-a}{n} \). Here, \( a = 1 \), \( b = 2 \). So \( \Delta x = \frac{2-1}{n} = \frac{1}{n} \).
03

Choose Rectangle Heights Using Midpoint Rule

We use the Midpoint Rule, where each rectangle's height is determined by the function value at the midpoint of the subinterval. The midpoint \( x_i \) for each subinterval is \( x_i = 1 + (i - 0.5) \Delta x \) for \( i = 1, 2, ..., n \).
04

Calculate Riemann Sum

For each value of \( n \), calculate the Riemann sum \( R = \sum_{i=1}^{n} f(x_i) \Delta x \), where \( f(x) = \frac{1}{x} \). Use the midpoints calculated in the previous step.
05

Evaluate for n=4

For \( n=4 \), calculate \( \Delta x = \frac{1}{4} = 0.25 \). Midpoints are at \( x_i = 1.125, 1.375, 1.625, 1.875 \). Calculate heights \( f(x_i) \) and sum the areas: \( R_4 = \sum \frac{1}{x_i} \times 0.25 \approx 0.6953 \).
06

Evaluate for n=10

For \( n=10 \), \( \Delta x = 0.1 \). Midpoints at \( x_i = 1.05, 1.15, ... , 1.95 \). Calculate \( R_{10} = \sum \frac{1}{x_i} \times 0.1 \approx 0.697 \).
07

Evaluate for n=20

For \( n=20 \), \( \Delta x = 0.05 \). List midpoints \( x_i = 1.025, 1.075, ..., 1.975 \). Calculate \( R_{20} = \sum \frac{1}{x_i} \times 0.05 \approx 0.694 \).
08

Evaluate for n=50

For \( n=50 \), \( \Delta x = 0.02 \). Midpoints are from \( x_i = 1.01 \) to \( 1.99 \) at every 0.02. Calculate \( R_{50} = \sum \frac{1}{x_i} \times 0.02 \approx 0.6935 \).
09

Draw Rectangles for Visualization (if possible)

Using a CAS (Computer Algebra System) or similar tool, draw rectangles at each of the evaluated \( n \) values to visualize how the approximation improves as \( n \) increases.
10

Conclusion

As \( n \) increases, the Riemann sum \( R_n \) becomes closer to the integral's actual value of 0.693. Thus, increasing \( n \) improves the approximation accuracy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Midpoint Rule
The Midpoint Rule is a technique used to approximate the integral of a function, which essentially means finding the area under a curve. This method uses the value of the function at the midpoint of each subinterval to determine the height of the rectangles. It's like estimating average temperature by checking at noon鈥攎idway through the day.

The basic steps to apply the Midpoint Rule include:
  • Dividing the interval into equal parts, or subintervals.
  • Calculating the midpoint of each subinterval.
  • Computing the function's value at each midpoint to find the rectangle's height.
  • Summing the areas of these rectangles to estimate the integral.
Using midpoints helps create a better balance in height estimates, offering a reliable integration approximation.
Integral Approximation
Integral approximation is a way of estimating the area under a curve when calculating the integral directly is challenging. Instead of trying to measure the entire area precisely, approximation methods like the Riemann sum are used for a close estimate.

Approximations become particularly helpful:
  • When the function is complex, and integration is not straightforward.
  • In digital computations where exact integral values are not necessary, but close approximations are sufficient.
  • For visualizing the problem, like when using rectangle approximations in graphs.
By approximating, calculations can be simplified, making complex problems manageable, while still providing insight into the real value of the integral.
Rectangle Method
Also known as Riemann sums, the Rectangle Method involves estimating the area under a curve using rectangles. Each subinterval on the x-axis forms the base of a rectangle, while the height is determined based on the specific rule used, such as the left endpoint, right endpoint, or midpoint (our focus).

In practical terms:
  • The width of each rectangle (\( \Delta x \)) is calculated by dividing the total interval length by the number of subintervals (\( n \)).
  • The area of each rectangle is the width times the height (function value at chosen point).
  • The sum of these areas gives an approximate measure of the integral.
This method helps conceptualize integration and prepares students for complex calculus concepts by breaking down the problem into simple, repetitive calculations.
Subintervals
In the context of Riemann sums and integral approximation, subintervals are small, manageable slices of the entire interval over which a function is being integrated. Imagine slicing a loaf of bread鈥攅ach slice is a subinterval. As you increase the number of slices, they become thinner, providing a more detailed picture.

Key characteristics of subintervals include:
  • The number of subintervals (\( n \)) directly affects the accuracy of the approximation.
  • A higher number of subintervals increases precision but also requires more computation.
  • All subintervals are typically of equal length, determined by dividing the original interval size by the total number of subintervals.
Using subintervals allows for step-by-step calculations that build toward a complete approximation, improving accuracy as the number of subintervals increases.

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Most popular questions from this chapter

Find the areas of the regions enclosed by the lines and curves in Exercises \(51-58 .\) $$ x=2 y^{2}, \quad x=0, \quad \text { and } \quad y=3 $$

Suppose the area of the region between the graph of a positive continuous function \(f\) and the \(x\) -axis from \(x=a\) to \(x=b\) is 4 square units. Find the area between the curves \(y=f(x)\) and \(y=2 f(x)\) from \(x=a\) to \(x=b\)

In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ f(x)=-3 x^{2}-1 \quad \text { on } \quad[0,1] $$

Suppose that \(f\) has a positive derivative for all values of \(x\) and that \(f(1)=0 .\) Which of the following statements must be true of the function $$g(x)=\int_{0}^{x} f(t) d t ?$$ Give reasons for your answers. \(\begin{array}{l}{\text { a. } g \text { is a differentiable function of } x \text { . }} \\ {\text { b. } g \text { is a continuous function of } x \text { . }} \\ {\text { c. The graph of } g \text { has a horizontal tangent at } x=1 \text { . }} \\ {\text { d. } g \text { has a local maximum at } x=1 \text { . }}\\\\{\text { e. } g \text { has a local minimum at } x=1.} \\\ {\text { f. The graph of } g \text { has an inflection point at } x=1.} \\\ {\text { g. The graph of } d g / d x \text { crosses the } x \text { -axis at } x=1}.\end{array}\)

In Exercises \(95-98,\) use a CAS to perform the following steps: $$ \begin{array}{l}{\text { a. Plot the functions over the given interval. }} \\\ {\text { b. Partition the interval into } n=100,200, \text { and } 1000 \text { subintervals }} \\ {\text { of equal length, and evaluate the function at the midpoint }} \\ {\text { of each subinterval. }}\\\ {\text { c. compute the average value of the function values generated in }} \\\ {\text { part (b). }} \\ {\text { d. Solve the equation } f(x)=(\text { average value) for } x \text { using the average}} \\ {\text { value calculated in part (c) for the } n=1000 \text { partitioning. }}\end{array} $$ $$ f(x)=x \sin \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right] $$
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