/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Graph each function \(f(x)\) ove... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 36

Graph each function \(f(x)\) over the given interval. Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum \(\Sigma_{k=1}^{4} f\left(c_{k}\right) \Delta x_{k},\) given that \(c_{k}\) is the (a) left-hand endpoint, (b) righthand endpoint, (c) midpoint of the \(k\) th subinterval. (Make a separate sketch for each set of rectangles.) $$f(x)=\sin x+1, \quad[-\pi, \pi]$$

Short Answer

Expert verified
Partition [-蟺, 蟺] into four subintervals. Draw rectangles for left, right, and midpoint Riemann sums for f(x)=sin x+1.

Step by step solution

01

Determine the Interval and Partition

The function given is \( f(x) = \sin x + 1 \) over the interval \([-\pi, \pi]\). The total length of the interval is \( \pi - (-\pi) = 2\pi \). We partition this interval into four subintervals of equal length, which makes each subinterval have a length of \( \Delta x = \frac{2\pi}{4} = \frac{\pi}{2} \).
02

Calculate Subintervals

Given the interval \([-\pi, \pi]\) and \( \Delta x = \frac{\pi}{2} \), the subintervals are:- \([-\pi, -\frac{\pi}{2}]\)- \([-\frac{\pi}{2}, 0]\)- \([0, \frac{\pi}{2}]\)- \([\frac{\pi}{2}, \pi]\)
03

Sketch the Graph of f(x)

Plot \( f(x) = \sin x + 1 \) over the interval \([-\pi, \pi]\). This is a sine wave shifted up by 1 unit.
04

Left-Hand Endpoint Rectangles

For each subinterval, use the left endpoint for the function value,hence calculating: - \( f(-\pi) = 0 \) for \([-\pi, -\frac{\pi}{2}]\)- \( f(-\frac{\pi}{2}) = 1 \) for \([-\frac{\pi}{2}, 0]\)- \( f(0) = 1 \) for \([0, \frac{\pi}{2}]\)- \( f(\frac{\pi}{2}) = 2 \) for \([\frac{\pi}{2}, \pi]\).Draw rectangles using these heights, with width \(\Delta x = \frac{\pi}{2}\).
05

Right-Hand Endpoint Rectangles

For each subinterval, use the right endpoint for the function value, hence calculating:- \( f(-\frac{\pi}{2}) = 1 \) for \([-\pi, -\frac{\pi}{2}]\)- \( f(0) = 1 \) for \([-\frac{\pi}{2}, 0]\)- \( f(\frac{\pi}{2}) = 2 \) for \([0, \frac{\pi}{2}]\)- \( f(\pi) = 1 \) for \([\frac{\pi}{2}, \pi]\).Draw rectangles using these heights, with width \(\Delta x = \frac{\pi}{2}\).
06

Midpoint Rectangles

For each subinterval, use the midpoint for the function value, calculate:- Midpoint of \([-\pi, -\frac{\pi}{2}]\) is \(-\frac{3\pi}{4}\), \(f(-\frac{3\pi}{4})\approx 0.29\)- Midpoint of \([-\frac{\pi}{2}, 0]\) is \(-\frac{\pi}{4}\), \(f(-\frac{\pi}{4})\approx 1.29\)- Midpoint of \([0, \frac{\pi}{2}]\) is \(\frac{\pi}{4}\), \(f(\frac{\pi}{4})\approx 1.71\)- Midpoint of \([\frac{\pi}{2}, \pi]\) is \(\frac{3\pi}{4}\), \(f(\frac{3\pi}{4})\approx 1.71\).Draw rectangles using these heights, with width \(\Delta x = \frac{\pi}{2}\).

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Key Concepts

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

Left Endpoint
The left endpoint method for a Riemann sum involves using the leftmost point within each subinterval as the representative point, or height, for constructing rectangles under a curve. In our exercise with the function \( f(x) = \sin x + 1 \), over the interval \([-\pi, \pi]\), we must divide the interval into four equal parts, each of length \(\Delta x = \frac{\pi}{2}\).
  • First subinterval \([-\pi, -\frac{\pi}{2}]\): height is \( f(-\pi) = 0 \)
  • Second subinterval \([-\frac{\pi}{2}, 0]\): height is \( f(-\frac{\pi}{2}) = 1 \)
  • Third subinterval \([0, \frac{\pi}{2}]\): height is \( f(0) = 1 \)
  • Fourth subinterval \([\frac{\pi}{2}, \pi]\): height is \( f(\frac{\pi}{2}) = 2 \)
The left endpoint approach provides an approximation of the area under the curve by consistently underestimating or overestimating it, depending on whether the function is increasing or decreasing.
Right Endpoint
The right endpoint method uses the rightmost point within each subinterval as the descriptor for the rectangle's height. Continuing with the exercise, for \( f(x) = \sin x + 1 \), we retain the same four subintervals, each with length \(\Delta x = \frac{\pi}{2}\).
  • First subinterval \([-\pi, -\frac{\pi}{2}]\): height is \( f(-\frac{\pi}{2}) = 1 \)
  • Second subinterval \([-\frac{\pi}{2}, 0]\): height is \( f(0) = 1 \)
  • Third subinterval \([0, \frac{\pi}{2}]\): height is \( f(\frac{\pi}{2}) = 2 \)
  • Fourth subinterval \([\frac{\pi}{2}, \pi]\): height is \( f(\pi) = 1 \)
The right endpoint approach can also either underestimate or overestimate the true area, depending on the function's behavior across the interval.
Midpoint
For many, the midpoint method strikes a balance when estimating the area under a curve using rectangles. By selecting the midpoint of each subinterval as the height's reference point, the approximation can more accurately reflect the true area. Using \( f(x) = \sin x + 1 \), consider our subintervals:
  • Midpoint of \([-\pi, -\frac{\pi}{2}]\) is \(-\frac{3\pi}{4}\), with height \( f(-\frac{3\pi}{4}) \approx 0.29 \)
  • Midpoint of \([-\frac{\pi}{2}, 0]\) is \(-\frac{\pi}{4}\), with height \( f(-\frac{\pi}{4}) \approx 1.29 \)
  • Midpoint of \([0, \frac{\pi}{2}]\) is \(\frac{\pi}{4}\), with height \( f(\frac{\pi}{4}) \approx 1.71 \)
  • Midpoint of \([\frac{\pi}{2}, \pi]\) is \(\frac{3\pi}{4}\), with height \( f(\frac{3\pi}{4}) \approx 1.71 \)
The midpoint method often yields a sum that more closely matches the actual integral, making it a preferable choice in many scenarios.
Subintervals
To compute a Riemann sum, you divide the main interval into smaller, equal-sized subintervals. The choice of these subintervals determines the outcome of your approximation. In the exercise, covering the interval \([-\pi, \pi]\), we divide it into four subintervals, each with a length of \(\Delta x = \frac{\pi}{2}\).
  • The breaking points form the subintervals \([-\pi, -\frac{\pi}{2}], [-\frac{\pi}{2}, 0], [0, \frac{\pi}{2}], [\frac{\pi}{2}, \pi]\)
Choosing how many subintervals you use affects the precision of the Riemann sum. More subintervals generally lead to a more accurate approximation but increase the complexity of calculations.
Trigonometric Functions
Trigonometric functions often appear in calculus, presenting unique challenges and opportunities for integration. The function \( f(x) = \sin x + 1 \) shifts the sine wave up by one unit, ensuring positivity over \([-\pi, \pi]\). Understanding properties of trigonometric functions like symmetry, periodicity, and amplitude can help in sketching the graph and predicting behavior.
  • \( \sin x \) is periodic with a period of \( 2\pi \)
  • \( \sin x + 1 \) adjusts the baseline, making the minimum value 0 and the maximum value 2 over the interval
These characteristics can simplify the process of applying Riemann sums or definite integrals by providing predictable patterns to work with.

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Most popular questions from this chapter

Find the total area between the region and the \(x\) -axis. $$y=x^{3}-3 x^{2}+2 x, \quad 0 \leq x \leq 2$$

Find the areas of the regions enclosed by the curves in Exercises \(81-84\). $$4 x^{2}+y=4 \text { and } x^{4}-y=1$$

In Exercises \(119-122,\) you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$f(x)=\frac{x^{3}}{3}-\frac{x^{2}}{2}-2 x+\frac{1}{3}, \quad g(x)=x-1$$

In Exercises \(119-122,\) you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$f(x)=x^{2} \cos x, \quad g(x)=x^{3}-x$$

Let \(F(x)=\int_{a}^{u(x)} f(t) d t\) for the specified \(a, u,\) and f. Use a CAS to perform the following steps and answer the questions posed. a. Find the domain of \(F\). b. Calculate \(F^{\prime}(x)\) and determine its zeros. For what points in its domain is \(F\) increasing? Decreasing? c. Calculate \(F^{\prime \prime}(x)\) and determine its zero. Identify the local extrema and the points of inflection of \(F .\) d. Using the information from parts (a)-(c), draw a rough handsketch of \(y=F(x)\) over its domain. Then graph \(F(x)\) on your CAS to support your sketch. $$a=1, \quad u(x)=x^{2}, \quad f(x)=\sqrt{1-x^{2}}$$
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