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Chapter 5: Problem 36

In Exercises \(33-36,\) graph each function \(f(x)\) over the given interval.Partition the interval into four sub intervals 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) right-hand endpoint, (c) midpoint of the \(k\) the sub interval. (Make a separate sketch for each set of rectangles.) $$f(x)=\sin x+1, \quad[-\pi, \pi]$$

Short Answer

Expert verified
Riemann sums: Left = \(\frac{3\pi}{2}\), Right = \(2\pi\), Midpoint = \(2\pi\). Sketch graph and rectangles accordingly.

Step by step solution

01

Determine Subinterval Length

The interval \([-\pi, \pi]\) is to be divided into four subintervals of equal length. To find the length of each subinterval, calculate the total interval length and divide by 4: \[\Delta x = \frac{\pi - (-\pi)}{4} = \frac{2\pi}{4} = \frac{\pi}{2}.\] Thus, each subinterval has a length of \(\frac{\pi}{2}\).
02

Define Subintervals and Endpoints

Identify the endpoints of each subinterval: 1. First subinterval: \([-\pi, -\frac{\pi}{2}]\)2. Second subinterval: \([-\frac{\pi}{2}, 0]\)3. Third subinterval: \([0, \frac{\pi}{2}]\)4. Fourth subinterval: \([\frac{\pi}{2}, \pi]\)
03

Calculate Left-Hand Riemann Sums

For the left-hand Riemann sum, use the left endpoints of the subintervals:1. For \([-\pi, -\frac{\pi}{2}]\), use \(-\pi\): \(f(-\pi) = \sin(-\pi) + 1 = 0+1 = 1\).2. For \([-\frac{\pi}{2}, 0]\), use \(-\frac{\pi}{2}\): \(f(-\frac{\pi}{2}) = \sin(-\frac{\pi}{2}) + 1 = -1+1 = 0\).3. For \([0, \frac{\pi}{2}]\), use \(0\): \(f(0) = \sin(0) + 1 = 0+1 = 1\).4. For \([\frac{\pi}{2}, \pi]\), use \(\frac{\pi}{2}\): \(f(\frac{\pi}{2}) = \sin(\frac{\pi}{2})+1 = 1+1=2\). The Riemann sum is: \[\Sigma_{k=1}^{4} f(c_{k}) \Delta x = 1 \cdot \frac{\pi}{2} + 0 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} + 2 \cdot \frac{\pi}{2} = \frac{3\pi}{2}.\]
04

Calculate Right-Hand Riemann Sums

For the right-hand Riemann sum, use the right endpoints of the subintervals:1. For \([-\pi, -\frac{\pi}{2}]\), use \(-\frac{\pi}{2}\): \(f(-\frac{\pi}{2}) = \sin(-\frac{\pi}{2}) + 1 = -1+1 = 0\).2. For \([-\frac{\pi}{2}, 0]\), use \(0\): \(f(0) = \sin(0) + 1 = 0+1 = 1\).3. For \([0, \frac{\pi}{2}]\), use \(\frac{\pi}{2}\): \(f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + 1 = 1+1 = 2\).4. For \([\frac{\pi}{2}, \pi]\), use \(\pi\): \(f(\pi) = \sin(\pi) + 1 = 0+1 = 1\). The Riemann sum is: \[\Sigma_{k=1}^{4} f(c_{k}) \Delta x = 0 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} + 2 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} = \frac{4\pi}{2} = 2\pi.\]
05

Calculate Midpoint Riemann Sums

For the midpoint Riemann sum, use midpoints of the subintervals:1. For \([-\pi, -\frac{\pi}{2}]\), use \(-\frac{3\pi}{4}\): \(f(-\frac{3\pi}{4}) = \sin(-\frac{3\pi}{4})+1 = -\frac{\sqrt{2}}{2} + 1\).2. For \([-\frac{\pi}{2}, 0]\), use \(-\frac{\pi}{4}\): \(f(-\frac{\pi}{4}) = \sin(-\frac{\pi}{4})+1 = -\frac{\sqrt{2}}{2} + 1\).3. For \([0, \frac{\pi}{2}]\), use \(\frac{\pi}{4}\): \(f(\frac{\pi}{4}) = \sin(\frac{\pi}{4})+1 = \frac{\sqrt{2}}{2} + 1\).4. For \([\frac{\pi}{2}, \pi]\), use \(\frac{3\pi}{4}\): \(f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4})+1 = \frac{\sqrt{2}}{2} + 1\). The Riemann sum is: \[\Sigma_{k=1}^{4} f(c_{k}) \Delta x = \left(-\frac{\sqrt{2}}{2} + 1\right) \frac{\pi}{2} + \left(-\frac{\sqrt{2}}{2} + 1\right) \frac{\pi}{2} + \left(\frac{\sqrt{2}}{2} + 1\right) \frac{\pi}{2} + \left(\frac{\sqrt{2}}{2} + 1\right) \frac{\pi}{2}\right)\]which simplifies to \[= 2\pi.\]
06

Draw the Graph

Sketch the function \(f(x) = \sin x + 1\) over the interval \([-\pi, \pi]\). Using different colors, draw the rectangles for each Riemann sum calculated: left-hand, right-hand, and midpoint. Note that the function is a vertical shift of the sine function one unit upward.

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Key Concepts

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

Left-Hand Riemann Sum
Riemann sums are a fundamental concept in calculus used to approximate the area under a curve. The idea is to divide the area into simpler shapes, usually rectangles, and sum their areas. The left-hand Riemann sum specifically uses the left endpoint of each subinterval to calculate the height of the rectangles.

When calculating a left-hand Riemann sum, you begin by determining the length of each subinterval. In our example, the interval \([-\pi, \pi]\) is divided into four subintervals, each with a length of \(\frac{\pi}{2}\).

The left-hand Riemann sum for the function \(f(x) = \sin x + 1\) involves using the left endpoints of each subinterval to evaluate the function:
  • First interval: \([-\pi, -\frac{\pi}{2}]\), evaluate at \(-\pi\), \(f(-\pi) = 1\).
  • Second interval: \([-\frac{\pi}{2}, 0]\), evaluate at \(-\frac{\pi}{2}\), \(f(-\frac{\pi}{2}) = 0\).
  • Third interval: \([0, \frac{\pi}{2}]\), evaluate at \(0\), \(f(0) = 1\).
  • Fourth interval: \([\frac{\pi}{2}, \pi]\), evaluate at \(\frac{\pi}{2}\), \(f(\frac{\pi}{2}) = 2\).
Adding these up while multiplying by the interval length \(\frac{\pi}{2}\), we obtain the left-hand Riemann sum of \(\frac{3\pi}{2}\). This sum provides an approximation, which may be underestimated compared to the true area, especially if the function is increasing in the interval.
Right-Hand Riemann Sum
In contrast to the left-hand approach, the right-hand Riemann sum uses the right endpoint of each subinterval to determine the heights of the rectangles. This can yield a different approximation than the left-hand method, especially when the function behavior changes over the interval.

For the interval \([-\pi, \pi]\) divided into subintervals of equal length \(\frac{\pi}{2}\), here is how we find the right-hand Riemann sum for \(f(x) = \sin x + 1\):
  • First interval: \([-\pi, -\frac{\pi}{2}]\), evaluate at \(-\frac{\pi}{2}\), \(f(-\frac{\pi}{2}) = 0\).
  • Second interval: \([-\frac{\pi}{2}, 0]\), evaluate at \(0\), \(f(0) = 1\).
  • Third interval: \([0, \frac{\pi}{2}]\), evaluate at \(\frac{\pi}{2}\), \(f(\frac{\pi}{2}) = 2\).
  • Fourth interval: \([\frac{\pi}{2}, \pi]\), evaluate at \(\pi\), \(f(\pi) = 1\).
The total right-hand Riemann sum equals the sum of these products: \[0 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} + 2 \cdot \frac{\pi}{2} + 1 \cdot \frac{\pi}{2} = 2\pi.\]
This estimation might overestimate the true area if the function is decreasing. Like left-hand Riemann sums, right-hand Riemann sums provide an approximation of the integral, useful for understanding the behavior of integrals.
Midpoint Riemann Sum
The midpoint Riemann sum provides another alternative for approximating the area under a curve. It uses the midpoint of each subinterval to calculate the height of the rectangles. This method often yields a more accurate approximation than either left or right-hand Riemann sums, especially over symmetric intervals.

With our interval \([-\pi, \pi]\), divided into subintervals \(\frac{\pi}{2}\) long, use midpoints to evaluate \(f(x) = \sin x + 1\):
  • For \([-\pi, -\frac{\pi}{2}]\), use \(-\frac{3\pi}{4}\), \(f(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} + 1\).
  • For \([-\frac{\pi}{2}, 0]\), use \(-\frac{\pi}{4}\), \(f(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} + 1\).
  • For \([0, \frac{\pi}{2}]\), use \(\frac{\pi}{4}\), \(f(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + 1\).
  • For \([\frac{\pi}{2}, \pi]\), use \(\frac{3\pi}{4}\), \(f(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} + 1\).
When summed, this midpoint approach evaluates to:\[ \left( -\frac{\sqrt{2}}{2} + 1 \right) \cdot \frac{\pi}{2} + \left( -\frac{\sqrt{2}}{2} + 1 \right) \cdot \frac{\pi}{2} + \left( \frac{\sqrt{2}}{2} + 1 \right) \cdot \frac{\pi}{2} + \left( \frac{\sqrt{2}}{2} + 1 \right) \cdot \frac{\pi}{2} = 2\pi.\]By capturing the function values at the midpoints, this sum takes into account both rises and falls within each subinterval, potentially balancing out overestimations and underestimations.

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Most popular questions from this chapter

In Exercises \(75-78\) , let \(F(x)=\int_{a}^{x} f(t) d t\) for the specified function \(f\) and interval \([a, b] .\) Use a CAS to perform the following steps and answer the questions posed. \begin{equation} \begin{array}{l}{\text { a. Plot the functions } f \text { and } F \text { together over }[a, b] \text { . }} \\ {\text { b. Solve the equation } F^{\prime}(x)=0 . \text { What can you see to be true about }} \\ {\text { the graphs of } f \text { and } F \text { at points where } F^{\prime}(x)=0 \text { . Is your observation }} \\ {\text { borne out by Part } 1 \text { of the Fundamental Theorem coupled }} \\ {\text { with information provided by the first derivative? Explain your }} \\ {\text { answer. }}\\\\{\text { c. Over what intervals (approximately) is the function } F \text { increasing }} \\\ {\text { and decreasing? What is true about } f \text { over those intervals? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { d. Calculate the derivative } f^{\prime} \text { and plot it together with } F . \text { What can }} \\ {\text { you see to be true about the graph of } F \text { at points where } f^{\prime}(x)=0 ?} \\\ {\text { Is your observation borne out by Part } 1 \text { of the Fundamental Theorem? }} \\ {\text { Explain your answer. }}\end{array} \end{equation} $$f(x)=2 x^{4}-17 x^{3}+46 x^{2}-43 x+12,\left[0, \frac{9}{2}\right]$$

Evaluate the integrals in Exercises 53 and 54 $$ \int \frac{(2 r-1) \cos \sqrt{3(2 r-1)^{2}+6}}{\sqrt{3(2 r-1)^{2}+6}} d r $$

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. $$ \int_{0}^{1}\left(x^{2}+1\right) d x=\frac{4}{3} $$

It would be nice if average values of integrable functions obeyed the following rules on an interval \([a, b] .\) $$ \begin{array}{l}{\text { a. } \operatorname{av}(f+g)=\operatorname{av}(f)+\operatorname{av}(g)} \\ {\text { b. } \operatorname{av}(k f)=k \operatorname{av}(f) \quad(\text { any number } k)} \\ {\text { c. } \operatorname{av}(f) \leq \operatorname{av}(g) \text { if } f(x) \leq g(x) \quad \text { on }[a, b]} \\ {\text { Do these rules ever hold? Give reasons for your answers. }}\end{array} $$

Upper and lower sums for increasing functions $$ \begin{array}{l}{\text { a. Suppose the graph of a continuous function } f(x) \text { rises steadily }} \\ {\text { as } x \text { moves from left to right across an interval }[a, b] . \text { Let } P} \\ {\text { be a partition of }[a, b] \text { into } n \text { subintervals of equal length }} \\ {\Delta x=(b-a) / n . \text { Show by referring to the accompanying figure }}\\\ {\text { that the difference between the upper and lower sums for }} \\ {f \text { on this partition can be represented graphically as the area }} \\\ {\text { of a rectangle } R \text { whose dimensions are }[f(b)-f(a)] \text { by } \Delta x \text { . }} \\ {\text { (Hint: The difference } U-L \text { is the sum of areas of rectangles }}\\\\{\text { whose diagonals } Q_{0} Q_{1}, Q_{1} Q_{2}, \ldots, Q_{n-1} Q_{n} \text { lie approximately }} \\ {\text { along the curve. There is no overlapping when these rectan- }} \\ {\text { gles are shifted horizontally onto } R . \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { b. Suppose that instead of being equal, the lengths } \Delta x_{k} \text { of the }} \\ {\text { subintervals of the partition of }[a, b] \text { vary in size. Show that }}\end{array} $$ $$ U-L \leq|f(b)-f(a)| \Delta x_{\max } $$ $$ \begin{array}{l}{\text { where } \Delta x_{\max } \text { is the norm of } P, \text { and hence that } \lim _{ \| P | \rightarrow 0}} \\\ {(U-L)=0}\end{array} $$
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