Problem 36 - University Calculus: Early Transcendentals

Question

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.) $$f(x)=\sin x+1, \quad[-\pi, \pi]$$

Step-by-Step Solution

Answer

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

Step 1: 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 \( \Delta x = \frac{2\pi}{4} = \frac{\pi}{2} \).

Step 2: Calculate Subintervals

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

Step 3: 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.

Step 4: 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 \(\Delta x = \frac{\pi}{2}\).

Step 5: 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 \(\Delta x = \frac{\pi}{2}\).

Step 6: 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 \(\Delta x = \frac{\pi}{2}\).

Key Concepts

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.