Question

Evaluate the Riemann sums \(L_{4}\) and \(R_{4}\) for the following functions over the specified interval. Compare your answer with the exact answer, when possible, or use a calculator to determine the answer. \(y=x^{2} \sin x\) over \([0, \pi]\)

Short answer

L_4 and R_4 help approximate the integral using Riemann sums.

Step-by-step solution

Determine the interval and partition

The interval given is \([0, \pi]\). We need to partition this interval into 4 equal subintervals because we're calculating \ L_{4} \ and \ R_{4} \.The width of each subinterval is \ \, \Delta x = \frac{b-a}{n} = \frac{\pi - 0}{4} = \frac{\pi}{4} \, \.

Compute Left Riemann Sum \(L_4\)

For the left Riemann sum \ L_4 \, we use the left endpoints of each subinterval. The left endpoints are \ x_0 = 0, x_1 = \frac{\pi}{4}, x_2 = \frac{\pi}{2}, x_3 = \frac{3\pi}{4} \.\[L_4 = \Delta x \left[f(x_0) + f(x_1) + f(x_2) + f(x_3)\right]\]\[L_4 = \frac{\pi}{4} \left[0^2 \sin(0) + \left(\frac{\pi}{4}\right)^2 \sin\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + \left(\frac{3\pi}{4}\right)^2 \sin\left(\frac{3\pi}{4}\right)\right]\]\[L_4 = \frac{\pi}{4} \left[0 + \frac{\pi^2}{16}\frac{\sqrt{2}}{2} + \frac{\pi^2}{4} \cdot 1 + \frac{9\pi^2}{16}\frac{\sqrt{2}}{2}\right]\]

Key concepts

Left Riemann Sum

The Left Riemann Sum is a method used in calculus to approximate the area under a curve. It involves taking several rectangles and summing their areas to estimate this region. For each subinterval of the curve, we take the left endpoint to determine the height of the rectangle. This height is then multiplied by the width of the subinterval, giving the area of each rectangle.

In the problem, for the function \(y = x^2 \sin x\) over the interval \([0, \pi]\), we divided the interval into four equal parts, giving us subintervals of width \(\Delta x = \frac{\pi}{4}\). The left endpoints of these subintervals are \(0, \frac{\pi}{4}, \frac{\pi}{2},\) and \(\frac{3\pi}{4}\).

To find the Left Riemann Sum \(L_4\), we calculate the value of the function \(f(x)\) at these left endpoints and multiply each by the subinterval width:

• At \(x_0 = 0\), \(f(x_0) = 0^2 \sin(0) = 0\)
• At \(x_1 = \frac{\pi}{4}\), \(f(x_1) = \left(\frac{\pi}{4}\right)^2 \sin\left(\frac{\pi}{4}\right)\)
• At \(x_2 = \frac{\pi}{2}\), \(f(x_2) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right)\)
• At \(x_3 = \frac{3\pi}{4}\), \(f(x_3) = \left(\frac{3\pi}{4}\right)^2 \sin\left(\frac{3\pi}{4}\right)\)

Finally, we add these products together to get the approximate area.

Right Riemann Sum

The Right Riemann Sum, similar to the Left Riemann Sum, is another technique used to estimate the area under a curve. Instead of the left endpoint, the right endpoint of each subinterval is used to determine the height of the rectangle. This method often yields a different approximation than the Left Riemann Sum, especially if the function is not constant.

In our exercise for the function \(y = x^2 \sin x\) over \([0, \pi]\), the right endpoints are considered. These endpoints will be \(\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4},\) and \(\pi\).

To compute the Right Riemann Sum \(R_4\), we evaluate the function at these right endpoints and multiply by \(\Delta x = \frac{\pi}{4}\):

• At \(x_1 = \frac{\pi}{4}\), \(f(x_1) = \left(\frac{\pi}{4}\right)^2 \sin\left(\frac{\pi}{4}\right)\)
• At \(x_2 = \frac{\pi}{2}\), \(f(x_2) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right)\)
• At \(x_3 = \frac{3\pi}{4}\), \(f(x_3) = \left(\frac{3\pi}{4}\right)^2 \sin\left(\frac{3\pi}{4}\right)\)
• At \(x_4 = \pi\), \(f(x_4) = \pi^2 \sin(\pi) = 0\)

We sum these computations to obtain the total approximate area using the Right Riemann Sum.

Interval Partitioning

Interval partitioning is the process of dividing an interval into smaller subintervals. When using Riemann sums, this technique is crucial because the sum involves calculating areas over these subintervals. The number of partitions, denoted by \(n\), and their widths, \(\Delta x\), play a significant role in obtaining a more accurate approximation.

In terms of calculating Riemann sums, partitioning helps in managing the calculations easily by setting consistent dimensions for the rectangles used in the approximation.

For our exercise, we are given the interval \([0, \pi]\) and must partition it into four equal parts because we're finding \(L_4\) and \(R_4\). This division results in subintervals of width \(\Delta x = \frac{b-a}{n} = \frac{\pi}{4}\).

With interval partitioning, we are able to define the subintervals:

• First subinterval: \([0, \frac{\pi}{4}]\)
• Second subinterval: \([\frac{\pi}{4}, \frac{\pi}{2}]\)
• Third subinterval: \([\frac{\pi}{2}, \frac{3\pi}{4}]\)
• Fourth subinterval: \([\frac{3\pi}{4}, \pi]\)

This partitioning allows us to use a structured approach for both the Left and Right Riemann Sums to approximate the integral.