Question

Left and right Riemann sums Complete the following steps for the given function, interval, and value of \(n.\) a. Sketch the graph of the function on the given interval. b. Calculate \(\Delta x\) and the grid points \(x_{0}, x_{1}, \ldots, x_{n^{*}}.\) c. Illustrate the left and right Riemann sums. Then determine which Riemann sum underestimates and which sum overestimates the area under the curve. d. Calculate the left and right Riemann sums. $$f(x)=2 x^{2} \text { on }[1,6] ; n=5$$

Short answer

Answer: The left Riemann sum is \(110\) and the right Riemann sum is \(190\).

Step-by-step solution

Sketch the graph of the function

To sketch the graph of the function \(f(x)=2x^2\) on the interval \([1,6]\), you will need to identify the behavior of the function in this interval. To do this, plug in the values of \(x\) from \(1\) to \(6\) into \(f(x)\) and plot the resulting points. Since it's a parabola, it's easy to see that the curve will be increasing on the entire interval.

Calculate \(\Delta x\) and the grid points

We can calculate \(\Delta x\) by using the formula \(\Delta x = \frac{b-a}{n}\), where \(a\) and \(b\) are the endpoints of the interval and \(n\) is the number of subintervals. In this case, we have \(a=1\), \(b=6\), and \(n=5\). So, \(\Delta x = \frac{6-1}{5} = 1\). The grid points are \(x_0=1\), \(x_1=2\), \(x_2=3\), \(x_3=4\), \(x_4=5\), and \(x_5=6\).

Illustrate the left and right Riemann sums

To illustrate the left Riemann sum, we will use the left-hand endpoints of the subintervals as the approximating points. The left Riemann sum will be the sum of the areas of rectangles with heights given by the function values. The same approach will be used for the right Riemann sum, but with the right-hand endpoints of the subintervals. Since the function is increasing, the left Riemann sum will underestimate the area under the curve, while the right Riemann sum will overestimate it.

Calculate the left and right Riemann sums

To calculate the left Riemann sum, first find the function values at the left-hand endpoints: \(f(1)\), \(f(2)\), \(f(3)\), \(f(4)\), and \(f(5)\). Then multiply each function value by \(\Delta x\) and sum them up: $$L_n = \Delta x \sum_{i=0}^{n-1} f(x_i) = 1 \left( 2(1)^2 + 2(2)^2 + 2(3)^2 + 2(4)^2 + 2(5)^2 \right) = 110$$ Similarly, calculate the right Riemann sum by finding the function values at the right-hand endpoints: \(f(2)\), \(f(3)\), \(f(4)\), \(f(5)\), and \(f(6)\). Then multiply each function value by \(\Delta x\) and sum them up: $$R_n = \Delta x \sum_{i=1}^{n} f(x_i) = 1 \left( 2(2)^2 + 2(3)^2 + 2(4)^2 + 2(5)^2 + 2(6)^2 \right) = 190$$ So the left Riemann sum is \(110\) and the right Riemann sum is \(190\).

Key concepts

Left Riemann Sum

The left Riemann sum is one of the methods to approximate the area under a curve over a given interval. This method uses the "left-hand" endpoints of each subinterval to estimate the height of each rectangle. Essentially, imagine you have an increasing curve, and you want to fill the area with rectangles. For the left Riemann sum, the height of each rectangle is determined by the value of the function at the beginning of each subinterval.

Here's how you can calculate it step-by-step:

• Divide the interval \([a, b]\) into \(n\) equal subintervals. This is done by determining \(\Delta x = \frac{b-a}{n}\).
• Identify the grid points, which are the boundaries of your subintervals. This will include \([x_0, x_1, ..., x_n]\).
• For each subinterval, calculate the function value at its left endpoint.
• The left Riemann sum \(L_n\) is then \(\Delta x\) multiplied by the sum of all these function values: \(L_n = \Delta x \sum_{i=0}^{n-1} f(x_i)\).

It's crucial to remember that if the function is increasing, the left Riemann sum will typically underestimate the true area under the curve, as it takes the shortest height from each subinterval.

Right Riemann Sum

Another method to find an approximation of the area under a curve is the right Riemann sum. This one is quite similar to the left Riemann sum, but instead, it uses the right-hand endpoints of the subintervals to determine the height of each rectangle. This means the top-right corner of each rectangle touches the curve rather than the top-left.

To calculate the right Riemann sum:

• As with the left sum, divide the interval \([a, b]\) into \(n\) equal parts using \(\Delta x = \frac{b-a}{n}\).
• Identify the grid points, focusing on \([x_1, x_2, ..., x_n]\) for calculation.
• Calculate the function's value at the right endpoints of each subinterval.
• Multiply \(\Delta x\) by the sum of these values to get your right Riemann sum: \(R_n = \Delta x \sum_{i=1}^{n} f(x_i)\).

For an increasing function, the right Riemann sum will usually overestimate the area under the curve because it uses the highest point of each part as the rectangle's height.

Area Under the Curve

Finding the area under a curve is a fundamental concept in calculus, closely tied to integration. When you're given a continuous function, and you want to determine the space it encloses with the x-axis over a certain interval, you're essentially looking for this area. Riemann sums provide one way to approximate this area by breaking it into manageable pieces.

Using Riemann sums:

• Riemann sums offer a series of approximations that get closer to the true area as the number of rectangles (subintervals) increases.
• The left and right Riemann sums are two particular methods, each with unique usefulness depending on whether the function is increasing or decreasing.
• For many functions, increasing the number of subintervals \(n\) reduces the error in the approximation, leading both sums to converge on the actual integral of the function.

Overall, Riemann sums introduce the concept of integration by providing a simple way to sum up small areas to approximate a total area. These sums form the foundation for understanding more advanced calculus such as definite integrals.