Question

Approximating definite integrals Complete the following steps for the given integral and the given value of \(n .\) a. Sketch the graph of the integrand on the interval of integration. b. Calculate \(\Delta x\) and the grid points \(x_{0}, x_{1}, \ldots, x_{n},\) assuming a regular partition. c. Calculate the left and right Riemann sums for the given value of \(n .\) d. Determine which Riemann sum (left or right) underestimates the value of the definite integral and which overestimates the value of the definite integral. $$\int_{1}^{7} \frac{1}{x} d x ; n=6$$

Short answer

Answer: The left Riemann sum overestimates the definite integral of the function \(f(x) = \frac{1}{x}\) from 1 to 7 using 6 partitions, and the right Riemann sum underestimates it.

Step-by-step solution

Sketch the graph of the integrand on the interval of integration

To sketch the graph of the integrand function \(f(x) = \frac{1}{x}\), we will graph the function on the domain \([1, 7]\). The graph looks like a continuous, decreasing curve with a vertical asymptote at x = 0 (not included in the domain).

Calculate \(\Delta x\) and grid points \(x_0, x_1, ..., x_n\)

To calculate \(\Delta x\), we will use the formula \(\Delta x = \frac{b - a}{n}\) where \(a = 1\), \(b = 7\), and \(n = 6\): $$\Delta x = \frac{7 - 1}{6} = 1$$ Now, let's compute the grid points for the interval [1,7]: $$x_0 = 1, x_1 = 2, x_2 = 3, x_3 = 4, x_4 = 5, x_5 = 6, x_6 = 7$$

Calculate the left and right Riemann sums

For the left Riemann sum, we will use the values of the function at the left endpoints of the partition intervals, and for the right Riemann sum, we will use the values at the right endpoints. Left Riemann sum: \(L_n = \Delta x[f(x_0) + f(x_1) + f(x_2)+ f(x_3) + f(x_4) + f(x_5)]\) $$L_n = 1\left[\frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right] = 2.45$$ Right Riemann sum: \(R_n = \Delta x[f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5)+ f(x_6)]\) $$R_n = 1\left[\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right] = 1.97$$

Determine which Riemann sum underestimates and overestimates the definite integral

The integrand function \(f(x) = \frac{1}{x}\) is decreasing on the interval [1, 7]. Thus, the left Riemann sum will overestimate the definite integral's value since the rectangle's height is greater than the function's value at the right endpoint of each interval. On the other hand, the right Riemann sum will underestimate the definite integral because the rectangle's height is smaller than the function's value at the left endpoint of each interval. So, the left Riemann sum overestimates the integral and the right Riemann sum underestimates it.

Key concepts

Riemann Sums

Riemann sums are a powerful tool for understanding definite integrals. They allow us to approximate the area under a curve by summing up the areas of simpler, easily calculable shapes, like rectangles. This process essentially provides an estimate of the integral, which becomes more accurate as we use more rectangles. In our exercise, we explored both left and right Riemann sums. These two types vary based on whether the left or right endpoint of the interval is used to calculate the height of the rectangle. This small difference can impact whether the sum overestimates or underestimates the actual integral value.
Riemann sums are very useful because they lay the foundation for the concept of limits and integrals. By partitioning a space into smaller pieces, and summing them up, we begin to see the idea of an integral taking shape.

Approximating Integrals

Approximating integrals is crucial when an exact calculation is either cumbersome or impossible. One of the simplest ways to approximate an integral is by using Riemann sums, as detailed in the exercise. By adding up rectangles under the curve, we obtain an estimate of the area, and hence, the value of the integral.
Approximating comes into play especially when you deal with functions that are not easily integrable using standard calculus techniques. While the approximation might not always be exact, it provides a valuable estimate that becomes more precise as you increase the number of partitions. Additionally, using approximations help in gaining a better understanding of how integrals behave with different functions and intervals.

Integrand Function

The integrand function is the function we are integrating, which in this case was given by \( f(x) = \frac{1}{x} \). Understanding the behavior of the integrand function is essential before proceeding with approximation techniques like Riemann sums.
The function \( \frac{1}{x} \) is continuous and decreasing on the interval \([1, 7]\). This behavior affects whether the Riemann sums overestimate or underestimate the integral. Because the function decreases, the left sum, which uses the left endpoint of each interval, overestimates the area, while the right sum underestimates it. Visualizing the function first by sketching the graph, as in step 1 of the exercise, helps grasp how the integrand impacts the area estimation under the curve.

Partition Intervals

Partitioning the interval of integration is a critical step in forming a Riemann sum. This involves dividing the interval \( [a, b] \) into \(n\) equal subintervals, as was done in our exercise from \(1\) to \(7\) with \(n = 6\). Each partition then provides a segment called a grid point, and the length of each subinterval is called \( \Delta x \).
In our example, \( \Delta x \) was calculated as \(\frac{b - a}{n}\), resulting in a size of \(1\) for each partition. Correct partitioning is vital as it influences the accuracy of the Riemann sum approximation. More partitions often lead to a more accurate approximation as they provide a better fit of the rectangles to the contour of the curve.