Problem 57 - Calculus Early Transcendentals

Question

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. $$\int_{1}^{7} \frac{1}{x} d x ; n=6$$

Step-by-Step Solution

Answer

Answer: The left Riemann sum overestimates the value of the definite integral.

Step 1: Sketch the graph of the integrand

The integrand is \(\frac{1}{x}\), which is a typical reciprocal function. Its graph is asymptotic to the x-axis and y-axis, and decreasing in the first quadrant. Since the integral is defined on the interval [1,7], we will focus on this domain in our sketch. The graph will decrease smoothly and monotonically as x increases from 1 to 7.

Step 2: Calculate \(\Delta x\) and the grid points

To find the width of each subinterval, \(\Delta x\), we can use the formula: $$\Delta x = \frac{b - a}{n}$$ where \(a\) is the left endpoint, \(b\) is the right endpoint, and \(n\) is the number of subintervals. In this case, \(a = 1\), \(b = 7\), and \(n = 6\). So, $$\Delta x = \frac{7 - 1}{6} = 1$$ Now we can find the grid points (the points that divide the interval into equal parts). We use \(x_i=a + i \times \Delta x\), with \(i=0,1,\ldots,n\): $$x_0 = 1$$ $$x_1 = 1 + 1 = 2$$ $$x_2 = 2 + 1 = 3$$ $$x_3 = 3 + 1 = 4$$ $$x_4 = 4 + 1 = 5$$ $$x_5 = 5 + 1 = 6$$ $$x_6 = 6 + 1 = 7$$ So the grid points are \(x_0=1, x_1=2, x_2=3, x_3=4, x_4=5, x_5=6, x_6=7\).

Step 3: Calculate the left and right Riemann sums

To find the left Riemann sum, we use the left endpoints of each subinterval to evaluate the function. The left Riemann sum is defined as: $$L_n = \Delta x \sum_{i=0}^{n-1} f(x_i)$$ For this function and partition, $$L_n = 1\left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) = 1\left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right)$$ To find the right Riemann sum, we use the right endpoints of each subinterval to evaluate the function. The right Riemann sum is defined as: $$R_n = \Delta x \sum_{i=1}^{n} f(x_i)$$ For this function and partition, $$R_n = 1\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \right)$$

Step 4: Determine underestimation or overestimation of the Riemann sums

Since the graph of \(\frac{1}{x}\) is decreasing on the interval [1,7], the left Riemann sum will overestimate the value of the definite integral (as it will include the extra areas formed by the left endpoints of each subinterval), while the right Riemann sum will underestimate the value of the definite integral (since it does not capture the areas above the rectangles formed by the right endpoints).

Key Concepts

Definite Integral

A definite integral represents the area under the curve of a function on a specified interval. In mathematical terms, it is the integral of a function with upper and lower limits, denoted by \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration. The definite integral provides a quantitative measure of accumulated value. It is fundamental in calculus as it connects the concept of accumulation with area, providing insights into areas bounded by curves. In our exercise, the integral \( \int_{1}^{7} \frac{1}{x} \, dx \) seeks to find this area under the reciprocal function between \(x = 1\) and \(x=7\). The solution to this integral can also inform us about the total accumulated change from starting point \(x=1\) to ending point \(x=7\). Understanding definite integrals is essential for grasping more advanced calculus topics like the Fundamental Theorem of Calculus.

Reciprocal Function

The reciprocal function, represented as \( f(x) = \frac{1}{x} \), plays a significant role in understanding inverse relationships. Here, each value of \(x\) has a "reciprocal" or opposite value, which provides interesting properties. For positive \(x\), the function is always positive and approaches zero as \(x\) increases, indicating a decrease. Its graph is asymptotic towards the x-axis and y-axis, implying that it never touches these axes but gets infinitely close. In our given integral for \( f(x) = \frac{1}{x} \), the function displays these properties over the interval from 1 to 7, creating a shape resembling a hyperbola. Recognizing the behavior and characteristics of reciprocal functions helps students understand other significant aspects of mathematics like behavior at infinity, asymptotic limits, and part of calculus concerning inverse operations.

Decreasing Function

A decreasing function is one where the function values decline as the input values (usually x-values on the x-axis) increase. This is precisely what \( f(x) = \frac{1}{x} \) exhibits between 1 and 7. As you move from left to right along the x-axis in this range, the y-values (or function values) continually drop. There is a clear

• Inverse relationship
• Decrease in function value with increase in input
• Absence of critical points (remains smooth)

This property plays a key role in determining the behavior of Riemann sums. For decreasing functions like \( \frac{1}{x} \), the left Riemann sum tends to overestimate the definite integral's actual value, as the rectangles surpass the curve at each left endpoint. Conversely, the right Riemann sum underestimates, since the curve lies above the rectangles drawn at right endpoints. Understanding the concept of decreasing functions is integral to mastering integral calculus, especially in estimating areas with Riemann sums.

Grid Points

Grid points are essential components when calculating Riemann sums, as they divide the interval into equal segments or subintervals. For the integral \( \int_{1}^{7} \frac{1}{x} \, dx \) with \(n = 6\), the interval is partitioned into six equal parts. By using the formula \( \Delta x = \frac{b-a}{n} \), where \(a\) and \(b\) are limits of integration, each subinterval width is 1. The grid points obtained are \(x_0 = 1\), \(x_1 = 2\), \(x_2 = 3\), up to \(x_6 = 7\), forming a clear partitioning along the x-axis. These points delineate where the function is sampled and evaluated in Riemann sum calculations.
Grid points provide the basis for every rectangle in a Riemann sum estimation, and without them, it would be challenging to approximate the areas under curves effectively. They are fundamental in numerical integration techniques, especially when analytical solutions are challenging or impossible to derive. Understanding grid points offers a practical approach to calculus, bridging the gap between pure theory and its application.