Problem 12 - A Graphical Approach to Precalculus with Limits

Question

Approximate the area under the graph of \(f(x)\) and above the \(x\) -axis, using each of the following methods with \(n=4\). (a) Use left endpoints. (b) Use right endpoints. (c) Average the answers in parts ( \(a\) ) and ( \(b\) ). (d) Use midpoints. $$f(x)=\frac{2}{x} \text { from } x=1 \text { to } x=9$$

Step-by-Step Solution

Answer

Approximations: Left = 7.36, Right = 3.16, Average = 5.26, Midpoint = 4.16.

Step 1: Calculate the Width of Each Subinterval

The interval from \(x = 1\) to \(x = 9\) is divided into \(n = 4\) equal subintervals. The width \(\Delta x\) of each subinterval is given by: \[\Delta x = \frac{(9 - 1)}{4} = 2\]

Step 2: Left Endpoint Riemann Sum

For left endpoints, the four subintervals are \([1, 3], [3, 5], [5, 7], [7, 9]\).The left endpoints are \(x_0 = 1, x_1 = 3, x_2 = 5, x_3 = 7\).Calculate the sum:\[\sum_{i=0}^{3} f(x_i) \cdot \Delta x = \left(f(1) + f(3) + f(5) + f(7)\right) \cdot 2 = \left(\frac{2}{1} + \frac{2}{3} + \frac{2}{5} + \frac{2}{7}\right) \cdot 2\]Evaluating the individual functions:\[= \left(2 + \frac{2}{3} + \frac{2}{5} + \frac{2}{7}\right) \cdot 2\]\[= (2 + 0.67 + 0.4 + 0.29) \cdot 2 = 7.36\]

Step 3: Right Endpoint Riemann Sum

For right endpoints, the four subintervals are \([1, 3], [3, 5], [5, 7], [7, 9]\).The right endpoints are \(x_1 = 3, x_2 = 5, x_3 = 7, x_4 = 9\).Calculate the sum:\[\sum_{i=1}^{4} f(x_i) \cdot \Delta x = \left(f(3) + f(5) + f(7) + f(9)\right) \cdot 2 = \left(\frac{2}{3} + \frac{2}{5} + \frac{2}{7} + \frac{2}{9}\right) \cdot 2\]Evaluating the individual functions:\[= \left(0.67 + 0.4 + 0.29 + 0.22\right) \cdot 2\]\[= 3.16\]

Step 4: Average of Left and Right Endpoint Approximations

Compute the average of the left and right endpoint sums.\[\text{Average} = \frac{7.36 + 3.16}{2} = 5.26\]

Step 5: Midpoint Riemann Sum

For the midpoint approximation, the midpoints of each subinterval are calculated: The subintervals are \([1, 3], [3, 5], [5, 7], [7, 9]\), and their midpoints are \(x_0 = 2, x_1 = 4, x_2 = 6, x_3 = 8\).Calculate the midpoint sum:\[\sum f(x_i^*) \cdot \Delta x = \left(f(2) + f(4) + f(6) + f(8)\right) \cdot 2 = \left(\frac{2}{2} + \frac{2}{4} + \frac{2}{6} + \frac{2}{8}\right) \cdot 2\]Simplifying further:\[= \left(1 + 0.5 + 0.33 + 0.25\right) \cdot 2 = 4.16\]

Key Concepts

Left Endpoint Method

The Left Endpoint Method is an approach to approximate the area under a curve by summing up rectangles whose heights are determined by the left endpoints of subintervals. In simpler terms, you start from the leftmost point of each subinterval, evaluate the function there, and use it as the height of your rectangle. This method creates rectangles that may underestimate or overestimate the area, depending on whether the function is increasing or decreasing in that interval.

For example, if you're working within the interval from 1 to 9 with 4 subintervals, you calculate the function values at the left endpoints, e.g., 1, 3, 5, and 7 for function \(f(x)=\frac{2}{x}\). Multiply each result by the width of the subinterval to find the approximate area. By doing this:

• The function at 1 gives you 2, at 3 it gives about 0.67, and so on.
• Sum these values and multiply the total by the width of each subinterval (2), resulting in 7.36 as an approximation of the area.

This method is straightforward but can lack accuracy depending on the function's shape.

Right Endpoint Method

The Right Endpoint Method is similar to the Left Endpoint Method but as the name signifies, utilizes the function's value at the right endpoint of each subinterval. This shift in evaluating point can significantly change the estimated area, particularly when the function behaves differently at the end of intervals.

For the interval from 1 to 9 split into 4 subintervals, you use the points 3, 5, 7, and 9 to determine your rectangle heights for \(f(x)=\frac{2}{x}\). Here,

• The function at 3 returns about 0.67 and keeps decreasing as 5, 7, and 9 are considered.
• The resulting sum is then multiplied by the interval width (2), providing an estimate of 3.16 for the area.

This method can also either overestimate or underestimate based on the function's behavior, leading to different results than the left endpoint approach.

Midpoint Rule

The Midpoint Rule offers a balanced alternative to Left and Right Endpoint Methods by using the midpoint in each subinterval to determine the rectangle's height. This approach can often provide a more accurate approximation of the area under a curve by reducing the error associated with only using either end of the subinterval.

In the given interval from 1 to 9, with 4 subdivisions, the midpoints are 2, 4, 6, and 8. For the function \(f(x) = \frac{2}{x}\):

• At these midpoints, the function values are 1, 0.5, 0.33, and 0.25, respectively.
• Add these values up and multiply by the subinterval width (2), producing a total area approximation of 4.16.

This method slightly mitigates the limitations of using just ends and can be a more reliable estimate in many scenarios.

Area Under a Curve

The concept of finding the Area Under a Curve is essentially about finding the integral of a function within a specific range. In the field of calculus, it represents a fundamental way to measure the "accumulation" of quantities, like distance or volume, which are otherwise difficult to evaluate from a basic graph.

In practice, calculating this area offers insights into a variety of real-world problems by providing definite answers to what might otherwise be an infinite list of points in a curve.

Whether using Riemann Sums or other numerical integration techniques, breaking the problem into manageable pieces like rectangles—as done through methods like Left Endpoint, Right Endpoint, or Midpoint—helps approximate these areas effectively.

Ultimately, understanding these methods builds a solid foundation for tackling more advanced calculus problems where exact calculation of an integral isn't possible or practical.