Problem 3 - Thomas Calculus

Question

In Exercises \(1-4,\) use finite approximations to estimate the area under the graph of the function using $$\begin{array}{l}{\text { a. a lower sum with two rectangles of equal width. }} \\ {\text { b. a lower sum with four rectangles of equal width. }} \\\ {\text { c. an upper sum with two rectangles of equal width. }} \\ {\text { d. an upper sum with four rectangles of equal width. }}\end{array}$$ \(f(x)=1 / x\) between \(x=1\) and \(x=5\)

Step-by-Step Solution

Answer

a) \(\frac{8}{3}\), b) \(\frac{25}{12}\), c) \(\frac{16}{15}\), d) \(\frac{77}{60}\).

Step 1: Determine the interval and width of subintervals

We are asked to find the area under the curve of the function \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \). The interval is \([1,5]\) and its length is 4. For parts a and c, we divide this into 2 equal subintervals, each of width \( \Delta x = \frac{4}{2} = 2 \). For parts b and d, we divide this into 4 equal subintervals, each of width \( \Delta x = \frac{4}{4} = 1 \).

Step 2: Lower sum with two rectangles (a)

For two rectangles, take the left endpoint of each subinterval to find the height. The subintervals are \([1, 3]\) and \([3, 5]\). The heights are \( f(1) = 1 \) and \( f(3) = \frac{1}{3} \). The lower sum is: \[ (2)(1) + (2)\left(\frac{1}{3}\right) = 2 + \frac{2}{3} = \frac{8}{3}. \]

Step 3: Lower sum with four rectangles (b)

For four rectangles, take the left endpoint to find the heights. Subintervals: \([1, 2]\), \([2, 3]\), \([3, 4]\), \([4, 5]\). Heights are \( f(1) = 1 \), \( f(2) = \frac{1}{2} \), \( f(3) = \frac{1}{3} \), \( f(4) = \frac{1}{4} \). Lower sum: \[ (1)(1) + (1)\left(\frac{1}{2}\right) + (1)\left(\frac{1}{3}\right) + (1)\left(\frac{1}{4}\right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12}. \]

Step 4: Upper sum with two rectangles (c)

For two rectangles, take the right endpoint to find the height. Subintervals: \([1, 3]\) and \([3, 5]\). Heights are \( f(3) = \frac{1}{3} \) and \( f(5) = \frac{1}{5} \). Upper sum: \[ (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{5}\right) = \frac{2}{3} + \frac{2}{5} = \frac{16}{15}. \]

Step 5: Upper sum with four rectangles (d)

For four rectangles, take the right endpoint to find the heights. Subintervals: \([1, 2]\), \([2, 3]\), \([3, 4]\), \([4, 5]\). Heights are \( f(2) = \frac{1}{2} \), \( f(3) = \frac{1}{3} \), \( f(4) = \frac{1}{4} \), \( f(5) = \frac{1}{5} \). Upper sum: \[ (1)\left(\frac{1}{2}\right) + (1)\left(\frac{1}{3}\right) + (1)\left(\frac{1}{4}\right) + (1)\left(\frac{1}{5}\right) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{77}{60}. \]

Key Concepts

Finite Approximations

Finite approximations are a method used to estimate the area under a curve. The idea is to divide the area into simple geometric shapes whose areas are easy to calculate, like rectangles. The more rectangles used, the more accurate the approximation of the area. This approach is especially useful for understanding integral calculus. By using finite approximations, one can explore the concept of the definite integral. The process involves selecting key points on the interval and aligning them with the curve to form rectangles. These approximations come in two primary forms: lower sums and upper sums. Together these methods show how close we can get to the actual area under a curve by using a set number of simple shapes.

Lower Sum

The lower sum method estimates the area under a curve by forming rectangles that sit below the curve. Each rectangle’s height is determined by the value of the function at the left endpoint of the subinterval. This often results in an underestimation. Here’s why:

• We calculate the height of each rectangle using the function value at the left endpoint of each subinterval.
• This usually means the rectangles can miss out on areas where the curve rises above the left endpoint value.

For example, in estimating the area under the curve of the function \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \), we break down the interval into subintervals. The heights of these rectangles are calculated at the left end, forming what is known as a ‘lower sum’. The lower the number and width of the subintervals, the more you'll likely underestimate the area compared to using an infinite number of infinitely small widths, which defines an integral.

Upper Sum

The upper sum works similarly to the lower sum but instead aims to overestimate the area. Conversely to the lower sum, we use the value of the function at the right endpoint of each subinterval to determine the height of each rectangle. This choice usually results in overestimation because:

• The rectangles crafted often exceed the height of the curve, capturing more area than what the curve alone may encompass between endpoints.
• This ensures a buffer over the curve from the right side.

For the function scenario \( f(x) = \frac{1}{x} \), forming upper sums involves taking the height at the right endpoints of subintervals as you slice the main interval \([1,5]\). As a rule of thumb, the more rectangles you use, with more precise heights at each interval point, the upper sum will more closely approximate the actual integral area.

Subintervals

Subintervals are sections into which a larger interval is divided for the purpose of approximation. The original interval in this case is \([1, 5]\), and by breaking this interval into either two or four sections (based on the task addition), we form the basis for creating geometrical shapes like rectangles.

• Each subinterval is chosen such that its width is equal, simplifying the math and consistency.
• A two-rectangle approximation will have wider subintervals than a four-rectangle approach, affecting precision.

Deciding on the number of subintervals is a crucial step. More subintervals usually translate to a better approximation of the integral, though may involve more computational steps. As subinterval width decreases and their number increases, the finite sums converge towards the true area under the curve, echoing the foundational idea of integrals in calculus.