Problem 12 - Calculus Volume 1

Question

Let \(L_{n}\) denote the left-endpoint sum using \(n\) subintervals and let \(R_{n}\) denote the corresponding right-endpoint sum. In the following exercises, compute the indicated left and right sums for the given functions on the indicated interval. $$L_{4} \text { for } f(x)=\frac{1}{x-1} \text { on }[2,3]$$

Step-by-Step Solution

Answer

The left-endpoint sum \(L_4\) is approximately 0.7595.

Step 1: Determine the Width of Each Subinterval

Given the interval \([2, 3]\) and \(n=4\) subintervals, calculate the width of each subinterval, denoted by \(\Delta x\). Compute it using the formula: \[\Delta x = \frac{b-a}{n} = \frac{3-2}{4} = \frac{1}{4}.\]

Step 2: Calculate the Left-endpoint Values

For the left-endpoint sum, identify the x-values at the left of each subinterval using the formula \(x_i = a + i\Delta x\) where \(i = 0, 1, 2, 3\). Therefore, the x-values are as follows:- \(x_0 = 2 + 0 \times \frac{1}{4} = 2\)- \(x_1 = 2 + 1 \times \frac{1}{4} = 2.25\)- \(x_2 = 2 + 2 \times \frac{1}{4} = 2.5\)- \(x_3 = 2 + 3 \times \frac{1}{4} = 2.75\)

Step 3: Evaluate the Function at Each Left-endpoint

Evaluate \(f(x) = \frac{1}{x-1}\) for each of the left-endpoints computed:- \(f(2) = \frac{1}{2-1} = 1\)- \(f(2.25) = \frac{1}{2.25-1} = \frac{1}{1.25} = 0.8\)- \(f(2.5) = \frac{1}{2.5-1} = \frac{1}{1.5} = \frac{2}{3} = 0.6667\)- \(f(2.75) = \frac{1}{2.75-1} = \frac{1}{1.75} \approx 0.5714\)

Step 4: Calculate the Left-endpoint Sum

Combine the function evaluations with the computed interval width, \(\Delta x = \frac{1}{4}\). The sum is then:\[L_4 = \Delta x \times \left[ f(2) + f(2.25) + f(2.5) + f(2.75) \right]\]\[= \frac{1}{4} \times (1 + 0.8 + 0.6667 + 0.5714)\]\[= \frac{1}{4} \times 3.0381 = 0.7595\].

Key Concepts

Understanding Left-Endpoint Sum

The left-endpoint sum is a method used to estimate the area under a curve over a specific interval. It does this by summing the areas of rectangles where the height of each rectangle is the value of the function at the left endpoint of each subinterval. In this calculation: - We divide our interval into equal parts, known as subintervals. - We determine the left-endpoint of each subinterval, which is essentially the first value in each set partitioned in the interval. For instance, if you have a function defined from [2, 3] broken into 4 subintervals, you are using the left-end of each subinterval to compute the function value. This means the heights of our rectangles in this example are determined by the function evaluated at 2, 2.25, 2.5, and 2.75. It is a simple but powerful tool in calculus, giving insight into the function's behavior without requiring the computation of complex integrals directly.

The Process of Function Evaluation

Function evaluation involves calculating the value of a function at specific points. In the context of Riemann sums, you're finding the function values at the left endpoints of your subintervals.In our example, we use the function: \(f(x) = \frac{1}{x-1}\) When evaluating, substitute each x-value of your left-endpoint into this equation to find the height of each rectangle:

• For \(x = 2\), \(f(2) = \frac{1}{2 - 1} = 1\)
• For \(x = 2.25\), \(f(2.25) = \frac{1}{2.25 - 1} = 0.8\)
• For \(x = 2.5\), \(f(2.5) = \frac{1}{2.5 - 1} = 0.6667\)
• For \(x = 2.75\), \(f(2.75) = \frac{1}{2.75 - 1} = 0.5714\)

Ignoring the intricacy of the function itself, evaluating it at strategic points helps conceptualize the net area or approximation of it for certain bounds.

Calculating Subinterval Width

Subinterval width, denoted by \(\Delta x\), is a crucial part of calculating Riemann sums. It's what divides your total interval into smaller, manageable segments and is calculated by the formula:\[\Delta x = \frac{b-a}{n}\]where \(b\) is the upper limit and \(a\) the lower limit of your interval, while \(n\) is the number of subintervals you want.For the interval \([2, 3]\) divided into 4 parts, this results in a subinterval width:\(\Delta x = \frac{3 - 2}{4} = \frac{1}{4}\)Understanding this width is vital because it determines how many left or right endpoints you will evaluate, and therefore, how many times you will calculate your function. Accurate division into subintervals connects directly with the precision of your Riemann sum approximation, linking each step towards the function's comprehensive area approximation.