Problem 14 - 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. $$L_{6} \text { for } f(x)=\frac{1}{x(x-1)} \text { on }[2,5]$$

Step-by-Step Solution

Answer

The left-endpoint sum \(L_6\) is \(\frac{499}{1260}.\)

Step 1: Determine the Width of Subintervals

We have 6 subintervals on the interval \([2, 5]\). To find the width \(\Delta x\) of each subinterval, use the formula: \[ \Delta x = \frac{{b-a}}{n} \] where \(a = 2\), \(b = 5\), and \(n = 6\). Thus, \( \Delta x = \frac{5-2}{6} = \frac{1}{2}. \)

Step 2: Identify the Left-Endpoints

For a left-endpoint sum \(L_6\), the endpoints of the subintervals will start from \(x_0 = 2\) and continue by adding \(\Delta x\) until we reach just before the last subinterval's end: \(x_0, x_1, x_2, x_3, x_4, x_5\). Calculate the left-endpoints: \(x_i = 2 + i\cdot\frac{1}{2} \) for \(i = 0, 1, 2, 3, 4, 5\): they are 2, 2.5, 3, 3.5, 4, and 4.5.

Step 3: Evaluate the Function at Each Left-Endpoint

Evaluate the function \(f(x) = \frac{1}{x(x-1)}\) at each left-endpoint: \[f(2) = \frac{1}{2(1)} = \frac{1}{2},\] \[f(2.5) = \frac{1}{2.5(1.5)} = \frac{2}{15},\] \[f(3) = \frac{1}{3(2)} = \frac{1}{6},\] \[f(3.5) = \frac{1}{3.5(2.5)} = \frac{4}{35},\] \[f(4) = \frac{1}{4(3)} = \frac{1}{12},\] \[f(4.5) = \frac{1}{4.5(3.5)} = \frac{2}{63}.\]

Step 4: Compute the Left-Endpoint Sum

The left-endpoint sum \(L_6\) is given by \(L_6 = \Delta x \sum_{i=0}^{5} f(x_i)\). Substitute the function values and \(\Delta x\): \[ L_6 = \frac{1}{2}\left( \frac{1}{2} + \frac{2}{15} + \frac{1}{6} + \frac{4}{35} + \frac{1}{12} + \frac{2}{63} \right).\] Calculate the sum: \[ L_6 = \frac{1}{2} \left( \frac{435 + 112 + 210 + 96 + 105 + 40}{1260} \right) = \frac{998}{2520}. \] Simplify the fraction: \( L_6 = \frac{499}{1260}.\)

Key Concepts

Understanding Left-Endpoint Sums

In numerical integration, left-endpoint sums provide an approximate way to integrate a function when finding the exact integral is difficult. It involves partitioning the interval

• Let's say we have an interval \[a, b\] divided into \ n \ subintervals.
• The left-endpoint sum, denoted as \ L_n \, sums up the areas of rectangles. Each rectangle's height is the function value at the left endpoint of each subinterval.
• The width of each of these rectangles is \ \Delta x \, the length of each subinterval.

For example, when dealing with the function \(f(x) = \frac{1}{x(x-1)}\) over the interval \[2, 5\] with 6 subintervals, we first divide the interval into equal parts. Here, \ \Delta x = \frac{1}{2} \. Each left endpoint is calculated as \[x_i = a + i\cdot\Delta x\] for \ i = 0, 1, 2, ..., n-1 \. This method lets us compute an approximate value of an integral by summing up the areas of these rectangles.This approach can sometimes underestimate the actual area, especially when the function is increasing over the interval because each rectangle's top-left corner touches the function line.

Right-Endpoint Sums Explained

Right-endpoint sums are similar to left-endpoint sums, but with a slight twist. Instead of using the left edge of each subinterval to determine the rectangle's height, we use the right edge.

• Here, the height of each rectangle is the function value at the right endpoint of each subinterval.
• This means that the right endpoint of a subinterval \[x_{i+1}\] is used to compute the height for the corresponding rectangle.

Just like with left-endpoint sums, these rectangles have a width \ \Delta x \, representing the division of the main interval. For our sample function, points such as 2.5, 3, 3.5, 4, 4.5, and 5 would represent these right endpoints.A right-endpoint sum gives a better approximation when the function is decreasing across the interval. In contrast, it can overestimate the area when the function increases. These nuances show how both left and right-endpoint sums can provide different approximations, highlighting limitations in calculating true integrals.

Details on Subintervals

Subintervals play a crucial role in any numerical integration method, providing the fundamental structure within which these methods operate.

• The main interval \[a, b\] is partitioned into smaller, equally spaced segments.
• The number of subintervals, \ n \, dictates how fine these partitions will be.
• The width \ \Delta x \ of each subinterval is computed as \[\Delta x = \frac{b-a}{n}\].

For instance, given the interval \[2, 5\] divided by 6 subintervals, each segment will have a width of \ \frac{1}{2} \. Subintervals help simplify the somewhat complex behavior of a function into manageable bits, allowing us to use simple shapes like rectangles to estimate the area.By increasing the number of subintervals, the approximation generally becomes more accurate, since each rectangle better aligns with the actual curve of the function. However, this also increases the calculations required. Balancing accuracy and computation is key, and understanding subintervals is central to mastering numerical integration techniques.