Problem 17 - 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. $$L_{4} \text { for } \frac{1}{x^{2}+1} \text { on }[-2,2]$$

Step-by-Step Solution

Answer

\(L_4 = \frac{16}{5}\).

Step 1: Identify the Function and Interval

The function given is \(f(x) = \frac{1}{x^2 + 1}\). We are asked to find \(L_4\), the left-endpoint sum with 4 subintervals on the interval \([-2, 2]\).

Step 2: Determine the Subinterval Width

The interval \([-2, 2]\) has a total length of \(2 - (-2) = 4\). With 4 subintervals, the width of each subinterval is \(\Delta x = \frac{4}{4} = 1\).

Step 3: Establish the Left Endpoint Values

For 4 subintervals, the left endpoints will be at \(x_0 = -2\), \(x_1 = -1\), \(x_2 = 0\), and \(x_3 = 1\).

Step 4: Calculate Function Values at Left Endpoints

Evaluate the function at each left endpoint: - \(f(x_0) = f(-2) = \frac{1}{(-2)^2 + 1} = \frac{1}{5}\) - \(f(x_1) = f(-1) = \frac{1}{(-1)^2 + 1} = \frac{1}{2}\) - \(f(x_2) = f(0) = \frac{1}{0^2 + 1} = 1\) - \(f(x_3) = f(1) = \frac{1}{1^2 + 1} = \frac{1}{2}\)

Step 5: Apply the Left-Endpoint Sum Formula

The left-endpoint sum \(L_4\) is given by \[L_4 = \Delta x \left(f(x_0) + f(x_1) + f(x_2) + f(x_3)\right) \]Substituting the values, \(L_4 = 1 \left(\frac{1}{5} + \frac{1}{2} + 1 + \frac{1}{2}\right)\).

Step 6: Simplify the Expression

Simplify the expression obtained in Step 5: Calculate the sum:\[ \frac{1}{5} + \frac{1}{2} + 1 + \frac{1}{2} = \frac{1}{5} + 2\times\frac{1}{2} + 1\]Convert to a common denominator:\[= \frac{1}{5} + \frac{5}{5} + \frac{10}{5} = \frac{16}{5}\]Thus, \(L_4 = 1 \times \frac{16}{5} = \frac{16}{5}\).

Key Concepts

left-endpoint sum

In calculus, the concept of a left-endpoint sum is utilized to approximate the area under a curve, which is a fundamental part of integral calculus. Using Riemann sums, the left-endpoint sum specifically focuses on evaluating a function at the left endpoint of each subinterval. This approach aids in estimating the total area beneath the graph of a function.

• The left-endpoint choice is systematic: rather than using complicated or random points, the start of each subinterval is picked. This simplifies calculations and provides a consistent method for approximation.
• In a practical sense, given a function and an interval divided into smaller sections, the height of the rectangle representing each section's contribution to the area is based on the function's value at the start of that section.

This method ensures that when you calculate the total sum, you are making a coherent approximation of the integral over the interval.
As the number of subintervals increases, the approximation becomes more accurate, serving as a stepping stone to understanding the concept of integration more deeply.

definite integral approximation

The concept of a definite integral approximation involves using various techniques to estimate the integral over a specified interval. In simple terms, it's about calculating the total area under a curve from one point to another, which is an essential aspect of calculus. Riemann sums, including the left-endpoint sum, provide a practical framework for this estimation.

• The approximation process involves dividing the interval into subintervals. These provide clear sections where calculations can take place.
• The sum of the areas of shapes, often rectangles, that fit under the curve within each subinterval approximates the total integrated area.

Different methods, such as left-endpoint, right-endpoint, and midpoint rules, offer varying levels of accuracy depending on the function and the number of subintervals.
The aim with any approximation method, including the left-endpoint sum, is to come closer to the exact integral value by refining the number and size of these subintervals.

subintervals in calculus

Subintervals in calculus are segments into which a whole interval is divided, serving as the basis for evaluating Riemann sums. Understanding subintervals is crucial when applying techniques to estimate the area under curves.

• The number and width of subintervals (\( \Delta x \)) are determined by how finely you wish to approximate the integral. More subintervals typically yield a more precise approximation.
• The width is calculated by dividing the total length of the interval by the number of subintervals, establishing a constant measure to base calculations on.

Subintervals play a pivotal role because they break down the task into manageable chunks, allowing you to methodically approach the complex shape of a curve with simpler, uniform calculations.
As you become familiar with this concept, you’ll find that it not only aids in Riemann sums but also in setting the stage for understanding more sophisticated integration techniques.

function evaluation at endpoints

Function evaluation at endpoints is a vital step in calculating Riemann sums, particularly with methods like the left-endpoint sum. It involves determining the function's value at the start or end of each subinterval, fundamentally impacting the outcome of the approximation.

• For the left-endpoint sum, evaluations occur at the beginnings of subintervals, utilizing these values to determine the height of rectangles under the curve.
• This step-by-step evaluation at systematic points aids in ensuring consistency and accuracy in the approximation process.

The chosen points—left, right, or middle—affect the resultant calculation of the Riemann sum and the ultimate approximation of the integral.
Understanding and being meticulous with function evaluations at these endpoints helps lay the groundwork for more complex numerical methods and exact integrations, enhancing your ability to solve calculus problems effectively.