Problem 27 - Calculus Volume 1

Question

Express the following endpoint sums in sigma notation but do not evaluate them. \(R_{100}\) for \(\ln x\) on \([1, e]\)

Step-by-Step Solution

Answer

\(\sum_{i=1}^{100} \ln(1 + i\cdot\frac{e-1}{100}) \cdot \frac{e-1}{100}\)

Step 1: Identify the Function and Interval

We are given the function \( \ln x \) on the interval \([1, e]\). The partition involves breaking this interval into smaller subintervals for the Riemann sum.

Step 2: Determine the Partition

Since the interval is \([1, e]\) and we are using \(100\) subintervals, the width \(\Delta x\) of each subinterval is given by: \[ \Delta x = \frac{e - 1}{100} \]

Step 3: Determine the Endpoints of Subintervals

For these subintervals, the endpoints \(x_i\) can be expressed in terms of the subinterval index \(i\):\[ x_i = 1 + i \cdot \Delta x, \space i = 0, 1, 2, ..., 100 \]

Step 4: Express the Riemann Sum in Sigma Notation

The Riemann sum \( R_{100} \) for upper endpoints is represented in sigma notation as:\[ \sum_{i=1}^{100} \ln(x_i) \cdot \Delta x \] Substitute \( x_i \) and \( \Delta x \) into the sum:\[ \sum_{i=1}^{100} \ln(1 + i\cdot\frac{e-1}{100}) \cdot \frac{e-1}{100} \]

Key Concepts

Riemann sum

The Riemann sum is a method in calculus used to approximate the total area under a curve, which represents the integral of a function over a particular interval. Imagine dividing the area under a curve into many small, manageable rectangle-like shapes, then adding up their areas. This sum of the areas of these shapes provides an approximation to the integral.
This technique is helpful when a function is complex and difficult to integrate analytically or when you're looking to estimate it numerically. There are different ways to choose the points at which a function is evaluated within each subinterval, leading to various types of Riemann sums, such as the Left, Right, and Midpoint Riemann sums.
In our case, the task involves using the upper endpoints of the subintervals to calculate the Riemann sum, which provides one of several ways to approach such problems.

natural logarithm function

The natural logarithm, denoted by \( \ln x \), is a mathematical function that tells us how many times we multiply the number \( e \) (approximately 2.71828) to get a particular number \( x \). The natural logarithm is the inverse of the exponential function, making it highly significant in various fields of science and engineering.
In this exercise, we work with the natural logarithm function over the interval \[ [1, e] \], meaning we analyze the behavior of \( \ln x \) from 1 to \( e \).
The natural logarithm is particularly useful because its derivative at any point \( x \) is exactly \( 1/x \), providing a straightforward way to relate the rate of change of a quantity to the quantity itself.

partitioning intervals

Partitioning intervals involves dividing a function's interval into smaller subintervals. This is a key technique in calculating Riemann sums. By partitioning, we break down the problem into smaller, more manageable pieces.

• In this exercise, the interval \[ [1, e] \] is partitioned into \( 100 \) subintervals.
• The width \( \Delta x \) of each subinterval is calculated as \[ \frac{e - 1}{100} \]

This arrangement helps in systematically calculating the sum of areas under the curve using each subinterval's contribution.
Proper partitioning ensures each part of the interval contributes accurately to the approximation, bringing the calculated Riemann sum closer to the actual integral.

upper endpoint approximation

Upper endpoint approximation in the context of Riemann sums involves selecting the endpoint at the higher x-value within each subinterval to evaluate the function. This choice generally results in a sum that either overestimates or underestimates the integral, influenced by the function's behavior.

• For an increasing function, using the upper endpoint usually yields an overestimation.
• Conversely, for a decreasing function, it tends to give an underestimation.

In our scenario, where \( \ln x \) is used, the approximation would be calculated by taking each subinterval's greater endpoint's logarithm.
The final Riemann sum involved in this task is represented by: \[ \sum_{i=1}^{100} \ln\left(1 + i\cdot\frac{e-1}{100}\right) \cdot \frac{e-1}{100} \] This formula captures how the chosen upper endpoint contributes to each slice's approximate area under the logarithmic curve.