91Ó°ÊÓ

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities, like area under a curve. For a given function, it can help determine the total change over an interval. In our exercise, we aim to calculate the definite integral of the function \( f(x) = \frac{1}{x} \) over the interval \([1, \frac{9}{4}]\).
To find this, we perform the integration of the function over the specified limits. The integral of \( \frac{1}{x} \) is known to be the natural logarithm function, \( \ln|x| \). Thus, we evaluate:

• \( \int_{1}^{\frac{9}{4}} \frac{1}{x} \, dx \)
• This results in \( \ln \left| x \right| \) evaluated from \(1\) to \( \frac{9}{4} \).

By computing \( \ln \left( \frac{9}{4} \right) - \ln(1) \), we obtain a final value of approximately \( 0.8109 \). This value gives us an exact measure of the area under the curve of \( f(x) = \frac{1}{x} \) from 1 to \( \frac{9}{4} \). Understanding how to compute this is crucial in viewing the relationship between sums and integrals, also illustrated by the Riemann Sum approximation.

Subinterval width is a key concept when calculating the Riemann sum. It defines how we split our entire interval into smaller, manageable parts. This is essential for approximating an integral by a sum.
We use this formula to determine the width of each segment in a partition:

• \( \Delta x = \frac{b-a}{N} \)

Where \( b \) and \( a \) are the interval bounds, and \( N \) is the number of subintervals. In our exercise, for the interval \([1, \frac{9}{4}]\) with \( N = 5 \), the calculation is straightforward:

• \( \Delta x = \frac{\frac{9}{4} - 1}{5} = \frac{1}{4} \)

Each subinterval, therefore, has a consistent width of \( \frac{1}{4} \). Having equal subintervals allows for a simple and balanced approach, making calculations easier and more reliable when forming a Riemann Sum.

In evaluating a Riemann sum, choosing endpoints is crucial, and in our case, we use left endpoints. The left endpoints tell us where to evaluate the function in each subinterval.
This involves starting from the lower bound of our original interval \( a = 1 \), then progressing one subinterval width at a time. Here's how we find them:

• Start at 1 and add the subinterval width \( \Delta x = \frac{1}{4} \).
• This results in the left endpoints: \( 1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2 \).

These points are key to accurately estimating the area under our curve using rectangles of height equal to the function value at these points. By summing over these regions, we get a reasonable approximation of the definite integral.

Function evaluation is the process of calculating the function's output for specific inputs. Within the context of the Riemann sum, it's essential to decide the height of each rectangle used in the approximation.
For each left endpoint identified, we need to plug the endpoint value into the function \( f(x) = \frac{1}{x} \). Here’s how it works:

• At \( x = 1 \), \( f(1) = 1 \)
• At \( x = \frac{5}{4} \), \( f\left( \frac{5}{4} \right) = \frac{4}{5} \)
• At \( x = \frac{3}{2} \), \( f\left( \frac{3}{2} \right) = \frac{2}{3} \)
• At \( x = \frac{7}{4} \), \( f\left( \frac{7}{4} \right) = \frac{4}{7} \)
• At \( x = 2 \), \( f(2) = \frac{1}{2} \)

Each of these values determines the heights of rectangles spanning across each subinterval. When multiplied by the subinterval width and summed, they generate the Riemann sum \( \mathcal{R}(f, \mathcal{S}) = 0.4464 \), providing an effective means to approximate the integral.