Problem 10 - Calculus Single Variable

Question

For the given function \(f\), interval \(I\), and uniform partition of order \(N\) : a. Evaluate the Riemann sum \(\mathcal{R}(f, \mathcal{S})\) using the choice of points \(\mathcal{S}\) that consists of the right endpoints of the subintervals of \(I\). b. Evaluate the definite integral that \(\mathcal{R}(f, \mathcal{S})\) approximates. $$ f(x)=1 / x \quad \mathcal{S}=\\{1,5 / 4,3 / 2,7 / 4,2\\}, I=[3 / 4,2], N=5 $$

Step-by-Step Solution

Answer

The Riemann sum is approximately 0.5798, and the definite integral is approximately 0.9808.

Step 1: Understand the Problem

We need to calculate the Riemann sum using right endpoints over the interval \(I = \left[ \frac{3}{4}, 2 \right]\) with a partition order \(N=5\). The function is \(f(x) = \frac{1}{x}\).

Step 2: Calculate the Width of Each Subinterval

The interval \(I = \left[ \frac{3}{4}, 2 \right]\) is divided into \(N=5\) equal parts. The width of each subinterval is \( \Delta x = \frac{2 - \frac{3}{4}}{5} = \frac{5}{4 \times 5} = \frac{1}{4} \).

Step 3: Identify the Right Endpoints

A uniform partition means moving from \(\frac{3}{4}\) to \(2\) in steps of \(\frac{1}{4}\). Thus, the right endpoints are \(\left\{ 1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2 \right\}\).

Step 4: Calculate the Riemann Sum

For each subinterval, use the formula \(f(x_i^*) \cdot \Delta x \) where \(x_i^*\) are the right endpoints: \(\mathcal{R}(f, \mathcal{S}) = f(1) \cdot \frac{1}{4} + f\left(\frac{5}{4}\right) \cdot \frac{1}{4} + f\left(\frac{3}{2}\right) \cdot \frac{1}{4} + f\left(\frac{7}{4}\right) \cdot \frac{1}{4} + f(2) \cdot \frac{1}{4} \)This simplifies to:\( \mathcal{R}(f, \mathcal{S}) = \left(\frac{1}{1} + \frac{1}{\frac{5}{4}} + \frac{1}{\frac{3}{2}} + \frac{1}{\frac{7}{4}} + \frac{1}{2}\right) \cdot \frac{1}{4} \)Calculate each term to approximate the sum.

Step 5: Simplify the Calculation

Evaluate each term: \(f(1) = 1, \ \frac{1}{\frac{5}{4}} = \frac{4}{5}, \ \frac{1}{\frac{3}{2}} = \frac{2}{3}, \ \frac{1}{\frac{7}{4}} = \frac{4}{7}, \ \frac{1}{2} = \frac{1}{2} \). Add these and multiply by \(\frac{1}{4}\):\(\mathcal{R}(f, \mathcal{S}) = \frac{1}{4} \left(1 + \frac{4}{5} + \frac{2}{3} + \frac{4}{7} + \frac{1}{2}\right) \= \frac{1}{4} \left(70 + 56 + 40 + 24 + 35 \right) / 70\ \approx 0.5798\)

Step 6: Evaluate the Definite Integral

The integral of \(f(x) = \frac{1}{x}\) from \(a = \frac{3}{4}\) to \(b = 2\) is:\( \int_{\frac{3}{4}}^{2} \frac{1}{x} \, dx = [\ln|x|]_{\frac{3}{4}}^{2} = \ln(2) - \ln\left(\frac{3}{4}\right) \)Simplify:\( = \ln\left(\frac{2}{\frac{3}{4}}\right) = \ln\left(\frac{8}{3}\right) \).

Step 7: Compare Sum to Integral

Compare \(\mathcal{R}(f, \mathcal{S}) \approx 0.5798\) and \(\int_{\frac{3}{4}}^{2} \frac{1}{x} \, dx = \ln\left(\frac{8}{3}\right) \approx 0.9808\) to understand the approximation's accuracy.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, allowing us to calculate the area under a curve. In our exercise, we deal with the function \( f(x) = \frac{1}{x} \) over the interval \( \left[ \frac{3}{4}, 2 \right] \). The definite integral's role here is to provide the exact value of that area. In simpler terms, a definite integral helps us find the total accumulation of a quantity.The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, defining the interval on which we compute the integral. In our case, you would compute:

• Find the antiderivative of \( f(x) \).
• Evaluate this antiderivative at the endpoints, \( a = \frac{3}{4} \) and \( b = 2 \).
• Subtract the value of the antiderivative at \( a \) from its value at \( b \).

This process gives us the area under the curve, which in this example turns out to be \( \ln\left(\frac{8}{3}\right) \), offering a precise way to understand accumulation over the interval.

Partition Intervals

Partitioning an interval is all about dividing it into equal segments, which makes it easier to approximate areas under curves. In our problem, the interval \( \left[ \frac{3}{4}, 2 \right] \) is divided into 5 equal parts, known as subintervals.The width of each subinterval, denoted by \( \Delta x \), plays a crucial role in setting up a Riemann sum. Here, \( \Delta x = \frac{1}{4} \) because \( 2 - \frac{3}{4} = \frac{5}{4} \) gets divided by our 5 parts.

You can visualize partitioning like slicing a bread loaf into equal slices; each slice represents a subinterval. Calculating the area using these subintervals gives us an approximate value of our integral. More intervals generally mean a more accurate approximation.

Right Endpoints

In the Riemann sum method, right endpoints are used as sample points within each subinterval to calculate the sum. This means that for each segment of our partition, the function's value is taken at its rightmost point.For the given interval \( \left[ \frac{3}{4}, 2 \right] \), the right endpoints of the subintervals are \( \{1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2\} \). These points are crucial because they determine the height of the rectangles in the Riemann sum used to approximate the integral.Imagine each rectangle's right edge aligning perfectly with the function’s value at these endpoints. This is why they’re called right endpoints. Choosing right endpoints helps us easily compute and understand how functions behave across subintervals.

Mathematical Approximation

Mathematical approximation is a method used to estimate values when exact results are either impossible or unnecessary. Riemann sums are a classic example where approximation is used to determine the area under a curve by summing up areas of rectangles under the curve instead of computing the exact integral.This approach is useful when dealing with complex functions or when an integral is too difficult to compute directly. To perform this:

• Use a series of rectangles to approximate the area under the curve.
• The height of each rectangle is the function's value at the right endpoint of each subinterval.
• Multiply this height by the width of the subinterval, \( \Delta x \).

The sum \( \mathcal{R}(f, \mathcal{S}) \) we calculated demonstrates approximation for our function \( f(x) = \frac{1}{x} \), over \( \left[ \frac{3}{4}, 2 \right] \), resulting in \( 0.5798 \), which is less than the definite integral's value, showing typical underestimation in such collaborations.