Problem 16 - Calculus Single Variable

Question

Calculate the lower and upper Riemann sums for the given function \(f\), interval \(I\), and uniform partition of order 2 $$ f(x)=x+1 / x \quad I=[1 / 2,2] $$

Step-by-Step Solution

Answer

Lower Riemann sum: \( \frac{123}{40} \), Upper Riemann sum: \( \frac{15}{4} \).

Step 1: Define the Partition Points

The interval \( I = [\frac{1}{2}, 2] \) needs to be divided into a uniform partition of order 2. This means we need to split it into two equal subintervals. Let's calculate the width of each subinterval. The width, \( \Delta x \), is given by \( \frac{b-a}{n} \) where \( n = 2 \), \( a = \frac{1}{2} \), and \( b = 2 \). This results in \( \Delta x = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \). Thus, the partition points are \( x_0 = \frac{1}{2} \), \( x_1 = \frac{5}{4} \), and \( x_2 = 2 \).

Step 2: Calculate the Function Values at Partition Points

Evaluate the function \( f(x) = x + \frac{1}{x} \) at each partition point. For \( x_0 = \frac{1}{2} \), we get \( f(x_0) = \frac{1}{2} + 2 = \frac{5}{2} \). For \( x_1 = \frac{5}{4} \), we get \( f(x_1) = \frac{5}{4} + \frac{4}{5} = \frac{41}{20} \). For \( x_2 = 2 \), we get \( f(x_2) = 2 + \frac{1}{2} = \frac{5}{2} \).

Step 3: Calculate Lower Riemann Sum

For each subinterval, find the minimum value of \( f \), then multiply by \( \Delta x \). The two intervals are \([\frac{1}{2}, \frac{5}{4}]\) and \([\frac{5}{4}, 2]\). The minimum of \( f \) in the first subinterval \([\frac{1}{2}, \frac{5}{4}]\) is \( f(x_1) = \frac{41}{20} \) and in the second \([\frac{5}{4}, 2]\) is \( f(x_1) = \frac{41}{20} \). So, the lower Riemann sum is \( \Delta x \left(f(x_1) + f(x_1)\right) = \frac{3}{4}\left(\frac{41}{20} + \frac{41}{20}\right) = \frac{123}{40} \).

Step 4: Calculate Upper Riemann Sum

For each subinterval, find the maximum value of \( f \), then multiply by \( \Delta x \). The maximum of \( f \) in the first subinterval \([\frac{1}{2}, \frac{5}{4}]\) is \( f(x_0) = \frac{5}{2} \) and in the second \([\frac{5}{4}, 2]\) is \( f(x_2) = \frac{5}{2} \). Thus, the upper Riemann sum is \( \Delta x \left(f(x_0) + f(x_2)\right) = \frac{3}{4}\left(\frac{5}{2} + \frac{5}{2}\right) = \frac{15}{4} \).

Key Concepts

Partition Points

When working with Riemann Sums, one of the first tasks is to define the partition points of the interval. In our example, we are given the interval \( I = [\frac{1}{2}, 2] \) \ and need to partition it into 2 equal subintervals. This is what we call a 'partition of order 2'. When dividing the interval, the calculation for each subinterval's width, \( \Delta x \), is crucial. You calculate this width using the formula: \( \Delta x = \frac{b-a}{n} \), where \( n \) is the number of partitions, \( a \) is the starting value, and \( b \) is the ending value of the interval.For our interval \([\frac{1}{2}, 2]\), with \( n = 2 \), we calculated \( \Delta x = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \). This width divides the interval into the following partition points: \( x_0 = \frac{1}{2} \), \( x_1 = \frac{5}{4} \), and \( x_2 = 2 \). These points are crucial as they will be used to calculate the Riemann sums.

Function Evaluation

Evaluating the function at the partition points is the next step in finding Riemann Sums. We have the function \( f(x) = x + \frac{1}{x} \) that needs to be evaluated at each partition point calculated earlier.

• For \( x_0 = \frac{1}{2} \), we substitute into the function: \[ f(x_0) = \frac{1}{2} + 2 = \frac{5}{2} \]
• At \( x_1 = \frac{5}{4} \), we find: \[ f(x_1) = \frac{5}{4} + \frac{4}{5} = \frac{41}{20} \]
• For \( x_2 = 2 \), substitute to get: \[ f(x_2) = 2 + \frac{1}{2} = \frac{5}{2} \]

It's crucial to compute these accurately, as they're fundamental to determining both the Lower and Upper Riemann sums. Each of these function values corresponds to specific subintervals, helping to determine where the function is minimized or maximized.

Lower Riemann Sum

The concept of a Lower Riemann Sum involves finding the minimum function value over each subinterval and then multiplying it by the width \( \Delta x \). This gives an estimate of the area under the curve, which approaches the actual integral as the number of partitions increases.In our example, the two subintervals are \([\frac{1}{2}, \frac{5}{4}]\) and \([\frac{5}{4}, 2]\). For both intervals, the minimum value is found at \( x_1 = \frac{5}{4} \), where \( f(x_1) = \frac{41}{20} \). Thus, \[\text{Lower Riemann Sum} = \Delta x \times (f(x_1) + f(x_1)) = \frac{3}{4}(\frac{41}{20} + \frac{41}{20}) = \frac{123}{40} \]This sum provides an approximation from below of the area under the function graph over the interval.

Upper Riemann Sum

Calculating the Upper Riemann Sum involves identifying the maximum function value over each subinterval and multiplying it by the width \( \Delta x \). This sum provides an overestimate of the area under the curve.In our scenario, for the first subinterval \([\frac{1}{2}, \frac{5}{4}]\), the maximum function value is at \( x_0 = \frac{1}{2} \), and for the second \([\frac{5}{4}, 2]\), it is at \( x_2 = 2 \). In both cases, \( f(x) = \frac{5}{2} \) at these points. Thus, \[\text{Upper Riemann Sum} = \Delta x \times (f(x_0) + f(x_2)) = \frac{3}{4}(\frac{5}{2} + \frac{5}{2}) = \frac{15}{4} \]The Upper Riemann Sum estimates the total area from above, ensuring that it is always greater than or equal to the actual integral. As the number of subintervals increases, this approximated overestimation becomes closer to the true value of the integral.