Problem 47
Question
Calculate \(\mathcal{R}\left(f, \mathcal{L}_{N}\right),\) the Riemann sum \(\mathcal{R}(f, \mathcal{S})\) using the midpoint of each subinterval for the choice of points, and \(\mathcal{R}\left(f, \mathcal{U}_{N}\right)\). (You will notice that the inequalities of line \((5.2 .2)\) hold. \()\) $$ f(x)=1 /(1+x) \quad I=[1,3], N=2 $$
Step-by-Step Solution
Verified Answer
The midpoint Riemann sum is 0.6857 and the upper Riemann sum is 0.5833.
1Step 1: Determine the Subintervals
The interval \( I = [1, 3] \) is divided into \( N = 2 \) subintervals. Calculate the width of each subinterval \( \Delta x = \frac{3 - 1}{2} = 1 \). The subintervals are \([1, 2] \) and \([2, 3] \).
2Step 2: Calculate Midpoints of Subintervals
The midpoint of the first subinterval \([1, 2] \) is \( \frac{1 + 2}{2} = 1.5 \). The midpoint of the second subinterval \([2, 3] \) is \( \frac{2 + 3}{2} = 2.5 \).
3Step 3: Evaluate Function at Midpoints
Calculate \( f(x) = \frac{1}{1+x} \) at each midpoint: \( f(1.5) = \frac{1}{1+1.5} = \frac{1}{2.5} = 0.4 \) and \( f(2.5) = \frac{1}{1+2.5} = \frac{1}{3.5} \approx 0.2857 \).
4Step 4: Compute Midpoint Riemann Sum
The midpoint Riemann sum is \( \mathcal{R}(f, \mathcal{S}) = \Delta x \cdot (f(1.5) + f(2.5)) = 1 \times (0.4 + 0.2857) = 0.6857 \).
5Step 5: Evaluate Function at Endpoints for Upper Sum
For the upper sum, calculate \( f(x) \) at the right endpoint of each subinterval: \( f(2) = \frac{1}{1+2} = \frac{1}{3} \approx 0.3333 \) and \( f(3) = \frac{1}{1+3} = \frac{1}{4} = 0.25 \).
6Step 6: Compute Upper Riemann Sum
The upper Riemann sum is \( \mathcal{R}(f, \mathcal{U}_N) = \Delta x \cdot (f(2) + f(3)) = 1 \times (0.3333 + 0.25) = 0.5833 \).
Key Concepts
Midpoint Riemann SumUpper Riemann SumInterval Partitioning
Midpoint Riemann Sum
The midpoint Riemann sum is a method of approximating the area under a curve, offering a more accurate estimation than using just the left or right endpoints of subintervals. This approach involves choosing a point in each subinterval at the midpoint, calculating the function's value at this point, and then multiplying by the width of the subinterval. In our exercise, the interval \([1, 3]\) is divided into two subintervals with a width of 1. We use the midpoints 1.5 and 2.5 for the respective subintervals \([1, 2]\) and \([2, 3]\). Calculating the function values, we get \(f(1.5) = 0.4\) and \(f(2.5) \approx 0.2857\). Hence, the midpoint Riemann sum is the sum of these products: \[0.4 \times 1 + 0.2857 \times 1 = 0.6857\] for the total approximation.
For better accuracy, one could increase the number of subintervals, which would generally result in a sum that provides closer approximation to the actual integral.
For better accuracy, one could increase the number of subintervals, which would generally result in a sum that provides closer approximation to the actual integral.
Upper Riemann Sum
The Upper Riemann Sum differs from the midpoint method by evaluating the function at the right endpoint of each subinterval. This often results in an overestimate of the area under a curve, particularly for functions that are non-decreasing in the interval.
In this exercise, the function is evaluated at the right endpoints of the subintervals \(2\) and \(3\). Calculation results in \(f(2) \approx 0.3333\) and \(f(3) = 0.25\). The resulting upper Riemann sum is \[1 \times (0.3333 + 0.25) = 0.5833\].
Upper Riemann sums are particularly useful in providing bounds and can be compared with other Riemann sums or integral approximations for improved insights into how close an estimate might be to an actual value.
In this exercise, the function is evaluated at the right endpoints of the subintervals \(2\) and \(3\). Calculation results in \(f(2) \approx 0.3333\) and \(f(3) = 0.25\). The resulting upper Riemann sum is \[1 \times (0.3333 + 0.25) = 0.5833\].
Upper Riemann sums are particularly useful in providing bounds and can be compared with other Riemann sums or integral approximations for improved insights into how close an estimate might be to an actual value.
Interval Partitioning
Interval partitioning is a crucial concept when working with Riemann sums. The main idea is dividing the overall interval \(I = [1, 3]\) into smaller, manageable parts or subintervals.
The number of subintervals, which is defined by \(N\), affects the width of each, calculated as \[\Delta x = \frac{3 - 1}{N}\]. In our case \(N = 2\), hence \(\Delta x = 1\), producing subintervals \([1, 2]\) and \([2, 3]\).
Each of these subintervals then serves as a basis for different Riemann sum calculations - midpoint, upper, and others. As \(N\) increases, the partitions become narrower, and the Riemann sum approximations generally converge closer to the true integral of the function over \(I\). Partitioning is foundational, setting the stage for more refined analyses and improved integration approximations.
The number of subintervals, which is defined by \(N\), affects the width of each, calculated as \[\Delta x = \frac{3 - 1}{N}\]. In our case \(N = 2\), hence \(\Delta x = 1\), producing subintervals \([1, 2]\) and \([2, 3]\).
Each of these subintervals then serves as a basis for different Riemann sum calculations - midpoint, upper, and others. As \(N\) increases, the partitions become narrower, and the Riemann sum approximations generally converge closer to the true integral of the function over \(I\). Partitioning is foundational, setting the stage for more refined analyses and improved integration approximations.
Other exercises in this chapter
Problem 47
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