Problem 49 - Calculus Single Variable

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. \()\) $$ f(x)=\ln (1+x) \quad I=[0,3], N=3 $$

Step-by-Step Solution

Answer

Midpoint sum: \(\ln(1.5) + \ln(2.5) + \ln(3.5)\), Upper sum: \(\ln(2) + \ln(3) + \ln(4)\).

Step 1: Identify Subintervals and Midpoints

For the interval \([0, 3]\) and \(N=3\), we divide the interval into three equal subintervals: \([0, 1]\), \([1, 2]\), and \([2, 3]\). Calculate the midpoint for each subinterval: the midpoint of \([0, 1]\) is \(0.5\), for \([1, 2]\) is \(1.5\), and for \([2, 3]\) is \(2.5\).

Step 2: Apply the Function to Midpoints

Use the function \(f(x) = \ln(1 + x)\) with each of the midpoints found in Step 1. Thus, compute \(f(0.5) = \ln(1.5)\), \(f(1.5) = \ln(2.5)\), and \(f(2.5) = \ln(3.5)\).

Step 3: Calculate the Midpoint Riemann Sum

The width of each subinterval is \(\Delta x = 1\) (since \(\frac{3}{3} = 1\)). The Riemann sum using midpoints is calculated as \(\mathcal{R}(f, \mathcal{L}_N) = \Delta x \times (f(0.5) + f(1.5) + f(2.5))\). Simplifying this gives \(1 \times (\ln(1.5) + \ln(2.5) + \ln(3.5))\).

Step 4: Calculate Upper Sum using Right Endpoints

For the upper sum, evaluate the function at the right endpoints of each subinterval: \(f(1) = \ln(2)\), \(f(2) = \ln(3)\), and \(f(3) = \ln(4)\). The upper Riemann sum is \(\mathcal{R}(f, \mathcal{U}_N) = 1 \times (\ln(2) + \ln(3) + \ln(4))\).

Step 5: Compare Midpoint and Upper Sums

Now, compare the sums: the midpoint Riemann sum \(\mathcal{R}(f, \mathcal{L}_N)\) is \(\ln(1.5) + \ln(2.5) + \ln(3.5)\) and the upper sum \(\mathcal{R}(f, \mathcal{U}_N)\) is \(\ln(2) + \ln(3) + \ln(4)\).

Key Concepts

Midpoint Rule

The Midpoint Rule is an effective technique for approximating the integral of a function over a certain interval. It works by using the value of the function at the midpoint of each subinterval. In our exercise, we're looking at the interval

• From 0 to 3, which is divided into three equal subintervals:
• [0, 1], [1, 2], and [2, 3].

To apply the midpoint rule, we calculate the midpoint of each subinterval. These midpoints are 0.5, 1.5, and 2.5. Instead of calculating the function at every point, using midpoints helps give us a simpler approach that still captures the essence of the function's behavior over each section.

Subintervals

Subintervals play a crucial role in Riemann sums and are essentially smaller sections of the main interval over which we are considering the integral. For the function

• \( f(x) = \ln(1+x) \),
• with the interval [0, 3]
• that is divided into 3 subintervals of equal lengths.

Each subinterval here has a length of 1, calculated as the difference between the start and end of one divided by the number of subintervals, \( \Delta x = \frac{3 - 0}{3} = 1 \). The midpoints of these subintervals are used to evaluate the function, demonstrating how subintervals help us focus on small, manageable parts.

Logarithmic Function

Logarithmic functions, such as

• \( f(x) = \ln(1+x) \) in our example,

are special functions that involve logarithms. In the context of Riemann sums, these functions are assessed at specific points to estimate areas under curves. Logarithms have unique growth behaviors, and understanding these nuances is helpful for realizing why approximations of their integrals are important. By evaluating the function at selected midpoints and endpoints of subintervals, we can better capture the pattern of growth and approach the integral of the function.

Riemann Sum Inequality

Riemann Sum Inequality refers to the mathematical concept that the sums calculated using midpoints and endpoints neither overestimate nor underestimate consistently. In analyzing functions like

• \( f(x) = \ln(1+x) \),
• using values at the right endpoints results in upper sums,
• while using midpoints gives another level of approximation.

The Inequality helps us understand where the midpoint sum stands in relation to upper and lower approximations, revealing how close or far off our estimate is from the true numerical integral. Being aware of this concept allows for more precise applications of integral approximation techniques.