Problem 40 - University Calculus: Early Transcendentals

Question

Find a formula for the Riemann sum obtained by dividing the interval \([a, b]\) into \(n\) equal subintervals and using the right-hand endpoint for each \(c_{k} .\) Then take a limit of these sums as \(n \rightarrow \infty\) to calculate the area under the curve over \([a, b]\). \(f(x)=2 x\) over the interval [0,3]

Step-by-Step Solution

Answer

The area under the curve from 0 to 3 is 9.

Step 1: Determine the Width of Each Subinterval

First, we need to divide the interval ([a, b]) into n equal subintervals. The interval in question is [0, 3]. So we calculate the width of each subinterval (\(\Delta x\)) using the formula \(\Delta x = \frac{b-a}{n}\). For this particular problem, \(a = 0, b = 3\), so \(\Delta x = \frac{3-0}{n} = \frac{3}{n}\).

Step 2: Determine the Right-Hand Endpoint Formula

The right-hand endpoint for each subinterval is determined by using the formula \(c_k = a + k \Delta x\), where \(k\) represents the subinterval index. Since \(a = 0\), and \(\Delta x = \frac{3}{n}\), we find that the right endpoint is \(c_k = 0 + k \cdot \frac{3}{n} = \frac{3k}{n}\).

Step 3: Construct the Riemann Sum

The Riemann sum using the right endpoints is given by \(S_n = \sum_{k=1}^{n} f(c_k) \Delta x\). Substituting the given function \(f(x) = 2x\) and the expressions for \(c_k\) and \(\Delta x\), we have \(S_n = \sum_{k=1}^{n} 2\left(\frac{3k}{n}\right) \cdot \frac{3}{n}\), which simplifies to \(S_n = \sum_{k=1}^{n} 2\left(\frac{3k}{n}\right) \cdot \frac{3}{n} = \frac{18}{n^2} \sum_{k=1}^{n} k\).

Step 4: Simplify the Sum

The sum of the first n natural numbers is given by the formula \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\). Plugging this into our Riemann sum gives \(S_n = \frac{18}{n^2} \cdot \frac{n(n+1)}{2}\). Simplifying, we have \(S_n = \frac{18(n+1)}{2n}\), which is \(S_n = 9\left(1 + \frac{1}{n}\right)\).

Step 5: Calculate the Limit as n Approaches Infinity

To find the area under the curve, we take the limit of \(S_n\) as \(n\to\infty\). This is given by \[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} 9\left(1 + \frac{1}{n}\right) = 9 \times 1 = 9. \] Thus, the area under the curve from 0 to 3 is 9.

Key Concepts

Subinterval

When dealing with Riemann sums, the concept of "subinterval" plays a central role. Essentially, to approximate the area under a curve, we divide the entire interval \([a, b]\) into smaller, equal-length partitions called subintervals. In this scenario, where we want to find the area from \([0, 3]\), we divide this interval into \(n\) equal parts. Each subinterval has a width, often denoted as \(\Delta x\), and is calculated using the formula \(\Delta x = \frac{b-a}{n}\).

• For our example, since \(a = 0\) and \(b = 3\), then \(\Delta x = \frac{3}{n}\).
• These subintervals allow us to systematically and evenly partition the area we are working with, setting the stage for calculating the Riemann sum.

Breaking up the interval into subintervals is a crucial first step because it sets up a framework that lets us apply mathematical operations incrementally and consistently across the interval.
Understanding subintervals helps students grasp how to distribute areas, how each portion contributes to the whole, and why this division is essential for Riemann sums.

Right-Hand Endpoint

When using the Riemann Sum, choosing a point within each subinterval is key. One common choice is the "right-hand endpoint". In simple terms, this means we evaluate the function at the rightmost point of each subinterval. If our interval is \([a, b]\), and it's divided into \(n\) subintervals, the right-hand endpoint for the k-th subinterval could be noted as \(c_k\).
For our interval and function:

• Calculate the right-hand endpoint using \(c_k = a + k\Delta x\).
• Substituting \(a = 0\), gives \(c_k = \frac{3k}{n}\).
• This helps in setting up the Riemann sum, where \(c_k\) is used to calculate the function value.

Right-hand endpoints often simplify computations and help better align with integral calculus, as these points mimic approaching the exact area under a curve.

Limit of a Sum

The "Limit of a Sum" in Riemann sums bridges the discrete world of sums and the continuous realm of areas under curves. As we manage more subintervals, specifically when \(n\) goes to infinity, Riemann sums become integrals from calculus.
The process involves:

• Setting up the Riemann sum: \(S_n = \sum_{k=1}^{n} f(c_k) \Delta x\)
• As \(n\) increases, each subinterval \(\Delta x\) shrinks, leading \(S_n\) to approach the true area.
• To find the exact area, compute \[ \lim_{n \to \infty} S_n \] using calculus or algebra.

For our example, as \(n\) approaches infinity, the Riemann sum transforms into an area of 9 units under the curve. This step connects finite sums to precise integrals, helping students appreciate the elegance and precision of calculus.