Problem 45 - 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 \([a, b]\). \(f(x)=2 x^{3}\) over the interval [0,1]

Step-by-Step Solution

Answer

The area under the curve is \( \frac{1}{2} \).

Step 1: Determine Subinterval Width

The width of each subinterval \( \Delta x \) is calculated as the total interval length divided by the number of subintervals: \[ \Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}. \]

Step 2: Identify Right-Hand Endpoints

The right-hand endpoint of each subinterval is given by \( x_k = a + k\Delta x \). For the interval \([0, 1]\), this becomes: \[ x_k = 0 + k\left(\frac{1}{n}\right) = \frac{k}{n}, \] where \( k \) ranges from 1 to \( n \).

Step 3: Calculate Function Value at Right-Hand Endpoints

Substitute the expression for \( x_k \) into the function to find \( f(x_k) \):\[ f\left(\frac{k}{n}\right) = 2\left(\frac{k}{n}\right)^3 = \frac{2k^3}{n^3}. \]

Step 4: Write the Riemann Sum Formula

The Riemann sum using the right-hand endpoints is given by:\[ S = \sum_{k=1}^{n}f(x_k)\Delta x = \sum_{k=1}^{n}\frac{2k^3}{n^3} \times \frac{1}{n}.\] Simplifying, we get:\[ S = \frac{2}{n^4} \sum_{k=1}^{n} k^3. \]

Step 5: Use a Formula for Summation

To evaluate the sum \( \sum_{k=1}^{n} k^3 \), use the formula:\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \] Substituting this into our expression for \( S \), we have:\[ S = \frac{2}{n^4} \times \left(\frac{n(n+1)}{2}\right)^2 = \frac{2}{n^4} \times \frac{n^2(n+1)^2}{4}. \]

Step 6: Simplify the Riemann Sum Formula

Simplifying the expression from Step 5:\[ S = \frac{2n^2(n+1)^2}{4n^4} = \frac{(n+1)^2}{2n^2}. \]

Step 7: Evaluate the Limit as n Goes to Infinity

To find the area, take the limit of the Riemann sum as \( n \rightarrow \infty \):\[ \lim_{n \rightarrow \infty} S = \lim_{n \rightarrow \infty}\frac{(n+1)^2}{2n^2}. \] Expanding and simplifying gives:\[ = \frac{n^2 + 2n + 1}{2n^2} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2}. \] As \( n \rightarrow \infty \), the terms \( \frac{2}{n} \) and \( \frac{1}{n^2} \) approach 0, leading to:\[ \frac{1}{2}. \]

Key Concepts

Subinterval Width

When dividing an interval into a number of equal parts, each smaller section is called a subinterval. The width of these subintervals, often denoted by \( \Delta x \), can be found using a simple formula. This is calculated by taking the length of the entire interval and dividing it by the total number of subintervals. For example, in the interval \([0, 1]\), the width for \(n\) subintervals would be \( \Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n} \). This width \( \Delta x \) is essential as it defines how finely you are slicing the interval to approximate the area under the curve.

Right-Hand Endpoint

In integrating functions approximately, we can choose various points within each subinterval to sample the function. Using the right-hand endpoint is a common choice. To find the right-hand endpoint of a subinterval, you start from the left end and add the width \( \Delta x \) times the interval index \( k \). This results in the formula \( x_k = a + k\Delta x \).
For our specific example of the interval \([0, 1]\), and function \(2x^3\), this becomes \( x_k = 0 + k \left(\frac{1}{n}\right) = \frac{k}{n} \). The index \( k \) ranges from 1 to \( n \), ensuring that the points cover the entire interval at exact right-hand spots.

Summation Formula

A crucial part of solving Riemann sums is combining the individual function values at these sample points using summation. To apply this method, plug the right-hand endpoint into the function. Then multiply it by the width of the subinterval \( \Delta x \). This gives the Riemann sum formula \( S = \sum_{k=1}^{n}f(x_k)\Delta x \).
In our exercise, the function value at the right-hand endpoint is \( f\left(\frac{k}{n}\right) = 2\left(\frac{k}{n}\right)^3 = \frac{2k^3}{n^3} \), leading to the sum \( S = \frac{2}{n^4} \sum_{k=1}^{n} k^3 \). A known formula for summing cubes is used here: \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \). Substituting into the Riemann sum helps simplify it to a manageable form.

Limit of a Sequence

As we refine our approximation, increasing the number \( n \) of subintervals makes the intervals narrower, leading to a limit—the true area under the curve. For this reason, we take the limit of the simplified form of the Riemann sum as \( n \rightarrow \infty \).

• Start with the simplified Riemann sum \( S = \frac{(n+1)^2}{2n^2} \).
• Break it down to \( \frac{n^2 + 2n + 1}{2n^2} \), which separates into \( \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2} \).
• Recognize that as \( n \rightarrow \infty \), both \( \frac{2}{n} \) and \( \frac{1}{n^2} \) tend to 0.

This leaves us with \( \frac{1}{2} \). Thus, the limit of this sequence gives us the exact area under the curve for the interval, confirming the precision of calculus in measuring area.