Problem 39 - 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 \([a, b]\). \(f(x)=1-x^{2}\) over the interval [0,1]

Step-by-Step Solution

Answer

The area under the curve is \(\frac{2}{3}\).

Step 1: Define the Width of Subintervals

First, divide the interval \([a, b]\) into \(n\) equal subintervals. If \(a = 0\) and \(b = 1\), the width of each subinterval, \(\Delta x\), is given by \(\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}\). Thus, each subinterval has a width of \(\frac{1}{n}\).

Step 2: Determine Right-Hand Endpoints

For right-hand endpoints, we choose \(c_k = a + k \Delta x = 0 + k \cdot \frac{1}{n} = \frac{k}{n}\) for each \(k = 1, 2, \ldots, n\).

Step 3: Write the Riemann Sum

The Riemann sum using right-hand endpoints is \[S_n = \sum_{k=1}^{n} f\left( \frac{k}{n} \right) \Delta x = \sum_{k=1}^{n} \left(1 - \left(\frac{k}{n}\right)^2 \right) \cdot \frac{1}{n}.\]This simplifies to \[S_n = \sum_{k=1}^{n} \left( \frac{1}{n} - \frac{k^2}{n^3} \right).\]

Step 4: Simplify the Riemann Sum

The sum can be separated into two simpler sums:\[S_n = \sum_{k=1}^{n} \frac{1}{n} - \sum_{k=1}^{n} \frac{k^2}{n^3}.\]The first sum is simply \(1\) since \(\sum_{k=1}^{n} \frac{1}{n} = n \cdot \frac{1}{n}\). The second sum, using the formula \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\), becomes \[S_n = 1 - \frac{n(n+1)(2n+1)}{6n^3}.\]

Step 5: Compute the Limit of the Riemann Sum

Take the limit of \(S_n\) as \(n \rightarrow \infty\):\[\lim_{n \to \infty} \left( 1 - \frac{n(n+1)(2n+1)}{6n^3} \right).\]Simplifying, we find:\[= 1 - \lim_{n \to \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = 1 - \frac{1}{3} = \frac{2}{3}.\]

Step 6: Final Result: Area Under the Curve

The area under the curve \(f(x) = 1 - x^2\) over the interval \([0, 1]\) is \(\frac{2}{3}\).

Key Concepts

Limit of a Function

When you're dealing with functions, the notion of a limit is essential. It helps us understand the behavior of a function as the input approaches a certain point. In the context of a Riemann sum, when we divide an interval into smaller and smaller subintervals, we are interested in what happens as the number of these subintervals, denoted as \(n\), becomes very large. This is where the limit comes in.

The limit of a function is a fundamental idea because it allows us to transition from discrete sums to continuous integrals. We express this as \(\lim_{n \to \infty}\). This notation indicates what value the sum \(S_n\) approaches as \(n\) becomes infinitely large.

This limit helps in calculating the total area under a curve, such as determining the precise area between a curve and the x-axis. By taking the limit, we effectively convert a Riemann sum into the concept of an integral, thereby defining the 'exact' area.

Integration

Integration is a powerful mathematical tool that allows us to find accumulative quantities such as areas under curves. When we calculate the limit of a Riemann sum, we are essentially performing an integration process. Integration is the reverse process of differentiation and helps in summing up parts to determine a whole.

The integral of a function \(f(x)\) over an interval \([a, b]\) gives us the accumulated value, often interpreted as the area. This is precisely what we did in the exercise by finding the limit of the sum of small areas. The resulting value, \(\frac{2}{3}\), represents this area under the curve \(f(x) = 1 - x^2\) on the interval \([0, 1]\). This continuous summation is the heart of integration and fundamental in calculus.

Definite Integral

A definite integral represents the signed area under a curve within a given interval \([a, b]\). It is denoted by \(\int_{a}^{b} f(x) \, dx\). The definite integral computes the net area by taking into account both areas above and below the x-axis, which may cancel out depending on their position.

In the context of our exercise, the definite integral of \(1 - x^2\) over \([0, 1]\) equals \(\frac{2}{3}\). This value is found by taking the limit of the Riemann sum, demonstrating how definite integrals and Riemann sums are intimately connected.

Importantly, the definite integral provides a way of finding exact areas, unlike approximations from finite sums. It's a cornerstone concept in analysis, granting us the tools to explore and quantify continuous spaces.