Problem 44 - 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 \([a, b]\). \(f(x)=3 x+2 x^{2}\) over the interval [0,1]

Step-by-Step Solution

Answer

The area under the curve is 2 units.

Step 1: Divide the Interval

The interval \([0, 1]\) is divided into \({n}\) equal subintervals. The width of each subinterval is calculated as \(\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}\).

Step 2: Identify the Right Endpoint

For each subinterval, identify the right endpoint \(c_k\). The \(k\)th subinterval ends at \(x_k = a + k\Delta x = 0 + k\frac{1}{n} = \frac{k}{n}\).

Step 3: Calculate Function Value at Right Endpoint

Substitute \(x_k\) into the function \(f(x) = 3x + 2x^2\). Thus, \(f\left(\frac{k}{n}\right) = 3\cdot\frac{k}{n} + 2\cdot\left(\frac{k}{n}\right)^2 = \frac{3k}{n} + \frac{2k^2}{n^2}\).

Step 4: Set Up Riemann Sum

The Riemann sum is \( S_n = \sum_{k=1}^{n} f\left(\frac{k}{n}\right) \Delta x = \sum_{k=1}^{n} \left( \frac{3k}{n} + \frac{2k^2}{n^2} \right) \cdot \frac{1}{n} \).

Step 5: Simplify the Riemann Sum

Simplify the terms in the sum: \[ S_n = \sum_{k=1}^{n} \left( \frac{3k}{n^2} + \frac{2k^2}{n^3} \right) = \frac{3}{n^2} \sum_{k=1}^{n} k + \frac{2}{n^3} \sum_{k=1}^{n} k^2 \].

Step 6: Evaluate Sum of Series

Use known formulas for sums:\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \].Substitute these into the Riemann sum formula.

Step 7: Substitute and Simplify

Substitute the sum formulas into the sum expression: \[ S_n = \frac{3}{n^2}\left(\frac{n(n+1)}{2}\right) + \frac{2}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right) \].Simplify the expression to get: \[ S_n = \frac{3(n+1)}{2n} + \frac{2(n+1)(2n+1)}{6n^2} \].

Step 8: Take the Limit as n Approaches Infinity

Find the limit as \(n \rightarrow \infty\) to get the area under the curve. Simplifying, \[ \lim_{{n \to \infty}} S_n = \lim_{{n \to \infty}} \left( \frac{3(n+1)}{2n} + \frac{2(n+1)(2n+1)}{6n^2} \right) = \lim_{{n \to \infty}} (\frac{3}{2} + \frac{1}{3}) = 2 \].

Key Concepts

Integral Calculus

Integral calculus is a fascinating branch of mathematics that focuses on the accumulation of quantities and the areas under and between curves. When you imagine the graph of a curve, such as a simple parabola or a more complicated waveform, it's integral calculus that allows us to "add up" the infinite number of tiny slices of area under that curve. The result is a single number, the integral, that represents the area.

There are various types of integrals, but in this context, we're focusing on definite integrals, which provide the area between the curve and the x-axis over a specified interval of x-values. The integrand, such as the expression \(3x + 2x^2\) in our given problem, is the function you're integrating. To solve such problems, we use various techniques, but they often relate back to evaluating how this summed area changes over small slices of the interval.

Definite Integral

A definite integral is used to calculate the area under a curve between two bounds, or limits, on the x-axis. For example, in our exercise, we are asked to find the area under the graph of the function \(f(x) = 3x + 2x^2\) from x = 0 to x = 1.

The process involves calculating the Riemann sum, a method that approximates the area by dividing it into several rectangles. The heights of these rectangles are determined by the value of the function at certain points within each subinterval. Here, we use the right endpoint method, where the function value at the right edge of each subinterval is used to calculate the rectangle's height.

By determining these areas and summing them up, we estimate the total area under the curve. The more subintervals we use, the closer the approximation to the true area. However, we still need to ensure accuracy by considering the limit process, which we'll discuss next.

Limit Process

The limit process is critical in integral calculus for achieving precision in our calculations. In the context of Riemann sums and definite integrals, the limit process refers to taking the number of subintervals, \(n\), to approach infinity. This means our rectangles become infinitely thin, and our approximation of the area gets more precise.

To illustrate, as \(n\) increases, our Riemann sum \(S_n\) turns into an integral. Mathematically, this is expressed as \(\lim_{{n \to \infty}} S_n\). Essentially, by taking the limit as \(n\) goes to infinity, we calculate the exact area under the curve for the given interval. In our problem, solving this limit gives us a finite value \(2\), which is indeed the exact area under the curve \(f(x) = 3x + 2x^2\) between x = 0 and x = 1.

Thus, this limit process is what bridges our approximations with precise, definite outcomes in integral calculus.