Problem 64 - Calculus Volume 1

Question

In the following exercises, given \(L_{n}\) or \(R_{n}\) as indicated,express their limits as \(n \rightarrow \infty\) as definite integrals, identifying the correct intervals. $$L_{n}=\frac{1}{n} \sum_{i=1}^{n} \frac{i-1}{n}$$

Step-by-Step Solution

Answer

The limit is \(\int_{0}^{1} x \, dx = 0.5\).

Step 1: Identify the Riemann Sum

The given expression \(L_{n}=\frac{1}{n} \sum_{i=1}^{n} \frac{i-1}{n}\) resembles a Riemann sum. This formula takes the form of a left Riemann sum, as it uses \(i-1\) in the numerator (the left endpoint).

Step 2: Determine the Function and Interval

The function in the sum is \(f\left(\frac{i-1}{n}\right) = \frac{i-1}{n}\). This indicates \(f(x) = x\). The interval in a left Riemann sum begins at 0 and ends at 1, since the sum index \(i-1\) starts from 0 when divided by \(n\), going up to \(n-1/n\).

Step 3: Express as a Definite Integral

As \(n \to \infty\), the left Riemann sum \(L_{n}\) converges to the definite integral. Thus, \(\lim_{{n \to \infty}} L_{n} = \int_{0}^{1} x \, dx\). This is because the sum approximates the area under \(f(x) = x\) from 0 to 1.

Step 4: Compute the Definite Integral

Calculate the integral \(\int_{0}^{1} x \, dx\). This integral computes to \([0.5]\), as the antiderivative of \(x\) is \(\frac{x^2}{2}\), and evaluating it from 0 to 1 gives \(\frac{1^2}{2} - \frac{0^2}{2} = 0.5\).

Key Concepts

Definite Integrals

Definite integrals represent the concept of finding the area under a curve, bounded by specific intervals on the x-axis. In the context of the given exercise, the definite integral is expressed as \( \int_{0}^{1} x \, dx \). This integral computes the total area under the linear function \( f(x) = x \) over the interval from 0 to 1.
A definite integral is not just about finding the area under a curve but also involves specific characteristics:

• The upper and lower bounds, which are 1 and 0 in this example.
• The function that we are integrating, \( f(x) = x \).

Calculating a definite integral like \( \int_{0}^{1} x \, dx \) means finding the antiderivative of \( x \), which is \( \frac{x^2}{2} \), and evaluating it at the bounds: substitute in 1 and subtract the result of substituting 0. This gives us \( \frac{1}{2} = 0.5 \). The result 0.5 represents the exact area under the curve between 0 and 1.

Limit of Riemann Sums

The concept of the limit of Riemann sums connects approximations to exact values. In calculus, Riemann sums allow us to approximate the area under a curve by summing up the areas of rectangles. As the number of rectangles \( n \) increases, the approximation becomes more accurate.
The term \( L_{n} = \frac{1}{n} \sum_{i=1}^{n} \frac{i-1}{n} \) refers to a Left Riemann Sum because each rectangle's height is determined by the function value at the left endpoint \( \frac{i-1}{n} \). This way, the sum approximates the true area under \( f(x) = x \) from 0 to 1.
When we consider the limit as \( n \to \infty \), the approximation improves, and the sum effectively becomes the definite integral \( \int_{0}^{1} x \, dx \). Understanding the transition from the discrete sum to the continuous integral is crucial in calculus as it illustrates the principle of integration as a method of finding areas under curves.

Calculus Problem Solving

Solving calculus problems often involves translating abstract mathematical notation into understandable concepts. The exercise makes use of a Riemann sum to introduce a fundamental idea in calculus - how sums can represent areas and how these sums transition into definite integrals as limits are applied.
In problem-solving:

• Identify the type of Riemann sum, Left or Right, based on the expression given.
• Recognize the function and interval involved. Here, the function \(f(x) = x\) over \([0,1]\) \.
• Convert the sum into an integral by taking the limit as \(n \to \infty\).
• Calculate the integral to get the precise area, which is the solution to the problem.

Mastering these steps ensures one can tackle more complex problems that involve integrating different functions over various intervals. By comprehending these methods, a student gains a deeper appreciation and grasp of calculus principles.