Problem 63 - Calculus Volume 1

Question

In the following exercises, express the limits as integrals. $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \cos ^{2}\left(2 \pi x_{i}^{*}\right) \Delta x \text { over }[0,1] $$

Step-by-Step Solution

Answer

\(\int_{0}^{1} \cos^2(2\pi x)\ dx\)

Step 1: Identifying the Problem Structure

We start with the given limit problem, which is a Riemann sum:\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \cos^2\left(2 \pi x_i^{*}\right) \Delta x\] over the interval \([0, 1]\). Our goal is to express this limit as a definite integral.

Step 2: Understanding Riemann Sums

The Riemann sum form \(\sum_{i=1}^{n} f(x_i^*) \Delta x\) approximates the integral \(\int_a^b f(x) \ dx\) as \(n \rightarrow \infty\), where \(x_i^*\) is a sample point in the subinterval \(\Delta x\), and \\(\Delta x = \frac{b-a}{n}\). In our case, \(a = 0\), \(b = 1\), \(f(x) = \cos^2(2\pi x)\).

Step 3: Defining the Definite Integral

Given that the Riemann sum represents the integral of \(f(x) = \cos^2(2\pi x)\) over the interval \([0, 1]\) as \(n\) approaches infinity, we can then express the limit as:\[\int_{0}^{1} \cos^2(2\pi x)\ dx\].

Step 4: Conclusion from the Steps

By understanding the relationship between the Riemann sum and the definite integral, we have rewritten the original limit problem as a definite integral \(\int_{0}^{1} \cos^2(2\pi x)\ dx\). This integral now accurately represents the limit.

Key Concepts

Riemann sum

The concept of a Riemann sum originates from the fundamental idea of calculus, which involves approximating the area under a curve. To begin with, the interval over which we want to compute the area (or integral) is divided into smaller subintervals, each with a length \(\Delta x = \frac{b-a}{n}\).
For each of these subintervals, a sample point, usually denoted as \(x_i^*\), is chosen. The function value at this sample point, \(f(x_i^*)\), is then multiplied by the length of the subinterval \(\Delta x\).
In essence, a Riemann sum is the sum of these products:\[ \sum_{i=1}^{n} f(x_i^*) \Delta x \]

• This sum approximates the total area under the curve \(f(x)\) over the interval \([a, b]\).
• As the number of subintervals \(n\) increases, the approximation becomes more accurate.
• The Riemann sum becomes a better representation of the integral \(\int_{a}^{b} f(x) \, dx\).

Riemann sums are foundational in understanding how integrals work, providing the stepping stones from discrete to continuous mathematics.

Definite integral

A definite integral is a fundamental concept in calculus used to compute the total accumulation of quantities, such as areas. It is defined over a specific interval \([a, b]\), where \(a\) and \(b\) are the bounds.
The definite integral of a function \(f(x)\) over \([a, b]\) is represented by:\[\int_{a}^{b} f(x) \, dx\]

• This integral provides a precise value for the accumulated area under the curve \(f(x)\) between \(x = a\) and \(x = b\).
• It relies on the concept of taking the limit of a Riemann sum as the number of subdivisions \(n\) approaches infinity, becoming more refined.
• The result of the definite integral is a number and it can be interpreted as a net area, considering the sign of \(f(x)\).

By transitioning from a Riemann sum to a definite integral, we're able to calculate the exact area rather than just an approximation.

Limit as integral

When we talk about the limit as an integral, we're referring to the transition from an approximate sum (the Riemann sum) to an exact value (the definite integral). The concept hinges on the fundamental theorem of calculus.
This theorem links the notion of a limit of a sum with that of an integral, stating that:\[\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_{a}^{b} f(x) \, dx\]
This equation shows how, as the number of subintervals \(n\) increases indefinitely, the Riemann sum converges to the definite integral.

• The Riemann sum gets increasingly closer to the true value of the total area under \(f(x)\) as more intervals are added.
• The limit ensures that all refinements to a Riemann sum are accounted for, leaving no margin for approximation error.
• This makes the limit as integral a key concept for transforming discrete sums into continuous measurements.

Understanding this progression from limit to integral helps illustrate how calculus transitions between discrete sums and continuous functions, providing powerful tools for analysis and solving complex problems.