Problem 31 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Determine whether the integral is convergent or divergent. Evaluate all convergent integrals. Be efficient. If \(\lim _{x \rightarrow \infty} \neq 0\), then \(\int_{a}^{\infty} f(x) d x\) is divergent. \(\int_{1}^{2} \frac{1}{x \ln x} d x\)

Step-by-Step Solution

Answer

The integral \(\int_{1}^{2} \frac{1}{x \ln x} d x\) is convergent and equals \(\ln 2\).

Step 1: Apply integration by substitution

Let \(u = \ln x\), then \(du = \frac{1}{x} dx\). This changes the integrand into \(\int_{1}^{2} \frac{1}{u} du\), which is much simpler to handle.

Step 2: Evaluate the integral

The antiderivative of \(\frac{1}{u}\) is \(\ln u\). So, \(\int_{1}^{2} \frac{1}{u} du = \ln u \Big|_{1}^{2} = \ln 2 - \ln 1 = \ln 2\).

Step 3: Apply the integral test

We need to check if \(\lim _{x \rightarrow \infty} f(x) = 0\) for our original function \(f(x) = \frac{1}{x \ln x}\). However, since our integral does not extend to infinity, this step is not needed. Therefore, the integral is automatically convergent.

Step 4: Recall results and conclude

The result of the integral is finite, \(\ln 2\), and by the fundamental theorem of calculus, the integral is convergent. Hence we confirmed that the integral \(\int_{1}^{2} \frac{1}{x \ln x} d x\) is, in fact, convergent

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify integrals and make them easier to solve. It involves changing the variable of integration to transform the integral into a simpler form. Think of it like "reverse engineering" your function to find an easier path to integrate.
For example, in the original exercise, substituting a new variable made the integration a lot easier. We used:

• Letting \( u = \ln x \)
• Realizing \( du = \frac{1}{x} dx \)

By substituting \( u \) and \( du \) into the integral, \( \int_{1}^{2} \frac{1}{x \ln x} dx \) became \( \int_{1}^{2} \frac{1}{u} du \).
This change of variables simplifies the process, converting a complex integral into a basic form, allowing us to find the answer more efficiently.

Convergence of Integrals

Determining whether an integral is convergent or divergent is crucial when solving integrals over an interval. A convergent integral suggests a finite area under the curve, while a divergent integral doesn't converge to a finite value.
In the given exercise, convergence is assessed within the finite interval \([1, 2]\), focusing on whether the actual defined limits provide a finite result. Because the bounds \(1\) and \(2\) don't extend to infinity, checking limit behavior isn't necessary.
However, when dealing with improper integrals, you'd check convergence by determining the integral over an infinite interval; if the limit approaches a non-zero finite number, the integral is convergent.

• Identify defines bounds (e.g., finite, infinite)
• Check if the function becomes undefined within the bounds
• Conclude based on mathematical calculations (e.g., \(\int_{1}^{\infty} f(x) dx\) requires limit checks)

Antiderivatives

Antiderivatives, also known as indefinite integrals, are essential to solving differential equations and finding functions whose derivatives are known. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).
In our exercise, we found the antiderivative for \( \frac{1}{u} \) as \( \ln u \). Applying this antiderivative helped compute the value of the definite integral from \( 1 \) to \( 2 \).

• Identify the basic form (e.g., \( \int \frac{1}{u} du = \ln u + C \))
• Solve definite integrals using limits (e.g., \( \ln u \bigg|_1^2 = \ln 2 - \ln 1 \))
• Ensure the calculated values further confirm the function type (convergent/divergent)

Antiderivatives are particularly helpful in evaluating simple integrals, providing a straightforward method to determine the total area under a curve within set bounds.