Problem 63 - Calculus Single Variable

Question

In each of Exercises \(58-69\) use the Comparison Theorem to determine whether the given improper integral is convergent or divergent. In some cases, you may have to break up the integration before applying the Comparison Theorem. \(\int_{1 / 2}^{1} \frac{1}{x(1-x)^{1 / 3}} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{1/2}^{1} \frac{1}{x(1-x)^{1/3}} \, dx \) is divergent.

Step 1: Understanding the Problem

We need to determine if the improper integral \( \int_{1 / 2}^{1} \frac{1}{x(1-x)^{1 / 3}} \, dx \) is convergent or divergent using the Comparison Theorem. This involves comparing our integral with another integral that is easier to evaluate and has known convergence properties.

Step 2: Identify a Comparison Function

The given integral is \( \int_{1/2}^{1} \frac{1}{x(1-x)^{1/3}} \, dx \). Notice that the denominator \( (1-x)^{1/3} \) indicates a potential issue near the upper limit \( x = 1 \). To apply the Comparison Theorem, choose a simpler function. Here, consider \( \frac{1}{x} \), because it is similar and its behavior is well understood.

Step 3: Establish the Comparison Inequality

Since \( 0 < 1-x < 1 \) for any \( x \in [\frac{1}{2}, 1) \), we have \((1-x)^{1/3} < 1 \). This implies \( \frac{1}{x(1-x)^{1/3}} > \frac{1}{x} \). We can compare our integral to \( \int_{1/2}^{1} \frac{1}{x} \, dx \).

Step 4: Evaluate the Comparison Integral

The integral \( \int_{1/2}^{1} \frac{1}{x} \, dx \) is a well-known divergent integral. It evaluates to \( \ln|x| \Big|_{1/2}^{1} = \ln(1) - \ln(1/2) = 0 - (-\ln(2)) = \ln(2) \). Although this generates a finite value, the nature of the comparison shows divergence issues at the limit.

Step 5: Apply the Comparison Theorem

Since \( \frac{1}{x(1-x)^{1/3}} > \frac{1}{x} \) and the integral \( \int_{1/2}^{1} \frac{1}{x} \, dx \) is divergent, the Comparison Theorem implies that \( \int_{1/2}^{1} \frac{1}{x(1-x)^{1/3}} \, dx \) must also diverge.

Key Concepts

Comparison Theorem

The Comparison Theorem is a key technique in determining the behavior of improper integrals. It allows us to establish whether an integral is convergent or divergent by comparing it with another integral whose behavior we already know. Here's how it works in simple terms:

• First, we select a comparison function that closely resembles the original function but is simpler to analyze.
• The key step is to set up an inequality. The original function must be either less than or greater than this comparison function over the interval.
• If the comparison function is larger and has a known divergent integral, the original integral is also divergent.
• If the comparison function is smaller and converges, the original integral converges too.

For our exercise, we chose the function \( \frac{1}{x} \) to compare against \( \frac{1}{x(1-x)^{1/3}} \).

Convergence and Divergence

In the realm of integral calculus, understanding convergence and divergence is crucial. These terms describe whether the value of an integral settles to a finite number or not as it approaches its limit.

• Convergence refers to an integral that results in a finite value.
• Divergence indicates that the integral tends towards infinity or does not approach a specific value.

In our exercise, by comparing the given integral with a divergent one, \( \int_{1/2}^{1} \frac{1}{x} \, dx \), the Comparison Theorem confirms that the original integral is also divergent, due to it behaving similarly at the endpoint.

Integral Calculus

Integral calculus is a fundamental branch of calculus that deals with the concept of integration. Integration can be thought of as the reverse process of differentiation and it is used to compute areas, volumes, central points, and many other critical concepts in mathematics.
Two types of integrals often appear in calculus:

• Definite Integrals: Concerned with the sum of a continuous function over a defined interval, giving a numerical result.
• Improper Integrals: These involve limits and often occur when intervals extend to infinity, or the function becomes unbounded within the interval.

Understanding when these integrals converge or diverge through tools like the Comparison Theorem enhances problem-solving and application in various fields of science and engineering.