Problem 69 - 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. \(\int_{1}^{3} \frac{1}{(x-1)^{5 / 3} \sqrt{3-x}} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{1}^{3} \frac{1}{(x-1)^{5 / 3} \sqrt{3-x}} \, dx \) is divergent.

Step 1: Analyze the integral

The given integral is \( \int_{1}^{3} \frac{1}{(x-1)^{5/3} \sqrt{3-x}} \, dx \). It is evident that the integral might be improper at \( x = 1 \) and \( x = 3 \). We will need to apply the Comparison Theorem to determine its convergence.

Step 2: Consider behavior at x = 1

As \( x \) approaches 1 from the right, \( \frac{1}{(x-1)^{5/3}} \) becomes very large. Compare this part with \( \frac{1}{(x-1)^{5/3}} \). An integral that diverges is \( \int_{1}^{b} \frac{1}{(x-1)^{5/3}} \, dx \), for any \( b > 1 \), which diverges because the power \( -5/3 < -1 \).

Step 3: Consider behavior at x = 3

As \( x \) approaches 3 from the left, \( \frac{1}{\sqrt{3-x}} \) becomes large. This term is comparable to \( \frac{1}{\sqrt{3-x}} \). An integral that converges is \( \int_{a}^{3} \frac{1}{\sqrt{3-x}} \,dx \), for any \( a < 3 \), which converges because \( -1/2 > -1 \).

Step 4: Apply the Comparison Theorem

Compare the important terms we identified in Steps 2 and 3. The term \( \frac{1}{(x-1)^{5/3}} \) has a dominant behavior at \( x=1 \), especially more significant than the diminishing \( \frac{1}{\sqrt{3-x}} \) at \( x=3 \). The divergence of the integral \( \int_{1}^{b} \frac{1}{(x-1)^{5/3}} \, dx \) suggests that the given integral is divergent overall.

Key Concepts

Improper Integrals

Improper integrals occur when the interval of integration is infinite or if the integrand has infinite discontinuities within the interval. In such cases, parts of the integral might not have finite limits, resulting in problematic computations. In the exercise, the integrand \( \frac{1}{(x-1)^{5/3} \sqrt{3-x}} \) has points of interest at both \( x = 1 \) and \( x = 3 \).
As \( x \) approaches these points, the behavior of the integral becomes crucial.

• At \( x = 1 \), \( \frac{1}{(x-1)^{5/3}} \) becomes excessively large as \( x \) approaches 1, signifying an improper behavior there.
• At \( x = 3 \), \( \frac{1}{\sqrt{3-x}} \) also leads to issues as \( x \) approaches 3 from the left.

Analyzing these behaviors separately helps us decide if the improper integral converges (yields a finite result) or diverges (yields an infinite result). Avoiding mistakes with improper integrals is essential in calculus, and understanding these problems in detail is the first step toward effectively handling more complex situations.

Convergence and Divergence

Understanding whether an improper integral converges or diverges is fundamental. Convergence indicates that the integral results in a specific value, whereas divergence suggests an infinite or undefined output. The Comparison Theorem is a potent tool to judge convergence or divergence. It involves comparing an unknown integral behavior with that of known integrals which have established convergence or divergence patterns.
In typical scenarios, we use known simple behaviors:

• For instance, \( \int_{1}^{b} \frac{1}{(x-1)^{5/3}} \, dx \) diverges because the exponent on \( (x-1) \) is less than -1, indicating that it doesn't converge to any finite number.


• Conversely, \( \int_{a}^{3} \frac{1}{\sqrt{3-x}} \, dx \) converges because its exponent is greater than -1, allowing it to sum up to a finite value.

By aligning the behavior of the integrals in question to comparable well-understood integrals, one can deduce their nature correctly, as shown in the step-by-step solution.

Calculus Integration Techniques

To effectively apply the Comparison Theorem or evaluate integrals like \( \int \), it's essential to be familiar with different integration techniques. These techniques include substitution, partial fraction decomposition, and integration by parts.
However, for improper integrals that are evaluated using the Comparison Theorem, one primarily relies on understanding the asymptotic behavior or growth at the limits of integration.
In our exercise, we didn't directly apply classical integration techniques to solve the problem. Instead, we scrutinized the integrand's behavior near the problematic points \( x = 1 \) and \( x = 3 \).

• Such analysis requires a firm grasp of limits and growth of functions like \( (x-1)^{-5/3} \) and \( (3-x)^{-1/2} \).
• These methods help to comprehend whether the terms lead toward infinity or stabilize, providing the necessary insight into convergence/divergence.

In short, while we utilize various integration techniques in calculus, mastering the behavior of functions around critical points remains vital for solving complex integrals effectively.