Problem 67 - 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. \(\int_{1}^{2} \frac{1}{\sqrt{x^{3}+x^{2}-2}} d x\)

Step-by-Step Solution

Answer

The integral is convergent.

Step 1: Analyze the behavior of the integrand

First, we need to understand how the integrand behaves as the variable approaches the limits of integration. We're looking at the function \( f(x) = \frac{1}{\sqrt{x^3 + x^2 - 2}} \) over the interval \([1, 2]\). By substituting values, we notice that the expression under the square root, \(x^3 + x^2 - 2\), is continuous and positive since \( x \geq 1 \), ensuring no division by zero or negative square roots. Thus, the function is defined over this interval.

Step 2: Find a Comparison Function

We will find a simpler function \( g(x) \) that we can compare with \( f(x) \). Since \( x^3 + x^2 - 2 \geq 0 \) for \( x \geq 1 \), a good candidate is to compare with \( g(x) = \frac{1}{\sqrt{x^3}} = \frac{1}{x^{3/2}} \). This function is simpler and can be analyzed easily for convergence or divergence.

Step 3: Apply the Comparison Theorem

To use the Comparison Theorem, we note that for \( x \geq 1 \), \[ \sqrt{x^3 + x^2 - 2} \geq \sqrt{x^3} = x^{3/2} \]Thus, it follows that \[ f(x) = \frac{1}{\sqrt{x^3 + x^2 - 2}} \leq \frac{1}{x^{3/2}} = g(x) \]The Comparison Theorem tells us that if \( \int_{1}^{2} g(x) \, dx \) converges, then \( \int_{1}^{2} f(x) \, dx \) also converges.

Step 4: Evaluate the Integral of the Comparison Function

Evaluate \( \int_{1}^{2} \frac{1}{x^{3/2}} \, dx \). This requires integrating \( x^{-3/2} \), which is done by finding the antiderivative:\[ \int x^{-3/2} \, dx = \frac{x^{-1/2}}{-1/2} = -2x^{-1/2} \]Evaluating from 1 to 2 gives:\[-2x^{-1/2} \bigg|_{1}^{2} = -2(2)^{-1/2} + 2(1)^{-1/2} = -2 \cdot \frac{1}{\sqrt{2}} + 2 \cdot 1 \]\[ = -\sqrt{2} + 2 \]This is a finite number, so the integral of \( \frac{1}{x^{3/2}} \) over \([1, 2]\) converges.

Step 5: Conclusion Based on Comparison Theorem

Since the comparison integral \( \int_{1}^{2} \frac{1}{x^{3/2}} \, dx \) converges and we have \( f(x) \leq g(x) \), it follows by the Comparison Theorem that the original integral \( \int_{1}^{2} \frac{1}{\sqrt{x^3 + x^2 - 2}} \, dx \) also converges.

Key Concepts

Understanding Improper Integrals

An improper integral involves integrals with infinite limits or integrands that approach infinity at one or more points within the interval of integration.
These can be tricky because traditional methods, which rely on finite limits and well-behaved functions, do not apply directly.
Improper integrals require special attention to ensure they are well-defined and could be finite or infinite:

• If an improper integral evaluates to a finite value, it is said to converge.
• Conversely, if it equals infinity, it diverges.

To analyze these integrals, it's crucial to look at the behavior of the integrand and check the interval of integration.
In the exercise, the integral is defined on a finite interval; thus, the focus is on the integrand behavior, not the limits.

Exploring Convergence and Divergence

Determining whether an improper integral converges or diverges requires understanding its behavior.
A useful method is using the Comparison Theorem. This theorem works by comparing the integral of interest to a known function that has a clear convergence behavior.

• If the given integrand is smaller than a converging comparison function, the original integral converges.
• If it is larger than a diverging comparison function, the original integral diverges.

It's like having a benchmark of behavior to guide us in making conclusions about our integral. In this exercise, we derive convergence of the integral by comparing it to another function known to converge.

Finding Antiderivatives

The antiderivative is the reverse process of differentiation, simple integration, crucial for evaluating definite integrals.
To find an antiderivative, you need to recognize the function type and use integration techniques to simplify.
In our step-by-step solution, the antiderivative of the comparison function, \( \frac{1}{x^{3/2}} \), is found using power rule integration: \[ \int x^{-3/2} \, dx = \frac{x^{-1/2}}{-1/2} = -2x^{-1/2} \]This transformation helps determine the value of the integral over selected bounds, allowing us to conclude its convergence effectively.

Conducting Integrand Analysis

Analyzing an integrand is crucial in determining the integral's behavior, particularly with improper integrals.
A deep dive into how the function behaves within the interval of integration provides insight.
In this exercise, \[ f(x) = \frac{1}{\sqrt{x^3 + x^2 - 2}} \] was assessed for values over \([1, 2]\). During this analysis, it became clear that the function is continuous and positive, indicating it is well-defined over the interval.
Such analysis helps in choosing an appropriate comparison function effectively, which is vital for leveraging the Comparison Theorem to determine convergence.