Problem 47 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{1}^{\infty} \frac{d x}{x^{3}+1} $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{\infty} \frac{dx}{x^3 + 1} \) converges by the Direct Comparison Test.

Step 1: Identify the Type of Integral

First, we identify whether the given integral \[ \int_{1}^{\infty} \frac{dx}{x^3 + 1} \]is an improper integral. It is improper because it has an infinite upper limit of integration.

Step 2: Choose a Comparison Function

For the Direct Comparison Test, choose a simpler comparison function. Notice that\[ \frac{1}{x^3 + 1} < \frac{1}{x^3}. \]The function \( \frac{1}{x^3} \) is simpler and suitable for comparison as it has a known convergence behavior.

Step 3: Test Convergence of the Comparison Function

Calculate the integral of the simpler function:\[ \int_{1}^{\infty} \frac{dx}{x^3} = \left. -\frac{1}{2x^2} \right|_{1}^{\infty} = 0 - \left(-\frac{1}{2(1)^2}\right) = \frac{1}{2}. \]Since the above integral converges, \( \int_{1}^{\infty} \frac{dx}{x^3} \) converges.

Step 4: Apply the Direct Comparison Test

Since \( \frac{1}{x^3 + 1} \) is less than \( \frac{1}{x^3} \) for \( x \geq 1 \) and \( \int_{1}^{\infty} \frac{dx}{x^3} \) converges, by the Direct Comparison Test, \( \int_{1}^{\infty} \frac{dx}{x^3 + 1} \) also converges.

Key Concepts

Improper Integrals

An improper integral is an integral where one or both of the limits are infinite, or where the integrand becomes infinite within the limits of integration. This makes evaluation a bit trickier than a regular integral. In our case, the integral \[\int_{1}^{\infty} \frac{dx}{x^3 + 1} \] has an infinite upper limit, classifying it as an improper integral.To determine if such integrals are convergent (finite) or divergent (not finite), we often use comparison tests. Improper integrals are crucial in Calculus because they emerge in various applications, like in physics for calculating quantities extending to infinity, or in probability theory.

Comparison Test

The Comparison Test is a handy tool for evaluating the convergence of improper integrals. It involves comparing your complex function to a simpler one (comparison function) with known convergence behavior. This helps in determining the behavior of the original function's integral. Here's how it works:

• Identify a simpler function which bounds your original function.
• Ensure both functions have finite limits over the same interval.
• Evaluate the simpler function's integral.

If the simpler function's integral converges, and your original function is smaller or equal, then your original integral is likely convergent.

Convergence

Convergence, in the context of improper integrals, means that the integral evaluates to a finite value. It's a way of verifying that despite extending to infinity, the area under the curve of the function is not infinite.To check for convergence in improper integrals, you must:

• Identify if the integral is improper due to infinite limits or discontinuities.
• Use tests like Direct Comparison Test or Limit Comparison Test.
• Evaluate the integral of a simpler function for comparison that we know converges.

In our example, we saw that another integral, \( \int_{1}^{\infty} \frac{dx}{x^3} \), converges to \( \frac{1}{2} \). Since it's larger than our original function, by comparison, the original integral converges too.

Direct Comparison Test

The Direct Comparison Test is a robust way to determine convergence of an improper integral. By finding a comparison function that bounds your original function from above or below, you can establish convergence or divergence.For the given exercise, we used:\[\int_{1}^{\infty} \frac{dx}{x^3}\]as our comparison function. We know for certain that this converges, as calculated:\[\int_{1}^{\infty} \frac{dx}{x^3} = \frac{1}{2}\]Since \( \frac{1}{x^3 + 1} < \frac{1}{x^3} \) for \( x \geq 1 \), and \( \int_{1}^{\infty} \frac{dx}{x^3} \) converges, our original integral converges by the Direct Comparison Test. This test simplifies checking convergence, especially when dealing with complex functions.