Problem 53 - 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. $$ \int_{1}^{\infty} \frac{\sqrt{x+1}}{x^{2}} d x $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{\infty} \frac{\sqrt{x+1}}{x^{2}} \, dx \) converges by the Direct Comparison Test.

Step 1: Analyze the Function

Consider the function \( f(x) = \frac{\sqrt{x+1}}{x^2} \). As \( x \to \infty \), the behavior of the function approximates \( \frac{\sqrt{x}}{x^2} = \frac{1}{x^{3/2}} \). This suggests we should compare it to the simpler integral \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \).

Step 2: Comparison Integral Setup

The integral \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \) is known to converge; it is a p-integral with \( p = \frac{3}{2} > 1 \). Therefore, we can use the Comparison Test to determine convergence.

Step 3: Direct Comparison Test

We need to show that \( \frac{\sqrt{x+1}}{x^2} \leq \frac{1}{x^{3/2}} \) for sufficiently large x. Notice that \( \sqrt{x+1} < \sqrt{2x} \) for all \( x \geq 1 \), so \( \frac{\sqrt{x+1}}{x^2} < \frac{\sqrt{2x}}{x^2} = \frac{\sqrt{2}}{x^{3/2}} \). Thus, \( \int_{1}^{\infty} \frac{\sqrt{x+1}}{x^2} \, dx \) converges by the Direct Comparison Test.

Step 4: Calculate the Known Comparison Integral

Let's confirm the convergence of the comparison integral for completeness. Compute \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \). This equals \[ \lim_{b \to \infty} \left[ -2x^{-1/2} \right]_1^b = \lim_{b \to \infty} \left( -2b^{-1/2} + 2 \right) = 2. \] Thus, it converges.

Key Concepts

Comparison Test

The Comparison Test is a method used to determine if an improper integral converges or diverges. This is especially useful when the integral has complex expressions that are difficult to integrate directly. The idea is to compare the given integral with another integral whose behavior is known. The Comparison Test is based on the following principle:

• If a function you are integrating, say \( f(x) \), is less than or equal to another known function \( g(x) \) for all \( x \) in the given interval, and the integral of \( g(x) \) from 1 to infinity converges, then the integral of \( f(x) \) also converges.
• Similarly, if \( f(x) \) is greater than or equal to \( g(x) \) and the integral of \( g(x) \) diverges, then the integral of \( f(x) \) diverges as well.

The test is simple and elegant because we don't need to evaluate the potentially complex integral directly. Instead, we utilize a known simpler integral that is easier to evaluate.

Direct Comparison Test

The Direct Comparison Test is a specific application of the Comparison Test. Here's how it works in practical terms:
For a given function \( f(x) \), choose a simpler function \( g(x) \) such that you know its integral behavior over the same interval. The main steps include:

• Verify \( 0 \leq f(x) \leq g(x) \) for all \( x \) in your desired interval, which is usually from 1 to infinity.
• If \( \int_{1}^{\infty} g(x) \,dx \) is known to converge, then \( \int_{1}^{\infty} f(x) \,dx \) also converges.
• If \( \int_{1}^{\infty} g(x) \,dx \) diverges, then \( \int_{1}^{\infty} f(x) \,dx \) diverges too.

In our exercise, we used \( \frac{\sqrt{2}}{x^{3/2}} \) as \( g(x) \) to show that \( \frac{\sqrt{x+1}}{x^2} \leq \frac{\sqrt{2}}{x^{3/2}} \) once \( x \geq 1 \). Since the integral of \( g(x) \) converges, our original integral also converges.

Improper Integrals

Improper integrals involve integrals where one (or both) of the limits of integration extend to infinity, or where the integrand becomes infinite within the interval of integration.
There are two main types:

• Infinite Limits: Where the integral bounds extend to infinity, such as \( \int_{1}^{\infty} f(x) \,dx \).
• Discontinuous Integrands: Where the function becomes undefined at some point in the interval.

In our case, the issue is the infinite upper bound \( \infty \). To determine the convergence of such an integral, tests like the Comparison Test are often employed. The integral \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \) is an example of how we determine whether the integral of \( \frac{\sqrt{x+1}}{x^2} \) converges or diverges.

Limit Comparison Test

The Limit Comparison Test is a variation of the Comparison Test that can be useful when direct comparison is difficult. This test is helpful in cases where the behavior of the function isn't easily bounded directly by a simpler function.
Here’s how it's generally applied:

• Choose a simpler function \( g(x) \) that has similar asymptotic behavior as \( f(x) \) when \( x \to \infty \).
• Compute the limit \( L = \lim_{x \to \infty} \frac{f(x)}{g(x)} \).
• If \( 0 < L < \infty \), then both \( \int_{1}^{\infty} f(x) \,dx \) and \( \int_{1}^{\infty} g(x) \,dx \) converge or diverge together.

This test is particularly advantageous when the simpler function used in the Direct Comparison Test is less straightforward to find. However, in the original step-by-step solution, we did not employ this test as the Direct Comparison Test was sufficient and suitable.