Problem 40 - Calculus for Biology and Medicine

Question

Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{6}}} d x $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{6}}} \, dx \) converges by comparison with \( \int_{1}^{\infty} \frac{1}{x^3} \, dx \).

Step 1: Identify the Integrand

We start with the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{6}}} \, dx \). The integrand is \( f(x) = \frac{1}{\sqrt{1+x^{6}}} \). Our goal is to compare this function to a simpler function to determine if the integral converges.

Step 2: Develop the Comparison Function

For large values of \(x\), the term \(x^6\) dominates over the constant 1 in the expression \(\sqrt{1 + x^6}\). Thus, a comparison function \( g(x) = \frac{1}{x^3} \) is suitable since for large \(x\), \( \sqrt{x^6} = x^3 \) closely approximates \( \sqrt{1 + x^6} \).

Step 3: Establish the Comparison Inequality

Next, we verify the inequality relationship. For \( x \geq 1 \), \( \sqrt{1 + x^6} > x^3 \), which implies \( \frac{1}{\sqrt{1 + x^6}} < \frac{1}{x^3} \). Therefore, \( f(x) < g(x) \) for all \( x \geq 1 \).

Step 4: Determine the Convergence of the Comparison Function

We analyze the integral of \( g(x) = \frac{1}{x^3} \). Determine if \( \int_{1}^{\infty} \frac{1}{x^3} \, dx \) converges. Compute \( \int_{1}^{\infty} \frac{1}{x^3} \, dx = \left[ -\frac{1}{2x^2} \right]_{1}^{\infty} = 0 - \left(-\frac{1}{2} \right) = \frac{1}{2} \). Consequently, this integral converges.

Step 5: Apply the Comparison Test

The integral of \( \frac{1}{x^3} \) converges, and \( \frac{1}{\sqrt{1 + x^6}} < \frac{1}{x^3} \) for all \( x \geq 1 \). By the Comparison Test, \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{6}}} \, dx \) also converges.

Key Concepts

Comparison Function

When dealing with improper integrals, one effective strategy for determining convergence is to use a comparison function. A comparison function is a simpler, related function whose integral we already know to converge or diverge. The goal is to compare the given function with this simpler function. If the given function is dominated by a convergent comparison function, then the original function's integral is also convergent.

In our specific example, the integrand is \( f(x) = \frac{1}{\sqrt{1 + x^6}} \). We are tasked with finding a suitable comparison function. We approximate \( \sqrt{1 + x^6} \) by focusing on the dominant term for large \( x \), which is \( x^3 \). This gives us the comparison function \( g(x) = \frac{1}{x^3} \).

Here’s how it works:

• For large \( x \), \( 1 + x^6 \) is close to \( x^6 \), making \( \sqrt{1 + x^6} \) close to \( x^3 \).
• Hence, \( f(x) \) is smaller than \( g(x) \) for large \( x \), simplifying the comparison.

By leveraging such a comparison, establishing the nature of the original integral becomes more straightforward.

Improper Integrals

Improper integrals are integrals that extend over an infinite interval or have unbounded integrands. In simple terms, they are integrals where either the limits of integration or the function values approach infinity.

This specific exercise involves the improper integral \( \int_{1}^{\infty} \frac{1}{\sqrt{1 + x^6}} \, dx \). The key challenge here is to determine whether such an integral converges (produces a real number) or diverges (grows without bound).

For improper integrals like \( \int_{1}^{\infty} \frac{1}{x^3} \, dx \), we can compute it directly. We find that the result converges to \( \frac{1}{2} \). Comparing our original integral to this determines its behavior.

• If the integral of a related function over the same interval converges, the integral of the given function may converge too.
• Use methods such as the Basic Comparison Test or Limit Comparison Test to analyze these integrals effectively.

This approach simplifies evaluating improper integrals in complex functions.

Limit Comparison Test

The Limit Comparison Test is a powerful method for determining the convergence of an improper integral. It is especially useful when the functions involved are complex, and finding a precise comparison is difficult. This test works by comparing the ratios of the functions to known convergent or divergent integrals.

In our scenario with \( f(x) = \frac{1}{\sqrt{1+x^6}} \) and the comparison function \( g(x) = \frac{1}{x^3} \), we determine if:
\[ \lim_{{x \to \infty}} \frac{f(x)}{g(x)} = L eq 0 \]

For example:

• If \( L \) is finite and non-zero, either both integrals \( \int f(x) \) and \( \int g(x) \) converge, or both diverge.
• If \( L = 0 \) and \( \int g(x) \) converges, then \( \int f(x) \) also converges.

This is essentially what happens here, as we find \( \lim_{{x \to \infty}} \frac{1/\sqrt{1 + x^6}}{1/x^3} = 1 \), which shows related behavior between the integrals. Thus, since \( \int g(x) \, dx \) converges, so does \( \int f(x) \, dx \), confirming convergence using this versatile technique.