Problem 43 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral \(\int_{1}^{\infty} \frac{1}{\sqrt{1+x}} dx\) is divergent.

Step 1: Determine a Simplified Comparison Function

The given integrand is \(\frac{1}{\sqrt{1+x}}\). For large values of \(x\), \(1+x\) behaves like \(x\). Thus, \(\sqrt{1+x} \approx \sqrt{x}\). Simplify the function for comparison as \(\frac{1}{\sqrt{x}}\).

Step 2: Formulate the Comparison Integral

Now, consider the comparison integral \(\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx\). This is a standard comparison to check for convergence.

Step 3: Evaluate the Comparison Integral

To evaluate \(\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx\), rewrite it as \(\int_{1}^{\infty} x^{-\frac{1}{2}} \, dx\). The antiderivative of \(x^{-\frac{1}{2}}\) is \(2x^{\frac{1}{2}}\). Evaluate from 1 to infinity:\[\lim_{b\to\infty} \left[2x^{\frac{1}{2}} \right]_1^b = \lim_{b\to\infty} \left(2b^{\frac{1}{2}} - 2\right) = \infty\]Since the integral diverges, \(\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx\) is divergent.

Step 4: Apply the Comparison Test

Since \(\frac{1}{\sqrt{1+x}} \leq \frac{1}{\sqrt{x}}\) for all \(x \geq 1\) and the comparison integral \(\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx\) is divergent, by the Comparison Test, \(\int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx\) is also divergent.

Key Concepts

Convergence of an Integral

Convergence is a key concept in understanding whether an improper integral has a finite value or not. When we talk about convergence in integrals, we essentially want to know if we evaluate the integral over an infinite interval or where the function approaches infinity, will it yield a finite number? If it does, the integral is said to be convergent; if not, it is divergent. In the case of improper integrals, these typically involve infinite limits or unbounded integrands. Thus, testing for convergence often involves using techniques or tests that can handle such infinities.When examining the convergence of our examples such as \[\int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx\], karrying out a comparison with a simpler function can reveal whether this integral converges or diverges. Using a function that is easier to integrate, opens up determining the behavior of the original integral by comparison.

Understanding Improper Integrals

Improper integrals extend the concept of definite integrals to cases with infinite limits or discontinuous integrands. These occur when the interval of integration is infinite, such as \(\int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx\), or the integrand becomes infinite within the interval. Roughly speaking, it's like asking for the total area under a curve that stretches to infinity or where the height of the curve becomes infinitely large. The key is to analyze whether this total area converges to a finite number.To evaluate an improper integral:

• Transform the infinite bound into a limit, e.g., \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{\sqrt{x}} \, dx \).
• If the limit approaches a finite number, the integral converges.
• If not, the integral diverges or becomes infinite.

This process helps determine whether stacking infinite amounts of small area additions results in something finite or infinite.

Integral Comparison Technique

The Integral Comparison Test is a practical method used to determine the convergence or divergence of an improper integral by comparing it to a known benchmark function. The principle is straightforward:

• Choose a simpler function of known behavior for comparison, often a function with a simpler denominator.
• Ensure the inequality holds, where the integrand is less than or equal to this comparison function for a given range.
• If the simpler integral diverges, and our function stays below, the original one diverges too.
• If the simpler integral converges and the original integrand is greater than or equal, the original converges.

In the problem at hand: Given \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx \), we compared it to the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \). Since \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) diverges and \( \frac{1}{\sqrt{1+x}} \leq \frac{1}{\sqrt{x}} \), by comparison, the original integral also diverges. This technique helps us make complex integral evaluation feasible by relying on known results.