Problem 40 - Calculus for Biology and Medicine

Question

(a) Show that $$0 \leq \frac{1}{\sqrt{x+x^{4}}} \leq \frac{1}{x^{2}}$$ for \(x>0\). (b) Use your result in (a) to show that $$\int_{1}^{\infty} \frac{1}{\sqrt{x+x^{4}}} d x$$ is convergent.

Step-by-Step Solution

Answer

(a) Both inequalities hold for all \( x > 0 \). (b) The integral converges.

Step 1: Establish the Range of the Function

For the given problem, we need to show that \( 0 \leq \frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2} \) when \( x > 0 \). The expression inside the square root, \( x + x^4 \), is always positive for \( x > 0 \). Hence, \( \frac{1}{\sqrt{x+x^4}} \) is always positive, which means \( \frac{1}{\sqrt{x+x^4}} \geq 0 \).

Step 2: Compare \( \frac{1}{\sqrt{x+x^4}} \) to \( \frac{1}{x^2} \)

Rewrite the inequality \( \frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2} \). This means showing \( \sqrt{x+x^4} \geq x^2 \). Simplify: \[ x + x^4 \geq x^4 \] \[ x \geq 0 \].Thus, since both sides are positive and \( x \geq 0 \) for \( x > 0 \), the inequality holds true. Therefore \( 0 \leq \frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2} \).

Step 3: Determine Convergence of the Integral

Using the comparison test for convergence, we compare the integral of \( \frac{1}{\sqrt{x+x^4}} \) to the integral of \( \frac{1}{x^2} \) from 1 to infinity.The integral of \( \frac{1}{x^2} \) is \[ \int_{1}^{\infty} \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_{1}^{\infty} = 1 \].Since the integral of \( \frac{1}{x^2} \) converges and \( \frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2} \), the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+x^4}} dx \) also converges by the Comparison Test.

Key Concepts

Integral Convergence

In calculus, the convergence of an integral is a vital concept, as it helps us determine if the area under a curve is finite or infinite. When dealing with improper integrals, which extend to infinity, convergence becomes crucial. To determine if \[\int_{1}^{\infty} \frac{1}{\sqrt{x+x^4}} \, dx\]converges, we need to apply mathematical techniques that assess whether the total area is finite. If the integral results in a finite number, it converges, meaning the total area under the curve can be measured.
Convergence is often assessed using comparison with integrals of functions that are already known to converge or diverge. This concept supports understanding the behavior of functions as they stretch towards infinity, providing important insights in both theoretical and applied mathematics.

Inequality

Inequalities are mathematical expressions showing the relationship between two values or functions which aren't necessarily equal. In this exercise, we were asked to demonstrate\[0 \leq \frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2}\]for \(x > 0\). Breaking this down:

• \(0 \leq \frac{1}{\sqrt{x+x^4}}\): Since \(x + x^4\) is positive for all \(x > 0\), \(\frac{1}{\sqrt{x+x^4}}\) also remains positive, satisfying the inequality.
• \(\frac{1}{\sqrt{x+x^4}} \leq \frac{1}{x^2}\): This suggests that \(\sqrt{x+x^4} \geq x^2\). When simplified, \(x + x^4\) being greater than or equal to \(x^4\) implies \(x \geq 0\), which is true for any positive \(x\).

Inequalities like this allow us to compare functions and determine bounds, which is particularly useful for understanding convergence in calculus. They lay the groundwork for more complex analysis by ensuring a function's behavior is within predictable limits.

Comparison Test

The comparison test is a powerful method used to determine the convergence or divergence of an improper integral by comparing it to another integral whose behavior (convergence or divergence) is already known. In this exercise, we utilized the comparison test to establish the convergence of\[\int_{1}^{\infty} \frac{1}{\sqrt{x+x^4}} \, dx\]by contrasting it with\[\int_{1}^{\infty} \frac{1}{x^2} \, dx\]which is a well-known convergent integral.
The key steps in the comparison test include:

• Identify a simpler function \(g(x)\) such that \(f(x) \leq g(x)\) for the interval in question.
• If \(\int_{a}^{\infty} g(x) \, dx\) converges, then \(\int_{a}^{\infty} f(x) \, dx\) must also converge.

This process simplifies the work, as already solved cases are used to infer the behavior of more complex functions. By ensuring that our function was always less than a known convergent function, we applied the comparison test successfully to show convergence. It's a handy tool for mathematicians dealing with infinite series or functions.