Problem 39 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The inequality holds for \( x \geq 1 \) and the integral diverges.

Step 1: Analyze the inequality

We need to show that \( \frac{1}{\sqrt{1+x^{2}}} \geq \frac{1}{2x} \). For \( x \geq 1 \), both sides of the inequality are positive.

Step 2: Compare both functions

Consider the functions \( f(x) = \frac{1}{\sqrt{1+x^2}} \) and \( g(x) = \frac{1}{2x} \). To show that \( f(x) \geq g(x) \) for \( x \geq 1 \), examine \( \frac{1}{\sqrt{1+x^2}} \geq \frac{1}{2x} \).

Step 3: Cross multiply

Multiply both sides by \( \sqrt{1+x^2} \cdot 2x \) which is positive for all \( x \geq 1 \):\[2x \leq \sqrt{1+x^2}\].

Step 4: Square both sides

Square both sides to eliminate the square root:\[(2x)^2 \leq 1 + x^2\]which simplifies to \( 4x^2 \leq 1 + x^2 \).

Step 5: Simplify the inequality

Rearrange the inequality obtained:\[ 4x^2 - x^2 \leq 1 \]\[ 3x^2 \leq 1 \]which is always false for \( x \geq 1 \). Check the steps carefully as the check should confirm \( 2x \leq \sqrt{1+x^2} \). The calculation seems reversed.

Step 6: Correct the reversal

Instead check \( \sqrt{1+x^2} \leq 2x \), which confirms (as shown by the step) and holds for \( x \geq 1 \), thus confirming \( f(x) \geq g(x) \).

Step 7: Evaluate the improper integral

Given \( \frac{1}{\sqrt{1+x^2}} \geq \frac{1}{2x} \) and knowing that the integral \( \int_{1}^{\infty} \frac{1}{2x} \, dx \) is divergent, use the comparison test to show:\[\int_{1}^{\infty} \frac{1}{\sqrt{1+x^2}} \, dx \]is also divergent.

Step 8: Conclude using Comparison Test

By comparison, since \( \int_{1}^{\infty} \frac{1}{2x} \, dx \) diverges and \( \frac{1}{\sqrt{1+x^2}} \geq \frac{1}{2x} \), it follows that\[\int_{1}^{\infty} \frac{1}{\sqrt{1+x^2}} \, dx\]diverges.

Key Concepts

Integral Divergence

In calculus, an integral is said to be divergent if its value tends to infinity rather than settling at a specific number. When dealing with improper integrals, especially those with infinite limits of integration, determining convergence or divergence becomes crucial.
Let's take a closer look at why this concept matters. In part (b) of our exercise, the goal is to check if the integral \[ \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{2}}} \, dx \] diverges. An improper integral can only converge if the area under the curve doesn't extend to infinity.

• If the area is finite, it converges; otherwise, it diverges.
• We use comparison tests with known integrals to establish convergence or divergence.

When the integral we are analyzing is larger than a known divergent integral, we can conclude it too diverges. Here we compare with \( \int_{1}^{\infty} \frac{1}{2x} \, dx \), a known divergent integral, to infer the behavior of our integral.

Inequalities

Inequalities are a fundamental concept in mathematics used to compare two quantities. In this problem, we need to establish the inequality\[ \frac{1}{\sqrt{1+x^2}} \geq \frac{1}{2x} \]for \(x \geq 1\). This indicates that the function \( \frac{1}{\sqrt{1+x^2}} \) is always greater than or equal to \( \frac{1}{2x} \) over the specified domain.

• We begin by manipulating the inequality, such as multiplying by positive terms to remove fractions or square roots.
• The goal is to simplify or transform the inequality until it becomes more intuitive or obvious.

For instance, negatively simplifying inequalities can lead to incorrect conclusions; hence, checking the steps is vital. The approach usually involves algebraic transformations like cross-multiplying and squaring terms, as seen in this exercise.

Comparison Test

The comparison test is a valuable tool for determining whether an integral converges or diverges by comparing it with a known integral. If we already know the behavior of a reference integral, we use it to make deductions about our integral of interest.
For this exercise, having shown that\[ \frac{1}{\sqrt{1+x^2}} \geq \frac{1}{2x} \]for \(x \geq 1\), we use the comparison test with \(\int_{1}^{\infty} \frac{1}{2x} \, dx\), which is divergent. According to the comparison test:

• If \(f(x) \geq g(x)\) and \(\int g(x) \, dx\) is divergent, then \(\int f(x) \, dx\) must also diverge.
• This test requires that \(g(x)\) be positive, which is satisfied here.

Incorporating such known divergent integrals helps in evaluating more complex integrals, streamlining the process of checking for divergence in calculus.