Problem 58 - Calculus Single Variable

Question

In each of Exercises \(58-69\) use the Comparison Theorem to determine whether the given improper integral is convergent or divergent. In some cases, you may have to break up the integration before applying the Comparison Theorem. \(\int_{0}^{\pi / 2} \frac{\cos (x)}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The integral converges.

Step 1: Analyze the Integral

We need to determine if the integral \( \int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{x}} \, dx \) is convergent or divergent. The function has a potential problem at \( x = 0 \) due to the \( \frac{1}{\sqrt{x}} \) term making the denominator zero.

Step 2: Choose a Comparison Function

To use the Comparison Theorem, we select a simpler function to compare with. Notice that \( \cos(x) \leq 1 \) for all \( x \), which implies \( \frac{\cos(x)}{\sqrt{x}} \leq \frac{1}{\sqrt{x}} \) on \( (0, \frac{\pi}{2}) \). We will compare our integral to \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x}} \, dx \).

Step 3: Evaluate the Comparison Integral

The integral \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x}} \, dx \) can be computed as: \[ \int_{0}^{\pi / 2} x^{-1/2} \, dx = \left. 2x^{1/2} \right|_0^{\pi/2} = 2\sqrt{\pi/2} - 2\sqrt{0} = 2\sqrt{\pi/2}. \] This integral is finite, indicating convergent behavior as \( x \to 0 \).

Step 4: Apply the Comparison Theorem

Since \( \frac{\cos(x)}{\sqrt{x}} \leq \frac{1}{\sqrt{x}} \), and we have shown that \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x}} \, dx \) converges, the Comparison Theorem tells us that \( \int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{x}} \, dx \) must also converge.

Key Concepts

Understanding Improper Integrals

Improper integrals are a key topic in calculus that help us deal with integrals that have infinite limits or discontinuous integrands. In the integral given in the original problem, we focus on an improper integral because the function \( \frac{\cos(x)}{\sqrt{x}} \) becomes problematic at \( x = 0 \). This is due to the term \( \frac{1}{\sqrt{x}} \) which becomes undefined as \( x \) approaches zero. To evaluate such integrals, we must assess the behavior as the approach the problematic points. In this exercise, the challenge is to manage how the integral behaves near zero. The use of comparison theorems helps determine convergence, allowing us to compare to a known integral for better understanding.

Convergence Explained

The convergence of an integral is about whether it has a finite value. In this exercise, we test the integral's behavior at the lower limit \( x = 0 \).
We used the Comparison Theorem to compare \( \int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{x}} \, dx \) with a simpler integral that we already understand: \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x}} \, dx \). By showing that this latter integral does not diverge as \( x \rightarrow 0 \), proving it is finite, we establish the convergence of our original integral.
The concept of convergence is essential for understanding many problems in calculus and is used to determine whether integrals have meaningful, finite results.

Exploring Trigonometric Functions

Trigonometric functions like \( \cos(x) \) are vital in calculus. Their properties make them ideal for integrals and comparisons.
In this problem, \( \cos(x) \) behaves predictably between 0 and \( \frac{\pi}{2} \), providing values between 0 and 1. This range helps us create an upper bound using \( \frac{1}{\sqrt{x}} \). By knowing \( \cos(x) \leq 1 \), we simplified the problem, making it easier to check the convergence using comparison.
Trigonometric functions often simplify complex integrals due to well-known limits, identities, and derivatives, thus making them integral to calculus.

Integral Calculus Fundamentals

Integral calculus allows us to evaluate the cumulative sum of infinite infinitesimal values, often through techniques that simplify complex problems. By using methods like the Comparison Theorem, we streamline solving improper integrals.
In the given problem, adopting a known, simpler function \( \frac{1}{\sqrt{x}} \) for comparison, allowed us to sidestep computational difficulties. This approach highlights a major strength of integral calculus: using known results to validate and solve more challenging problems.
Integral calculus not only helps in evaluating areas under curves but also finds significant applications in physics, engineering, and other scientific domains, showcasing its wide-reaching significance.