Problem 64 - 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. \(\int_{0}^{\pi / 2} \frac{1}{\sqrt{x}+\sin (x)} d x\)

Step-by-Step Solution

Answer

The integral is divergent by the Comparison Theorem.

Step 1: Identify the Improper Integral

We need to examine the integral \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x} + \sin(x)} \, dx \). We know it is improper because the integral's lower limit, \(x = 0\), makes the denominator approach zero, indicating a potential issue at this point.

Step 2: Define the Comparison Function

To use the Comparison Theorem, we need a function that is easier to integrate and can serve as a comparison. Here, \(\sin(x)\) is bounded from 0 (at \(x = 0\)) to 1 (at \(x = \pi/2\)), so we consider the function \(g(x) = \frac{1}{\sqrt{x}}\), which will be used to compare with the given integrand.

Step 3: Comparison Analysis Validation

We check if \( \frac{1}{\sqrt{x} + \sin(x)} \leq \frac{1}{\sqrt{x}} \) over the interval \([0, \pi / 2]\). Since \(\sin(x)\) adds a non-negative value to \(\sqrt{x}\), the inequality holds as the denominator of our integral is larger, thus making the term smaller.

Step 4: Evaluate the Comparison Integral

Evaluate \(\int_{0}^{\pi/2} \frac{1}{\sqrt{x}} \, dx \), which is known to be divergent. The integral of \(\frac{1}{\sqrt{x}}\) is \(2\sqrt{x}\), and we take the limit as \(x \to 0\), resulting in a divergent integral.

Step 5: Conclude Using the Comparison Theorem

Since \(\int_{0}^{\pi/2} \frac{1}{\sqrt{x}} \, dx\) is divergent and \( \frac{1}{\sqrt{x} + \sin(x)} \leq \frac{1}{\sqrt{x}} \), by the Comparison Theorem, \(\int_{0}^{\pi/2} \frac{1}{\sqrt{x} + \sin(x)} \, dx\) is also divergent.

Key Concepts

Understanding Improper Integrals

Improper integrals arise when we calculate the integral over an interval where the function is unbounded, or the limits of integration involve infinity or a point where the function is undefined. This requires special consideration since normal integration techniques cannot handle these issues directly.
For instance, consider the integral \( \int_{0}^{\pi / 2} \frac{1}{\sqrt{x} + \sin(x)} \, dx \) from our exercise. The issue at hand occurs at the lower limit \(x = 0\), where the denominator \(\sqrt{x} + \sin(x)\) approaches zero, suggesting the function tends to infinity as \(x\) approaches the lower limit. This is a classic example of an improper integral.
Instead of traditional methods, we need tools like limits to carefully resolve these cases, effectively completing the integration by examining the behavior as it approaches the troubling points.

Convergence and Divergence Explained

Understanding whether an improper integral converges or diverges is crucial:

• A convergent improper integral means the integral evaluates to a finite number despite its initial complications.
• A divergent improper integral means the integral evaluates to infinity or does not have a well-defined finite value.

For our example, we apply the Comparison Theorem, which is particularly useful in these situations. We compare the given function \( \frac{1}{\sqrt{x} + \sin(x)} \) with the simpler function \( \frac{1}{\sqrt{x}} \). Since the integral of \( \frac{1}{\sqrt{x}} \) from 0 to \(\pi/2\) diverges, and our original function is always smaller or equal, it leads us to conclude that our initial integral is also divergent.

Integration Techniques: The Comparison Theorem

The Comparison Theorem is a powerful tool used in determining the convergence or divergence of an improper integral. This theorem involves comparing the function in question to another function whose integral behavior is already known or easier to ascertain.
When applying the Comparison Theorem, the steps are as follows:

• Choose a function\( g(x) \) which is simpler and has a known integral behavior that acts as a benchmark for comparison.
• Check if \( f(x) \leq g(x) \) holds over the interval of interest. If it does and \( \int g(x) \, dx \) diverges, \( \int f(x) \, dx \) is also divergent. Similarly, if \( \int g(x) \, dx \) converges, so does \( \int f(x) \, dx \).

In our exercise, we successfully used the function \( \frac{1}{\sqrt{x}} \) to determine the behavior of the given integral. Since \( \int_{0}^{\pi/2} \frac{1}{\sqrt{x}} \, dx \) diverges, and it's greater than or equal to our original function, it confirms that \( \int_{0}^{\pi/2} \frac{1}{\sqrt{x} + \sin(x)} \, dx \) is also divergent, providing a clearer understanding of the integral's nature.