Problem 40 - Calculus Single Variable

Question

In each of Exercises \(31-40\), determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{1} \frac{1}{\sqrt{x} \sqrt{1-x}} d x\)

Step-by-Step Solution

Answer

The integral converges and equals \( \pi \).

Step 1: Understand the Type of Integral

The integral \( \int_{0}^{1} \frac{1}{\sqrt{x} \sqrt{1-x}} \, dx \) needs to be evaluated from 0 to 1. This is an improper integral because the integrand \( \frac{1}{\sqrt{x} \sqrt{1-x}} \) becomes undefined as it approaches the endpoints 0 and 1.

Step 2: Determine the Behavior Near Each Endpoint

We first examine the behavior of the integrand near each endpoint. As \( x \to 0^+ \), \( \frac{1}{\sqrt{x}} \to \infty \). Similarly, as \( x \to 1^- \), \( \frac{1}{\sqrt{1-x}} \to \infty \). This indicates potential divergence at both ends.

Step 3: Split the Integral at the Midpoint

To handle both singularities, split the integral: \[I = \int_{0}^{0.5} \frac{1}{\sqrt{x} \sqrt{1-x}} \, dx + \int_{0.5}^{1} \frac{1}{\sqrt{x} \sqrt{1-x}} \, dx.\]Each part needs to be examined separately for convergence.

Step 4: Apply Substitution for Evaluation

To evaluate these, use the substitution \( x = \sin^2(\theta) \). Then \( dx = 2\sin(\theta)\cos(\theta) d\theta \) and the integral bounds transform to \( \theta = 0 \) to \( \pi/4 \) and \( \pi/4 \) to \( \pi/2 \). The integral becomes:\[2 \int_{0}^{\pi/2} d\theta = 2\left[ \theta \right]_{0}^{\pi/2} = \pi.\]

Step 5: Evaluate Each Part Separately

The substitution effectively re-evaluates the whole integral as one piece from 0 to \( \pi/2 \) due to symmetry. Each separate integral computation results in \( \pi/2 \), confirming it converges.

Step 6: Confirm Convergence and Calculate Result

Both integrals converge and collectively give a result of \( \pi/2 + \pi/2 = \pi \). Hence, the original improper integral converges.

Key Concepts

Convergence of Integrals

Improper integrals involve limits, often because the function does not behave well at the boundaries of integration. In the problem we have, the integral \[ \int_{0}^{1} \frac{1}{\sqrt{x} \sqrt{1-x}} \, dx \]is improper due to the behavior of the function at both endpoints 0 and 1.To determine convergence:

• Analyze the behavior of the function as it approaches the limits. In our case, \( \frac{1}{\sqrt{x}} \to \infty \) as \( x \to 0^+ \), and \( \frac{1}{\sqrt{1-x}} \to \infty \) as \( x \to 1^- \).
• Check if both sides converge individually by splitting the integral. Convergence requires each part to produce a finite value.

If both integrals show convergence through their evaluations, it indicates that the original integral is convergent. That's the fundamental idea for convergence in improper integrals.

Substitution Method in Integration

The substitution method is a powerful tool in calculus used to simplify complex integrals. The core idea is to transform the integral into a simpler form by changing variables.For the given problem, substituting \( x = \sin^2(\theta) \) helps in two ways:

• It simplifies the limits of integration, transforming them to \( \theta = 0 \) to \( \pi/2 \).
• It converts the function into a form that's easier to integrate. \[ dx = 2\sin(\theta)\cos(\theta) d\theta \]

This substitution not only helps in evaluating specific parts of the integral more easily but also allows us to utilize symmetry in the problem, ultimately simplifying the computational process. This method shows how intricately related trigonometric identities and calculus concepts can be used to tackle seemingly difficult integrals.

Behavior of Integrands at Endpoints

Understanding how a function behaves at the endpoints of an integral is crucial in determining whether an integral is improper and if it converges.1. **Endpoint Analysis**: - At \( x = 0 \), \( \frac{1}{\sqrt{x}} \) tends towards infinity which indicates a vertical asymptote at the starting point. - As \( x \to 1 \), \( \frac{1}{\sqrt{1-x}} \) also tends towards infinity, signaling a similar asymptotic behavior.2. **Implications**: - Such behaviors suggest the potential for divergence. But that’s not always the case; it depends on whether these infinities balance out over the whole interval.By carefully evaluating the effects at these endpoints using methods like substitution or splitting at a midpoint, we can assess integrals that might initially seem divergent but are indeed convergent. This approach helps us determine the correct behavior and ensures clarity in evaluating such improper integrals.