Problem 59 - 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. \(\int_{0}^{1} \frac{e^{x}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The integral is divergent.

Step 1: Identify the Type of Improper Integral

The given integral \( \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} \, dx \) is improper because the integrand \( \frac{e^{x}}{\sqrt{x}} \) has a vertical asymptote at \( x = 0 \). This means the function is not defined when \( x = 0 \).

Step 2: Choose a Comparison Function

For the integral \( \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} \, dx \), consider the integrand \( \frac{e^{x}}{\sqrt{x}} \). For \( x \) near 0, \( e^{x} \approx 1 \), so we can use \( \frac{1}{\sqrt{x}} \) as a comparison function since it is simpler.

Step 3: Compare Integrals Using the Comparison Theorem

By the Comparison Theorem, if \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) is known to be convergent or divergent, and \( \frac{e^{x}}{\sqrt{x}} \leq C\frac{1}{\sqrt{x}} \) for some constant \( C \) on \( (0, 1] \), then \( \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} \, dx \) takes on the same property.

Step 4: Evaluate Simple Comparison Integral

To evaluate \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \), find:\[\text{Let } u = \sqrt{x} \Rightarrow du = \frac{1}{2\sqrt{x}}dx \Rightarrow dx = 2u du.\]So,\[\int \frac{1}{\sqrt{x}} \, dx = 2 \int 1 \, du = 2u = 2\sqrt{x}.\]Evaluate from 0 to 1 gives \[ 2 \sqrt{x} \bigg|_{0}^{1} = 2(1) - 2(0) = 2. \] This suggests divergence due to the lower limit's behavior.

Step 5: Analyze The Comparison

Since \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) is divergent, we compare \( \frac{e^{x}}{\sqrt{x}} \leq C\frac{1}{\sqrt{x}} \) for \( e^{x} \) bounded near \( x = 0 \). This confirms that the original integral \( \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} \, dx \) must also be divergent.

Key Concepts

Improper Integrals

Improper integrals involve an infinite limit or an integrand that becomes infinite over the interval of integration. When dealing with such integrals, traditional techniques won't work because the function tends to become undefined or infinite at some point, typically at the bounds. In the case of our exercise, the integrand \( \frac{e^{x}}{\sqrt{x}} \) becomes problematic at \( x = 0 \) because \( \frac{1}{\sqrt{x}} \rightarrow \infty \) as \( x \rightarrow 0^+ \). This results in what we call a vertical asymptote at the lower end of the integration. To manage these scenarios, we often need to use special theorems and techniques, such as the Comparison Theorem, to understand the behavior of the integral. By comparing our difficult integral to a simpler one that we know more about, we can determine whether the initial integral converges or diverges.

Integral Convergence

Integral convergence relates to understanding whether the area under a curve that extends infinitely or near an asymptote sums up to a finite number or grows indefinitely.

• A convergent integral indicates that the total area described by the integrand approaches a finite limit.
• A divergent integral means that the area is infinite or doesn't settle to a specific value.

For improper integrals, patterns of behavior are crucial. Using tools like the Comparison Theorem helps decide convergence. In our exercise, we compare \( \int_{0}^{1} \frac{e^{x}}{\sqrt{x}} \, dx \) with \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \). We find the simpler integral diverges. Hence, by comparison, our given integral must also be divergent. This method grants a clearer understanding of whether these more complex integrals settle to something finite or not.

Vertical Asymptote

A vertical asymptote appears in a graph where the function heads towards infinity at a specific value. This occurs because at that point, the function becomes undefined or ill-behaved. In the problem's context, the function \( \frac{e^{x}}{\sqrt{x}} \) has a vertical asymptote at \( x = 0 \) since \( \frac{1}{\sqrt{x}} \) increases endlessly as \( x \) approaches zero.

• Asymptotes often indicate a boundary for analysis, where the function's behavior needs thorough examination.
• Vertical asymptotes create challenges for integration because the function doesn't have a fixed, finite area.

Recognizing an asymptote invites careful application of tests and theorems to avoid pitfalls in outright calculations. Establishing how the function behaves as it nears this point helps decide on convergence or divergence effectively.