Problem 40 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral \(\int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} dx\) converges by the Direct Comparison Test.

Step 1: Analyze the Integral

We need to determine whether the integral \( \int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \ dx \) converges. Since this involves an exponential function divided by a root, we can consider the behavior near the boundary points. As \(x\) approaches 0 from the right, the term \(\frac{1}{\sqrt{x}}\) becomes significant in determining the behavior of the integral.

Step 2: Choose Appropriate Comparison Function

For small values of \(x\), the exponential part \(e^{-\sqrt{x}}\) tends towards 1. Therefore, near \(x = 0\), the integrand \(\frac{e^{-\sqrt{x}}}{\sqrt{x}}\) behaves similarly to \(\frac{1}{\sqrt{x}}\). We can consider the simpler function \(\frac{1}{\sqrt{x}}\) for comparison since the integral \(\int_{0}^{1} \frac{1}{\sqrt{x}} \ dx\) is known to converge.

Step 3: Apply Direct Comparison Test

Since our function \(\frac{e^{-\sqrt{x}}}{\sqrt{x}}\) is less than \(\frac{1}{\sqrt{x}}\) on \([0, 1)\), and \(\int_{0}^{1} \frac{1}{\sqrt{x}} \ dx\) converges (as it's equivalent to \(\int_{0}^{1} x^{-1/2} \ dx\) which equals \(2\)), by the Direct Comparison Test, \(\int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \ dx\) also converges.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a powerful tool for determining whether an integral converges or diverges. To use this test effectively, we need to identify a simpler, well-understood function to compare to our given function. If our original function is always smaller than this simpler one, and we know that the simpler function's integral converges, then the original integral also converges.

To apply the Direct Comparison Test, follow these steps:

• Identify the function you want to test for convergence.
• Choose a comparison function that is easier to integrate.
• Check that your original function is less than or equal to this comparison function over the domain of interest.
• Determine whether the integral of the comparison function converges.

In our original exercise, we compared the function \( \frac{e^{-\sqrt{x}}}{\sqrt{x}} \) with \( \frac{1}{\sqrt{x}} \) and found that the smaller function's integral converges. Hence, by the Direct Comparison Test, the given integral converges as well.

Limit Comparison Test

The Limit Comparison Test allows us to determine the convergence or divergence of an integral when the Direct Comparison Test might not be straightforward or applicable. This test involves comparing the given function to another function, but using a limit to draw conclusions about convergence.

When using the Limit Comparison Test, the steps involved are:

• Select another function, \( b(x) \), that is similar to your function, \( f(x) \).
• Calculate the limit: \( \lim_{{x \to c}} \frac{f(x)}{b(x)} \), where \( c \) is the point of interest (usually 0, \( \infty \), etc.).
• If the limit is a positive, finite number, then both functions behave similarly. Therefore, if \( \int b(x) \ dx \) converges, so does \( \int f(x) \ dx \), and vice versa.

This test is especially useful when dealing with complex functions where intuition about inequalities is not clear.

exponential functions

Exponential functions are represented mathematically as \( e^x \), where \( e \) is Euler's number, approximately 2.71828. These functions are notable for their rates of increase or decrease and can significantly impact the behavior of an integral.

In our problem, the integrand \( \frac{e^{-\sqrt{x}}}{\sqrt{x}} \) includes \( e^{-\sqrt{x}} \). As \( x \) approaches 0, the exponential part, \( e^{-\sqrt{x}} \), approaches 1, simplifying the function's comparison to \( \frac{1}{\sqrt{x}} \).

Understanding the behavior of exponential functions with negative exponents helps to predict the decay rate and convergence nature. In this case, they slow down the growth of the function, thus aiding in convergence.

improper integrals

Improper integrals occur when the integral's limits of integration involve infinity or the integrand becomes infinite within the domain of integration. Such integrals require special consideration to ascertain convergence.

Our integral, \( \int_{0}^{1} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \ dx \), is improper because the integrand \( \frac{1}{\sqrt{x}} \) becomes infinite as \( x \) approaches 0.

To evaluate an improper integral, we often use limits to handle the infinity:

• Rewrite the integral with a limit that approaches the point of infinity or discontinuity.
• Evaluate whether the result is finite, indicating convergence.

In this example, the limit as \( x \to 0 \) ensures that we handle the point where the function becomes large, leading to a more rigorous calculation of convergence.