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Chapter 8: Problem 48

In Exercises \(35-68\) , 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. $$\int_{0}^{2} \frac{d x}{1-x}$$

Short Answer

Expert verified
The integral diverges due to the vertical asymptote at \( x = 1 \).

Step by step solution

01

Analyze the Integral

The given integral is \( \int_{0}^{2} \frac{dx}{1-x} \). This function \( \frac{1}{1-x} \) has a vertical asymptote at \( x = 1 \) within the integration limits [0, 2]. Thus, the integral is improper.
02

Split the Integral

Because of the asymptote at \( x = 1 \), we split the integral into two parts: \( \int_{0}^{1} \frac{dx}{1-x} \) and \( \int_{1}^{2} \frac{dx}{1-x} \). Evaluate each integral separately to determine convergence.
03

Evaluate the First Integral

Consider \( \int_{0}^{1} \frac{dx}{1-x} \). Let \( u = 1-x \), then \( du = -dx \) and the limits change accordingly: \( x = 0 \to u = 1 \) and \( x = 1 \to u = 0 \). The integral becomes \( -\int_{1}^{0} \frac{du}{u} = \int_{0}^{1} \frac{du}{u} \). This integral evaluates to \( \ln|u| \bigg|_0^1 \), which diverges as it approaches infinity when \( u \to 0^+ \).
04

Evaluate the Second Integral

Consider \( \int_{1}^{2} \frac{dx}{1-x} \). Setup \( u = 1-x \), then \( du = -dx \), and the limits change: \( x = 1 \to u = 0 \) and \( x = 2 \to u = -1 \). The integral becomes \( -\int_{0}^{-1} \frac{du}{u} = \int_{-1}^{0} \frac{du}{u} \). However, evaluating this also implies \( \ln|u| \bigg|_{-1}^0 \), which diverges as \( u \to 0^- \).
05

Conclusion on Convergence

Since both integrals diverge, the original integral \( \int_{0}^{2} \frac{dx}{1-x} \) diverges. Thus, the improper integral does not converge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence Tests
Improper integrals like \( \int_{0}^{2} \frac{dx}{1-x} \) involve functions that have unbounded behavior in their range of integration or extend over an infinite interval. To determine whether such integrals converge or diverge, convergence tests are essential.

There are several types of convergence tests available, each serving different scenarios:
  • Direct Comparison Test: Compares the given function with another function that is known to converge or diverge.
  • Limit Comparison Test: Uses the limit of a ratio of the given function and a known function to determine convergence.
Before applying these tests, split the integral if there's a discontinuity, as seen at the asymptote \(x = 1\) in this exercise. This method helps in analyzing the behavior separately around the point of discontinuity.
Limit Comparison Test
The Limit Comparison Test is used when direct comparison is not straightforward, but you can still compare the given function to another function. For this test, suppose you have two functions, \( f(x) \) and \( g(x) \), both positive, and you're considering the integral of \( f(x) \). To use the Limit Comparison Test, you calculate the limit:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} \]

If this limit is a positive finite number, both integrals \( \int f(x) \) and \( \int g(x) \) either both converge or both diverge.

This comparison is most effective when choosing \( g(x) \) such that its convergence or divergence is already known. However, in the exercise given, determining a suitable \( g(x) \) may not be apparent immediately, making this test less practical for the problem at hand.
Direct Comparison Test
The Direct Comparison Test is a handy tool when a suitable function for comparison is available. It involves comparing the integral of a function \( f(x) \) with another function \( g(x) \), where integrating \( g(x) \) is manageable and its convergence behavior is known.

Here鈥檚 how the Direct Comparison Test works:
  • If \( f(x) \leq g(x) \) and \( \int g(x) \) converges, then \( \int f(x) \) also converges.
  • If \( f(x) \geq g(x) \) and \( \int g(x) \) diverges, then \( \int f(x) \) also diverges.
For our exercise, if you can identify a function \( g(x) = \frac{1}{x} \) or similar, it simplifies analysis, but as shown, both portions around the asymptote independently diverge. Therefore, further comparison won鈥檛 alter this integral's divergent nature.

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Most popular questions from this chapter

$$ \begin{array}{c}{\text { a. Use a CAS to evaluate }} \\ {\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x} \\ {\text { where } n \text { is an arbitrary positive integer. Does your CAS find }} \\ {\text { the result? }}\end{array} $$$$ \begin{array}{l}{\text { b. In succession, find the integral when } n=1,2,3,5, \text { and } 7 .} \\ {\text { Comment on the complexity of the results. }}\end{array} $$$$ \begin{array}{l}{\text { c. Now substitute } x=(\pi / 2)-u \text { and add the new and old }} \\ {\text { integrals. What is the value of }} \\\ {\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x ?} \\\ {\text { This exercise illustrates how a little mathematical ingenuity }} \\\ {\text { Solves a problem not immediately amenable to solution by a }} \\\ {\text { CAS. }}\end{array} $$

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