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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. $$ \int_{0}^{\pi / 2} \tan \theta d \theta $$

Short Answer

Expert verified
The integral \(\int_{0}^{\pi/2} \tan \theta \, d\theta\) diverges.

Step by step solution

01

Recognize the Problem Type

The given problem \(\int_{0}^{\pi / 2} \tan \theta \, d \theta\) requires determining whether the integral converges or diverges. Because it is an improper integral due to the asymptote of \(\tan \theta\) at \(\theta = \frac{\pi}{2}\), we need to use a convergence test.
02

Express the Integral in a Suitable Form

Rewrite the integral by recognizing that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Thus, the integral becomes \(\int_{0}^{\pi/2} \frac{\sin \theta}{\cos \theta} \, d \theta\).
03

Apply a Substitution to Simplify

Let \(u = \cos \theta\), then \(du = -\sin \theta \, d\theta\) which implies \(\sin \theta \, d\theta = -du\). Change the limits, when \(\theta = 0\), \(u = \cos(0) = 1\) and when \(\theta = \pi/2\), \(u = \cos(\pi/2) = 0\). Substitute to get the integral \(-\int_{1}^{0} \frac{1}{u} \, du\). This can be rewritten as \(\int_{0}^{1} \frac{1}{u} \, du\).
04

Evaluate the Integral

The integral \(\int_{0}^{1} \frac{1}{u} \, du\) is \(\lim_{b \to 0^+} \left[ \ln |u| \right]_b^1 = \lim_{b \to 0^+} (\ln 1 - \ln b) = \lim_{b \to 0^+} -\ln b\).
05

Determine Convergence or Divergence

As \(b\) approaches 0 from the right, \(-\ln b\) approaches \(+\infty\). Therefore, the original integral \(\int_{0}^{\pi/2} \tan \theta \, d\theta\) diverges.

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Key Concepts

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

Convergence Tests
When faced with an improper integral like \(\int_{0}^{\pi / 2} \tan \theta \, d \theta\), convergence tests help us determine if the integral converges to a finite value or diverges to infinity. These tests offer mathematical strategies to analyze the behavior of functions as they approach certain limits, especially when standard calculus approaches can't be applied directly due to undefined behavior such as asymptotes or infinity limits.

Convergence tests can generally be categorized into several types: the Direct Comparison Test, the Limit Comparison Test, and others designed for specific situations. The choice of test usually depends on the nature of the function under the integral and the limits involved. By selecting an appropriate test, we can diagnose the integral's convergence status accurately.

Ultimately, convergence tests provide us with a systematic way to handle integrals that appear complex or undefined at boundaries, ensuring our results are logical and sound.
Direct Comparison Test
The Direct Comparison Test is one of the most fundamental tools in analyzing improper integrals. It involves comparing the given integral with a simpler integral whose behavior we already know, either as convergent or divergent. By establishing a logical relationship between two functions, we can draw conclusions about their integrals.

Consider an example integral, \(\int_{a}^{b} f(x) \, dx\), and let's say we compare it with \(\int_{a}^{b} g(x) \, dx\). If \(0 \leq f(x) \leq g(x)\) for all \(x\) in the interval \([a, b]\), and \(\int_{a}^{b} g(x) \, dx\) is finite, then \(\int_{a}^{b} f(x) \, dx\) must also converge. Conversely, if \(f(x) \geq g(x) \geq 0\) and \(\int_{a}^{b} g(x) \, dx\) diverges, then \(\int_{a}^{b} f(x) \, dx\) diverges too.

This comparison strategy helps break down complex integral rules to manageable portions by relying on inequalities and previously established results about similar functions.
Limit Comparison Test
The Limit Comparison Test is a powerful technique especially effective when the Direct Comparison Test proves challenging due to closely matched functions. This test involves taking the limit of the ratio of two functions as they approach an endpoint of the interval in question.

Suppose we have two functions, \(f(x)\) and \(g(x)\), and their corresponding integrals over the same interval. The Limit Comparison Test states that if:
1. \(f(x) \geq 0\) and \(g(x) \geq 0\) eventually for all \(x\) in the interval,
2. \(\lim_{{x \to c}} \frac{f(x)}{g(x)} = L\), where \(L\) is a positive finite number,

then both \(\int f(x) \, dx\) and \(\int g(x) \, dx\) will either both converge or both diverge. This method focuses on the asymptotic behavior of the functions as they approach potential points of discontinuity or infinity, making it highly effective for evaluating challenging integrals.
Divergent Integrals
Divergent integrals arise when an integral does not settle to a finite value as we evaluate across its entire interval. Essentially, the limit of the integral extends to infinity, either due to unbounded behavior of the integrand or the limits of integration themselves

In mathematical terms, if the integral of a function from a point \(a\) to a point \(b\) results in infinity, we label it as divergent. The integral \(\int_{0}^{\pi/2} \tan \theta \, d \theta\) is a perfect example: due to the asymptote of \(\tan \theta\) at \(\theta = \frac{\pi}{2}\), the function shoots up to infinity, causing the integral to diverge.

Understanding divergence is crucial for determining when solutions do not exist in a traditional sense, and often necessitates regulating the problematic behavior using mathematical tools like comparison tests to make meaningful assessments.

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Most popular questions from this chapter

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises \(53-56\) about the \(x\) -axis. \(y=x+\sin 2 x, \quad-2 \pi / 3 \leq x \leq 2 \pi / 3 \quad\) (the curve in Section \(4.4,\) Exercise 5\()\)

Estimating the value of a convergent improper integral whose domain is infinite a. Show that $$\int_{3}^{\infty} e^{-3 x} d x=\frac{1}{3} e^{-9}<0.000042$$ and hence that \(\int_{3}^{\infty} e^{-x^{2}} d x<0.000042 .\) Explain why this means that \(\int_{0}^{\infty} e^{-x^{2}} d x\) can be replaced by \(\int_{0}^{3} e^{-x^{2}} d x\) without introducing an error of magnitude greater than 0.000042 . b. Evaluate \(\int_{0}^{3} e^{-x^{2}} d x\) numerically.

Usable values of the sine-integral function The sine-integral function, $$ \operatorname{Si}(x)=\int_{0}^{x} \frac{\sin t}{t} d t $$ is one of the many functions in engineering whose formulas cannot be simplified. There is no elementary formula for the antiderivative of \((\sin 1) / t\) . The values of \(\mathrm{Si}(x),\) however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is $$ f(t)=\left\\{\begin{array}{cl}{\frac{\sin t}{t},} & {t \neq 0} \\ {1,} & {t=0}\end{array}\right. $$ the continuous extension of \((\sin t) / t\) to the interval \([0, x] .\) The function has derivatives of all orders at every point of its domain. Its graph is smooth, and you can expect good results from Simpson's Rule. a. Use the fact that \(\left|f^{(4)}\right| \leq 1\) on \([0, \pi / 2]\) to give an upper bound for the error that will occur if $$ \operatorname{Si}\left(\frac{\pi}{2}\right)=\int_{0}^{\pi / 2} \frac{\sin t}{t} d t $$ is estimated by Simpson's Rule with \(n=4\) . b. Estimate \(\operatorname{Si}(\pi / 2)\) by Simpson's Rule with \(n=4\) c. Express the error bound you found in part (a) as a percentage of the value you found in part (b).

Evaluate the integrals in Exercises \(1-28\). $$ \int_{0}^{2} \frac{d x}{8+2 x^{2}} $$

Evaluate the integrals in Exercises \(1-28\). $$ \int_{-2}^{2} \frac{d x}{4+x^{2}} $$

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