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Chapter 7: Problem 30

Evaluate the integrals that converge. $$ \int_{1}^{+\infty} \frac{d x}{x \sqrt{x^{2}-1}} $$

Short Answer

Expert verified
The integral converges by comparison with \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \).

Step by step solution

01

Analyze the Integral for Convergence

The given integral is an improper integral because it has an infinite limit of integration. In general, we want to evaluate whether it converges or diverges by comparing it to a known convergent or divergent integral.
02

Comparison with a Known Integral

Consider the function \[ f(x) = \frac{1}{x \sqrt{x^2 - 1}}. \]For large values of \(x\), \(\sqrt{x^2 - 1} \approx x\), so \[ f(x) \approx \frac{1}{x^2}. \]The integral \[ \int_{1}^{\infty} \frac{1}{x^2} \, dx \]is a known convergent integral. Thus, we can use comparison to infer the behavior of the given integral.
03

Evaluate the Comparison Integral

Calculate \[ \int_{1}^{\infty} \frac{1}{x^2} \, dx, \]which converges:\[ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{1}^{\infty} = 0 + 1 = 1. \]
04

Conclusion on Convergence

Because \( f(x) = \frac{1}{x \sqrt{x^2 - 1}} \approx \frac{1}{x^2} \) and the integral \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges, the given integral \( \int_{1}^{\infty} \frac{1}{x \sqrt{x^2 - 1}} \, dx \) also converges by the comparison test.

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Key Concepts

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

Comparison Test
When dealing with improper integrals, the comparison test is an essential tool.
This test helps us determine whether an integral converges (finishes at a value) or diverges (goes to infinity).
For this, we compare our integral with another integral that is easier to evaluate.

Here's how the comparison test works:
  • Find a comparative function:For our function \( f(x) = \frac{1}{x \sqrt{x^2 - 1}} \), we need to find a simpler function \( g(x) \) that is known either to converge or diverge.
  • Check the behavior at infinity:If \( g(x) = \frac{1}{x^2} \), we see that as \( x \) becomes very large, \( \sqrt{x^2 - 1} \approx x \), leading \( f(x) \) to approximate \( \frac{1}{x^2} \).
  • Use the known result:The integral \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges, and because \( f(x) \leq g(x) \) for all large \( x \), the convergence of \( \int_{1}^{\infty} g(x) \, dx \) implies the convergence of \( \int_{1}^{\infty} f(x) \, dx \) as well.
This way, we can safely conclude the convergence of more complex integrals without directly solving them.
Convergence
Convergence in the context of integrals refers to the idea that the area under the curve of a function on a given interval leads to a finite value.
For improper integrals, we often deal with functions that go to infinity, so evaluating convergence is crucial.

When approaching convergence:
  • Start by identifying the type of integral: Improper integrals have one or more infinite limits or unbounded behaviors within the interval of integration.
  • Test for convergence: Use methods such as the comparison test, the ratio test, and others tailored to the particular form of the integral.
  • Opt for simplicity: Whenever possible, simplify your function's behavior through approximation methods to match it with a known benchmark integral.
Determining convergence helps to ascertain whether a particular integral contributes to a finite total area. It's about knowing where to draw the line between endless sums and a bounded result.
Integral Calculus
Integral calculus is a branch of mathematical analysis dealing with integrals and their properties.
It provides tools for solving problems related to areas under curves, volumes, central points, and more.

Here are some key concepts in integral calculus:
  • Definite Integrals: Integral calculus helps find the area under a curve in a specific interval. This is represented as \( \int_{a}^{b} f(x) \, dx \).
  • Indefinite Integrals: Unlike definite integrals that provide a number, indefinite integrals represent a family of functions and include the concept of an 'unknown constant' often denoted by \( C \).
  • Improper Integrals: These are integrals with infinite bounds or unbounded integrands. The example \( \int_{1}^{\infty} \frac{dx}{x \sqrt{x^2-1}} \) is improper due to the upper limit being infinity.
Integral calculus is a fascinating area, as it allows us to explore the nuances of changing quantities across complex, continuous functions.

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Most popular questions from this chapter

Use any method to solve for \(x\). $$ \int_{1}^{x} \frac{1}{t \sqrt{2 t-1}} d t=1, x>\frac{1}{2} $$

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{x}{\sqrt{x+1}} d x $$

A transform is a formula that converts or "transforms" one function into another. Transforms are used in applications to convert a difficult problem into an easier problem whose solution can then be used to solve the original difficult problem. The Laplace transform of a function \(f(t)\), which plays an important role in the study of differential equations, is denoted by \(\mathscr{L}\\{f(t)\\}\) and is defined by $$\mathscr{L}\\{f(t)\\}=\int_{0}^{+\infty} e^{-s t} f(t) d t$$ In this formula \(s\) is treated as a constant in the integration process; thus, the Laplace transform has the effect of transforming \(f(t)\) into a function of \(s .\) Use this formula in these exercises. In each part, find the Laplace transform. (a) \(f(t)=t, s>0\) (b) \(f(t)=t^{2}, s>0\) (c) \(f(t)=\left\\{\begin{array}{ll}0, & t<3 \\ 1, & t \geq 3\end{array}, s>0\right.\)

Approximate the integral using Simpson's rule \(S_{10}\) and compare your answer to that produced by a calculating utility with a numerical integration capability. Express your answers to at least four decimal places. $$ \int_{0}^{1} \cos \left(x^{2}\right) d x $$

In each part, try to evaluate the integral exactly with a CAS. If your result is not a simple numerical answer, then use the CAS to find a numerical approximation of the integral. (a) \(\int_{-\infty}^{+\infty} \frac{1}{x^{8}+x+1} d x\) (b) \(\int_{0}^{+\infty} \frac{1}{\sqrt{1+x^{3}}} d x\) (c) \(\int_{1}^{+\infty} \frac{\ln x}{e^{x}} d x\) (d) \(\int_{1}^{+\infty} \frac{\sin x}{x^{2}} d x\)
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