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Define the terms converges and diverges when working with improper integrals.

Short Answer

Expert verified
In the context of improper integrals, the term 'converges' refers to a situation where the values approach a finite number as the variable approaches infinity. On the other hand, 'diverges' refers to situations where the values don't tend towards any specific finite number as the variable approaches infinity.

Step by step solution

01

Defining 'Converges'

In the context of improper integrals, a series or function is said to 'converge' if the values approach a finite number as the variable approaches infinity (positive or negative). Mathematically, this can be represented as \(\lim_{{x \to \infty}} f(x) = L\), where L is a finite number. This means that as x increases or decreases without bound, the function f(x) approaches but never reaches the value L.
02

Defining 'Diverges'

A series or function 'diverges' when its values don't tend towards a specific finite value as the variable approaches infinity. This usually happens in two scenarios. Firstly, when as x approaches infinity, f(x) also approaches infinity or negative infinity, represented as \(\lim_{{x \to \infty}} f(x) = \infty\) or \(\lim_{{x \to \infty}} f(x) = -\infty\). Secondly, when the limit as x approaches infinity does not exist, meaning the function doesn't tend towards any specific number (finite or infinite).
03

Applying to Improper Integrals

The concepts of convergence and divergence apply to improper integrals in a similar way. An improper integral converges when the definite integral from a to infinity, \(\int_a^{\infty} f(x) dx\), or from negative infinity to a, \(\int_{-\infty}^a f(x) dx\), exists and equals a finite number. If the integral does not exist or equals infinity, then the improper integral diverges.

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Most popular questions from this chapter

In L'Hôpital's 1696 calculus textbook, he illustrated his rule using the limit of the function \(f(x)=\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}\) as \(x\) approaches \(a, a>0 .\) Find this limit.

Determine all values of \(p\) for which the improper integral converges. $$ \int_{1}^{\infty} \frac{1}{x^{p}} d x $$

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=\cos a t $$

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=\sinh a t $$

(b) Use the result of part (a) to find the equation of the path of the weight. Use a graphing utility to graph the path and compare it with the figure. (c) Find any vertical asymptotes of the graph in part (b). (d) When the person has reached the point (0,12) , how far has the weight moved?A person moves from the origin along the positive \(y\) -axis pulling a weight at the end of a 12 -meter rope (see figure). Initially, the weight is located at the point (12,0) . (a) Show that the slope of the tangent line of the path of the weight is $$ \frac{d y}{d x}=-\frac{\sqrt{144-x^{2}}}{x} $$

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