Integral Question: Deriving Formula for $\int|f(x)|dx$

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Discussion Overview

The discussion revolves around deriving a formula for the integral of the absolute value of a continuous function, specifically the expression $\int |f(x)| dx$. Participants explore different approaches to handle the integral, including piecewise integration based on the sign of the function.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asks for a formula or method to derive the integral $\int |f(x)| dx$, noting a connection to the signum function for odd-order integrations.
  • Another participant suggests that the integral can be expressed with two solutions, $F(x)$ and $-F(X)$, where $F(X)$ is the antiderivative of $f(x)$, depending on the sign of $f(x)$.
  • There is a clarification that $|f(x)| = f(x)$ if $f(x) \geq 0$ and $|f(x)| = -f(x)$ if $f(x) < 0$, indicating the need for piecewise integration.
  • One participant humorously comments on needing reading glasses, suggesting a light-hearted tone amidst the technical discussion.
  • A later reply points out that a previous participant was not repeating answers but was using different notations, highlighting a potential misunderstanding in the conversation.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation and approach to the integral, with no consensus reached on a definitive method or formula.

Contextual Notes

Participants discuss the integral in the context of piecewise functions, but there are unresolved aspects regarding the specific conditions under which the piecewise definitions apply.

Paul
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Hello everyone! Can anyone tell me a formula (or a way to derive) this integral?
[tex]\int|f(x)|dx[/tex]
where [tex]f(x)[/tex] is a real, continuous function of x in the vector space [tex]C^\infty[/tex]. So far, all I've figured out is that odd-order integrations are related to the signum function.
Thanks!
 
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Derive or find the area under |f(x)|?

|f(x)| = f(x), if x >=0; -f(x), if x<0

so &Int; |f(x)| dx has 2 solutions: F(x), and -F(X), where F(X) is the antiderivative of f(x).
 
PrudensOptimus said:
Derive or find the area under |f(x)|?

|f(x)| = f(x), if x >=0; -f(x), if x<0

so &Int; |f(x)| dx has 2 solutions: F(x), and -F(X), where F(X) is the antiderivative of f(x).

No, |f(x)| = f(x) if f(x) >= 0 and -f(x) if f(x) < 0.

I recommend breaking up the integral into separate domains as I've indicated and integrating piecewise.
 
Tide said:
No, |f(x)| = f(x) if f(x) >= 0 and -f(x) if f(x) < 0.

I recommend breaking up the integral into separate domains as I've indicated and integrating piecewise.
Stop repeating my answers :p
 
PrudensOptimus said:
Stop repeating my answers :p

LoL! Man, I've just GOTTA get some reading glasses! :bugeye:
 
He wasn't repeating your answers. Prudens used x, while Tide correctly used f(x).
 
what i really meant was f(x)... but i was thinking about beer.
 

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