Calculating the Area Under an Inexisting Curve?

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Homework Help Overview

The discussion revolves around the concept of calculating the area under a curve that is not defined over the entire interval of integration, specifically focusing on the function arctan(1/sqrt(x^2-1). Participants explore the implications of integrating a function with a restricted domain and the meaning of area in such contexts.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the validity of integrating a function that is not continuous over the interval. Some suggest that if a curve is not defined, the area may be considered meaningless. Others explore the idea of complex interpretations of the integral.

Discussion Status

The discussion is active, with participants examining different interpretations of integration and the conditions under which a function can be considered integrable. There is a recognition of the complexities involved in defining integrals for functions with discontinuities.

Contextual Notes

Participants note that the function in question is not defined on part of the interval [0, 10], raising questions about the assumptions underlying the integration process. There is also mention of different definitions of integration, such as Riemann-Stieltjes versus Lebesgue, which may influence the discussion.

alingy1
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Hi,
I always wondered,
what is the area under an inexisting curve.

That arctan(1/sqroot(x^2-1)) for example. Its domain does not include from -1 to 1.

If I take its integral from 0 to 10, what answer should I get?
 
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alingy1 said:
Hi,
I always wondered,
what is the area under an inexisting curve.
If the curve isn't defined, there's no region. Hence, area is meaningless.
alingy1 said:
That arctan(1/sqroot(x^2-1)) for example. Its domain does not include from -1 to 1.

If I take its integral from 0 to 10, what answer should I get?
Whatever answer you get will almost certainly be wrong. For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating. Your arctan function isn't defined on part of the interval [0, 10]. If you were to ignore these requirements for integration, whatever antiderivative you get won't be defined at 0.
 
You could interpret it by allowing y to be complex. Whether that leads to a meaningful integral I'm not sure. I tried it for x2+y2=1, x from 1 to 2. I got terms like ##i \ln(2+\sqrt 3)##.
 
Mark44 said:
For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating.
Hmmm... you sure about that?
 
skiller said:
Mark44 said:
For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating.
Hmmm... you sure about that?
Good catch. Even functions that would ordinarily be considered fairly pathological can be integrable. E.g. f(x) = 1 if x rational, 0 otherwise.
Depends partly on what definition of integration is being used, as in Riemann-Stieltjes v. Lebesgue.
 
haruspex said:
Good catch.
No, I didn't think it was a particularly great catch. As I'm sure you know, there are extremely simple functions that have discontinuities and yet are integrable. A simple step function, for one.
 
skiller said:
No, I didn't think it was a particularly great catch.
I just excusing myself for not having read Mark44's post fully enough to have caught it:wink:.
 

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