The area under a discontinuous & integrable function

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Does it make sense to talk about the area of the region {(x,y)|x[itex]\in[/itex][a,b];y[itex]\in[/itex][0,f(x)]} for a positive function f defined on an interval [a,b], where a,b[itex]\in[/itex]ℝ and f is integrable on that interval, even if the function is not continuous?
 
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  • #2
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It depends on what you mean by area. It is common to define this area as, in fact, the integral of f from a to b. In that case it makes perfect sense to speak of the area under an integrable, discontinuous function.
 
  • #3
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It depends on what you mean by area. It is common to define this area as, in fact, the integral of f from a to b. In that case it makes perfect sense to speak of the area under an integrable, discontinuous function.
Do you all agree with this?

My teacher says that the notion of area can be used only for bounded regions and we can say that the integral form a to b is that area only if the function is continuous, because the surface is limited by the lines x=a, x=b, Ox and the graph of f and, if f is discontinuous, then its graph will not wrap the upper part of the region completely.

http://img207.imageshack.us/img207/5329/imagu.png [Broken]

She also says that open disks don’t have area.

On the other hand, I think that the existence of area depends only on the set of points, and, following the definition on Wikipedia (go to Formal Definition, the last point), the set {(x,y)|x[itex]\in[/itex][a,b];y[itex]\in[/itex][0,f(x)]} should have area.
 
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  • #4
Char. Limit
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If the set of discontinuous points in an interval is finite, then the function can be integrated over that interval, I believe.
 
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  • #5
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If the set of discontinuous points in an interval is finite,
It is, because the function is integrable.
then the function can be integrated over that interval.
It can, for the same reason.

The problem is whether we can talk about the area of that surface or not.
 
  • #6
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The problem is whether we can talk about the area of that surface or not.
Yes, we talk about area there. The area is just defined as the integral [itex]\int_a^b |f(x)|dx[/itex]. There is no reason not to call that the area.

And open disks DO have an area. I don't get where your teacher is getting all that nonsense.
 
  • #7
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Yes, we talk about area there. The area is just defined as the integral [itex]\int_a^b |f(x)|dx[/itex]. There is no reason not to call that the area.

And open disks DO have an area. I don't get where your teacher is getting all that nonsense.
Thank you.
 
  • #8
mathwonk
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well maybe your teacher is defining area in only a restricted sense, to make your life simpler. ask him/her.
 

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