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The area under a discontinuous & integrable function

  1. Mar 12, 2012 #1
    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?
     
    Last edited: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2
    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.
     
  4. Mar 12, 2012 #3
    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.
     
    Last edited by a moderator: May 5, 2017
  5. Mar 12, 2012 #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.
     
    Last edited: Mar 12, 2012
  6. Mar 12, 2012 #5
    It is, because the function is integrable.
    It can, for the same reason.

    The problem is whether we can talk about the area of that surface or not.
     
  7. Mar 12, 2012 #6

    micromass

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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.
     
  8. Mar 12, 2012 #7
    Thank you.
     
  9. Mar 12, 2012 #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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