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Area and Definite Integration

  1. Jul 12, 2009 #1
    What is the proof that proves:

    A = the definite integral of a function with upper limit a and lower limit b. ?
     
  2. jcsd
  3. Jul 12, 2009 #2
    What definition would you like to use for "area" ?
     
  4. Jul 12, 2009 #3
    The one that is used most often.
     
  5. Jul 12, 2009 #4

    HallsofIvy

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    You see a problem with giving you a proof that the integral gives the area if you can't give a general definition of area, don't you?

    "Area" of a two dimensional set is a function that assigns to a set, X, a number, A(X), such that.
    1) If X is a rectangle with side lengths h and w, its area is A(X)= hw.
    2) If X is a subset of Y, then [itex]A(X)\le A(Y)[/itex].
    3) if X and Y are have no points in common, except possibly on their boundaries, then [itex]A(X\cup Y)= A(x)+ A(y)[/itex].

    Those are the properties that any notion of "area" should have. Of course to make it a true definition of "area" we would have to show that, for any set X, there exist a specific A(X)- and that's not true. There exist "non-measurable" sets. In terms of two-dimensional sets, that says that no matter how you tried to "define" area, there will be sets for which "area" cannot be defined (not that they have 0 area, that no number, not even 0, can be assigned to their area will violate some of tose conditions). Given that, and that we can prove that the integral, when it exists, satisfies those three conditions, the definition of area that "is used most often" is the integral.
     
  6. Jul 12, 2009 #5
    Apparently, the definition of area is too sophisticated to be given in any standard calculus textbook.

    I am unaware of most advanced mathematical notation and would like to know what the U-symbol means. Also, is a two-dimensional set a set of ordered pairs (x,y) where X may be a region? What are non-measurable sets?

    I suppose I am rather looking for a (more precise) explanation.
     
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