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Definite Integration proof

  1. Dec 6, 2007 #1
    Hey guys! I'm looking for a rigorous proof which shows that the definite integral does indeed give us the area under a curve or which proves how is a definite integral a summation process.
    I already know that the any given function is the derivative of the function of its area but most of the books show this by considering y=x and the then finding the expression of the area by using the area of triangle and comparing it with the result obtained by integration.

    Thanks in advance for any help you guys can offer. :)

    Abdullah

    (PS: Sorry for my bad English)
     
  2. jcsd
  3. Dec 6, 2007 #2

    arildno

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    For starters, study the construction of partitions and Riemann sums.
     
  4. Dec 6, 2007 #3
    i would suggest studying upper and lower sums. The integral IS a summation process, that is how the integral is constructed - for that no proof is required. If you study the actual theory of the integral (i.e what a partition is , what a lower and upper integral is and what it means for a function to be integrable on a interval) the geometric interpretation becomes obvious.
     
  5. Dec 6, 2007 #4
    I struggled with this in high school....

    In high school I struggled with being able to "see" why the definite integral gave the area under the curve. My high school teacher had no idea how to get me to see it was true. I spent many hours in my car thinking about this. (I had a long commute to work.)

    Assume that there is some function A(x) for a given f(x). Also assume that one of the properties of this function is that A(d) - A(c), where c & d are real constants and d > c, yields the area under the curve of f(x). Now, lets try to find a small area between (x, x+h).

    Using the function A(x), we find that the area would be represented by the following:

    [tex]A(x+h) - A(x)[/tex]

    If we use f(x) to estimate the area, then we have:

    [tex]{f(x)}{h}[/tex]

    These two equations should be approximately equal.

    [tex]A(x+h) - A(x) = {f(x)}{h}[/tex]

    [tex]\frac{A(x+h) - A(x)}{h} = f(x)[/tex]

    Now we take the limit as h -> 0 and we come up with a relationship between f(x) and A(x). We can see that f(x) is the derivative of this so called function. However at this point, we can only find the area when the width is really small. If we take the infinite sums of these really small rectangles, then we can see that the inner A(x)'s will cancel out and you're only left with the A(d) - A(c).

    I hope this helps. I'm sure that this is far from a rigorous proof, but it helped me "believe" that the definite integral does yield the area under the curve.

    Ken
     
  6. Dec 6, 2007 #5

    HallsofIvy

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    That actually is a proof of the "second" Fundamental Theorem of Calculus- that the derivative of the integral (the "area" function) is the original function.

    I think what the original poster was referring to was the fact that the definite integral (defined as the limit of the Riemann sums) does in fact give the area.

    To do that, for every positive integer, n, divide the interval from x= a to x= b into n parts. Within each of those divisions, choose [itex]x*_n[/itex] such that [itex]f(x*_n)[/itex] is the maximum value on the (closed) interval. Choose [itex]x_{*n}[/itex] such that [itex]f(x_{*n})[/itex] is the minimum value on the (closed) interval. It is easy to see that the rectangle formed by taking its height to be [itex]f(x*_n)[/itex] contains the area between the curve and the x-axis so that the Riemann sum is greater than the actual area. It is easy to see that the rectangle formed by taking its height to be [itex]f(x_{*n})[/itex] is contained in the area between the curve and the x-axis and so the Riemann sum is less than the actual area. The actual area is always trapped between the two sums so, as long as the function is integrable and the two sums give the same limit, that limit must be the area.
     
  7. Dec 7, 2007 #6

    Gib Z

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    I always thought the "area" underneath a function was defined by the definite integral =(
     
  8. Dec 7, 2007 #7

    HallsofIvy

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    Well, yes and no! If we are going to call that the "area" we would like to have some evidence that it has properties like those of "area" in elementary geometry. The interesting thing about the proof that I outlined above, is that we don't need to give a detailed definition of "area" for general sets in R2. We do require some basic properties:
    Area is a function that assigns to sets a real number, satisfying:
    If U is a subset of V then [itex]Area(U)\le Area(V)[/itex].
    If U and V area disjoint sets (except possible for boundary points) then [itex]Area(U \cup V)= Area(U)+ Area(V)[/itex].
    The area function is "invariant" under translation. That is if we start with set U, set up a coordinate system, and form set V by adding the same number to every x coordinate and adding the same number to every y coordinate, then Area(U)= Area(V).
    If U is a square with side length x then the area of U is x2.

    I think those area sufficient to show, as above, that the area "under a curve" is always "trapped" between the "outer measure" and "inner measure" and so, if in the limit those coincide, they must coincide at the area.

    By the way, it is fairly easily provable that, for any "area" function satisfying those properties, there must exist some sets to which it cannot be applied- that do not have "area".
     
  9. Dec 8, 2007 #8

    Gib Z

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    Thanks for that Halls =]
     
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