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How definite integration works?

  1. Nov 14, 2012 #1
    Basically my question is, how do we know that the antiderivitave of a function represents the area under that curve?
    If you look graphically, you can see that the area of f(x)=x from 0 to 1 is 1/2 which agrees with (x^2)/2 but why?
  2. jcsd
  3. Nov 14, 2012 #2
    We know that because we have proven it. The statement is exactly the fundamental theorem of calculus. The proof of that theorem tells us why we can calculate areas with the inverse of derivatives.

    The fact is certainly not trivial and is in fact quite a deep theorem.
    Last edited by a moderator: Nov 15, 2012
  4. Nov 15, 2012 #3


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    Let y= f(x) be continuous function such that [itex]f(x)\ge 0[/itex] for all [itex]a\le x[/itex]. Define F(x) to be the area under the graph from a to x. Then F(x+[itex]\Delta[/itex]x) is the area under the graph from a to x+ [itex]\Delta[/itex]x. F(x+ [itex]\Delta[/itex]x)- F(x) is the area under the graph from x to x+ [itex]\Delta[/itex]x. Because f is continuous (this is the "deep" part!), there exist some x', [itex]x\le x'\le x+ \Delta x[/itex], such that [itex]f(x')\Delta x[/itex] is equal to that area. That is, [itex]F(x+ \Delta x)- F(x)= f(x')\Delta x[/itex] so that
    [tex]\frac{F(x+ \Delta x)- F(x)}{\Delta x}= f(x')[/tex]

    Now, x' is always between x and x+ [itex]\Delta[/itex] x so if we take the limit as [itex]\Delta[/itex]x goes to 0, f(x') goes to f(x). That is:
    [tex]\lim_{\Delta x\to 0}\frac{F(x+ \Delta x)- F(x)}{\Delta x}=\frac{dF}{dx}= f(x)[/tex]

    That is, the area really is given by an anti-derivative. The interesting thing is that to do this- and so be able to calculate the are of some very complicated sets- we don't have to give a very detailed definition of "area". All that is needed about area is
    1) The area of a set is a non-negative number
    2) If sets A and B are disjoint (except possibe on their boundaries) the area of A U B is the area of A plus the area of B
    3) The area of a rectangle is "height times width".
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