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So why does the integral represent an infinite sum?

  1. Jul 22, 2013 #1
    In an earlier post i was shown how to represent an integral as an infinite sum. So why is the anti derivative a summation by definition? For example, the derivative dy/dx is found by f(x+h)-f(x)/h.
     
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  3. Jul 22, 2013 #2

    Zondrina

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    I assume you mean the Riemann integral. Integrals are a sum of infinite rectangles taken over an interval. Each of these rectangles has an area and we use those areas to approximate the integral ( presuming we are talking about definite integrals ).

    The Riemann sum, is that sum of the areas of those rectangles, and as we let n→∞ ( we chop the rectangles finely enough so that they cover the area completely ), the sum converges to the same value you will get from the definite integral. Hence why you can write :

    ##lim_{n→∞} \sum^{n}_{i=0} f(x_{i}^{*}) Δx_i = \int_{a}^{b} f(x) dx##
     
  4. Jul 22, 2013 #3

    LeonhardEuler

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    The answer to that question is basically the fundamental theorem of calculus.

    To show enough to make this plausible, imagine you break the interval [a,b] into pieces. Call the pieces [a,x1], [x1,x2], [x2,x3] ... [xN-1,xN], [xN,b].
    Assume they are equally large intervals and that xi+1-xi=h. Now you have a function F(x) and its derivative is F'(x)=f(x).

    To show that the summation definition of the integral is in a sense the opposite of the derivative, we need to show that
    [tex]\int_{a}^{b}f(x)dx=\int_{a}^{b}F'(x)dx = F(b)-F(a)[/tex]
    Use the definition of the derivative as an approximation (becomes exact as h goes to 0):
    [tex]F'(x_i)\approx\frac{F(x_{i+1})-F(x_i)}{h}[/tex]
    Now look at the summation definition of the integral (again, only becomes exactly the integral as h goes to 0). Also, for convenience, let a=x0 and b=xN+1.
    [tex]\int_{a}^{b}F'(x)dx =\sum_{i=0}^{N}F'(x_i)*(x_{i+1}-x_i)[/tex]
    Remember that xi+1-xi=h, so
    [tex]\int_{a}^{b}F'(x)dx =\sum_{i=0}^{N}F'(x_i)*h[/tex]
    [tex]\approx\sum_{i=0}^{N}\frac{F(x_{i+1})-F(x_i)}{h}*h[/tex]
    [tex]=\sum_{i=0}^{N}F(x_{i+1})-F(x_i)[/tex]
    This type of sum is called a 'telescoping sum'. Watch what happens when you write the first few terms:
    F(x1) -F(x0)+F(x2) -F(x1)+F(x3) -F(x2)+F(x4) -F(x3)
    Notice that every term cancels another one except -F(x0) and +F(x4). No matter how many terms you add, all of them will cancel except the first and last. So that integral is
    [tex]\int_{a}^{b}F'(x)dx = F(x_{N+1})-F(x_0)=F(b)-F(a)[/tex]

    So hopefully that shows how they are opposites.
     
  5. Jul 22, 2013 #4
    I thought the 'theoretical' answer to that question was a millenium problem (Riemann Hypothesis). I might be wrong though.
     
  6. Jul 22, 2013 #5

    LeonhardEuler

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    The Riemann Hypothesis is something else. It relates to the zeroes of the Riemann zeta function. This problem was solved in the 1600's.
     
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