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Converting an Integral to a Rieman Sum

  1. Aug 18, 2012 #1
    I know I should know this, but how would one convert a typical integral into a Rieman Sum?

    0n sinx + x dx for whatever n.

    for example.
     
  2. jcsd
  3. Aug 18, 2012 #2

    chiro

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    This function happens to be Riemann Integrable with closed form anti-derivatives: try using the fundamental theorem of calculus to get a closed form expression in terms of G(n) - G(0) where G(x) is the appropriate anti-derivative.
     
  4. Aug 18, 2012 #3

    Well, since the function [itex]\,f(x)=\sin x + x\,[/itex] is continuous everywhere, it is Riemann integrable in any finite

    interval, and we can choose any partition for it we want, for example the partition
    [tex]x_0=0\,,\,x_1=\frac{n}{k}\,,\,x_2=\frac{2n}{k},...,x_k=\frac{kn}{k}=n[/tex]
    for the interval [itex]\,[0,n]\,[/itex] , thus

    [tex]\int_0^n (\sin x +x)dx=\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k \left( \sin \frac{in}{k}+\frac{in}{k} \right)[/tex]

    DonAntonio
     
  5. Aug 19, 2012 #4

    Bacle2

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    To add to Don Antonio's comment:

    1) Partition your domain of integration [a,b] into a collection a=x0,

    x1,....,xn=b .

    2)Select a point xi* in each (xi-1,xi).

    3)Form the sum Ʃi=1,..,Nf(xi*)(xi-xi-1)

    In your case, f(xi*)=xi*+sin(xi*)
     
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