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Limit of a Summation.

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{1}{n}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)[/tex]





    3. The attempt at a solution
    I tried to solve it simply.
    [tex]\frac{1}{n}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)=\int_{0}^{1}f(a+(b-a)x)dx[/tex]

    [tex]=f(b)-f(a)[/tex]
     
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 10, 2007 #2
    Sorry,
    [tex]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)=\int_{0}^{1}f(a+(b-a)x)dx[/tex]
     
  4. Oct 10, 2007 #3

    HallsofIvy

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    Very good. Now integrate by letting u= a+ (b-a)x. What is du? Also, be careful about the limits of integration.
     
  5. Oct 10, 2007 #4
    [tex]\int_{a}^{b}\frac{f(u)du}{(b-a)}=\frac{1}{b-a}(f(b)-f(a))[/tex]
    Yes?
     
  6. Oct 10, 2007 #5

    Dick

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    Yes, to the left hand side. No, to the right. Unless the 'f' on the right stands for the antiderivative of the 'f' on the left. In which case you should say so and use a different symbol.
     
  7. Oct 10, 2007 #6
    Than ,actually,answer remains just
    [tex]\frac{1}{b-a}\int_{a}^{b}f(u)du[/tex]
     
  8. Oct 10, 2007 #7

    Dick

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    Sure the summation is just a definition of a riemann integral.
     
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