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Can someone explain this proccess for expressing limits as an integral?

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    My professor gave us this problem:

    Convert the limit to an integral:


    [tex]\lim_{n\rightarrow inf} \frac{1}{n}(\sin(\pi/2) + \sin(\pi/2) + ... + \sin(\pi))[/tex]
    he said it was right endpoint from [0,π]
    then he set Δx = [itex]\frac{\pi}{n}[/itex]
    then

    [tex] \lim_{n\rightarrow inf} \frac{1}{\pi}(\frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{i\pi}{n}))[/tex]

    [tex] = \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx[/tex]

    [tex] = \int\limits_{0}^{1}\sin{x}\, dx [/tex]

    I just don't understand how he found Δx as being π/n or how he munipulated the formula to get the limits for the inetgral.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 8, 2011 #2

    I like Serena

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    Hi NWeid1! :smile:


    It appears that your first formula contains a number of typos, since I think it should match the summation.


    Do you know how a (Riemann) integral is actually defined?
    In this case [itex]\int\limits_{0}^{\pi}\sin{x}\, dx[/itex]?
     
    Last edited: Dec 8, 2011
  4. Dec 8, 2011 #3
    Well this is exactly what I have in my notes, and I know the answer is [tex] = \int\limits_{0}^{1}\sin{x}\, dx [/tex]

    But I was just writing it down when he was teaching it, I never really understood the process.
     
  5. Dec 8, 2011 #4

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  6. Dec 8, 2011 #5
    You mean familiar with the process of converting an integral to a limit? I learned it but that is kind of what I'm asking. The only notes I have on this is what I have above, which I can't understand by the example.
     
  7. Dec 9, 2011 #6

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    Well, do you know that an integral is approximated by inscribed rectangles?

    Sorry for not explaining more, but I have no clue what you know and do not know.
     
    Last edited: Dec 9, 2011
  8. Dec 9, 2011 #7

    HallsofIvy

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    No. Based on what you say below, you want
    [tex]\lim_{n\to \inf} (\frac{1}{n} sin(\pi/n)+ sin(2\pi/n)+ ...+ sin(n\pi/n))[/tex]

    Yes, you are dividing the interval from 0 to [itex]\pi[/itex] into n equal length sub-intervals so each has length [itex]\pi/n[/itex]. The two endpoints of the first sub-interval are 0 and [itex]\pi/n[/itex], the two endpoinst of the second sub-interval are [itex]\pi/n[/itex] and [itex]2\pi/n[/itex], etc.

    So you are approximating the value of sin(x) inside the "i"th sub-interval by its value at the right endpoint, [itex]sin(i\pi/n)[/itex]. And you are approximating the area under the graph of y= sin(x) inside each sub-interval by the area of the rectangle having height [itex]sin(i\pi/n)[/itex] and base [itex]\pi/n[/itex]

    Now you make that approximation exact by taking the limit- this limit of the sum will be equal to the area under y= sin(x).

    and, of course, the area under the graph is the integral.

    The sum goes from [itex]0\pi/n= 0[/itex] to [itex]n\pi/n= \pi[/itex]. That's how he got the limits of integration. And the difference bertween two consecutive value, say [itex]2\pi/n- \pi/n[/itex], or [itex]n\pi/n- (m-1)\pi/n)[/itex] is [itex]\Delta x= \pi/n[/itex] so you are dividing the length, [itex]\pi[/itex] into n equal parts.

     
    Last edited: Dec 9, 2011
  9. Dec 9, 2011 #8

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    Heh! What HoI said! :smile:
    (@Hoi: you have way too much spare time! ;)


    Just one addition.
    You write:
    [tex] = \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx
    = \int\limits_{0}^{1}\sin{x}\, dx[/tex]
    But this is not true, which you will see if you evaluate the integral.
     
  10. Dec 9, 2011 #9
    Oh, yeah I learned all the riemann sums.
     
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