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Question about the Riemann-Integral definiton.

  1. Mar 29, 2013 #1
    Hello, I have a question about the definiton of the Riemann integral. I will explain my question with a problem.

    It says that the Riemann Integral on[a,b] exists, if for any epsilon this holds:
    |Riemannsum(f(x) on [a,b] )- Integralvalue|< epsilon, for any partion over [a,b] where the norm of the partion is less than delta.

    Now comes my problem, lets say I want to prove that the Riemann-Integral of x on[a,b] is
    (b^2-a^2)/2.

    I would then choose a partion (b-a)/n, and show that for a given epsilon, I can make n as big as I want, so that every Riemann sum of this partion goes to the desired value. I would do this by first making the Riemann sum as small as possible with a given n, by choosing the left value at each subinterval, and then as big as I can for a given n, by choosing the right value at each subinterval.


    Now comes my question, does this really prove that x is Riemann-integrable over [a,b], the problem is when we look at the definiton. It says that it is supposed to hold for every partion where the norm goes to zero. But I have only proved it for a partion where each subinterval as equal length (b-a)/n. Isn't this a really big problem with the definition. I mean you can choose that the norm of the partion goes to zero with very many partions, how does one solve this problem?
     
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  3. Mar 30, 2013 #2

    HallsofIvy

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    You say "I would do this by first making the Riemann sum as small as possible with a given n, by choosing the left value at each subinterval, and then as big as I can for a given n, by choosing the right value at each subinterval", which is the crucial point. If we call the sum you get using the left value "Ln" and choosing the right value "Rn", then using any other partition, to get "Mn", we must have [itex]L_n\le M_n\le R_n[/itex]. If the function is "integrable" then the limits of Ln and Rn must be the same so that Mn, trapped between the two, must also converge to the same limit.
     
  4. Mar 30, 2013 #3
    Thanks, but how do you show that for any other partion it must lie between those two values? Maybe it is trivial, but I do not see it.
    I guess I can accept if for the function x, but I would also like to see it for a another random function.
    That is, can we prove that if:

    We have a given partion where the norm of the partion goes to zero. For instance a partion where we have equal subinterval lengths, (b-a)/n.
    Then for any epsilon, we can get a delta, so that if the norm of the partions is less than delta.
    Then we have |Riemannsum - Integralvalue| < epsolon.

    Now can we prove that if what I said holds for the partions with equal subinterval lengths, then this implies that the epsilon-delta conditions will also hold for every other partion we can construct in a way where we can control the norm of the partion?
     
    Last edited: Mar 30, 2013
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