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Riemann Integral Identification from Sum

  1. May 11, 2009 #1
    Hi There Everyone

    I am studying undergraduate calculus in first year. My question regards the rules for identifying a limit sum as a Riemann sum and therefore a definite integral. The book we are using says that when choosing [tex] \inline \large c_{i} [/tex] for some [tex] f(x) [/tex], if [tex] \inline \large x_{i - 1} < c_{i} < x_{i} [/tex], then the sum is indeed a Riemann sum for [tex] f(x) [/tex] over an interval.

    Allow me to include an example afterwhich I will pose my question in clarity.

    ========================

    Express the limit [tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} [/tex] as a definite integral

    Solution:

    We want to interpret the sum as a Riemann sum for [tex] f(x)=(1+x)^{1/3} [/tex]. The factor 2/n suggests that the interval of integration has length 2 and is partitioned into n equal subintervals, each of length 2/n. Let [tex] \inline \large c_{i} = (2i - 1)/n [/tex] for i = 1, 2, 3, ..., n. As [tex] n \to \infty, c_{1} = 1/n \to 0 [/tex] and [tex] c_{n} = (2n -1)n \to 2 [/tex]. Thus, the interval is [0, 2], and the points of the partition are [tex] x_{i} = 2i/n [/tex]. Observe that [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] for each i, so that the sum is indeed a Riemann sum for [tex] f(x) [/tex] over [0, 2]. Since f is continuous on that interval, it is integrable there, and

    [tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} = \int_{0}^{2} (1+x)^{1/3} dx[/tex]

    ==========================

    My question is, is it necessary for the condition [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] be met for the sum to be converted to a definite integral. The reason I ask is that as [tex] n \to \infty [/tex] adding or subtracting any constant from the index does not change the sum at infinity in any case because if we choose [tex] c_{i} = (2i- 20000000)/n [/tex] the big number vanishes when [tex] n \to \infty [/tex] so that the sum is the same what ever the big number is. But then [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] is no longer true and by the definition of a Riemann sum, [tex] c_{i} [/tex] must lie with in the subinterval [tex] [x_{i-1}, x_{i} ][/tex]
     
  2. jcsd
  3. May 11, 2009 #2

    mathman

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    You are correct in that changing it the way you described will lead to the correct result, as long as the number is fixed (not dependent on n). As n gets larger, more of the terms will obey condition on ci, and the remaining terms will become vanishingly small. The question is - why bother?
     
  4. May 12, 2009 #3
    Hi Mathman

    Thank you very much for your response.

    I agree. But also, our text book says that for it to be defined as a general riemann sum that [tex] c_{i} [/tex] must be within [tex]
    [x_{i-1}, x_{i} ]
    [/tex]

    The following two limits produce the same answer but the last one does not satisfy the need for [tex] c_{i} [/tex] to be with in the partition subinterval. Does that mean I am wrong in assuming that [tex] c_{i} [/tex] must lie with in the subinterval for it to be a Riemann sum? Or is it that it does not need to be a Riemann sum to calculate the area or function as a definite integral? Or am I no where near the mark?:

    [tex]
    \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3}
    [/tex]

    [tex]
    c_{i} = \frac{2i - 1}{n} ,\
    \Delta x_{i} = \frac{2}{n},\
    x_{i} = \frac{2i}{n}
    x_{i-1} = \frac{2(i-1)}{n},\

    x_{i-1} < c_{i} < x_{i}
    [/tex]

    condition for Riemann sum is satisfied

    [tex]
    \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 100000}{n}\right)^{1/3}
    [/tex]

    [tex]
    c_{i} = \frac{2i - 100000}{n},\
    \Delta x_{i} = \frac{2}{n},\
    x_{i} = \frac{2i}{n}
    x_{i-1} = \frac{2(i-1)}{n},\

    x_{i-1} < c_{i} < x_{i}
    [/tex]

    condition not satisfied.
     
  5. May 12, 2009 #4

    jbunniii

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    Let's look at just the part in parentheses. In the first example, it is

    [tex]\left(1 + \frac{2i-1}{n}\right)[/tex]

    and "plugging" in the endpoints of the sum, namely [tex]i = 1[/tex] and [tex]i = n[/tex] gives (1 + 1/n) and (3 - 1/n), which is consistent with your integral bounds for each n.

    In the second example, you have

    [tex]\left(1 + \frac{2i - 100000}{n}\right)[/tex]

    andif we again plug in the endpoints [tex]i = 1[/tex] and [tex]i = n[/tex] , we get 1-(99998/n) and 3-(100000/n).

    Thus for each n, you are covering a very different interval, starting way over at [-99997,-99997], (of zero length?!) and while this interval eventually slides over or "converges" to [1,3] and, because of the continuity of (1+x)^1/3, in the limit you might even get the right answer (though I'm not fully convinced of that - you would have to argue this rigorously), this is by no means a Riemann sum, nor could you expect this to work in general.

    For example, what if the exponent had been 1/2 instead of 1/3? Then most of the terms of the sum wouldn't even be defined!
     
  6. May 14, 2009 #5
    Definition of a Definite Integral

    Hi

    Subject Change to: Definition of a definite Integral

    Thank you for your response. I've been thinking about your post and I understand now that the Riemann sum itself can never be defined if [tex]
    c_{i}
    [/tex] does not lie within [tex]

    [x_{i-1}, x_{i} ]

    [/tex]. However, the textbook that we are using says that a definite integral is defined by the limiting case of a Riemann sum. But by the fact that you can reach the definite integral by the limit of a sum that is not a Riemann sum, the conclusion must be that the limiting case of a Riemann sum is not a definitive definition for the definite integral.
     
    Last edited: May 14, 2009
  7. May 14, 2009 #6

    jbunniii

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    Re: Definition of a Definite Integral

    That much is certainly true. There are a number of different definitions for the definite integral. (An author named Frank Burk has even written a book called "A Garden of Integrals" that discusses a number of them.) Some definitions are equivalent, whereas others are more powerful than the Riemann integral because they can integrate all Riemann-integrable functions and then some. (And, a fact of key importance, they produce the same answer as the Riemann integral for those functions that are Riemann integrable.)

    The Lebesgue integral, for example, is usually used in analysis instead of the Riemann integral, not only to be able to integrate more functions, but because more manipulations involving the exchange of limit and integral are valid. It also generalizes to abstract spaces, making it a natural choice for probability theory, for example.

    Then there's the Henstock-Kurzweil ("gauge") integral which on the surface appears to be a trivial modification of the Riemann definition in which, loosely speaking, the [tex]\delta[/tex] that specifies the maximum rectangle width in the Riemann sum is allowed to vary across the interval of integration. If my understanding is right, this simple change produces an integral that is even more powerful than the Lebesgue integral.
     
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