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Limit of the Integral Sum

  1. Aug 7, 2005 #1


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    To integrate functions as the limit of an integral sum, how can we know which way to take the partition points in the interval?

    For example, in
    [tex] \int_{a}^{b} x dx [/tex]

    I can take the partition points as

    [tex] x_o = a [/tex]
    [tex] x_1 = a + \delta x [/tex]
    [tex] x_k = a+ k\delta x[/tex]
    where [tex] \delta x = \frac{b-a}{n} [/tex]

    So that the sum is [tex] \sum_{k=1}^{n} f(x_{k-1}) \delta x [/tex]

    But to integrate
    [tex] \int_{a}^{b} \sqrt{x} dx [/tex]

    If I take the partition points as above the sum will be
    [tex] (\delta x)(\sqrt{a} + \sqrt{a+ \delta x} + ... ) [/tex]

    which I cannot find.

    I can solve the question if I take the partition points as
    [tex] x_0 = a [/tex]
    [tex] x_1 = aq [/tex]
    [tex] x_k=aq^k [/tex]

    Where [tex] q=(\frac{b}{a})^(1/n) [/tex].
    {The idea to take it this way was given as a hint in the book}

    So, is there any other specific manner in which I should spilt the partition points and if so is there a general method in which I can know how to take the values of x_0, x_1, x_2 to solve the problem?
  2. jcsd
  3. Aug 7, 2005 #2

    matt grime

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    the choice of partition depends upon the function. experience is the thing you need and an educated guess. Ideally you want to choose the partition such that the resulting series is one you can sum. egn if tou wre doing sqrt(x) from 0 to 1 then picking the partition i^2/n^2 as i goes from 0 to n is good cos the square root spits out i/n.
    Last edited: Aug 7, 2005
  4. Aug 7, 2005 #3


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    Are you talking about a numerical integration? In that case, there are two things you can do:(1) take the points equally spaced so the calculations are easier or (2) take the points closer together where the function changes rapidly (i.e. where the derivative is large) so that the integration is more accurate.

    Of course, if you are talking about the Riemann sums used to define the integral, then it doesn't matter. As long as the function is integrable, any partition will give the same answer.
  5. Aug 7, 2005 #4

    matt grime

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    Ah, but proving that something is riemann integrable by the criterion of integrability requiers you to chose the partition family intellignetly, in my experience of teaching analysis. actually to be honest, for any example in the course or the exam there was always a strongly suggested hint.

    the approximate rules is pick a partiiton so that the length of the interval and the max.min value spat out are such that when you multiply them together you are addding up r^k from 1 to n times some constant and then appeal to seom result about summing these series.

    of course the sensilbe person simply cites that any continuous or montone function is riemann integrable.
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