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How to prove this?

  1. May 1, 2006 #1
    How can it be shown that the limit of the Riemann sum is equal to definite integral? [tex]\int_{a}^{b}f(x) \, dx = \lim_{n\rightarrow\infty} \frac{b-a}{n}\sum_{k=1}^{n}f\left( a+\frac{b-a}{n}k\right) [/tex]
     
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  3. May 1, 2006 #2

    matt grime

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    The limit of the Riemann sum, if it exists, is by definition (or at least the criterion of integrabilty) the integral.
     
  4. May 1, 2006 #3

    HallsofIvy

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    Well, first of all, it's not! Unless f happens to be "Riemann integrable".

    So of course the answer depends on the definition of "Riemann integrable" and "definite integral".

    Let f be defined and non-negative on the interval from a to b. partition the interval [a, b] into n subintervals. For each i choose a number xi* in the ith interval. Let [itex]\Delta x_i[/itex] be the length of the ith interval. Then
    [tex]\sum_{i=1}^n f(x_i^*)\Delta x_i[/tex]
    is the Riemann sum for those choices.

    If f is bounded on [a,b] then it is easy to show that the set of all such Riemann sums, for a given n, has an upper bound (the lower bound is trivially 0) and so has a greatest lower bound and least upper bound, Un and Vn. f is said to be "Riemann integrable" on [a,b] if and only if [itex]\lim_{n\rightarrow\infty} U_n= \lim_{n\rightarrow\infty} V_n[/itex].

    If f is Riemann integrable, one way of defining [itex]\int_a^b f(x)dx[/itex] is as that common value. With that definition, there is nothing to prove- just that the sum you give is, for each n, a Riemann sum and so the limit must be the integral.

    Another way to define the definite integral is as "area bounded by x= a, y= f(x), x= b, and y= 0". To prove your equality, choose each xi* to be, first, the x in that interval that gives the largest value for f(xi*) and, secondly, the x in the interval that gives the smallest value of f. Calling the first sum Mn and the second mn it is trivial that [itex]m_n \le area \le M_n[/itex]. Since that is true for all n, if f is Riemann integrable, so that those to sums have a common limit, by the "pinching theorem", that common limit must be the area under the curve.
     
  5. May 1, 2006 #4

    shmoe

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    Which would raise the question, how do you define "area bounded by x= a, y= f(x), x= b, and y= 0" without using a definite integral (or something essentially equivalent)?
     
  6. May 1, 2006 #5

    HallsofIvy

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    Yes, it would. The answer to that question is:

    If A is a bounded set in R2, then there exist numbers a, b, c, d such that A is contained in the interval [itex]a\le x \le b[/itex], [itex]c\le y\le d[/itex]. Divide each side into n intervals (so that the whole area is divided into n2 small rectangles). We say that such a small rectangle is "interior to A" if every point of it (except possibly boundary points) is in A, that such a small rectangle is "exterior to A" if every point of it (except possible boundary points) is NOT in A, and that such a small rectangle is "bounding A" if there are some points (other than boundary points) of the rectangle in A and some not in A.
    The "inner n-th measure" of A is the sum of the areas of all "interior" rectangles and the "outer n-th measure" of A is the sum of all "interior" and "bounding" rectangles of A. It is easy to show that the sequence of "inner n-th measures" of A is an increasing sequence having the area of the original rectangle as upper bound and so converges (to the "inner measure" M- of A). It is easy to show that the sequence of "outer n-th measures" of A is a decreasing sequence having lower bound 0 and so converges (to the "outer measure" M+ of A). Finally, it is easy to show that, for all n, the "inner n-th measure" of A is less than or equal to the "outer n-th measure" of A and so M-<= M+. A is said to be "measurable" if and only if those two limits are equal and, in that case, the common value is the area of A.
     
    Last edited: May 1, 2006
  7. May 1, 2006 #6
    Ok, it makes sense (although I can't say I understand all of it :)). However, I was looking for a simple mathematical solution. I came up with this:

    [tex] \lim_{n\rightarrow\infty} \frac{b-a}{n}\sum_{k=1}^{n}f\left( a+\frac{b-a}{n}k\right)=(f(a+n\frac{b-a}{n})+f(a+(n-1)\frac{b-a}{n})+\mbox{...}+f(a+\frac{b-a}{n}))\frac{b-a}{n}=[/tex]


    [tex]
    (\frac{F(a+n\frac{b-a}{n})-F(a+(n-1)\frac{b-a}{n})}{\frac{b-a}{n}}+\frac{F(a+(n-1\frac{b-a}{n})-F(a+(n-2)\frac{b-a}{n})}{\frac{b-a}{n}}+\mbox{...}+\frac{F(a+\frac{b-a}{n})-F(a)}{\frac{b-a}{n}})\frac{b-a}{n}=[/tex]


    [tex]
    \frac{F(a+n\frac{b-a}{n})-F(a)}{\frac{b-a}{n}}\frac{b-a}{n}=F(b)-F(a)=
    \int_{a}^{b}f(x) \, dx [/tex]

    where [tex] F\prime(x)=f(x)[/tex]

    Is this mathematically correct?
    (of course, n always approaches infinity)
     
    Last edited: May 1, 2006
  8. May 1, 2006 #7

    shmoe

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    Have you ever seen a text that went this route? I mean essentially considering a function on the real line as defining a region in the plane, and defining the integral in terms of our previously defined area.

    This seems at least philosophically different from the usual riemann sum definition for an integral (though of course yielding the same results where both make sense). The rectangles that usually accompany this version to me are a justification in the other direction- why it is reasonable to call the thing definined by the integral an "area" (along with it matching the "usual" definitions we have for area of simple shapes).
     
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