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Math Help - Help Summing Difficult Function

  1. #1
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    Help Summing Difficult Function

    Hey everyone, this is my first thread, and I hope it will give me a good experience on this site.

    Anyway, onto the question:

    I have been trying to figure out a way to sum this:

    \lim_{n\to\infty}\frac{r}{n}\sum_{i=0}^n\sqrt{r^2-x_i^2}\Delta x
    Where
    x_i=\frac{ri}{n}
    and
    \Delta x=\frac{r}{n}

    The next step is
    \lim_{n\to\infty}\sum_{i=0}^n(\sqrt{r^2-\frac{r^2i^2}{n^2}})(\frac{r}{n})

    I did some manipulating, and got:
    \lim_{n\to\infty}\frac{r}{n}\sum_{i=0}^n\sqrt{r^2-\frac{r^2i^2}{n^2}}

    All the help I have gotten has told me that the answer is zero, but that can't be possible, because I started with the area under the curve of the basic unit circle in the first quadrant,
    y=\sqrt{r^2-x^2}
    and I have not made any mistakes so far.

    So how can I come up with a non-zero answer for this?

    EDIT: I want to solve without having to use the fundamental theorem of calculus, and just saying that it is equal to \frac{\pi r^2}{4}

    EDIT 2: Just like for
    \sum_{i=0}^n i=\frac{n(n+1)}{2},
    is there a way I can do that sort of method the problem I have presented?
    Last edited by wolfman29; September 12th 2009 at 01:24 PM.
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  2. #2
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    Quote Originally Posted by wolfman29 View Post
    Anyway, onto the question:
    I have been trying to figure out a way to sum this:
    \lim_{n\to\infty}\frac{r}{n}\sum_{i=0}^n\sqrt{r^2-x_i^2}\Delta x
    Where
    x_i=\frac{ri}{n}
    and
    \Delta x=\frac{r}{n}
    \lim_{n\to\infty}\frac{r}{n}\sum_{i=0}^n\sqrt{r^2-x_i^2}\Delta x =\int_0^r {\sqrt {r^2  - x^2 } dx}  = \frac{{\pi r^2 }}{4}
    It is the area of one-fourth a circle.
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  3. #3
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    Well, yes, I know that. But what I am trying to do is trying to solve it assuming I don't know the Fundamental Theorem of Calculus (which I do, but this is just a curiosity).

    Is there a way to solve this without just saying, "Oh, by definition, it's the area of a quarter of a circle, and thus it is (pi*r^2)/4"?
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  4. #4
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    Quote Originally Posted by wolfman29 View Post
    Well, yes, I know that. But what I am trying to do is trying to solve it assuming I don't know the Fundamental Theorem of Calculus (which I do, but this is just a curiosity).

    Is there a way to solve this without just saying, "Oh, by definition, it's the area of a quarter of a circle, and thus it is (pi*r^2)/4"?
    You are missing the point.
    You simply must recognize that as an approximating sum for that integral.
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  5. #5
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    I understand that. But just like for
    \sum_{i=0}^n i=\frac{n(n+1)}{2},
    is there a way I can do that sort of method the problem I have presented?
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