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Math Help - riemann sum

  1. #1
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    riemann sum

    Write down the Riemann sum for int_2^3 2x dx using N equal subintervals, simplify, and find the limit as N approaches infinity.

    I am not sure if I set the sum up properly and I am having trouble with the simplification. I don't really know what it is I am trying to isolate... N? i?

    I am using the formula:

    (b-a)/N Sigma_i=1^N [ f(a + i(b-a / N))]

    so I have:

    1/N Sigma_i=1 ^ N [ 2(2 + i(1/N))]

    So where do I go from here? And also, when I take the limit as N approaches infinity, what do I do with terms including i?

    Thanks!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by littlejodo View Post
    Write down the Riemann sum for int_2^3 2x dx using N equal subintervals, simplify, and find the limit as N approaches infinity.

    I am not sure if I set the sum up properly and I am having trouble with the simplification. I don't really know what it is I am trying to isolate... N? i?

    I am using the formula:

    (b-a)/N Sigma_i=1^N [ f(a + i(b-a / N))]

    so I have:

    1/N Sigma_i=1 ^ N [ 2(2 + i(1/N))]

    So where do I go from here? And also, when I take the limit as N approaches infinity, what do I do with terms including i?

    Thanks!
    Assume that f is integrable over [a,b] then if we define \Delta x=\frac{b-a}{n} and M_i=a+\Delta x\cdot i then

    \int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f\left(M_i\right) \Delta x. So in your case

    \Delta x=\frac{3-2}{n}=\frac{1}{n}
    f\left(M_i\right)=f\left(2+\frac{i}{n}\right)=2\le  ft(2+\frac{i}{n}\right)

    So then your integral is equal to

    \begin{aligned}\int_2^3 2x~dx&=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}\  left\{2+\frac{i}{n}\right\}\\<br />
&=\frac{2}{n}\left\{2n+\frac{n+1}{2}\right\}{\colo  r{red}\star}\\<br />
&=4+1\\<br />
&=5\end{aligned}

    \color{red}\star was gotten by the formulas \sum_{i=1}^{n}c=cn and \sum_{i=1}^{n}i=\frac{n(n+1)}{2}
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  3. #3
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    Very helpful! Thank you!

    So, is it safe to say that in taking the limit as N --> inf I can treat each N as a 1? I'm trying to see how taking the limit leads to 4 + 1...

    I had thought that when the variable was in the denominator that term would equal 0...

    The only way I see it leading to 4 + 1 is to have N = 1 but I'm just not sure if that's the correct thing to do. Clearly, I can see that the limit is 5 by looking at graph, but I was hoping to understand the algebra behind the concept. Thanks!
    Last edited by littlejodo; December 13th 2008 at 09:20 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by littlejodo View Post
    Very helpful! Thank you!

    So, is it safe to say that in taking the limit as N --> inf I can treat each N as a 1? I'm trying to see how taking the limit leads to 4 + 1...

    I had thought that when the variable was in the denominator that term would equal 0...

    The only way I see it leading to 4 + 1 is to have N = 1 but I'm just not sure if that's the correct thing to do. Clearly, I can see that the limit is 5 by looking at graph, but I was hoping to understand the algebra behind the concept. Thanks!
    Word to the wise, I think most MHFers look at a response once usually. So if you need to ask a question don't edit your response make a new one if you had asked your previous one a while ago.

    \frac{2}{n}\left\{2n+\frac{n+1}{2}\right\}=4+\frac  {n+1}{n}

    Now taking the limit of this we get \lim\left\{4+\frac{n+1}{n}\right\}=4+1=5

    I assume you know that \lim\frac{n+1}{n}=1 right?
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