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Math Help - Riemann sum

  1. #1
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    Riemann sum

    Use the Archimedes-Riemann Theorem to show that for 0 \leq a < b , we have  \int ^b_a xdx = \frac {b^2-a^2}{2}

    I'm trying to find my sum but I can't seem to remember how to do it, any hints ,please?
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  2. #2
    Member Nacho's Avatar
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    <br />
\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left\{ {\frac{{\left( {b - a} \right)k}}<br />
{n}} \right\}}  \cdot \frac{{b - a}}<br />
{n} = \int_a^b x dx<br />

    This is your question?
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  3. #3
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    Yes, thank you.

    But now I'm still stuck... How would I go on with this problem?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Nacho View Post
    <br />
\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left\{ {\frac{{\left( {b - a} \right)k}}<br />
{n}} \right\}} \cdot \frac{{b - a}}<br />
{n} = \int_a^b x dx<br />

    This is your question?
    Yes, thank you.

    But now I'm still stuck... How would I go on with this problem?
    \begin{aligned}\lim_{n\to\infty}\sum_{k=1}^{n}\lef  t\{{\color{red}a}+\frac{(b-a)k}{n}\right\}\cdot\frac{b-a}{n}&=\lim_{n\to\infty}\sum_{k=1}^{\infty}\left\{  \frac{a(b-a)}{n}+\frac{(b-a)^2k}{n^2}\right\}\\<br />
&=\lim_{n\to\infty}\left\{a(b-a)+\frac{(b-a)^2n(n+1)}{2n^2}\right\}\quad{\color{red}\star}\\<br />
&=ab-a^2+\frac{b^2}{2}-ab+\frac{a^2}{2}\\<br />
&=\frac{b^2-a^2}{2}\end{aligned}

    {\color{red}\star}~\sum_{k=1}^{n}k=\frac{n(n+1)}{2  }~~\sum_{k=1}^{n}c=cn
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