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Thread: Riemann sum

  1. #1
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    Riemann sum

    Consider the function $\displaystyle f(x) = \frac {x^2}{3} - 9$
    In this problem you will calculate $\displaystyle \int_{0}^{2} ( \frac {x^2}{3} - 9) \,dx$ by using the definition

    $\displaystyle \int_{a}^{b} f(x) \,dx = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} f(x_i) \Delta x \right]$

    The summation inside the brackets is $\displaystyle R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

    Calculate $\displaystyle R_n$ for $\displaystyle f(x) = \frac {x^2}{3} - 9$
    on the interval [0, 2]

    $\displaystyle R_n = $

    $\displaystyle \lim_{n \to \infty} R_n = $

    i have no clue on how to do this one.
    Last edited by viet; Nov 21st 2006 at 09:55 PM. Reason: typo
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    Consider the function $\displaystyle f(x) = \frac {x^2}{3} - 9$
    In this problem you will calculate $\displaystyle \int_{0}^{2} ( \frac {x^2}{3} - 9) \,dx$ by using the definition

    $\displaystyle \int_{a}^{b} f(x) \,dx = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} f(x_i) \Delta x \right]$

    The summation inside the brackets is $\displaystyle R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

    Calculate $\displaystyle R_n$ for $\displaystyle f(x) = \frac {x^2}{3} - 9$
    on the interval [0, 2]

    $\displaystyle R_n = $

    $\displaystyle \lim_{n \to \infty} R_n = $

    i have no clue on how to do this one.
    First the interval has n points, so $\displaystyle \Delta x = \frac{2 - 0}{n} = \frac{2}{n}$. Thus each $\displaystyle x_i = 0 + i \cdot \frac{2}{n} = i \cdot \frac{2}{n}$

    So:
    $\displaystyle R_n = \sum_{i = 1}^n \left ( \frac{x_i^2}{3} - 9 \right ) \frac{2}{n} = \sum_{i = 1}^n \left ( \frac{ \left ( i \cdot \frac{2}{n} \right )^2 }{3} - 9 \right ) \frac{2}{n}$

    $\displaystyle R_n = \sum_{i = 1}^n \left ( \frac{8i^2}{3n^3} - \frac{18}{n} \right ) = \sum_{i = 1}^n \frac{8i^2}{3n^3} - \sum_{i = 1}^n \frac{18}{n}$

    $\displaystyle R_n = \frac{8}{3n^3} \sum_{i = 1}^n i^2 - \frac{18}{n} \sum_{i = 1}^n 1$

    $\displaystyle R_n = \frac{8}{3n^3} \frac{n(n+1)(2n+1)}{6} - \frac{18}{n} n$

    $\displaystyle R_n = \frac{4(2n^2 + 3n + 1)}{9n^2} - 18$

    Now:
    $\displaystyle \lim_{n \to \infty}R_n = \lim_{n \to \infty} \left ( \frac{4(2n^2 + 3n + 1)}{9n^2} - 18 \right )$

    = $\displaystyle \lim_{n \to \infty} \frac{4}{9} \left ( 2 + 3 \frac{1}{n} + \frac{1}{n^2} \right ) - 18$

    = $\displaystyle \frac{4}{9} \cdot 2 - 18 = \frac{8}{9} - 18 = \frac{-154}{9}$

    -Dan
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  3. #3
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    Right end point: $\displaystyle x_{k}=a+k{\Delta}x$.....[1]

    Subdivide the interval [0,2] into n equal parts, then each will have length:

    $\displaystyle {\Delta}x=\frac{2-0}{n}=\frac{2}{n}$

    From [1]:

    $\displaystyle x_{k}=\frac{2k}{n}$

    The kth rectangle has area:

    $\displaystyle f(x_{k}){\Delta}x=\left(\frac{(\frac{2k}{n})^{2}}{ 3}-9\right){\Delta}x$

    =$\displaystyle \left(\frac{4k^{2}}{3n^{2}}-9\right)(\frac{2}{n})$

    =$\displaystyle \frac{8k^{2}}{3n^{3}}-\frac{18}{n}$

    Sum up the rectangles:

    $\displaystyle \frac{8}{3n^{3}}\sum_{k=1}^{n}k^{2}-\sum_{k=1}^{n}\frac{18}{n}$

    Remember, the sum of $\displaystyle k^{2}=\frac{n(n+1)(2n+1)}{6}$

    Sub into our summation:

    $\displaystyle \frac{8}{3n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}-18$

    =$\displaystyle \frac{-2(77n^{2}-6n-2)}{9n^{2}}$

    Now take the limit:

    $\displaystyle \lim_{n\to\infty}\frac{-2(77n^{2}-6n-2)}{9n^{2}}$

    $\displaystyle \lim_{n\to\infty}\frac{4}{3n}+\lim_{n\to\infty}\fr ac{4}{9n^{2}}-\underbrace{\frac{154}{9}}$

    See the limit and, consequently, your integral?.

    EDIT: Ol' Topsquark beat me, but after all that I ain't deleting it
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    EDIT: Ol' Topsquark beat me, but after all that I ain't deleting it
    I don't blame you. I'd've done the same thing.

    -Dan
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