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Math Help - Riemann sum

  1. #1
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    Riemann sum

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

      \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 R_n which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

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

    R_n =

    \lim_{n \to \infty} R_n =

    i have no clue on how to do this one.
    Last edited by viet; November 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 f(x) = \frac {x^2}{3} - 9
    In this problem you will calculate \int_{0}^{2} ( \frac {x^2}{3} - 9) \,dx by using the definition

      \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 R_n which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

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

    R_n =

    \lim_{n \to \infty} R_n =

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

    So:
    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}

    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}

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

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

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

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

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

    =  \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: x_{k}=a+k{\Delta}x.....[1]

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

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

    From [1]:

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

    The kth rectangle has area:

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

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

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

    Sum up the rectangles:

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

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

    Sub into our summation:

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

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

    Now take the limit:

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

    \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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