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Math Help - Finding Riemann sum

  1. #1
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    Finding Riemann sum

    Let f(x) = 3x(4 - x)

    Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).
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  2. #2
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    Quote Originally Posted by TsAmE View Post
    Let f(x) = 3x(4 - x)

    Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).
    The interval has length 2 units. If you divide it into n subintervals, then the length of each subinterval is \frac{2}{n}. This is the length of each rectangle.

    The width of each rectangle will be f(x_i) = 3x_i(4 - x_i) for i = 1, 2, 3, \dots, n (since you are taking the right hand endpoints you don't use x_0).

    So that means each rectangle's area is

    \frac{2}{n}\left[3x_i(4 - x_i)\right]

     = \frac{6x_i(4 - x_i)}{n}.


    The total area therefore is

    \sum_{i = 1}^{n}\frac{6x_i(4 - x_i)}{n}


    By making n \to \infty, the area of the rectangles converges on the area underneath the curve.
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  3. #3
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    Quote Originally Posted by TsAmE View Post
    Let f(x) = 3x(4 - x)

    Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).
    Each subinterval has length

    \Delta x = \frac{3-1}{n}=\frac{2}{n}

    The right endpoint choices for x_1^*,...,x_n^* is given by:

    x^*_k=x_k=a+k \Delta x

    x_k^*= 1+k \frac{2}{n} =\frac{2n+4k}{n^2}

    Thus

    \sum ^n_{k=1} f(x^*_k)\Delta x = \sum ^n_{k=1} 3x_k^*(4 - x^*_k) \Delta x

    = \sum ^n_{k=1} 3( \frac{2n+4k}{n^2} ) (4- \frac{2n+4k}{n^2}) \frac{2}{n}

    This is the required expression for the Riemann sum:

    A= \lim_{(\Delta x \to \infty)} \sum ^n_{k=1} \left[ 3( \frac{2n+4k}{n^2} ) (4- \frac{2n+4k}{n^2}) \frac{2}{n} \right]
    Last edited by Roam; May 7th 2010 at 06:16 PM.
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  4. #4
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    x_k^*= 1+k \frac{2}{n} =\frac{2n+4k}{n^2}

    How did you get: \frac{2n+4k}{n^2} ?
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