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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 $\displaystyle 2$ units. If you divide it into $\displaystyle n$ subintervals, then the length of each subinterval is $\displaystyle \frac{2}{n}$. This is the length of each rectangle.

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

    So that means each rectangle's area is

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

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


    The total area therefore is

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


    By making $\displaystyle n \to \infty$, the area of the rectangles converges on the area underneath the curve.
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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

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

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

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

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

    Thus

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

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

    This is the required expression for the Riemann sum:

    $\displaystyle 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 05:16 PM.
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  4. #4
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    $\displaystyle x_k^*= 1+k \frac{2}{n} =\frac{2n+4k}{n^2}$

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