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Math Help - Area of a region using the limit process

  1. #1
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    Area of a region using the limit process

    The equation is y=-2x+3 and the points are [0,1]

    i simplified and got

    lim n as it approaches infiniy of
    -2/n^2 * n(n+1)/2 + 3/n *n

    the answer is supposed to be 2 but i dont understand how to get the limits to equal 2. I cant get all the n to cancel out
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  2. #2
    Senior Member nikhil's Avatar
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    y=-2x+3
    on integration we get RHS as
    =-2((x^2)/2)+3x +c
    =3x-x^2+c
    now put limits from 0 to 1
    so area obtained will be
    =3(1)-1+c-[3(0)-0+c]
    =3-1
    =2 sq.units
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  3. #3
    Eater of Worlds
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    {\Delta}x=\frac{1-0}{n}=\frac{1}{n}

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

    f(x_{k})==2(\frac{k}{n})+3

    f(x_{k}){\Delta}x=\left[-2(\frac{k}{n})+3\right]\cdot \frac{1}{n}

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

    But, \sum_{k=1}^{n}k=\frac{n(n+1)}{2}

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

    3-(\frac{1}{n}+1)

    Limit:

    \lim_{n\to {\infty}}\left(2-\frac{1}{n}\right)=2
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