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Math Help - definite integral answer different from Riemann Sum calc

  1. #1
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    definite integral answer different from Riemann Sum calc

    Hi everybody,

    I have to calculate a definite integral using the oh so necessary Riemann Sum.
    I first calculated the integral with the fundamental theorem as a check.
    The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

    When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

    I started by finding delta x = 1/n and xi = i/n
    then I set up the equation to be:
    lim of n->inf Sigma i=1 to n of f(i/n) 1/n

    Isn't that correct?
    I then solved it and got 1 as the answer.

    If anybody could help me out that would be great.

    Thx
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rexexdesign View Post
    Hi everybody,

    I have to calculate a definite integral using the oh so necessary Riemann Sum.
    I first calculated the integral with the fundamental theorem as a check.
    The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

    When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

    I started by finding delta x = 1/n and xi = i/n
    then I set up the equation to be:
    lim of n->inf Sigma i=1 to n of f(i/n) 1/n

    Isn't that correct?
    I then solved it and got 1 as the answer.

    If anybody could help me out that would be great.

    Thx

    \lim_{n\to\infty}\sum_{x=1}^{n}f(M_i)\Delta{x_i}

    so \Delta{x_i}=\frac{2-1}{n}


    f(M_i)=2+3\bigg(1+\frac{x}{n}\bigg)^2=5+\frac{6x}{  n}+\frac{3x^2}{n^2}

    So \frac{1}{n}\sum_{x=1}^{\infty}\bigg[5+\frac{6x}{n}+\frac{6x^2}{n^2}\bigg]=\frac{1}{n}\bigg[5n+\frac{6n(n+1)}{2n}+\frac{3n(n+1)(2n+1)}{6n^2}\b  igg]= \frac{1}{n}\bigg[5n+3n+\frac{3}{n}+n+\frac{3}{2}+\frac{1}{2n}\bigg]=5+3+\frac{3}{n^2}+1+\frac{3}{2n}+\frac{1}{2n^2}
    The limit of that as it goes to infinity is 9

    \lim_{n\to\infty}\sum_{x=1}^{n}\bigg[\bigg(2+3\bigg(1+\frac{x}{n}\bigg)^2\bigg)\bigg(\f  rac{1}{n}\bigg)\bigg]=9=\int_{1}^{2}\bigg[2+3x^2\bigg]dx


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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    Quote Originally Posted by rexexdesign View Post
    Hi everybody,

    I have to calculate a definite integral using the oh so necessary Riemann Sum.
    I first calculated the integral with the fundamental theorem as a check.
    The integral is from 1 to 2 of (2+3x^2)dx and it equals 9 according to my calculation.

    When I calculated with tv Riemann sum, it equaled 1, so I must have made a mistake.

    I started by finding delta x = 1/n and xi = i/n
    then I set up the equation to be:
    lim of n->inf Sigma i=1 to n of f(i/n) 1/n

    Isn't that correct?
    I then solved it and got 1 as the answer.

    If anybody could help me out that would be great.

    Thx
    Note that
    x_i=a+i\Delta x =1+\frac{i}{n}

    \lim_{n \to \infty}\sum_{i=1}^{n}\left( 2+3(1+\frac{i}{n})^2\right)\frac{1}{n}= \lim_{n \to \infty}\left( \frac{2}{n}\sum_{i=1}^{n}1+\frac{3}{n}\sum_{i=1}^{  n}\left( 1+\frac{2i}{n}+\frac{i^2}{n^2}\right)\right)

    You should be able to finish from here

    Edit Dang I type slow
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  4. #4
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    Thank you both for the quick answer. I guess I got messed up with xi being 1+i/n

    Thanks again
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