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Math Help - Finding exact area with the Definite Integral

  1. #1
    Bar0n janvdl's Avatar
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    Finding exact area with the Definite Integral

    Okay i was busy trying to figure these things out and then i got stuck.

    Now using the Riemann notation to calculate the area under the graph x^2 + 1 we get the following:

    R_{6} = \sum_{i = 1}^{6} [ f(x_{i}) \cdot ( \frac{3-0}{6})]

    (I calculated R_{6} = 14,375)

    They use rectangles under the graph to approximate the area. (6 rectangles and the x-axis value ranges from 0 to 3.)

    Then the book goes further and says:

    For any number of rectangles under the graph(Still ranging from 0 to 3 on the x-axis):

    R_{n} = \sum_{i = 1}^{n} [ f(x_{i}) \cdot ( \frac{3-0}{n})]

    At the end we arrive at the following:

    R_{n} = \frac{27}{n^3} \sum_{i = 1}^{n} i^2 + \frac{3}{n} \sum_{i = 1}^{n} 1


    Up to there i understand perfectly.

    Then from that we get

    R_{n} = \frac{27}{n^3} \left( \frac{ n(n+1)(2n+1) }{6} \right) + \frac{3}{n} (n)


    EDIT: posted by accident. (Would appreciate it if a mod would delete this)
    Last edited by janvdl; November 15th 2007 at 12:37 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Okay i was busy trying to figure these things out and then i got stuck.

    Now using the Riemann notation to calculate the area under the graph x^2 + 1 we get the following:

    R_{6} = \sum_{i = 1}^{6} [ f(x_{i}) \cdot ( \frac{3-0}{6})]

    (I calculated R_{6} = 14,375)

    They use rectangles under the graph to approximate the area. (6 rectangles and the x-axis value ranges from 0 to 3.)

    Then the book goes further and says:

    For any number of rectangles under the graph(Still ranging from 0 to 3 on the x-axis):

    R_{n} = \sum_{i = 1}^{n} [ f(x_{i}) \cdot ( \frac{3-0}{n})]

    At the end we arrive at the following:

    R_{n} = \frac{27}{n^3} \sum_{i = 1}^{n} i^2 + \frac{3}{n} \sum_{i = 1}^{n} 1


    Up to there i understand perfectly.

    Then from that we get

    R_{n} = \frac{27}{n^3} \left( \frac{ n(n+1)(2n+1) }{6} \right( + \frac{3}{n} (n)
    what is your question? how thy changed the sums to those formulas? those are identities. see the identities section here

    it should not be hard to find the proofs of these identities floating around the internet if you're interested
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