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Math Help - [SOLVED] Geometric Argument!

  1. #1
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    [SOLVED] Geometric Argument!

    Assume that 0< a < b <n infinity

    Use geometric argument to show that integral a to b x^2 dx = (b^3-a^3/3)

    I don't even understand what this is asking me - let alone a clue on how to attempt it.

    Please help!

    Nevermind - I figured this one out!!!!!!!!
    Last edited by an1985; May 3rd 2008 at 01:30 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by an1985 View Post
    Assume that 0< a < b <n infinity

    Use geometric argument to show that integral a to b x^2 dx = (b^3-a^3/3)

    I don't even understand what this is asking me - let alone a clue on how to attempt it.

    Please help!

    Nevermind - I figured this one out!!!!!!!!
    Can you tell us what the argument was?

    RonL
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  3. #3
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    Its a long one.

    Sn = (a/n)f(0) = (a/n)f(a/n) = (a/n)f(2a/n)......(a/n)f((n-1)a/n)
    = (a/n)0^2 + (a/n)(a^2/n^2) +.........(a/n)((n-1)^2a^2/n^2)
    = a^3[1^2 = 2^2......(n-1)2]

    Then you do this again but substitute a for b now to get the other side.

    then
    lim Sn = a^3/6(1-1/n)*1*(2 - 1/n) = a^3/6*1*1*2 = A^3/3

    Now you do this for same thing but insert b.

    Then you add a^3/3 + b^3/3 = (b^3 + a^3/3)


    Quote Originally Posted by CaptainBlack View Post
    Can you tell us what the argument was?

    RonL
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