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Math Help - Definition of Definite Integral

  1. #1
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    THanks so much for your help!!
    Last edited by aquaglass88; February 7th 2008 at 06:57 AM. Reason: delete
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  2. #2
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    Quote Originally Posted by aquaglass88 View Post
    Hi I am trying to find use the definition of the definite integral to prove that
    the integral of (x^2)dx in [a,b] = [(b^3) - (a^3)]/3

    I started out using the Reiman Sums except I ended up with
    lim as n approaches infinity [nab^2 - bab^2 + b^3 - na^3 + a^3 - ab^2]^2
    -------------------------------------------------------
    n^2

    I have no clue how that equals b cubed minus a cubed over 3 and I'm not sure what I did wrong if I did.

    THanks so much for your help!!
    Using right end-points, by definition:



    \displaystyle \int_a^b x^2 \, dx = \lim_{n \rightarrow \infty} \left( \frac{b-a}{n} \right) \sum_{i=1}^{\infty} f\left( a + \frac{(b-a)i}{n}\right) = \lim_{n \rightarrow \infty} \left( \frac{b-a}{n} \right) \sum_{i=1}^{\infty} \left( a + \frac{(b-a)i}{n} \right)^2



    Expand:

    \displaystyle = \lim_{n \rightarrow \infty} \left[ \frac{(b-a)}{n} \sum_{i=1}^{\infty} a^2 + \frac{2a(b-a)^2}{n^2} \sum_{i=1}^{\infty} i + \frac{(b-a)^3}{n^3} \sum_{i=1}^{\infty} i^2 \right]



    Substitute the well known sum of each series:


    \displaystyle = \lim_{n \rightarrow \infty} \left[ \frac{(b-a)}{n} (a^2n) + \frac{2a(b-a)^2}{n^2} \frac{n(n-1)}{2} + \frac{(b-a)^3}{n^3} \frac{n(n+1)(2n+1)}{6} \right]



    Simplify each term by cancelling common factors:

    \displaystyle = \lim_{n \rightarrow \infty} \left[ (b-a)a^2 + \frac{a(b-a)^2(n-1)}{n} + \frac{(b-a)^3 (n+1)(2n+1)}{6n^2} \right]



    Take the limit term by term:

    \displaystyle = (b-a)a^2 + a(b-a)^2 \lim_{n \rightarrow \infty}\frac{(n-1)}{n} + \frac{(b-a)^3}{6}\lim_{n \rightarrow \infty}\frac{(n+1)(2n+1)}{n^2}



    Evaluate the simple limits:

    \displaystyle = (b-a)a^2 + a(b-a)^2 (1) + \frac{(b-a)^3}{6} (2)


    \displaystyle = (b-a)a^2 + a(b-a)^2 + \frac{(b-a)^3}{3}

    Simplify:

    \displaystyle = \frac{b^3}{3} - \frac{a^3}{3}.
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