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Math Help - Method of Exhaustion - Archimedes's Method

  1. #1
    Member RedBarchetta's Avatar
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    Method of Exhaustion - Archimedes's Method

    Find the area under the curve using Archimedes's method for 2x+1 from zero to some number b.

    So we divide the area under the parabolic segment into rectangles. Where we divide the segment b into n parts where we have (b/n) sections and k=1,2,3,...,n.

    So the area of the rectangles are:



    .....but the answer should be +b not +b/n. Any help is appreciated. Thank you!
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    We will compute the areas two ways. One way through the left endpoints one way through the right endpoints. The left endpoints for f(x)=2x^2+1 are smaller than the true area of the parabola, is given by, \sum_{k=0}^{n-1} f\left( \frac{bk}{n} \right) \cdot \frac{b}{n}. And the right endponts are larger than the true area of the parabola, is given by, \sum_{k=1}^n f\left( \frac{bk}{n} \right) \cdot \frac{b}{n}.

    Thus, \sum_{k=0}^{n-1} \frac{2b^3k^2}{n^3} + \frac{b}{n} \leq A \leq \sum_{k=1}^n \frac{2b^3k^2}{n^3} + \frac{b}{n}.

    Now use the formula \sum_{k=0 \text{ or } 1}^m k^2 = \frac{m(m+1)(2m+1)}{6}.
    Last edited by ThePerfectHacker; January 1st 2009 at 05:32 PM.
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  3. #3
    Member RedBarchetta's Avatar
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    Shouldn't the first term contain b^3 and not b^2?
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    Quote Originally Posted by RedBarchetta View Post
    Shouldn't the first term contain b^3 and not b^2?
    It is fixed now.
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    Member RedBarchetta's Avatar
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    ....

    So do you evaluate that last summation by replacing m with n-1 and n respectively? I'm still not sure how this leads to the formula for the area.
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    Careful! \sum_{k=1}^n \frac{b}{n} = \frac{b}{n}+...+\frac{b}{n} = n\cdot \frac{b}{n} = b.

    After you do that take the limit n\to \infty
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  7. #7
    Member RedBarchetta's Avatar
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    I'm not sure if i'm following this....
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    Quote Originally Posted by RedBarchetta View Post
    I'm not sure if i'm following this....
    When you have \sum_{k=1}^n a_k it means a_1+a_2+...+a_n.

    Now if a_k = \tfrac{b}{n} then \sum_{k=1}^n a_k = \underbrace{\tfrac{b}{n}+...+\tfrac{b}{n}}_{n \text{ times }} = b

    EDIT: Fixed
    Last edited by ThePerfectHacker; January 2nd 2009 at 08:58 AM.
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    Typo - n times. Fixed ;-)
    Last edited by Jester; January 2nd 2009 at 12:32 PM. Reason: typo fixed
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