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Math Help - Trapezoidal rule proof

  1. #1
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    Trapezoidal rule proof

    Consider the function y=x^2 for x[0,1]. Show using the trapezoidal rule with n intervals that the integral 0 -> 1 of x^2 dx is equivalent to n(n+1)(2n+1)/6n^3 - 1/2n.

    I am having a lot of trouble arranging the trapezoidal rule into a sigma type sequence.

    My current working goes as far as
    integral of x^2 is equivilant to 1-0/2n [f(x0) +2(f(x1)+f(x2)+...+f(xn-1))+f(n)]

    All help is greatly appreciated.
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  2. #2
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    Let the width of a strip be h so nh = 1.

    Then f(x_1) = h^2, f(x_2) = (2h)^2, f(x_3) = (3h)^2, and so on.

    If you substitute these into your trapezium rule formula and remove h^2 as a common factor, the formula for the sum of squares

    1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n}{6}(n+1)(2n+1),

    can be used to simplify.

    Towards the end, remove h by writing it as 1/n.

    Incidently it's best to sum the series as

    1^2 + 2^2 + ... + (n-1)^2 = 1^2 + 2^2 + ... + n^2 - n^2.
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