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Math Help - Using Riemann Sum for intergrals

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    Using Riemann Sum for intergrals

    Evaluate the definite integral (x^3-x)dx with bounds 1 to 5 by the limit definition of a Riemann Sum.
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    Quote Originally Posted by Hibijibi View Post
    Evaluate the definite integral (x^3-x)dx with bounds 1 to 5 by the limit definition of a Riemann Sum.
    We break the interval into n steps:
    \Delta x = \frac{5 - 1}{n} = \frac{4}{n}

    We then form each little rectangle for the area. The area of the ith rectangle will be
    \Delta A = ( f(x_{i + 1}) - f(x_i) ) \Delta x
    where x_i = 1 + i \cdot \Delta x

    So this is
    \Delta A = \left ( \left [ \left ( 1 + (i + 1) \cdot \frac{4}{n} \right ) ^3 - \left ( 1 + (i + 1) \cdot \frac{4}{n} \right ) \right ] \right . \left . ~- \left [ \left ( 1 + i \cdot \frac{4}{n} \right ) ^3 - \left ( 1 + i \cdot \frac{4}{n} \right ) \right ] \right ) \cdot \frac{4}{n}

    So the estimate of the area will be
    \sum_{i = 0}^n \Delta A

    which is exact when taking the limit:
    A = \lim_{n \to \infty} \sum_{i = 0}^n \Delta A

    There's a lot of simplifying to do but the process is otherwise simple.

    -Dan
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