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Math Help - Integrating a linear polynomial approx

  1. #1
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    Integrating a linear polynomial approx

    hi,

    thanks for reading.can anyone help me out?
    see the attached file.i am working out for many hours how to integrate P1(t). does anyone know?
    how do you get -1/2hf(..... ) + 3/2hf(....) ?
    thats all i need to know!
    i tried integrating MANY times. it looks easy.so what am i not understanding?

    thanks all
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  2. #2
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    Hi

    The first integral is

    \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}

    Substituting t_{i+1} = t_i + h

    \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    The first integral is

    \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}

    Substituting t_{i+1} = t_i + h

    \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}


    aha...! thats where confusion happened... h can be anything =S
    a difference between one point and another. thanks! =D
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