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Thread: Question about Integral

  1. #1
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    Question about Integral

    Please see the attached file !!!!

    Thanks a lot!
    Attached Files Attached Files
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  2. #2
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    Quote Originally Posted by owenji81 View Post
    Please see the attached file !!!!

    Thanks a lot!
    Are those definite or indefinite integrals?

    RonL
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    indefinite integrals
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  4. #4
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    The problem is we cannot find anti-derivatives on closed intervals, we require that $\displaystyle |x|<.5$ in that case $\displaystyle f(x) = 1 \mbox{ on }-.5<x<.5$ and hence the anti-derivative is $\displaystyle x$ on $\displaystyle (-5,5)$ and a constant everywhere.

    So,
    $\displaystyle f(x) = \left\{ \begin{array}{c}C_1 \mbox{ for }x<-.5\\ x+C_2 \mbox{ for }-.5<x<.5 \\ C_3 \mbox{ for }x>.5 \end{array} \right\}$

    But $\displaystyle f$ needs to be continous differenciable at $\displaystyle x=\pm .5$ as well. So $\displaystyle C_1 = C_2 - .5$ and $\displaystyle C_3 = C_2+.5$ where $\displaystyle C_2$ is any arbitraty konstant.
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  5. #5
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    Quote Originally Posted by owenji81 View Post
    Please see the attached file !!!!

    Thanks a lot!
    Well I still don't like it so I will do the following instead:

    $\displaystyle
    c_K(x) = \int_{-\infty}^x K^2(u) du
    $

    Then:


    $\displaystyle c_K(x) = \left\{ \begin{array}{cc}
    0 &\mbox{ for }x<-0.5 \\
    x + 0.5& \mbox{ for }-0.5\le x \le 0.5 \\
    1& \mbox{ for }x>0.5 \end{array}\right.
    $

    and if you add an arbitary constant onto this $\displaystyle c_K(x)$ you will still have an anti-derivative of $\displaystyle K(x)$.

    If everything has worked out right with the extra arbitary constant this should be the same as the soution given by ImPerfectHacker.

    RonL
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