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Math Help - 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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  3. #3
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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 |x|<.5 in that case f(x) = 1 \mbox{ on }-.5<x<.5 and hence the anti-derivative is x on (-5,5) and a constant everywhere.

    So,
    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 f needs to be continous differenciable at x=\pm .5 as well. So  C_1 = C_2 - .5 and C_3 = C_2+.5 where 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:

    <br />
c_K(x) = \int_{-\infty}^x K^2(u) du<br />

    Then:


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

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