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Math Help - Integral inequality.

  1. #1
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    Integral inequality.

    Hi,

    I hope this is valid:
    Why does

    \int_a^b s_n \le x_1 \le \int_a^b t_n and \int_a^b s_n \le x_2 \le \int_a^b t_n

    multiply by -1 and add the result to the second give

    0 \le x_2-x_1 \le \int_a^b t_n - \int_a^b s_n

    and not

    \int_a^b s_n - \int_a^b t_n \le x_2-x_1 \le \int_a^b t_n - \int_a^b s_n

    Thanks
    Regards
    Craig.
    Last edited by craigmain; June 23rd 2009 at 09:27 AM. Reason: more complete.
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  2. #2
    Moo
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    Hello,

    Without further information, it's not possible to know that x_2-x_1\geq 0
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  3. #3
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    Hi,

    I thought as much. This is from Apostol. I am sure that I have missed something. for example, n \ge 1, and these integrals are the upper and lower step function bounds. It's part of the integral definition.

    I suppose because the upper step function is always above the lower step function that the difference must be more than zero.

    Maybe I need to read it more carefully.
    Last edited by craigmain; June 23rd 2009 at 09:59 PM. Reason: clearer
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