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Math Help - integration theorem!

  1. #1
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    integration theorem!

    i need some help or understanding or a book or material that explains this type of problem

    find the value of

     \int_{0.3}^{2.6} x^3 d(x + [x] )

    thanks
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  2. #2
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    Re: integration theorem!

    That is an example of a "Stieljes" integral. The original "Riemann" integral, \int_a^b f(x)dx, divides the x-axis into sub-intervals, [x_i, x_{i+1}], and then measures the "area" by f(x)\Delta x where \Delta x is just the length, x_{i+1}- x_i. The "Stieljes" integral, \int_a^b f(x)d\alpha(x), generalizes that by measuring each sub-interval by \alpha(x_{i+1})- \alpha(x_i) where \alpha(x) can be any increasing function of x. If \alpha(x) happens to be differentiable the Stieljes integral reduces to a Riemann integral: \int_a^b f(x)d\alpha= \int_a^b f(x)\alpha'(x)dx.

    But \alpha(x) does not have to be differentiable or even continuous. One application of the the Stieljes integral is to be able to write a sum as an integral: if \alpha(x)= \lfloor x\rfloor then \alpha(x_{i+1})- \alpha(x_i) is non-zero only for x_{i+1} and x_i to be on different sides of an integer: \int_0^5 f(x)d\lfloor x\rfloor= f(0)+ f(1)+ f(2)+ f(3)+ f(4)+ f(5).

    That, together with the fact that
    \int_{0.3}^{2.6} x^3d(x+ \lfloor x\rfloor)= \int_{0.3}^{2.6}x^3 dx+ \int_{0.3}^{2.6} x^3 d\lfloor x\rfloor
    should be enough for you to do this problem.

    (By the way, I have interpreted your "[x]" as the floor function. Latex has "lfloor" and "rfloor" to get \lfloor x\rfloor.
    Last edited by HallsofIvy; June 29th 2013 at 06:34 AM.
    Thanks from lawochekel, Shakarri and topsquark
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  3. #3
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    Re: integration theorem!

    do u carry out normal integration on

     \int_{0.3}^{2.6} x^3 dx

    thanks
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  4. #4
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    Re: integration theorem!

    Yes, of course.
    Thanks from lawochekel
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