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Math Help - integration by part theorem!

  1. #1
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    integration by part theorem!

    i need some clarification about the problem below;

     \int_{-1.2}^{3.9} xd[x]

    the only way i understand to solve the above problem is to apply integration by part

    that is

     \int_{-1.2}^{3.9} xd[x] + \int_{-1.2}^{3.9} |x|dx = f(b)g(a) - f(a)g(a)

    do i just substitute a as -1.2 and b as 3.9 and that's all?

    pls help here

    and again how do one disband modulus sign from integration function? asking for general purpose
    thanks
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  2. #2
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    Re: integration by part theorem!

    What do you mean by d[x]? Is that the "greatest integer function" or "floor function", \lfloor x\rfloor? If so that is a "Stieljes" integral, not a "Riemann" integral. \lfloor x \rfloor changes value only at integer values so the Stieljes integral is just the sum of the function being integrated at -1, 0, 1, 2, and 3. Since here the function is just "x", that would be -1+ 0+ 1+ 2+ 3= 5.

    I have no idea why you think you should just add a new integral to it! That certainly has nothing to do with "integration by parts".
    Thanks from lawochekel
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  3. #3
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    Re: integration by part theorem!

    Quote Originally Posted by lawochekel View Post
    i need some clarification about the problem below;

     \int_{-1.2}^{3.9} xd[x]

    the only way i understand to solve the above problem is to apply integration by part
    I think that you have misunderstood the parts rule.
    See this source.

    Also have a look at this.
    Last edited by Plato; July 2nd 2013 at 10:12 AM.
    Thanks from lawochekel
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  4. #4
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    Re: integration by part theorem!

    i want to be sure if i am improving on this, pls check the problem below i attempted

     \int_{-0.3}^{4.5} x^2d([x]) = \int_{-0.3}^{0} x^2d([x]) + \int_{0}^{2} x^2d([x]) + \int_{2}^{4} x^2d([x]) + \int_{4}^{4.5} x^2d([x])

     =[0-(-1)]0^2+[2-0]2^2+[4-2]4^2+[4-4](4.5)^2 = 0+8+32++0 = 40



    thanks
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