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  • Integral Question

    From example 7-8 of the Actex 2007 manual, you have to find the integral from 0 to 50 of k*(x^5)((100-2x)^6)dx. The manual uses the beta distribution (to find the mean of a pdf of X that is proportional to x^4(100-2x)^6 on the interval 0<x<50). Isn't there a way to do this without using the beta distribution and without having to multiply out (100 - 2x)^6?

    Thanks!!

  • #2
    Originally posted by mmmmgodiva View Post
    From example 7-8 of the Actex 2007 manual, you have to find the integral from 0 to 50 of k*(x^5)((100-2x)^6)dx. The manual uses the beta distribution (to find the mean of a pdf of X that is proportional to x^4(100-2x)^6 on the interval 0<x<50). Isn't there a way to do this without using the beta distribution and without having to multiply out (100 - 2x)^6?

    Thanks!!
    Sure. You can say u = 100-2x therefore x = (100-u)/2, then du/2 = dx and now you have integral of [(100-u)/2]^5 * u^6 * 1/2 du and now you have to expand out a 5th degree polynomial instead of a 6th degree one

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    • #3
      Is that really the best way to do it? Do you think I need to memorize the beta distribution for the exam?

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      • #4
        From what I heard, most if not all exotic distributions have been removed from 1/P. It does show up on 4/C but you are given a distribution table. In reality, it shouldn't take 6 minutes to expand and integrate that problem. Your Calculus abilities should be well beyond that.

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