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Math Help - Integral from my applied statistics class

  1. #1
    Member oldguynewstudent's Avatar
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    Integral from my applied statistics class

    I am brain dead at this point and I'm not sure how to do this integral. Any help would be appreciated.

    E(X)= (12/5)\intop_{0}^{\infty}x(1+\frac{x}{2.5})^{-7}dx

    The original was easy  k\intop_{0}^{\infty}(1+\frac{x}{2.5})^{-7}dx=\frac{-5k}{12}(1+\frac{x}{2.5})^{-6}\mid_{0}^{\infty}=\frac{5k}{12} This integral must equal 1 so k=12/5.

    Integrate by parts? Partial fractions? I just can't think how to approach this expected value problem.
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    Let u=1+\frac{x}{2.5} then 2.5du=dx.
    So we have (2.5u-2.5)(u^{-7})(2.5du) from u=1 to infinity.
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    Let u=1+\frac{x}{2.5} then 2.5du=dx.
    So we have (2.5u-2.5)(u^{-7})(2.5du) from u=1 to infinity.
    So for the variance where V(X)= (12/5)\intop_{0}^{\infty}x^2(1+\frac{x}{2.5})^{-7}dx

    Will the following work?

    Let u=1+\frac{x}{2.5}

    Then integrate ((2.5)^2u^2-(25/2)u+(25/4))(u^{-7})(2.5du) from u=1 to infinity.
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