Integral from my applied statistics class

**oldguynewstudent** · June 27th 2010, 09:32 AM

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.

**Plato** · June 27th 2010, 09:51 AM

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.

**oldguynewstudent** · June 27th 2010, 12:06 PM

Originally Posted by Plato

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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