# Thread: Integral from my applied statistics class

1. ## 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.

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

The original was easy$\displaystyle k\intop_{0}^{\infty}(1+\frac{x}{2.5})^{-7}dx=\frac{-5k}{12}(1+\frac{x}{2.5})^{-6}\mid_{0}^{\infty}=\frac{5k}{12}$ 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.

2. Let $\displaystyle u=1+\frac{x}{2.5}$ then $\displaystyle 2.5du=dx$.
So we have $\displaystyle (2.5u-2.5)(u^{-7})(2.5du)$ from $\displaystyle u=1$ to infinity.

3. Originally Posted by Plato
Let $\displaystyle u=1+\frac{x}{2.5}$ then $\displaystyle 2.5du=dx$.
So we have $\displaystyle (2.5u-2.5)(u^{-7})(2.5du)$ from $\displaystyle u=1$ to infinity.
So for the variance where V(X)=$\displaystyle (12/5)\intop_{0}^{\infty}x^2(1+\frac{x}{2.5})^{-7}dx$

Will the following work?

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

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