Question about limits: int [x e^(-px)] dx

[mooshupork34]

Below I did out an integral (from 0 to infinity)

∫x * e^(-px) dx

u=x, dv=e^(-px), du=1, v=(-1/p)e^(-px)

x*(-1/p)e^(px) - ∫ (-1/p)e^(-px)

I integrated the second integral using parts:
u=-1/p, dv = e^(-px), du = 0, v=(-1/p)e^(-px)

This resulted in a value of (1/p^2)e^(-px)

So the value of ∫x * e^(-px) dx turns out to be:

x*(-1/p)e^(-px) minus (1/p^2)e^(-px)

This factors to (-1/p)e^(-px) (x - (1/p)).

This is my question. According to the back of my book, applying the limits is supposed to give 1/p. Yet I'm getting 1/p^2 here. Is there an error in my calculations?

 

[arthur ohlsten]

I also get 1/p^2, so either there is a error in the text ,or we are both wrong.

int 0-->oo xe^(-px) dx

let u=x
then du=dx
let dv= e^-(px) dx
then v= [-1/p] e^-(px]

int 0-->oo xe^-(px) dx = -[x/p]e^-(px) + [1/p] int e^-[px] dx
int 0-->oo xe^-[px]=[ -x/p]e^-px -[1/p^2]e^-[px] evaluated at oo and 0
int 0-->xe^-[px]=-0-0+0+1/p^2 answer

Arthur