# need help! intergration

• Apr 18th 2008, 08:34 AM
winganger
need help! intergration
Y is a random variable that is exponentially distributed with parameter k=2.
Find E[Y].

Anyone know how to intergrate 2ye^(-2y) w.r.t. y from 0 to infinity? I have no idea on it.(Clapping)
• Apr 18th 2008, 08:38 AM
TheEmptySet
Quote:

Originally Posted by winganger
Y is a random variable that is exponentially distributed with parameter k=2.
Find E[Y].

Anyone know how to intergrate 2ye^(-2y) w.r.t. y from 0 to infinity? I have no idea on it.(Clapping)

by parts $\displaystyle u=y \mbox{ and }dv=e^{-2ydy}$

The final result should be

$\displaystyle 2 \int ye^{-2y}dy=2[\frac{-y}{2}e^{-2y}-\frac{1}{4}e^{-2y}]$

Good luck
• Apr 18th 2008, 08:56 AM
winganger
Quote:

Originally Posted by TheEmptySet
by parts $\displaystyle u=y \mbox{ and }dv=e^{-2ydy}$

The final result should be

$\displaystyle 2 \int ye^{-2y}dy=2[\frac{-y}{2}e^{-2y}-\frac{1}{4}e^{-2y}]$

Good luck

Thank you.
May I ask one more question, that is, what is the value when putting y=infinity into
2ye^(-2y) ?
• Apr 18th 2008, 06:24 PM
Mathstud28
Quote:

Originally Posted by winganger
Thank you.
May I ask one more question, that is, what is the value when putting y=infinity into
2ye^(-2y) ?

it is zero...since $\displaystyle \lim_{x \to 0}\frac{2x}{e^{2x}}=0$ you can acheive this limt due to L'hopitals rule or simple comparison of rates of increase