Integration ?e^(-?y)

[rbcc]

Hi,

I'm having trouble with this, can some one step me through it?

integrate from 0 to infinity y?e^(-?y) dy

thanks!

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \int_{0}^{\infty} y \lambda e^{-\lambda y} \ = \ \lim_{b\to\infty}\int_{0}^{b}\frac{y \lambda}{e^{\lambda y}}, \ \lambda \ > \ 0$

$\displaystyle Let \ u \ = \ y\lambda, \ then \ du \ = \ \lambda dy$

$\displaystyle Hence, \ \lim_{b\to\infty}\frac{1}{\lambda}\int_{0}^{b}\frac{u}{e^{u}}du \ = \ \lim_{b\to\infty}\frac{1}{\lambda}\bigg[\frac{-u-1}{e^{u}}\bigg]_{0}^{b}$

$\displaystyle = \ \lim_{b\to\infty}\bigg[\frac{-b}{\lambda e^{b}}-\frac{0}{\lambda}-\frac{1}{\lambda e^{b}}+\frac{1}{\lambda}\bigg]$

$\displaystyle = \ \lim_{b\to\infty} \bigg[\frac{-1}{\lambda e^{b}}-0-0+\frac{1}{\lambda}\bigg] \ = \ \frac{1}{\lambda}$

 

[BigGlenntheHeavy]

$\displaystyle After \ looking \ over \ your \ thread, \ I \ am \ assuming \ (knowing \ a \ mere \ assumption \ is \ no \ proof) \ that$

$\displaystyle you \ meant, \ to \ wit: \ \int_{0}^{\infty}\lambda e^{-\lambda y}dy. \ If \ I'm \ correct, \ we \ will \ proceed \ as \ follows:$

$\displaystyle \int_{0}^{\infty}\lambda e^{-\lambda y}dy \ = \ \lim_{b\to\infty}\lambda \int_{0}^{b}e^{-\lambda y}dy, \ \lambda \ > \ 0.$

$\displaystyle = \ \lambda \lim_{b\to\infty}-\frac{e^{-\lambda y}}{\lambda}\bigg]_{0}^{b}$

$\displaystyle = \ \lim_{b\to\infty}\frac{-1}{e^{\lambda y}}\bigg]_{0}^{b} \ = \ \lim_{b\to\infty}\bigg[\frac{-1}{e^{\lambda b}}-\frac{-1}{1}\bigg] \ = \ 1$

 

[rbcc]

I actually meant the first one (i made a mistake in the title) , thanks for all your help!