Integration ?e^(-?y)
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BigGlenntheHeavy
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\(\displaystyle f(x) \ = \ \int_{0}^{\infty} y \lambda e^{-\lambda y} \ = \ \lim_{b\to\infty}\int_{0}^{b}\frac{y \lambda}{e^{\lambda y}}, \ \lambda \ > \ 0\)
\(\displaystyle Let \ u \ = \ y\lambda, \ then \ du \ = \ \lambda dy\)
\(\displaystyle Hence, \ \lim_{b\to\infty}\frac{1}{\lambda}\int_{0}^{b}\frac{u}{e^{u}}du \ = \ \lim_{b\to\infty}\frac{1}{\lambda}\bigg[\frac{-u-1}{e^{u}}\bigg]_{0}^{b}\)
\(\displaystyle = \ \lim_{b\to\infty}\bigg[\frac{-b}{\lambda e^{b}}-\frac{0}{\lambda}-\frac{1}{\lambda e^{b}}+\frac{1}{\lambda}\bigg]\)
\(\displaystyle = \ \lim_{b\to\infty} \bigg[\frac{-1}{\lambda e^{b}}-0-0+\frac{1}{\lambda}\bigg] \ = \ \frac{1}{\lambda}\)
\(\displaystyle Let \ u \ = \ y\lambda, \ then \ du \ = \ \lambda dy\)
\(\displaystyle Hence, \ \lim_{b\to\infty}\frac{1}{\lambda}\int_{0}^{b}\frac{u}{e^{u}}du \ = \ \lim_{b\to\infty}\frac{1}{\lambda}\bigg[\frac{-u-1}{e^{u}}\bigg]_{0}^{b}\)
\(\displaystyle = \ \lim_{b\to\infty}\bigg[\frac{-b}{\lambda e^{b}}-\frac{0}{\lambda}-\frac{1}{\lambda e^{b}}+\frac{1}{\lambda}\bigg]\)
\(\displaystyle = \ \lim_{b\to\infty} \bigg[\frac{-1}{\lambda e^{b}}-0-0+\frac{1}{\lambda}\bigg] \ = \ \frac{1}{\lambda}\)
BigGlenntheHeavy
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\(\displaystyle After \ looking \ over \ your \ thread, \ I \ am \ assuming \ (knowing \ a \ mere \ assumption \ is \ no \ proof) \ that\)
\(\displaystyle you \ meant, \ to \ wit: \ \int_{0}^{\infty}\lambda e^{-\lambda y}dy. \ If \ I'm \ correct, \ we \ will \ proceed \ as \ follows:\)
\(\displaystyle \int_{0}^{\infty}\lambda e^{-\lambda y}dy \ = \ \lim_{b\to\infty}\lambda \int_{0}^{b}e^{-\lambda y}dy, \ \lambda \ > \ 0.\)
\(\displaystyle = \ \lambda \lim_{b\to\infty}-\frac{e^{-\lambda y}}{\lambda}\bigg]_{0}^{b}\)
\(\displaystyle = \ \lim_{b\to\infty}\frac{-1}{e^{\lambda y}}\bigg]_{0}^{b} \ = \ \lim_{b\to\infty}\bigg[\frac{-1}{e^{\lambda b}}-\frac{-1}{1}\bigg] \ = \ 1\)
\(\displaystyle you \ meant, \ to \ wit: \ \int_{0}^{\infty}\lambda e^{-\lambda y}dy. \ If \ I'm \ correct, \ we \ will \ proceed \ as \ follows:\)
\(\displaystyle \int_{0}^{\infty}\lambda e^{-\lambda y}dy \ = \ \lim_{b\to\infty}\lambda \int_{0}^{b}e^{-\lambda y}dy, \ \lambda \ > \ 0.\)
\(\displaystyle = \ \lambda \lim_{b\to\infty}-\frac{e^{-\lambda y}}{\lambda}\bigg]_{0}^{b}\)
\(\displaystyle = \ \lim_{b\to\infty}\frac{-1}{e^{\lambda y}}\bigg]_{0}^{b} \ = \ \lim_{b\to\infty}\bigg[\frac{-1}{e^{\lambda b}}-\frac{-1}{1}\bigg] \ = \ 1\)
, thanks for all your help!