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Integral! (related to gamma function maybe?) \[\int\limits_{0}^{X}\frac{\lambda^3}{2}t^2e^{-\lambda t}dt

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Here it is: \[\int\limits_{0}^{X}\frac{\lambda^3}{2}t^2e^{-\lambda t}dt\] I tried doing this trying to relate it to a gamma function but myupper bound is not infinity :S: \[=\frac{\lambda^3}{2}\int\limits_{0}^{X}t^2e^{-\lambda t}dt\] \[Let : u=\lambda t =>t=u/t\]\[So: du=\lambda dt\]\[Bounds: u=0=>t=0 ; u=x=>t=x/\lambda\] \[\frac{\lambda^2}{2}\int\limits_{0}^{X/\lambda}(\frac{u}{\lambda})^2e^{-u}du\]\[=\frac{1}{2}\int\limits_{0}^{X/\lambda}u^{3-1}e^{-u}du\]
why do you not just use tabular method
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Other answers:

it's a special type of by parts.. you can do by parts also
\[\int t^2e^{-yt}=\frac{-t^2e^{-yt}}{y}-\frac{2te^{-yt}}{y^2}-\frac{2e^{-yt}}{y^3}+c\]
Hmm interesting o_o!! As I was trying to do integration by parts, I realized I have to do it twice. Is this method related to doing int. by parts multiple times? same answer as mine only simplified and yes Tabular method is when you have something of this sort \[\int t^nf(t)dt\] you take the derivatives of the t to thepower until you get to zero giving them alternating signs and integrate the f(t) one more time than how many it takes the t power function goes to zero
look up tabular method for a derivation of this
Oh wow this is quite neat !! :) thank you so much
I can't believe they never taught this technique..

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