Here's the question you clicked on:
kirbykirby
Integral! (related to gamma function maybe?) \[\int\limits_{0}^{X}\frac{\lambda^3}{2}t^2e^{-\lambda t}dt
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
it's a special type of by parts.. you can do by parts also
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\[\int t^2e^{-yt}=\frac{-t^2e^{-yt}}{y}-\frac{2te^{-yt}}{y^2}-\frac{2e^{-yt}}{y^3}+c\]
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?
http://www.wolframalpha.com/input/?i=integral+of+t%5E2e%5E%7B-yt%7D 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..