kirbykirby
  • kirbykirby
Integral! (related to gamma function maybe?) \[\int\limits_{0}^{X}\frac{\lambda^3}{2}t^2e^{-\lambda t}dt
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

kirbykirby
  • kirbykirby
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: \[=\frac{\lambda^3}{2}\int\limits_{0}^{X}t^2e^{-\lambda t}dt\] \[Let : u=\lambda t =>t=u/t\]\[So: du=\lambda dt\]\[Bounds: u=0=>t=0 ; u=x=>t=x/\lambda\] \[\frac{\lambda^2}{2}\int\limits_{0}^{X/\lambda}(\frac{u}{\lambda})^2e^{-u}du\]\[=\frac{1}{2}\int\limits_{0}^{X/\lambda}u^{3-1}e^{-u}du\]
anonymous
  • anonymous
why do you not just use tabular method
kirbykirby
  • kirbykirby
What's that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
it's a special type of by parts.. you can do by parts also
anonymous
  • anonymous
|dw:1358916012425:dw|
anonymous
  • anonymous
|dw:1358916166540:dw|
anonymous
  • anonymous
\[\int t^2e^{-yt}=\frac{-t^2e^{-yt}}{y}-\frac{2te^{-yt}}{y^2}-\frac{2e^{-yt}}{y^3}+c\]
kirbykirby
  • kirbykirby
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
look up tabular method for a derivation of this
kirbykirby
  • kirbykirby
Oh wow this is quite neat !! :) thank you so much
kirbykirby
  • kirbykirby
I can't believe they never taught this technique..

Looking for something else?

Not the answer you are looking for? Search for more explanations.