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Christos
 2 years ago
integral of 1/(u^3) ?
Christos
 2 years ago
integral of 1/(u^3) ?

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large \int\limits \frac{ 1 }{ u^3 }du= \int\limits u^{3}du\] you know how to integrate that right.

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0to tell you the truth I already knew how to do it, I just think this exersise is wrong so I need to verify, look at the bottom http://screencast.com/t/XEJ9Dmj1S

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0my result: (1/2)*((1/18) + (1/72))

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0http://screencast.com/t/XEJ9Dmj1S this solution displays a different result

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0Just making sure I didnt miss anything lol

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0where did the \(\frac{1}{18}\) come from ?

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0the solution looks fine

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0show us what you did, did u change your bounds when you substituted?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0three squared is not equal to nine squared

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\[F(u)=\frac{1}{4u^2}\] \[F(3)F(6)=\frac{1}{4}\left(\frac{1}{9}\frac{1}{36}\right)\]

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0where did the 1/4 come from

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0there might be a probability that my solution and the other one are equal idk

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0the u  sub gave a factor of \(\frac{1}{2}\) and the anti derivative gave another \(\frac{1}{2}\)

Christos
 2 years ago
Best ResponseYou've already chosen the best response.0That was the trick, thank you
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