Christos
  • Christos
integral of 1/(u^3) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Christos
  • Christos
@zzr0ck3r u there?
Jhannybean
  • Jhannybean
\[\large \int\limits \frac{ 1 }{ u^3 }du= \int\limits u^{-3}du\] you know how to integrate that right.
Christos
  • Christos
to 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

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More answers

zzr0ck3r
  • zzr0ck3r
looks good
Christos
  • Christos
my result: (1/2)*((-1/18) + (1/72))
zzr0ck3r
  • zzr0ck3r
what is the problem?
Christos
  • Christos
http://screencast.com/t/XEJ9Dmj1S this solution displays a different result
Christos
  • Christos
Just making sure I didnt miss anything lol
anonymous
  • anonymous
where did the \(\frac{1}{18}\) come from ?
zzr0ck3r
  • zzr0ck3r
the solution looks fine
Christos
  • Christos
3^2 = 9*2 = 18
zzr0ck3r
  • zzr0ck3r
show us what you did, did u change your bounds when you substituted?
zzr0ck3r
  • zzr0ck3r
3^2 = 3*3 = 9
anonymous
  • anonymous
three squared is not equal to nine squared
Christos
  • Christos
9*2(denominator) = 18
zzr0ck3r
  • zzr0ck3r
18 = 2*3^2
Christos
  • Christos
Right
zzr0ck3r
  • zzr0ck3r
I m confused:)
Christos
  • Christos
;p
anonymous
  • anonymous
\[F(u)=-\frac{1}{4u^2}\] \[F(3)-F(6)=-\frac{1}{4}\left(\frac{1}{9}-\frac{1}{36}\right)\]
Christos
  • Christos
where did the 1/4 come from
Christos
  • Christos
there might be a probability that my solution and the other one are equal idk
Christos
  • Christos
:D
anonymous
  • anonymous
the u - sub gave a factor of \(\frac{1}{2}\) and the anti derivative gave another \(-\frac{1}{2}\)
Christos
  • Christos
aah!
Christos
  • Christos
hold on
Christos
  • Christos
That was the trick, thank you
anonymous
  • anonymous
yw

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