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