Question

integral of 1/(u^3) ?

Answer 1

@zzr0ck3r u there?

Answer 2

\[\large \int\limits \frac{ 1 }{ u^3 }du= \int\limits u^{-3}du\] you know how to integrate that right.

Answer 3

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

Answer 4

looks good

Answer 5

my result: (1/2)*((-1/18) + (1/72))

Answer 6

what is the problem?

Answer 7

http://screencast.com/t/XEJ9Dmj1S this solution displays a different result

Answer 8

Just making sure I didnt miss anything lol

Answer 9

where did the \(\frac{1}{18}\) come from ?

Answer 10

the solution looks fine

Answer 11

3^2 = 9*2 = 18

Answer 12

show us what you did, did u change your bounds when you substituted?

Answer 13

3^2 = 3*3 = 9

Answer 14

three squared is not equal to nine squared

Answer 15

9*2(denominator) = 18

Answer 16

18 = 2*3^2

Answer 17

Right

Answer 18

I m confused:)

Answer 19

;p

Answer 20

\[F(u)=-\frac{1}{4u^2}\]
\[F(3)-F(6)=-\frac{1}{4}\left(\frac{1}{9}-\frac{1}{36}\right)\]

Answer 21

where did the 1/4 come from

Answer 22

there might be a probability that my solution and the other one are equal idk

Answer 23

:D

Answer 24

the u - sub gave a factor of \(\frac{1}{2}\) and the anti derivative gave another \(-\frac{1}{2}\)

Answer 25

aah!

Answer 26

hold on

Answer 27

That was the trick, thank you

Answer 28

yw