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

1. Christos

@zzr0ck3r u there?

2. anonymous

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

3. 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

4. zzr0ck3r

looks good

5. Christos

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

6. zzr0ck3r

what is the problem?

7. Christos

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

8. Christos

Just making sure I didnt miss anything lol

9. anonymous

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

10. zzr0ck3r

the solution looks fine

11. Christos

3^2 = 9*2 = 18

12. zzr0ck3r

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

13. zzr0ck3r

3^2 = 3*3 = 9

14. anonymous

three squared is not equal to nine squared

15. Christos

9*2(denominator) = 18

16. zzr0ck3r

18 = 2*3^2

17. Christos

Right

18. zzr0ck3r

I m confused:)

19. Christos

;p

20. anonymous

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

21. Christos

where did the 1/4 come from

22. Christos

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

23. Christos

:D

24. anonymous

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

25. Christos

aah!

26. Christos

hold on

27. Christos

That was the trick, thank you

28. anonymous

yw