Problem: Definite integral by subtitution

[don.bandar]

Hi everyone, to day I have a problem in dealing with the following question:

and I hope that you help me

here is may question and my try:
[PLAIN]http://img268.imageshack.us/img268/9291/scan0004ov.jpg

wish you all the best

 

[fzero]

If u = 1 - r^3, then

r = (1-u)^{1/3} = \sqrt[3]{1-u}.

Your formula for du is actually correct, so why don't you just complete making the substitution and see if you can do the resulting integral?

 

[don.bandar]

hi dear,
i know that i can do the subtitution.But i want to know how -3du came?

if you could

 

[fzero]

hi dear,
i know that i can do the subtitution.But i want to know how -3du came?

if you could

I'm not sure what the question is. You explicitly computed that

9r^2 dr = - 3 du.

The only thing left to do is finish the computation.

 

[don.bandar]

Evluating Integrals:

evaluate the indefinite integral by using the given substitution to reduce the integral to standard form.

7. [PLAIN]http://img232.imageshack.us/img232/8840/kmnk.jpg


this is the question

i want to know how did -3du come?!

i'm sorry for asking too many questions

 

[Dickfore]

Try the subst:

<br /> u = 1 - r^{3}<br />

 

[fzero]

No need to repeat the statement of the problem. Can you try rewriting

\frac{9r^2 dr}{\sqrt{1-r^3}}

entirely in terms of u and du?