Factorising Integration

1. Apr 12, 2007

r_maths

2. Apr 12, 2007

Integral

Staff Emeritus
Just do the substitution.

in part b, let $k^3 = \rho ^2$

Solve the resulting quadratic.

3. Apr 12, 2007

r_maths

??? :S

4. Apr 12, 2007

Mystic998

k^3 = p^2 implies k^(3/2) = p.

5. Apr 12, 2007

r_maths

I see now, so silly. Thanks

6. Apr 13, 2007

r_maths

Still cannot get answer.

7. Apr 13, 2007

dextercioby

You've gotten "p", now you have to find "k".

8. Apr 13, 2007

r_maths

How is that going to lead me to the answer?
If I substitute "p" back to "k". I don't see how I can get 0.44

9. Apr 13, 2007

Mystic998

What is an expression for k in terms of p?

And you'll get an extraneous solution, but which one is right will be fairly clear when you get them. Look at bounds of your integration. Then look at the k's you get.

10. Apr 13, 2007

r_maths

11. Apr 13, 2007

HallsofIvy

Staff Emeritus
Do you see an error here?

12. Apr 13, 2007

r_maths

I hope...

I don't quite understand, how i'm I to know 0.29 instead of 1.71 was to be used?

Last edited: Apr 13, 2007
13. Apr 13, 2007

Mystic998

What is 1.79^(2/3)? Why is it that k clearly can't take on that value?

14. Apr 13, 2007

r_maths

Because of: ?

15. Apr 13, 2007

Mystic998

You've pretty much got the right idea. When you split the integral from 0 to 1 into an integral from 0 to k and an integral from k to 1, you tacitly assumed that k was between 0 and 1. Otherwise, what you did wouldn't be valid.

16. Apr 13, 2007

r_maths

Ok, thanks Mystic998 and everyone else who helped out.