# Factorising Integration

1. Apr 12, 2007

### r_maths

[Broken]

[Broken]

Last edited by a moderator: Apr 22, 2017 at 4:57 PM
2. Apr 12, 2007

### Integral

Staff Emeritus
Just do the substitution.

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

3. Apr 12, 2007

### r_maths

[Broken]

??? :S

Last edited by a moderator: Apr 22, 2017 at 4:57 PM
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

[Broken]

Last edited by a moderator: Apr 22, 2017 at 4:57 PM
7. Apr 13, 2007

### dextercioby

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

8. Apr 13, 2007

### r_maths

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

[Broken]

Last edited by a moderator: Apr 22, 2017 at 4:57 PM
11. Apr 13, 2007

### HallsofIvy

Staff Emeritus
Do you see an error here?

12. Apr 13, 2007

### r_maths

[Broken]

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 by a moderator: Apr 22, 2017 at 4:57 PM
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: ?[/PLAIN] [Broken]

Last edited by a moderator: Apr 22, 2017 at 4:57 PM
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.