# Factorising Integration

http://img62.imageshack.us/img62/1393/graph005tm7.png [Broken]

http://img49.imageshack.us/img49/2026/graph006td1.png [Broken]

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Integral
Staff Emeritus
Gold Member
Just do the substitution.

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

http://img267.imageshack.us/img267/7201/graph007pp4.png [Broken]

??? :S

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

I see now, so silly. Thanks

http://img142.imageshack.us/img142/7294/graph008qz9.png [Broken]

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

If I substitute "p" back to "k". I don't see how I can get 0.44

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.

http://img143.imageshack.us/img143/5615/graph009um6.png [Broken]

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HallsofIvy
Homework Helper
r maths said:
$$k=p^{\frac{3}{2}}$$
so
$$p= k^{\frac{3}{2}}$$
Do you see an error here?

http://img124.imageshack.us/img124/2865/graph011ux7.png [Broken]

I hope...

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

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

Because of: http://img66.imageshack.us/img66/6338/graph012xj9.png [Broken]?[/URL]

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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.

Ok, thanks Mystic998 and everyone else who helped out.