Change of Variables

KellyK said:
integral (2y-y^2) dy dx with dy from 0 to x and dx from 0 to 1
Click to expand...

\(\displaystyle \displaystyle{\int_0^1 \underbrace{\int_0^x (2y-y^2)dy}_{\text{do this integral first}} dx = \int_0^1 (x^2-\frac13 x^3) dx} \)

I'm very interested to know why and where you had difficulties to do this integral.
 
This is your integral?:

\(\displaystyle \displaystyle \int_{0}^{1}\int_{0}^{x}(2y-y^{2})^{\frac{2}{3}}dydx\)

If so, this is not easily done using elementary methods. In other words, it takes more advanced methods to evaluate

this.

Are you supposed to use a Jacobian?.
 
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Probably. Its been so long since I have done these integrals I have forgotten how. I was under the impression it could be done with a change of variables if I could find the right one.
 
Just so you know. It evaluates to 3/10

At least now you know what you're shooting for.
 
willmoore21 said:
This is a stab in the dark, but can you log it?
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actually, I spent I while on this,

you can do this:

Reverse the order of integration. It becomes
int[0,1]int[y,1]{[2y-y^2]^(2/3)}dxdy=int[0,1]{1-y}{[2y-y^2]^(2/3)}dy. Substitute u=2y-y^2 and the answer is 3/10.
 
when you change the order of integration how do you know to integrate x from y to 1?
 
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