# Change of variables problem

1. Nov 18, 2005

### jordanl122

let f be continuous on [0,1] and R be a triangular region with vertices (0,0), (1,0) and (0,1). Show:

the double integral over the region R of f(x+y)dxdy = the integral from 0 to 1 over u f(u)du

I recognize it is a change of variables problem but I'll be damned if I can create a set of functions u and v that yield the right jacobian. If anyone can help, its much appreciated.

2. Nov 18, 2005

### kleinwolf

I think it's obvious that u=x+y, v=x for example for v...this leads to |det(Jacob)|=1..then byconstruction of u and v, f(x+y)=f(u) which is independent of v, hence you can integrate over v (take care of the limit of integration, i was not able at first glance to be convinced of the u limit, but for v you get integral from 0 to 1 dv, which gives 1, and hence "disappears")...which hence lead an integral of the form integral from 0 to 1 over f(u) du...

3. Nov 19, 2005

### jordanl122

yeah that certainly would be a nice result if that were the answer to the problem but theree is a u out front and so a jacobian of 1 will not get the right results, but thanks anyway.

4. Nov 20, 2005

### kleinwolf

Oh, ok, then u could do something like : int{0,1}du[u f(u)]=F(1)-int{0,1}du[ F(u)] .where F(u)=int{0,u}ds[ f(s)]+C...by integration by parts,you have to adjust the integration constant c so that only the double integral remains....then use the first computation backward and get x+y as argument.........does it work like this better...?

Last edited: Nov 20, 2005