Double integral to simple integral

[aldrinkleys]

Hello. Can anyone help me, please?

R = { (x,y) $$\in$$ R² | 0 $$\leq$$ x $$\leq$$ 1, 0 $$\leq$$ y$$\leq 1-x$$}

f is continuous at [0,1]

Show that

$$\iint_$$_R f(x+y) dxdy = $$\int_{[0,1]}$$ u f(u) du

 

[tiny-tim]

hi aldrinkleys!

the obvious thing would be to change to coordinates one of which is u = x+y

 

[aldrinkleys]

I tried to it. But I don't know what to do after.

 

[tiny-tim]

when you tried it, what was your other variable?

 

[aldrinkleys]

What I did:

u = x+y
so
x = u-y
y = u-x

x $$\geq$$ 0
y $$\geq$$ 0
y $$\leq$$ 1-x

u-y $$\geq$$ 0 $$\rightarrow$$ u $$\geq$$ y
u-x $$\geq$$ 0 $$\rightarrow$$ u $$\geq$$ x
u <= 1

x $$\in$$ [0,1]
y $$\in$$ [0,1]

So

u $$\in$$ [0,1]

And I don't know what to do about the integral and the Jacobian, etc :(

what to you think about?

Ps: OMG, tex isn't working :(

 

[tiny-tim]

… and i don't know what to do abou the integral and the jacobian, etc :(

you need two variables:

x+y and x

or x+y and y

x+y and x-y …

make a choice, then find the new limits and the Jacobian! ​

 

[aldrinkleys]

:!) I'm so happy!

Thank you very much!