Double integral to simple integral

aldrinkleys
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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
 
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hi aldrinkleys! :smile:

the obvious thing would be to change to coordinates one of which is u = x+y :wink:
 
I tried to it. But I don't know what to do after.
 
when you tried it, what was your other variable?
 
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 :(
 
aldrinkleys said:
… 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! :smile:
 
:!) I'm so happy!

Thank you very much!
 
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