- #1
ehrenfest
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Homework Statement
Can someone explain to me the logic of this statement:
"Since the value of f (x, y) is unchanged when we swap x with y,
[tex]\int_0^1 \int_0^x f (x+y)dydx = 1/2 \int_0^1 \int_0^1 f (x+y)dydx.[/tex]"
Homework Equations
The Attempt at a Solution
[tex]\int_0^1 \int_0^x f (x+y)dydx = \int_0^1 \int_0^y f (y+x)dxdy[/tex]
But I do not think that is the same.