Triangle with Fubini

  • Thread starter Faust90
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  • #1
20
0
Hi,

I should Show the following:


D is subset of R^2 with the triangle (0,0),(1,0),(0,1). g is steady.

Integral_D g(x+y) dL^2(x,y)=Integral_0^1 g(t)*t*dt

my ansatz:

Integral_0^1(Integral_0^(1-x) g(x+y) dy) dx

With Substitution t=x+y

Integral_0^1(Integral_x^1 g(t) dt) dx

But now i dont Have any ideas how to go on:-(
 

Answers and Replies

  • #2
611
24
Hi,

I should show the following:


[itex]D \subseteq R^2[/itex] with the triangle (0,0),(1,0),(0,1). g is steady.

[itex]\int_D g(x+y) dL^2(x,y)=\int_0^1 g(t)\ t \ dt[/itex]

My answer:

[itex]\int_0^1(\int_0^{(1-x)} g(x+y) dy) dx[/itex]

With Substitution t=x+y

[itex]\int_0^1(\int_x^1 g(t) dt) dx[/itex]

But now I don't have any ideas how to go on :frown:
I cleared up your post, hope you don't mind. Or, in terms of your username,

"'Twere better nothing would begin.
Thus everything that that your terms, sin,
Destruction, evil represent—
That is my proper element.”
-Goethe​

My love of good literature aside, there are two things you need to remember to continue. Firstly, you are evaluating the a double integral, so you need to evaluate the innermost one first. Secondly, g is steady. What might that imply about its integral?
 
  • #3
20
0
Hi,

thanks for your answer :) One of my favorite quotations.
I think I got it. If somebody want, i can write the solution here.

Greetings
 

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