# The use of Fubini's theorem

1. Nov 24, 2005

### Bernoulli

Hi, I usually dont have any problems with Fubini's theorem, but there is something I just cant figure out. Let f be integrable, and a some positive constant. How do i apply the theorem to this integral:
$$\int_0^a\int_x^a \frac{1}{t}|f(t)|dtdx$$
Really; I know the answer is
$$\int_0^a \int_0^t \frac{1}{t}|f(t)|dxdt$$
but I just dont get it. To me this is not obvious (should it be?). Can someone explain this to me?

2. Nov 24, 2005

### Bernoulli

I guess I have made it more difficult then it really is. I just want to know why the second integration domain turns out like that. For simplicity put $$\frac{1}{t}|f(t)| = t$$ for example (and $$t\in (0,a)$$). Then the integration area becomes
$$\int_0^a\int_x^a t dtdx = \int_?^?\int_?^?t dxdt$$
Why?

3. Nov 24, 2005

### shmoe

Draw a picture in the x-t plane. Your region is bounded by the curves t=a, x=0 and x=t. For a fixed t, x ranges from 0 to t. t itself can range from 0 to a.

4. Nov 24, 2005

### Bernoulli

Oh, man... I must be tired :) The original problem was not posted like this. I did not realize that the 2D integration area was infact the upper triangle of the square [0,1]^2. I thought I was dealing with the hole square... Stupid me :)

Then of course it is easy.
Thanks