Double Integral of f(y) over region in xy plane I was given that $\int_{0}^{1} \int_{x}^{1-x}f(y)dydx$ is equal to a specific value for all integratable functions $f(y)$, I was able to work out that the value is $0$, but when I try to visualise this I just get confused. Is it possible I'm missing something? Can someone provide a nice visual explanation?
 A: 
So, I've drawn the region of integration in the $x-y$ plane. So, the bounds of the inner integral $ x$ and $ 1-x$ denote these two lines in the $x-y$ plane. If we were to imagine $f(y)$ as some density function (eg: mass density) then integrating it from $x$ to $ 1-x$ can be thought of as finding the mass of a vertical strip between the lines $ y=x $ and $ y=1-x$.
The mass of the strip varies vertically as a function of $y$ , that is the curve connects the black image connected to the green inner integral in my picture. We can think of the mass of the strip varying as we move up vertically along our vertical stirp.
Once we do the inner integral, we get this expression as @Ninad Munshi has shown:
$$ \int_{0}^1 g(1-x) - g(x) dx$$
So, this expression we can think of as finding the mass of vertical strip from the $y=1-x$ line to the x-axis, and chopping of the mass from the line $y=x$ to the x-axis. The outer integral can be thought of as adding up the vertical strips horizontally, to get the mass of bow-tie as Ninad put it.
$$ \int_0^1 g(1-x) dx - \int_0^1 g(x) dx$$
For the first integral, we can do a trick. Instead of adding up the mass of the vertical bar from left to right along the x-axis, we can add them from the right to left ( starting from the end). This gives:
$$ \int_0^1 g(x) dx - \int_0^1 g(x) dx$$
But now we see that the mass of the region is zero.. the mass between $x$ axis to the line $ y=x $ is negative to that of the mass between the line $ y=1-x$ to the line $ y=x$
A: If $f(y)$ is integrable, then it is the derivative of some other function $g(y)$ s.t. $g'(y) = f(y)$. This gives us
$$\int_0^1 \int_x^{1-x} g'(y)\:dydx = \int_0^1 g(1-x)-g(x)\:dx$$
Then split apart the integrals and use the variable interchange $x\leftrightarrow 1-x$ on the first integral
$$\int_0^1g(1-x)dx-\int_0^1g(x)dx = \int_0^1g(x)dx-\int_0^1g(x)dx = 0$$
