I have a uniform probability distribution with density function $f(x,y)$ such that
$$\int_0^2 \int_0^2 f(x,y)dy dx = 1$$
Now I know that $f(x,y)=\frac{1}{4}$ simply by considering the dimensions of a cuboid of volume 1.
But is it possible to determine what $f(x,y)$ is analytically rather than geometrically?
Since $f(x,y)$ is a uniform probability distribution we know that it is should be independent of $x$ and $y$ on it's support. That is, there exists a constant $c \in \mathbb{R}$ such that $$f(x,y)=c,$$ for all $x\in[0,2], y \in [0,2]$. By substituting in the double integral we have that $$1=\int^{2}_{0}\int^{2}_{0}f(x,y)dydx=\int^{2}_{0}\int^{2}_{0}c\,dydx=\int^{2}_{0}[cy]_0^2dx=\int^{2}_{0}2c\,dx=[2cx]_0^2=4c,$$ From the above we deduce that $c=\frac{1}{4}$.