The two dimensional random variable is uniformly distributed over the triangle $A(0,0), B(4,0), C(2,2)$
1) Calculate for all $(x,y) \in \mathbb{R}^²$ the respective value of the density function $f(x,y)$.
2) Calculate the probability $P(X \leq 3, Y \leq 1)$
for 1) I tried to use the formulas for the triangular distribution https://en.wikipedia.org/wiki/Triangular_distribution, however we are two dimensional here. In the "solution" it is stated, that one should solely calculate the area of the triangle to yield $f(x,y)$?! But I do not really understand this.
2) I know that $\int_{-\infty}^{1}\int_{-\infty}^{3}f(x,y) dx dy$, so I would need the result from (1) to calculate this.
The area inside the triangle is 4 so the joint density is the following
$$f_{XY}(x,y)=\frac{1}{4}\mathbb{1}_{(0;2)}(y)\mathbb{1}_{(y;y+4)}(x)$$
The knowlegde of this function in NOT necessary to answer to point 2)
In fact, to calculate the requested probability it is enough to look at the draw
And calculate the probabilty as the ratio of Purple area on the triangle area, say
$$\mathbb{P}[X \leq 3; Y \leq 1]=\frac{(\frac{1}{2}+2)}{4}=\frac{5}{8}$$
EDIT: answering to the latest comment of the O.P.
You can solve the same probability using the integral; of course you have always to integrate in the purlple area
$$\frac{1}{4}\int_0^1dy \int_y^3 dx=\frac{1}{4}\int_0^1[3-y]dy=\frac{5}{8}$$