# Transformation of Bivariate Random variables

I am having trouble with this question and feel like my supports are not entirely correct:

Given $$f_u(u)= 1_{[1,0]}(u)$$ and $$f_v(v)= 2v*1_{[1,0]}$$ with U and V being independent and $$U,V \in \mathbb{R}$$

with the Transformations $$X = U, Y = U-V$$

What is the density of $$f_{X,Y}(x,y)$$ AND what is the median of Y $$f_Y(y$$)

I have the joint density of $$f_{U,V}(u,v)= 2v*1_{{[1,0]^2}}(u,v)$$

then with the transformations of $$h_1(u,v)=u=x$$ and $$h_2(u,v)=u-v=y$$ with inverses of $$h_1^{-1}(x,y)=x=u$$ and $$h_2^{-1}(x,y)=x-y=v$$ & a Jacobian of 1

with supports $$\{(x,y)\in\mathbb{R}:x\in[0,1],y\in[x-1,x]\}$$

which gives me a Joint PDF for X,Y of $$f_{X,Y}(x,y)= 2(x-y)*1_{{[1,0]^2}}(x,y)$$

To get the marginal PDF of Y, I have integrated $$\int^1_0 f_{X,Y}(x,y)$$ to get $$f_Y(y)=1-2y$$ which seems incorrect, then further, attempting to find the Distribution Function of Y, I don't know if I should be integrating $$\int^x_{x-1}f_Y(y)$$