Marginal density and conditional density from joint density I am having trouble understanding how to solve this when the variables are not discrete.
Let the simultaneous density of the non-discrete stochastic variables (X,Y) be

I am then supposed to find marginal densities g(x) and h(y), conditional density f(y|x) and P(Y > 1|X=1/2)
I understand i should integrate the joint probability density function, but with what boundaries?
 A: The marginal pdf will be calculated over the area defined by a triangle as mentioned in the comments. The reason for it lies in the boundary constraints $0 < x < y < 2$, where the bivariate joint pdf is defined. To see this, we can mentally slide along the $x$ axis from $0$ to $2$, and see how at any given point, the $y$ axis will be past the bisecting line ($y = x$) on the $xy$ plane by the inequality $y > x$, and with a maximum of $2.$ 
I tried illustrating this with the following plot, which looks at the $z= \frac{1}{2}\,x\,y$ surface slightly from the top and front. The area we are going to integrate over when obtaining the marginal pdf's will be the light blue (crayon) triangle, which will cut through the overhead surface "in front of" the yellow plane:

And here is the view of the region where the pdf is defined (in blue) seen from above with the $z = 1/2\,x\,y$ surface in gray:

Therefore the marginal pdf of $Y$ will be found by integrating the joint pdf, $f(x,y)$ over the support of $Y$, corresponding to the triangle in the plot:
$f_Y(y) =\displaystyle\int_{x\,=\,0}^{x\,=\,y} f(x,y)\, dx = \displaystyle\int_0^y \frac{1}{2}\,x\,y\, dx = \frac{x^2}{4}\,y\,\Big|_{x\,=\,0}^{x\,=\,y}=\frac{y^3}{4}$, for $0<y<2.$
And the marginal of $X$ will be obtained by integrating away the $y$, keeping in mind, the the lower boundary of $Y$ is $X$:
$f_X(x) =\displaystyle\int_{y\,=\,x}^{y\,=\,2} f(x,y)\, dy = \displaystyle\int_x^2 \frac{1}{2}\,x\,y\, dy =x\, \frac{y^2}{4}\,\Big|_{y\,=\,x}^{y\,=\,2}=x-\frac{x^3}{4}$, for $0 < x < 2.$
The conditional pdf of $Y$ given $X=x$, $f_{Y|X}$ is given by 
$\large \frac{f(x,y)}{f_X(x)}=\frac{2\,y}{4-x^2}$. 
We know that the marginal of $X$ and the joint pdf. The conditional will have the same support, $0<x<y$ and $x<y<2.$
