Probability Help! (X,Y) ~ f(x,y) = 8xy $I_D(x,y)$ a) $f_X (x) =$ ?
b) $P( X + Y < \frac{1}{2}) =$ ?
c) $f_Y(y \,| \, X = \frac{3}{4}) =$ ?
d) $P( Y < \frac{1}{2} \, | \, X = \frac{3}{4}) = $ ?
Any help is greatly appreciated! Thanks!!
Here is my work so far...
a) To get marginal density of $x$, I need to integrate $f(x,y)$ once with respect to $y$. From the drawing, we see the appropriate bounds for integration is $0$ and $x$. So we have 
$∫8xydy$ from $0$ to $x$, yielding $4x^3$ for $0<x<1$. 
b) First we need to understand what exactly is $X + Y < \frac{1}{2}$. This the line between $(.5,0)$ and the endpoint where it intersects with $y=x$, so that I end up with an isosceles triangle. You can see that the appropriate double integral is 
$∫∫8xydydx$, over $(0,.5-x)$ for $y$ and $(0,.5)$ for $x$. Then split the integral into two parts to make it easier ?
c) $f(y|X)= f(x,y)/f(x)= 8xy/(4x^3)$ Then plug in $x = .75$ ?
d) Now that I have part c), I can integrate $y$ from $0$ to $.5$ ?


$\operatorname{\bf I}_D(x,y)$ is an indicator function; a characteristic equation that has the value of $1$ when the argument exists within the domain, and a value if $0$ when it does not.  Here the argument is $(x,y)$ and the domain is $D$.
This is sometimes written as: ${\large\bf 1}_D(x,y)$.
It is compact notation for the piecewise function:
$$\operatorname{\bf I}_D(x,y) = \begin{cases}1 & \text{if } 0\leq y \leq x\leq 1 \\ 0 & \text{elsewise}\end{cases}$$

a) $f_X (x) =\int 8xy\operatorname{\bf I}_D(x,y) \operatorname{d} y = \int_0^x 8xy \operatorname{d} y \operatorname{\bf I}_{x\in[0,1]}(x)= 4x^3 \operatorname{\bf I}_{x\in[0,1]}(x)$
b) $P( X + Y < \frac{1}{2}) =$ ? : Simply observe what portion of the domain $D$ is below the line $x=\frac 1 2 - y$?
d) $P( Y < \frac{1}{2} \mid X = \frac{3}{4}) = $ ? : Find the line segment at $x=\frac 34$ what portion lies below $y=\frac 1 2$.
