Probabilities and joint pdf I'm supposed to compute $P(Y>1/2)$ for the joint pdf
$$f_{X,Y}(x,y)=\begin{cases}12(y-x^2),&0<x<y<1\\0,&\text{elsewhere.}\end{cases}$$
and  figured that I can do this if I use the given information by rewriting it as $P(1/2 < Y < 1),$ which would allow me to solve the problem by integration.
$$12\int_{0.5}^1\left[\int_{0.5}^y\big(y-x^2\big)\,dx\right]dy=0.5625.$$
Firstly, I'm not sure if I've gotten the integral right. Secondly, I believe this solves for
$P(1/2 ≤ Y ≤ 1),$ not $P(1/2 < Y < 1).$ Can someone help me clarify this problem?

Your density function is not normalized, since
$$12\int_{0}^1\left[\int_0^y\big(y-x^2\big)\,dx\right]dy=3.$$
You need to divide by $3$ to get a normalized density function, which would be
$$f_{X,Y}(x,y)=\begin{cases}4(y-x^2),&0<x<y<1\\0,&\text{elsewhere.}\end{cases}$$
Your integral is incorrect, because $x$ has to be allowed to vary all the way from $0$ to $y,$ based on the density definition. To compute your desired probability, you need to calculate
$$4\int_{0.5}^1\left[\int_0^y\big(y-x^2\big)\,dx\right]dy=0.854167.$$
