Finding the probability of a region $D$ given a joint density

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Suppose $X$ and $Y$ have the joint density $$ f(x,y)= \begin{cases} 6y,\ \text{for} \ (x,y)\in D \\ 0 \ \text{otherwise} \end{cases} $$ where $D$ is the triangular region where $x > 0$, $y > 0$, and $x + y < 1$.

Find $P\left(Y \geq \frac{1}{2}(1-X)\right)$

Attempted solution:

I think I am getting the bounds wrong but I get

$$P\left(Y \geq \frac{1}{2}(1-X)\right) = \int_{0}^{1}\int_{0}^{(1-x)/2} 6y dydx$$

which equals $1/4$ but the answer is $3/4$. I drew the picture and I am really not sure what I am doing wrong.

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Judging by the limits of your integral, you're actually computing $P\left(Y \leq \frac{1}{2}(1-X)\right)$.