How do I find this probability,given the following joint distribution? $X$ and $Y$ are jointly distributed random variables having the pdf
$f(x,y) = x+y, $if  $0<x<1,0<y<1,$
$\            $$0,$ otherwise. 
I have to find $P(X+Y>0.5)$
Given the constraints, I solved to get the ranges of $x$ and $y$ as follows:
$0<y<0.5$ and $0.5-y<x<0.5$. (Is this correct?)
Then I solved the probability by double integrating the function $f(x,y)$ over this range, with respect to $x$ first and then with respect to $y$. 
I got the answer as $\frac{1}{12}$. But this is incorrect,according to the key. Where am I going wrong?
 A: When you integrate over $0<y<0.5$ and $0.5−y<x<0.5$, you integrates over the small triangle with vertices $(0.5, 0)$, $(0.5,0.5)$ and $(0, 0.5)$. And you need to integrate over all the unit square except triangle with vertices $(0,0)$, $(0.5, 0)$, and $(0, 0.5)$. 
It is convinient to find the probability of the opposite event: 
$$
\mathbb P(X+Y\leq 0.5) = \int_0^{0.5} \left(\int_0^{0.5-y}f(x,y)\, dx \right)dy = \frac{1}{24}.
$$
And then find $\mathbb P(X+Y >0.5)$ from this value.
A: 
Given the constraints, I solved to get the ranges of $x$ and $y$ as
  follows: $0<y<0.5$ and $0.5-y<x<0.5$. (Is this correct?)

Almost. It is $0<y<0.5$ and $0<x<0.5-y$. This is $P(X+Y<0.5)$. Then you have to subtract this probability from 1. The integral is
$$P(X+Y<0.5)=\int\limits_0^{0.5} \int\limits_0^{0.5-y} x+y \, dx \, dy$$
The inner integral: 
$$\int\limits_0^{0.5-y} x+y \, dx =\left[\frac12x^2+yx \right]_0^{0.5-y}=\frac12\cdot (0.5-y)^2+0.5y-y^2=\frac18-\frac{y}2+\frac{y^2}2+0.5y-y^2$$
$=\frac18-\frac{y^2}2$
Now the outer integral
$$\int\limits_0^{0.5} \frac18-\frac{y^2}2\, dy=...=\frac1{24}$$
Finally we have $P(X+Y>0.5)=1-P(X+Y<0.5)$
