Calculating joint probability problem This is the example from Casella and Berger 2nd page 146.
you have
$$f(x,y)=e^{-y}, 0<x<y< \infty $$
and you are interested in finding 
$$P(X+Y\ge1)$$
They solved this problem the following way
$$P(X+Y\ge1)=1-P(X+Y<1)=1-\int_0^{1/2}{\int_x^{1-x}{e^{-y}dydx}}\\=
1-\int_0^{1/2}{(e^{-x}-e^{-(1-x)}dx)}=2e^{-1/2}-e^{-1}$$
My question is why did they solve it this way?
1) Why can't I just solve $P(X+Y\ge1)$ rather than $1-P(X+Y<1)$?
I am confused because the example before this was solved different way. There
$f(x,y)=6xy^2$ and to solve $P(X+Y\ge1)$,
$$P(X+Y\ge1)=\int_0^1\int_{1-y}^16xy^2dxdy=9/10$$
2) Why is the limit of integral for x is 0 to 1/2? Where do I find this value?
They solve this using a graph, but is there way to justify their method without using a graph?
 A: Ans 1) The reason they used $1-P(X+Y<1)$ is that finding the probability of $P(X+Y<1)$ is much more easier than $P(X+Y\geq 1)$. To explain more, have a look at the following graph. The shaded red lines shows the area where $f(X,Y)$ is defined i.e. $0<x<y<\infty$. The region shown by vertical green lines is actually the event $X+Y\geq 1$ (and within the red lines). If you want to find $P(X+Y\geq 1)$, then you should take your integration over this green shaded area. the diagonal black lines indicates the event $X+Y<1$ (and within the red lines). You need to take your integration over this black diagonal lines if you want to find $P(X+Y<1)$. As you can see, we have a nice triangle defined the the black diagonal lines. But the green vertical lines is not that much well shaped and if you want to take your integration over this region, you should split it into 2-3 parts (a rectabgle, a triangle, etc...) which makes the computation difficult. Now let's answer the 2nd question:

Ans 2) First of all, $X=1/2$ is actually the intersection of $Y=X$, which is in the domain of $f(x,y)$, and $X+Y=1$. Now how do they find the boundaries of the integral in $\int_0^{1/2}\int_x^{1-x}d_yd_x$. First note that whenever you write $d_y$ it means that your $x$-coordinates are fixed, and you are just changing the $y$ coordinates. So what you do is to consider a vertical tiny rectangle like the blue one that you have in the following graph. 

The lower part of this rectangle goes from $Y=x$ to the upper part of this rectangle i.e. $Y=-x+1$ or $Y=1-x$. These two form the boundaries for the first integral. Now to find the boundaries for the 2nd integral you change your $X$ (that's why you have $d_x$) and move your rectangle to the left and right until you cover all the diagonal black lines. If you move it to the left, you will go until the line $X=0$ and if you move it to the right, you will go until the line $X=1/2$. These two form the boundaries for the 2nd integral. 
