Jtdylktuy

A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.

My attempt:
$$int_{-1}^1int_{-1}^{2-x} frac{1}{4},dy,dx$$
Did I set up the double integral right?

Comment: @herbsteinberg is correct that the problem as stated
is pointless, because the answer is obviously $1.$ The integral would be
$int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$

The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
Because the joint distribution of $(X,Y)$ on the square is uniform it
seems clear that $P(|X+Y|<1) = 3/4.$

If you want to use integral calculus, it is probably best to break
the integral into two parts, perhaps to the left and right of the vertical green
line.

Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
[A larger number of points, such as m = 10^6, would give a much closer
approximation, but an ugly plot.]

set.seed(1112);  m = 50000

x = runif(m, -1,1);  y = runif(m, -1,1)

plot(x,y, pch=".")

cond = (abs(x+y)<1)

points(x[cond],y[cond], pch=".", col="blue")

abline(h=0, col="green2", lwd=2);  abline(v=0, col="green2", lwd=2)

mean(cond)

[1] 0.7519  # aprx P(|X + Y| < 1) = 3/4

Note thet:
$$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$
See the graph:

If the question is to find $P(|X+Y|<1)$, then:
$$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$

See the graph: