Given the following density function $$ f_{(X, Y)}(x, y)=\frac{1}{10} e^{-0.5(y+3-x)}, \quad 5 \leq x \leq 10, \quad y \geq x-3 $$ I am asked to find $P(X<Y)$ but I dont understand the limits of $y$ in the solution: $$ P(X<Y)=\int_{x=5}^{10} \int_{y=x}^{\infty} 0.1 e^{-0.5(y+3-x)} d y d x=\left.0.1 \int_{x=5}^{10}(-2) e^{-0.5(y+3-x)}\right|_{x} ^{\infty} $$
why this is not $$ P(X<Y)=\int_{x=5}^{10} \int_{y=x-3}^{\infty} 0.1 e^{-0.5(y+3-x)} d y d x=...$$
Because the request is exactly $P(Y>X)$, that is the integral over the domain upper the line $X=Y$
see the picture below