Let $(X,Y)$ be a uniformly chosen point of a region $A \subset \mathbb{R}^2$
Given we have the following joint pdf:
$$ f(x,y) = \begin{cases} \dfrac{a}{ \text{area of}~ A} & (x,y)\in A \\ 0 & \text{else} \end{cases}$$
And Let $$A = {(x,y),|y| \leq x \cdot e^{-x}}$$
the provided solutions state that A is 2 and it computes
$$EX = \int_{0}^{\infty} \int_{-xe^{-x}}^{e^{-x}} \frac{x}{2} dydx = \int_{0}^{\infty} x^2 e^{-x} = 2$$
Not that doesn't make sense to me:
1) how is the area of $A = 2$ , when the graph of $|y| = x e^{-x}$ looks like this:
The area of $A$ is given by $$ \int_0^\infty xe^{-x}-(-xe^{-x})\, dx=2\int_0^\infty xe^{-x}\, dx=2. $$ To see that $\int_0^\infty xe^{-x}\, dx=1$ use integration by parts. Indeed, $$ \int_0^\infty xe^{-x}\, dx=-xe^{-x}]_{0}^\infty+\int_0^\infty e^{-x}\, dx=0+1=1. $$