Answer
$$
S=\left\{(x, y) | x \geqslant 0,0 \leqslant y \leqslant x e^{-x}\right\}
$$
The shaded area is the region of interest.
$$
\begin{split}
\text { Area } & = \int_{0}^{\infty} x e^{-x} d x \\
& =1
\end{split}
$$
Work Step by Step
$$
S=\left\{(x, y) | x \geqslant 0,0 \leqslant y \leqslant x e^{-x}\right\}
$$
The shaded area is the region of interest.
$$
\begin{aligned} \text { Area } &=\int_{0}^{\infty} x e^{-x} d x=\lim _{t \rightarrow \infty} \int_{0}^{t} x e^{-x} d x \\
&=\lim _{t \rightarrow \infty} \left[ -x e^{-x}+ \int_{0}^{t} e^{-x} d x\right]
\\
& \quad\quad\quad\left[\text { use integration by parts with } u=x \text { and } d v=e^{-x} d x \right] \\
&=\lim _{t \rightarrow \infty}\left[-x e^{-x}-e^{-x}\right]_{0}^{t} \quad \\ &=\lim _{t \rightarrow \infty}\left[\left(-t e^{-t}-e^{-t}\right)-(-1)\right] \\
&=\lim _{t \rightarrow \infty}\left[\left( -\frac{t}{e^{t}} -e^{-t}\right)-(-1)\right] \\
& \quad\quad\quad\left[\text {The fraction part is } \frac{\infty}{\infty} \text { so we can use L'Hospital's } \right] \\
&=0 -0+1=1. \end{aligned}
$$
