Answer
$1$ and $\dfrac{1}{4}$
Work Step by Step
We have: $\overline {x}=\dfrac{1}{A}\int_{0}^{\infty} xe^{-x} dx=\lim\limits_{a \to \infty}[-xe^{-x}-e^{-x}]_{0}^a=\lim\limits_{a \to \infty}[-ae^{-a}-(-e^{-a})]-(0-e^{-0})=0+1=1$
and $\overline {y}=\dfrac{1}{2A}\int_{0}^{\infty} (e^{-x})^2 dx=\dfrac{1}{2}\int_{0}^{\infty} e^{-2x} dx=\lim\limits_{a \to \infty} (1/2) \dfrac{-e^{-2a}}{2} -\dfrac{1}{2} (\dfrac{-e^{-2(0)}}{2})=0+\dfrac{1}{4}=\dfrac{1}{4}$
