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