## Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
Use the definition for improper integrals $I= \int_{0}^{\infty} \lim\limits_{b \to \infty} \int_0^b xe^{-x-2y} dx \space dy$ or, $= \int_{0}^{\infty} e^{-2y} \lim\limits_{b \to \infty} [-x e^{-x} -e^{-x}]_0^b dy$ or, $=\int_{0}^{\infty} e^{-2y} \lim\limits_{b \to \infty} [-b e^{-b} -e^{-b}+1] dy$ Thus, we have $I=\dfrac{1}{2} \lim\limits_{b \to \infty} (-e^{-2b}+1)=\dfrac{1}{2}$