Answer
Converges to 9.
Work Step by Step
$\displaystyle \int_{0}^{\infty}xe^{-x/3}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}xe^{-x/3}dx$
...Integration by parts
$\displaystyle \int xe^{-4x}dx=\int udv\quad\left[\begin{array}{ll}
u=x & dv=e^{-x/3}\\
du=dx & v=-3e^{-x/3}
\end{array}\right]=uv-\displaystyle \int vdu$
$=-3xe^{-x/3}+3\displaystyle \int e^{-x/3}dx$
$=-3xe^{-x/3}+3(-3)e^{-x/3}+C$
$=\displaystyle \lim_{b\rightarrow\infty}[(-3x-9)e^{-x/3}]_{0}^{b}$
$=\displaystyle \lim_{b\rightarrow\infty}[(-3b-9)e^{-b/3}-(-9)]$
$=9$
Converges to 9.
