Answer
Convergent.
Work Step by Step
$$
\int_{2}^{\infty} y e^{-3 y} d y
$$
Observe that the given integral is improper
$$
\begin{aligned}\int_{2}^{\infty} y e^{-3 y} d y &=\lim _{t \rightarrow \infty} \int_{2}^{t} y e^{-3 y} d y \\
& \quad\quad\quad\left[\text { use integration by parts} \right] \\
& \quad\quad\quad\left[\begin{array}{c}{u=y, \quad d v=e^{-3y} dy } \\ {d u=dy, \quad v=\frac{-1}{3}e^{-3y} \quad}\end{array}\right] \\
&= \lim _{t \rightarrow \infty}\left[ \frac{-1}{3} y e^{-3 y} +\frac{1}{3}\int_{2}^{t} e^{-3 y} dy \right]\\
&=\lim _{t \rightarrow \infty}\left[-\frac{1}{3} y e^{-3 y}-\frac{1}{9} e^{-3 y}\right]_{2}^{t} \\
&=\lim _{t \rightarrow \infty}\left[\left(-\frac{1}{3} t e^{-3 t}-\frac{1}{9} e^{-3 t}\right)-\left(-\frac{2}{3} e^{-6}-\frac{1}{9} e^{-6}\right)\right] \\
&=\lim _{t \rightarrow \infty}\left[\left(-\frac{1}{3} \frac{t}{e^{3 t}} -\frac{1}{9} e^{-3 t}\right)-\left(-\frac{2}{3} e^{-6}-\frac{1}{9} e^{-6}\right)\right] \\
& \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\
&=0-0+\frac{7}{9} e^{-6}\\
&=\frac{7}{9} e^{-6} .\\
\end{aligned}
$$
It follows that the integral
$$
\int_{2}^{\infty} y e^{-3 y} d y
$$
is convergent.
