Answer
$$\frac{1}{3 e^{12}}$$
Work Step by Step
\begin{aligned}
\int_{4}^{\infty} e^{-3 x} d x &=\lim _{R \rightarrow \infty} \int_{4}^{R} e^{-3 x} d x \\
&=\lim _{R \rightarrow \infty}\left.\frac{e^{-3 x}}{-3}\right|_{4} ^{R} \\
&=\lim _{R \rightarrow \infty} \left(\frac{1}{3 e^{12}}-\frac{1}{3 e^{3 R}} \right) \\
&=\frac{1}{3 e^{12}}
\end{aligned}
Converges.
