Answer
Convergent $\;,\;$ $\large\frac{1}{5}e^{-10}$
Work Step by Step
Let
\[I=\int_{2}^{\infty}e^{-5p}dp\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow \infty}\int_{2}^{t}e^{-5p}dp\;\;\;\ldots(2)\]
\[I=\lim_{t\rightarrow \infty}\left[-\frac{1}{5}e^{-5p}\right]_{2}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[-\frac{1}{5}e^{-5t}+\frac{1}{5}e^{-10}\right]\]
\[I=\frac{1}{5}e^{-10}\]
Since limit on R.H.S. of (2) exists
So given improper integral (1) is convergent
and \[I=\frac{1}{5}e^{-10}.\]
