Answer
Convergent $\;,\;$ $\large\frac{1}{\ln 2}$
Work Step by Step
Let
\[I=\int_{-\infty}^{0}2^{r}dr\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow -\infty}\int_{t}^{0}2^{r}dr\;\;\;\ldots(2)\]
\[I=\lim_{t\rightarrow -\infty}\left[\frac{2^r}{\ln 2}\right]_{t}^{0}\]
\[I=\lim_{t\rightarrow -\infty}\left[\frac{1}{\ln 2}-\frac{2^t}{\ln 2}\right]\]
\[I=\frac{1}{\ln 2}\]
Since limit on R.H.S. of (2) exists
So given improper integral (1) is convergent and \[I=\frac{1}{\ln 2}.\]
