Answer
$$\frac{1}{{2\ln 2}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {{2^{ - x}}} dx \cr
& {\text{definition of improper integral}} \cr
& \int_1^\infty {{2^{ - x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {{2^{ - x}}} dx \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( { - \frac{{{2^{ - x}}}}{{\ln 2}}} \right)} \right|_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{{{2^{ - b}}}}{{\ln 2}} + \frac{{{2^{ - 1}}}}{{\ln 2}}} \right) \cr
& {\text{simplify}} \cr
& = \frac{1}{{\ln 2}}\mathop {\lim }\limits_{b \to \infty } \left( {{2^{ - b}} - {2^{ - 1}}} \right) \cr
& {\text{evaluate the limit}} \cr
& = \frac{1}{{\ln 2}}\left( {{2^{ - \infty }} - {2^1}} \right) \cr
& = \frac{1}{{\ln 2}}\left( {0 - \frac{1}{2}} \right) \cr
& = \frac{1}{{2\ln 2}} \cr} $$
