Answer
$$\frac{1}{{2\ln 2}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{2^{ - \theta }}} d\theta \cr
& {\text{use substitution}}{\text{. Let }}u = - \theta,{\text{ so that }}du = - d\theta \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = 1,{\text{ }}u = - 1 \cr
& \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = 0 \cr
& {\text{Then}} \cr
& \int_0^1 {{2^{ - \theta }}} d\theta = \int_0^{ - 1} {{2^u}} \left( { - du} \right) \cr
& {\text{integrate}} \cr
& = - \int_0^{ - 1} {{2^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = - \left( {\frac{{{2^u}}}{{\ln 2}}} \right)_0^{ - 1} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - \left( {\frac{{{2^{ - 1}}}}{{\ln 2}}} \right) + \left( {\frac{{{2^0}}}{{\ln 2}}} \right) \cr
& {\text{simplifying}} \cr
& = - \left( {\frac{1}{{2\ln 2}}} \right) + \left( {\frac{1}{{\ln 2}}} \right) \cr
& = \frac{{ - 1 + 2}}{{2\ln 2}} \cr
& = \frac{1}{{2\ln 2}} \cr} $$
