Answer
$$ 1$$
Work Step by Step
$$\eqalign{
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx \cr
& {\text{use the formula }}\int_a^b {{e^{kx}}} dx = \left( {\frac{{{e^{kx}}}}{k}} \right)_a^b \cr
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx = \left( { - {e^{ - x}}} \right)_{ - \ln 2}^0 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx = - {e^{ - 0}} + {e^{ - \left( { - \ln 2} \right)}} \cr
& {\text{simplifying}} \cr
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx = - 1 + {e^{\ln 2}} \cr
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx = - 1 + 2 \cr
& \int_{ - \ln 2}^0 {{e^{ - x}}} dx = 1 \cr} $$
