Answer
$$\frac{{24}}{{\ln 5}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^0 {{5^{ - \theta }}} d\theta \cr
& {\text{use substitution}}{\text{. Let }}u = - \theta,{\text{ so that }}du = - d\theta \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = \cr
& \,\,\,\,\,\,{\text{If }}\theta = - 2,{\text{ }}u = 2 \cr
& {\text{Then}} \cr
& \int_{ - 2}^0 {{5^{ - \theta }}} d\theta = \int_2^0 {{5^u}} \left( { - du} \right) \cr
& = \int_0^2 {{5^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = \left( {\frac{{{5^u}}}{{\ln 5}}} \right)_0^2 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{5^2} - {5^0}}}{{\ln 5}} \cr
& {\text{simplifying}} \cr
& = \frac{{25 - 1}}{{\ln 5}} \cr
& = \frac{{24}}{{\ln 5}} \cr} $$
