Answer
$$ e - 1$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 1} {{e^{ - \left( {x + 1} \right)}}} dx \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = - \left( {x + 1} \right),{\text{ so that }}\frac{{du}}{{dx}} = - 1,\,\,\,dx = - du \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = - 1,{\text{ }}u = - \left( { - 1 + 1} \right) = 0 \cr
& \,\,\,\,\,\,{\text{If }}x = - 2,{\text{ }}u = - \left( { - 2 + 1} \right) = 1 \cr
& {\text{Then}} \cr
& \int_{ - 2}^{ - 1} {{e^{ - \left( {x + 1} \right)}}} dx = \int_1^0 {{e^u}} \left( { - du} \right) \cr
& {\text{integrate}} \cr
& = - \left( {{e^u}} \right)_1^0 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - \left( {{e^0} - {e^1}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = - \left( {1 - e} \right) \cr
& = e - 1 \cr} $$
