Answer
$$\int^{-1}_{-2}e^{-(x+1)}dx=e-1$$
Work Step by Step
$$A=\int^{-1}_{-2}e^{-(x+1)}dx$$
We set $u=-(x+1)$, which means $$du=-dx$$ $$dx=-du$$
For $x=-1$, $$u=-(-1+1)=0$$
For $x=-2$, $$u=-(-2+1)=1$$
Therefore, $$A=-\int^0_1e^udu$$ $$A=-e^u\Big]^0_1$$ $$A=-(e^0-e^1)$$ $$A=-(1-e)$$ $$A=e-1$$
