Answer
$$\frac{5}{2}-\frac{1}{e} \approx 2.132$$
Work Step by Step
$$
Question: \int_0^1 \int_1^2 (x+e^{-y}) dxdy $$
$Solution: $
$ =\int_0^1 (\frac{x^2}{2}+xe^{-y})_1^2dy$
$= \int_0^1 [(2+2e^{-y})-(\frac{1}{2}+e^{-y})]dy$
$= \int_0^1 (\frac{3}{2}+e^{-y})dy$
$= (\frac{3}{2}y - e^{-y})_0^1$
$= (\frac{3}{2}-\frac{1}{e})-(0-1)$
$= \frac{5}{2}-\frac{1}{e}$
$$Answer\approx2.132$$
$$\blacksquare$$
