Answer
$e^2 -2e+1 $
Work Step by Step
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A=\int_0^1 \int_{0}^{1} e^{x+y} dx dy\\=\int_0^1 [e^{x+y}]_0^1 \ dy \\=\int_0^1 [e^{1+y}-e^y] \ dy \\=[e^{y+1}-e^y]_0^1 \\=e^2 -e^1-(e^1-e^0) \\=e^2-e-e+1 \\=e^2 -2e+1 $
