Answer
$2e^{12}-\dfrac{e^{9}}{2}+\dfrac{e^5}{2}$
Work Step by Step
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A= \int_{2}^{4} \int_{y+1}^{12-y} e^{x+y} \ dx \ dy \\= \int_{2}^{4} [e^{x+y}]_{y+1}^{12-y} \ dy\\= \int_{2}^{4} (e^{12}-e^{2y+1}) \ dy \\= [e^{12} y-\dfrac{e^{2y+1}}{2}]_2^4\\=[e^{12} (4-2)-\dfrac{e^{2(4)+1}}{2}+\dfrac{e^{2(2)+1}}{2} \\= 2e^{12}-\dfrac{e^{9}}{2}+\dfrac{e^5}{2}$
