Answer
$\dfrac{1}{2} (e^{16}-17) \\ \approx 4.44 \times 10^{6}$
Work Step by Step
The domain $D$ is bounded by $y=x; y=4 $ and $x=0$
Now, we can work out with the given integral as follows:
$A=\iint_{D} y^2 e^{xy} dA=\int_{0}^{4} \int_{0}^{y} y^2 e^{xy} dx \ dy \\= \int_{0}^{4} [ye^{xy}]_{0}^{y} \ dy \\ = \int_{0}^{4} [ye^{y^2}-y] \ dy \\\dfrac{1}{2} \times [e^{y^2} -y^2]_0^4 \\ = \dfrac{1}{2} (e^{4^2} -(4)^2-(1-0)] \\= \dfrac{1}{2} (e^{16}-17) \\ \approx 4.44 \times 10^{6}$
