Answer
$\dfrac{1}{2}(e^2-3)$
Work Step by Step
Plug $u=y^2 \implies du=2y dy$
Re-arrange the given integral as follows:
$(\dfrac{1}{2}) \int_{0}^{2} \int_{0}^{1}2xye^{xy^2} dy dx=(\dfrac{1}{2}) \int_{0}^{2} \int_{0}^{1}xe^{ux} du dx$
This implies that
$(\dfrac{1}{2}) \int_{0}^{2} \int_{0}^{1}xe^{ux} du dx= (\dfrac{1}{2}) \int_{0}^{2}x(e^{ux}/x)]_{0}^{1} dx$
Hence, $(\dfrac{1}{2}) (e^x-x)_0^2=\dfrac{1}{2}(e^2-3)$
