Let $R$ be the region bounded by the graph of $y=x^2$, $x=0$ and $y=4$ in the first quadrant. Please compute the following double integral: $$\iint_R xe^{y^2} \ dx\ dy $$ Please round off your answer to $1$ decimal place.
My work is the following:$$\begin{align}\iint_Rxe^{y^2}dxdy&=\int_0^4\int_0^\sqrt{y}xe^{y^2}dxdy\\&=\int_0^4\left[\frac{x^2}{2}\right]_0^\sqrt{y}e^{y^2}dy\\&=\frac12\int_0^4ye^{y^2}dy\\&=\frac14\int_0^{16}e^tdt\\&=\frac14[e^t]_0^{16}\\&=\frac14[e^{16}-1]\\&=2221527.4\end{align}$$but it doesn't seem correct