Answer
$\frac{e-1}{4}$
Work Step by Step
Given: $\int_{0}^{1}\int_{\sqrt y}^{1}\frac{ye^{x^{2}}}{x^{3}}dxdy$
$\int_{0}^{1}\int_{\sqrt y}^{1}\frac{ye^{x^{2}}}{x^{3}}dxdy=\int_{0}^{1}\int_{0}^{x^{2}}\frac{ye^{x^{2}}}{x^{3}}dydx$
$=\int_{0}^{1}\frac{e^{x^{2}}}{x^{3}}\cdot \frac{y^{2}}{2}|_{0}^{x^{2}}dx$
$=\frac{1}{2}\int_{0}^{1}xe^{x^{2}}dx$
Put $t=x^{2}$ thus, $t=0$ when $x=0$
$dt=2xdx$ and $t=1$ when $x=1$
$=\frac{1}{2}\cdot\frac{1}{2}\int_{0}^{1}e^{t}dt$
$=\frac{1}{4}[e^{t}]|_{0}^{1}$
$=\frac{e-1}{4}$
