## Calculus 8th Edition

$\frac{e-1}{4}$
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 $x=1$ $=\frac{1}{2}\cdot\frac{1}{2}\int_{0}^{1}e^{t}dt$ $=\frac{1}{4}[e^{t}]|_{0}^{1}$ $=\frac{e-1}{4}$