. Let $I=\int_{0}^{1} \sqrt{x} e^{\sqrt{x}} d x$
Put $\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t$
$\therefore I=\int_{0}^{1} t e^{t}(2 t d t)$
$=2 \int_{0}^{1} t^{2} e^{t} d t$
$=2\left[t^{2} e^{t}-\int 2 t e^{t} d t\right]_{0}^{1}$
$=2\left[t^{2} e^{t}-2 t e^{t}+2 e^{t}\right]_{0}^{1}$
$=2\left[1 e^{1}-2 e^{1}+2 e^{1}-\left(0-0+2 e^{0}\right)\right]$
$=2[e-2]$