令 $u=\sqrt{x}$，则 $x=u^2$、$\mathrm{d}x=2u\mathrm{d}u$，积分限由 $0,4$ 变为 $0,2$。

因此

$$\begin{array}{rl}{\int }_0^4{\mathrm{e}}^{\sqrt{x}}\mathrm{d}x& =2{\int }_0^{2}u{\mathrm{e}}^u\mathrm{d}u\\ & =2[(u-1){\mathrm{e}}^u{]}_0^2\\ & =2{\mathrm{e}}^2+2.\end{array}$$