Ta có: $I=\int\limits_{0}^{a}{x.{{e}^{\frac{x}{2}}}dx}$. Đặt $\left\{ \begin{align}  & u=x \\  & dv={{e}^{\frac{x}{2}}}dx \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align}  & du=dx \\  & v=2.{{e}^{\frac{x}{2}}} \\ \end{align} \right.$

$\Rightarrow I=\left. 2x.{{e}^{\frac{x}{2}}} \right|_{0}^{a}-2\int\limits_{0}^{a}{{{e}^{\frac{x}{2}}}dx}=2a{{e}^{\frac{a}{2}}}-\left. 4.{{e}^{\frac{x}{2}}} \right|_{0}^{a}=2\left( a-2 \right){{e}^{\frac{a}{2}}}+4$

Theo đề ra ta có: $I=4\Leftrightarrow 2\left( a-2 \right){{e}^{\frac{a}{2}}}+4=4\Leftrightarrow a=2$