Ta có: \(I = \int\limits_0^a {x.{e^{\frac{x}{2}}}dx}\). Đặt: \(\left\{ \begin{array}{l} u = x\\ dv = {e^{\frac{x}{2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = 2.{e^{\frac{x}{2}}} \end{array} \right.\)
\(\Rightarrow I = \left. {2x.{e^{\frac{x}{2}}}} \right|_0^a - 2\int\limits_0^a {{e^{\frac{x}{2}}}dx} = 2a{e^{\frac{a}{2}}} - \left. {4.{e^{\frac{x}{2}}}} \right|_0^a = 2\left( {a - 2} \right){e^{\frac{a}{2}}} + 4\)
Theo đề ra ta có: \(I = 4 \Leftrightarrow 2\left( {a - 2} \right){e^{\frac{a}{2}}} + 4 = 4 \Leftrightarrow a = 2\)
