Tìm a sao cho \(I = \int\limits_0^a {x.{e^{\frac{x}{2}}}d{\rm{x}}} = 4\).

* Hướng dẫn giải

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:

\(I = 4 \Leftrightarrow 2\left( {a - 2} \right){e^{\frac{a}{2}}} + 4 = 4 \Leftrightarrow a = 2\)