Biết \(\int\limits_0^{\sqrt a } {(x - 1){e^{2{\rm{x}}}}d{\rm{x}} = \frac{{3 - {e^2}}}{4}} ;\,\,a > 0\) . Tính giá trị của \(a\)

* Hướng dẫn giải

\(I=\int\limits_0^{\sqrt a } {(x - 1){e^{2{\rm{x}}}}d{\rm{x}} = \frac{{3 - {e^2}}}{4}} ;\,\,a > 0\)

Đặt \(\left\{ \begin{array}{l}
u = x - 1\\
dv = {e^{2{\rm{x}}}}d{\rm{x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = d{\rm{x}}\\
v = \frac{1}{2}{e^{2{\rm{x}}}}
\end{array} \right.\)

\(\begin{array}{*{20}{l}}
{ \Rightarrow I = \left. {\frac{1}{2}\left( {x - 1} \right){e^{2x}}} \right|\begin{array}{*{20}{c}}
{\sqrt a }\\
0
\end{array} - \frac{1}{2}\int\limits_0^{\sqrt a } {{e^{2x}}} dx}\\
{ = \frac{1}{2}\left( {\sqrt a  - 1} \right).{e^{2\sqrt a }} + \frac{1}{2} - \frac{1}{4}.{e^{2\sqrt a }} + \frac{1}{4}}\\
{ = \frac{{\left( {2{e^{2\sqrt a }} - 3{e^{2\sqrt a }} + 3} \right)}}{4} = \frac{{3 - {e^2}}}{4}}\\
{ \Rightarrow a = 1}
\end{array}{\rm{ }}\)