Tích phân \(\int\limits_0^1 {\left( {x - 2} \right){e^{2x}}dx} \) bằng

* Hướng dẫn giải

Gọi \(I = \int\limits_0^1 {\left( {x - 2} \right){e^{2x}}dx} \)

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

Khi đó ta có: 

\(\begin{array}{l}I = \left. {\left( {x - 2} \right)\frac{{{e^{2x}}}}{2}} \right|_0^1 - \int\limits_0^1 {\frac{{{e^{2x}}}}{2}dx} \\\,\,\, =  - \frac{1}{2}{e^2} + 1 - \left. {\frac{{{e^{2x}}}}{4}} \right|_0^1\\\,\,\, =  - \frac{1}{2}{e^2} + 1 - \frac{{{e^2}}}{4} + \frac{1}{4}\\\,\,\, =  - \frac{3}{4}{e^2} + \frac{5}{4} = \frac{{5 - 3{e^2}}}{4}\end{array}\)