Tích phân \(\int\limits^1_0\left(x-2\right)e^{2x}\text{d}x\) bằng
\(\dfrac{5+3e^2}{4}\).\(\dfrac{-5-3e^2}{4}\).\(\dfrac{5-3e^2}{4}\).\(\dfrac{5-3e^2}{2}\).Hướng dẫn giải:Đặt \(\begin{cases}u=x-2\\v'=e^{2x}\end{cases}\) \(\Rightarrow\begin{cases}u'=1\\v=\dfrac{1}{2}e^{2x}\end{cases}\)
\(I=\int\limits^1_0\left(x-2\right)e^{2x}\text{d}x=\dfrac{1}{2}\left(x-2\right)e^{2x}|^1_0-\dfrac{1}{2}\int\limits^1_0e^{2x}\text{d}x\)
\(=\left(\dfrac{1}{2}\left(x-2\right)e^{2x}-\dfrac{1}{4}e^{2x}\right)|^1_0\)
\(=\left(\dfrac{1}{2}xe^{2x}-\dfrac{5}{4}e^{2x}\right)|^1_0\)
\(=\dfrac{1}{2}e^2-\dfrac{5}{4}e^2+\dfrac{5}{4}\)
\(=\dfrac{5-3e^2}{4}\).