$\quad I=  \displaystyle\int\limits_0^{\infty}xe^{-3x}dx$

$\Leftrightarrow I = \lim\limits_{t\to 0}\displaystyle\int\limits_0^{t}xe^{-3x}dx$

Đặt $\begin{cases}u = x\\dv = e^{-3x}dx\end{cases}\Rightarrow \begin{cases}du = dx\\v = -\dfrac13e^{-3x}\end{cases}$

Ta được:

$\quad I = \lim\limits_{t\to 0}\left(-\dfrac13xe^{-3x}\right)\Bigg|_0^t + \dfrac13\lim\limits_{t\to 0}\displaystyle\int\limits_0^t e^{-3x}dx$

$\Leftrightarrow I =\lim\limits_{t\to 0}\left(-\dfrac13te^{-3t}\right) - \dfrac19\lim\limits_{t\to 0}\left(e^{-3x}\right)\Bigg|_0^t$

$\Leftrightarrow I = - \dfrac19\lim\limits_{t\to 0}\left(e^{-3t}\right) + \dfrac19e^0$

$\Leftrightarrow I =\dfrac19$

I=∞∫0xe−3xdxI=∫0∞xe−3xdx

⇔I=limt→0t∫0xe−3xdx⇔I=limt→0∫0txe−3xdx

Đặt {u=xdv=e−3xdx⇒{du=dxv=−13e−3x{u=xdv=e−3xdx⇒{du=dxv=−13e−3x

Ta được:

I=limt→0(−13xe−3x)∣∣∣t0+13limt→0t∫0e−3xdxI=limt→0(−13xe−3x)|0t+13limt→0∫0te−3xdx

⇔I=limt→0(−13te−3t)−19limt→0(e−3x)∣∣∣t0⇔I=limt→0(−13te−3t)−19limt→0(e−3x)|0t

⇔I=−19limt→0(e−3t)+19e0⇔I=−19limt→0(e−3t)+19e0

⇔I=19