Cho hàm số \(y = f\left( x \right)\) biết \(f\left( 0 \right) = \dfrac{1}{2}\) và \(f'\left( x \right) = x{e^{{x^2}}}\) với mọi \(x \in \mathbb{R}\). Khi đó \(\int\limits_0^1 {xf\left( x \right)dx} \) bằng:

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = \dfrac{{{x^2} - 1}}{2}\end{array} \right.\).

\(\begin{array}{l} \Rightarrow \int\limits_0^1 {xf\left( x \right)dx}  = \left. {\dfrac{{{x^2} - 1}}{2}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {\dfrac{{{x^2} - 1}}{2}f'\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}f\left( 0 \right) - \dfrac{1}{2}\int\limits_0^1 {\left( {{x^2} - 1} \right).x{e^{{x^2}}}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}.\dfrac{1}{2} - \dfrac{1}{2}I = \dfrac{1}{4} - \dfrac{1}{2}I\end{array}\)

Đặt \(t = {x^2} \Rightarrow dt = 2xdx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 1\end{array} \right.\).

Khi đó ta có:

\(\begin{array}{l}I = \dfrac{1}{2}\int\limits_0^1 {\left( {t - 1} \right){e^t}dt}  = \dfrac{1}{2}\left( {\int\limits_0^1 {t{e^t}dt}  - \int\limits_0^1 {{e^t}dt} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {\left. {t.{e^t}} \right|_0^1 - \int\limits_0^1 {{e^t}dt}  - \int\limits_0^1 {{e^t}dt} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {e - \left. {2{e^t}} \right|_0^1} \right) = \dfrac{1}{2}\left( {e - 2e + 2} \right) = \dfrac{{2 - e}}{2}\end{array}\)

Vậy \(\int\limits_0^1 {xf\left( x \right)dx}  = \dfrac{1}{4} - \dfrac{1}{2}I = \dfrac{1}{4} - \dfrac{{2 - e}}{4} = \dfrac{{e - 1}}{4}\).