Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên \(\mathbb{R}\) thỏa mãn \(f\left( 0 \right) = 3\) và \(f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\,\,\forall x \in \mathbb{R}\). Tích phân \(\int\limits_0^2 {xf'\left( x \right)dx} \) bằng:

\(\int\limits_0^2 {xf'\left( x \right)dx}  = \int\limits_0^2 {xd\left( {f\left( x \right)} \right)}  = \left. {xf\left( x \right)} \right|_0^2 - \int\limits_0^2 {f\left( x \right)dx}  = 2f\left( 2 \right) - \int\limits_0^2 {f\left( x \right)dx} \)

Theo bài ra ta có:

\(f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\,\,\forall x \in \mathbb{R} \Rightarrow f\left( 0 \right) + f\left( 2 \right) = 2 \Rightarrow f\left( 2 \right) = 2 - f\left( 0 \right) =  - 1\).

\( \Rightarrow \int\limits_0^2 {xf'\left( x \right)dx}  =  - 2 - \int\limits_0^2 {f\left( x \right)dx}  =  - 2 - \int\limits_0^2 {f\left( t \right)dt} \).

Đặt \(t = 2 - x \Rightarrow dt =  - dx\). Đổi cận \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 2\\x = 2 \Rightarrow t = 0\end{array} \right.\).

\(\begin{array}{l} \Rightarrow \int\limits_0^2 {f\left( t \right)dx}  =  - \int\limits_2^0 {f\left( {2 - x} \right)dx}  = \int\limits_0^2 {f\left( {2 - x} \right)dx} \\ \Rightarrow \int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^2 {f\left( {2 - x} \right)dx} \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^2 {f\left( x \right)dx}  + \int\limits_0^2 {f\left( {2 - x} \right)dx} \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^2 {\left[ {f\left( x \right) + f\left( {2 - x} \right)} \right]dx} \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^2 {\left( {{x^2} - 2x + 2} \right)dx} \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx}  = \left. {\left( {\dfrac{{{x^3}}}{3} - {x^2} + 2x} \right)} \right|_0^2 = \dfrac{8}{3}\\ \Rightarrow \int\limits_0^2 {f\left( x \right)dx}  = \dfrac{4}{3}\end{array}\)  

Vậy \(\int\limits_0^2 {xf'\left( x \right)dx}  =  - 2 - \dfrac{4}{3} =  - \dfrac{{10}}{3}\).