Ta có: \(f\left( {2 - x} \right) + {x^2}f''\left( x \right) = 2x \Rightarrow f\left( 2 \right) = 0\).

Lại có: \(\int\limits_0^2 {f\left( {2 - x} \right){\rm{d}}x} + \int\limits_0^2 {{x^2}f''\left( x \right){\rm{d}}x} = \int\limits_0^2 {2x{\rm{d}}x} = \left. {{x^2}} \right|_0^2 = 4\).

Xét \({I_1} = \int\limits_0^2 {f\left( {2 - x} \right){\rm{d}}x} \). Đặt \(2 - x = t \Rightarrow {\rm{d}}x = - {\rm{d}}t\).

Ta có \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 2\\x = 2 \Rightarrow t = 0\end{array} \right.\).

Khi đó \[{I_1} = - \int\limits_2^0 {f\left( t \right)dt} = \int\limits_0^2 {f\left( t \right)dt} = \int\limits_0^2 {f\left( x \right)dx} \]. Đặt \(\left\{ {\begin{array}{*{20}{l}}{u = f\left( x \right)}\\{{\rm{d}}v = {\rm{d}}x}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{\rm{d}}u = f'\left( x \right){\rm{d}}x}\\{v = x}\end{array}} \right.} \right.\).

\( \Rightarrow {I_1} = \left. {\left[ {xf\left( x \right)} \right]} \right|_0^2 - \int_0^2 x f'\left( x \right){\rm{d}}x = 2f\left( 2 \right) - \int_0^2 x f'\left( x \right){\rm{d}}x = - \int_0^2 x f'\left( x \right){\rm{d}}x{\rm{. }}\)

Xét \({I_2} = \int\limits_0^2 {{x^2}f''\left( x \right){\rm{d}}x} \). Đặt \(\left\{ {\begin{array}{*{20}{l}}{u = {x^2}}\\{\;{\rm{d}}v = f''\left( x \right){\rm{d}}x}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{\rm{d}}u = 2x\;{\rm{d}}x}\\{v = f'\left( x \right)}\end{array}} \right.} \right.\).

\( \Rightarrow {I_2} = \left. {\left[ {{x^2}f'\left( x \right)} \right]} \right|_0^2 - \int_0^2 2 xf'\left( x \right){\rm{d}}x = 4f'\left( 2 \right) - 2\int_0^2 x f'\left( x \right){\rm{d}}x = 8 - 2\int_0^2 x f'\left( x \right){\rm{d}}x{\rm{. }}\)

Vậy \( - \int\limits_0^2 x f'\left( x \right){\rm{d}}x + 8 - 2\int\limits_0^2 x f'\left( x \right){\rm{d}}x = 4 \Rightarrow \int\limits_0^2 x f'\left( x \right){\rm{d}}x = \frac{4}{3}\).

Đáp án: \(\frac{4}{3}\).