Lời giải chi tiết:

Ta có: \(\int\limits_0^1 {{{\left( {f'\left( x \right)} \right)}^2}dx}  + \int\limits_0^1 {4f\left( x \right)}  = \int\limits_0^1 {\left( {8{x^2} + 4} \right)dx}  = \dfrac{{20}}{3}\).

Xét

\(\begin{array}{l}\int\limits_0^1 {4f\left( x \right)dx}  = \int\limits_0^1 {f\left( x \right)d\left( {4x} \right)}  = \left. {f\left( x \right).4x} \right|_0^1 - \int\limits_0^1 {4xf'\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4f\left( 1 \right) - 4\int\limits_0^1 {xf'\left( x \right)dx}  = 8 - 4\int\limits_0^1 {xf'\left( x \right)dx} \end{array}\)

\(\begin{array}{l} \Rightarrow \int\limits_0^1 {{{\left( {f'\left( x \right)} \right)}^2}dx}  + 8 - 4\int\limits_0^1 {xf'\left( x \right)dx}  = \dfrac{{20}}{3}\\ \Leftrightarrow \int\limits_0^1 {{{\left( {f'\left( x \right)} \right)}^2}dx}  - 4\int\limits_0^1 {xf'\left( x \right)dx}  + 4\int\limits_0^1 {{x^2}dx}  - 4\int\limits_0^1 {{x^2}dx}  = \dfrac{{20}}{3} - 8\\ \Leftrightarrow \int\limits_0^1 {\left[ {{{\left( {f'\left( x \right)} \right)}^2} - 4xf'\left( x \right) + 4{x^2}} \right]dx}  - \dfrac{4}{3} =  - \dfrac{4}{3}\\ \Leftrightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right) - 2x} \right]}^2}dx}  = 0 \Leftrightarrow f'\left( x \right) - 2x = 0 \Leftrightarrow f'\left( x \right) = 2x\\ \Leftrightarrow f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {2xdx}  = {x^2} + C\end{array}\)

Lại có \(f\left( 1 \right) = 2 \Leftrightarrow 1 + C = 2 \Leftrightarrow C = 1 \Rightarrow f\left( x \right) = {x^2} + 1\).

Vậy \(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\left( {{x^2} + 1} \right)dx}  = \dfrac{4}{3}\).

Chọn C