Câu hỏi 33

\(f\left( {4x} \right) = f\left( x \right) + 4{x^3} + 2x \Leftrightarrow f\left( {4x} \right) - f\left( x \right) = 4{x^3} + 2x\)

\( \Rightarrow f\left( x \right)\) có dạng \(f\left( x \right) = a{x^3} + b{x^2} + cx + d\)

Vì \(f\left( 0 \right) = 2 \Rightarrow d = 2\).

Ta có:

\(\begin{array}{l}f\left( {4x} \right) - f\left( x \right) = 4{x^3} + 2x\\ \Rightarrow \left( {64a{x^3} + 16b{x^2} + 4cx + 2} \right) - \left( {a{x^3} + b{x^2} + cx + 2} \right) = 4{x^3} + 2x\\ \Rightarrow 63a{x^3} + 15b{x^2} + 3cx = 4{x^3} + 2x\\ \Rightarrow \left\{ \begin{array}{l}63a = 4\\15b = 0\\3c = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{4}{{63}}\\b = 0\\c = \dfrac{2}{3}\end{array} \right.\\ \Rightarrow f\left( x \right) = \dfrac{4}{{63}}{x^3} + \dfrac{2}{3}x + 2\end{array}\)

Vậy \(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\left( {\dfrac{4}{{63}}{x^3} + \dfrac{2}{3}x + 2} \right)dx}  = \dfrac{{148}}{{63}}\).

Chọn A.