Hàm số \(f( x ) = a{x^3} + b{x^2} + cx + d\) có \(f( 0 ) = 2\) và \(f( {4x} - f( x )= 4{x^3} + 2x

Ta có: \(\left\{ \begin{array}{l}f\left( {4x} \right) - f\left( x \right) = 4{x^3} + 2x,\,\forall x \in \mathbb{R}\\f\left( 0 \right) = 2\end{array} \right.\)

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

Vậy \[I = \int\limits_0^1 {f\left( x \right)dx} = \int\limits_0^1 {\left( {\frac{4}{{63}}{x^3} + \frac{2}{3}x + 2} \right)dx} = \frac{{148}}{{63}} \approx 2,35\].