Ta có

\(f\left( x \right) = 3f\left( {\frac{x}{2}} \right) \Leftrightarrow \mathop \smallint \nolimits^ f\left( x \right)dx = \mathop \smallint \nolimits^ 3f\left( {\frac{x}{2}} \right)dx \Leftrightarrow \mathop \smallint \nolimits^ f\left( x \right)dx = \mathop \smallint \nolimits^ 6f\left( {\frac{x}{2}} \right)d\left( {\frac{x}{2}} \right) \Leftrightarrow F\left( x \right) = 6F\left( {\frac{x}{2}} \right) + C\).

Với \(x = 4 \Rightarrow F\left( 4 \right) = 6F\left( 2 \right) + C \Rightarrow 6F\left( 2 \right) + C = 1\) (1).

Với \(x = 8 \Rightarrow F\left( 8 \right) = 6F\left( 4 \right) + C \Rightarrow F\left( 8 \right) - C = 6\) (2).

Từ (1) và (2) ta có \(6F\left( 2 \right) + F\left( 8 \right) = 7\).

Ta có hệ \(\left\{ {\begin{array}{*{20}{c}}{6F\left( 2 \right) + F\left( 8 \right) = 7}\\{5F\left( 2 \right) + 2F\left( 8 \right) = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{F(2) = 2}\\{F\left( 8 \right) = - 5}\end{array}} \right.\)

Có .

Do đó ta chọn như sau