Cho hàm số \(y = f\left( x \right)\) có đạo hàm liên tục trên \(\mathbb{R}\) thỏa mãn \(x \cdot f\left( x \right) \cdot f'\left( x \right) = {f^2}\left( x \right) - x,\,\,\forall x \in \mathbb{R}\) và có \(f\left( 2 \right) = 1\). Tích phân \[\int\limits_0^2 {{f^2}\left( x \right)dx} \] bằng

Ta có: \(x \cdot f\left( x \right) \cdot f'\left( x \right) = {f^2}\left( x \right) - x \Leftrightarrow 2x \cdot f\left( x \right) \cdot f'\left( x \right) = 2f_2^2\left( x \right) - 2x\)

\[ \Leftrightarrow 2x \cdot f\left( x \right) \cdot f'\left( x \right) + {f^2}\left( x \right) = 3{f^2}\left( x \right) - 2x \Leftrightarrow \int\limits_0^2 {{{\left( {x,\,\,{f^2}\left( x \right)} \right)}^\prime }dx} = 3\int\limits_0^2 {{f^2}\left( x \right)dx} - \int\limits_0^2 {2xdx} \]

\(\left. { \Leftrightarrow \left( {x \cdot {f^2}\left( x \right)} \right)} \right|_0^2 = 3I - 4 \Leftrightarrow 2 = 3I - 4 \Leftrightarrow I = 2\). Chọn C.