Phương pháp giải:

Sử dụng các tính chất của tích phân:

\(\begin{array}{l}\int\limits_a^b {kf\left( x \right)dx}  = k\int\limits_a^b {f\left( x \right)dx} \,\,\,\left( {k \ne 0} \right)\\\int\limits_a^b {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx}  + \int\limits_c^b {f\left( x \right)dx} \\\int\limits_a^b {f\left( x \right)dx}  =  - \int\limits_b^a {f\left( x \right)dx} \\\int\limits_a^b {f\left( x \right)dx}  \pm \int\limits_a^b {g\left( x \right)dx}  = \int\limits_a^b {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} \end{array}\)

Lời giải chi tiết:

Ta có: \(\int\limits_0^1 {\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]dx = 5} \)

\(\begin{array}{l}\int\limits_0^1 {{{\left[ {f\left( x \right) + 1} \right]}^2}dx = 36}  \Leftrightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2f\left( x \right) + 1} \right]} dx = 36\\ \Rightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2f\left( x \right) + 1} \right]} dx - \int\limits_0^1 {\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]dx}  = 36 - 5\\ \Leftrightarrow \int\limits_0^1 {\left[ {3f\left( x \right) + 1} \right]dx}  = 31 \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + \int\limits_0^1 {dx}  = 31\\ \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + \left. x \right|_0^1 = 31 \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  + 1 = 31\\ \Leftrightarrow 3\int\limits_0^1 {f\left( x \right)dx}  = 30 \Leftrightarrow \int\limits_0^1 {f\left( x \right)dx}  = 10.\end{array}\)

Chọn D.