Nếu \[\mathop \smallint \limits_0^1 \left[ {{f^2}\left( x \right) - f\left( x \right)} \right]dx = 5\]và \[\mathop \smallint \limits_0^1 {\left[ {f\left( x \right) + 1} \right]^2}dx = 36\]thì \(\int\limits_0^1 {f\left( x \right)dx} \) bằng:

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

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

Đáp án cần chọn là: D