Nếu tích phân từ 0 đến 1[f2(x)-f(x)]dx=5 và

Đáp án D

Ta có: $\underset{0}{\overset{1}{\int }}{\left[f\left(x\right)+1\right]}^{2}dx=36\iff \underset{0}{\overset{1}{\int }}\left[f^2\left(x\right)+2f\left(x\right)+1\right]dx=36$

$\begin{array}{l}\Rightarrow \underset{0}{\overset{1}{\int }}\left[f^2\left(x\right)+2f\left(x\right)+1\right]dx-\underset{0}{\overset{1}{\int }}\left[f^2\left(x\right)-f\left(x\right)\right]dx=36-5\\ \iff \underset{0}{\overset{1}{\int }}\left[3f\left(x\right)+1\right]dx=31\\ \iff 3\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{0}{\overset{1}{\int }}dx=31\end{array}$ .$\iff 3\underset{0}{\overset{1}{\int }}f\left(x\right)dx+x\left|{}_0^1\right.=31\iff 3\underset{0}{\overset{1}{\int }}f\left(x\right)dx+1=31\iff 3\underset{0}{\overset{1}{\int }}f\left(x\right)dx=30\iff \underset{0}{\overset{1}{\int }}f\left(x\right)dx=10$