Cho hàm số f(x) có đạo hàm liên tục trên [0;1]

* Hướng dẫn giải

Đáp án A

Đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=\left(2x-1\right)dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=f\left(x\right)dx\\ v=x^2-x\end{array}\right.$

$\Rightarrow -\frac{1}{30}=\underset{0}{\overset{1}{\int }}\left(2x-1\right)f\left(x\right)dx=\left(x^2-x\right)f\left(x\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.-\underset{0}{\overset{1}{\int }}\left(x^2-x\right)f^{\text{'}}\left(x\right)dx=-\underset{0}{\overset{1}{\int }}\left(x^2-x\right)f^{\text{'}}\left(x\right)dx$

$\Rightarrow \underset{0}{\overset{1}{\int }}\left(x^2-x\right)f^{\text{'}}\left(x\right)dx=\frac{1}{30}$.

Ta có: $\underset{0}{\overset{1}{\int }}{\left(x^2-x\right)}^{2}dx=\underset{0}{\overset{1}{\int }}\left(x^4-2x^3+x^2\right)dx=\left(\frac{x^5}{5}-\frac{x^4}{2}+\frac{x^3}{3}\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.=\frac{1}{30}$.

Do đó, $\underset{0}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)-\left(x^2-x\right)\right]}^{2}dx=\underset{0}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)\right]}^{2}dx-2\underset{0}{\overset{1}{\int }}\left(x^2-x\right)f\left(x\right)dx+\underset{0}{\overset{1}{\int }}{\left(x^2-x\right)}^{2}dx=0$  

$\Rightarrow f^{\text{'}}\left(x\right)=x^2-x\Rightarrow f\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}+C$, mà f(0)=1 nên $C=1\Rightarrow f\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}+1$

Vậy $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}\left(\frac{x^3}{3}-\frac{x^2}{2}+1\right)dx=\left(\frac{x^4}{12}-\frac{x^3}{6}+x\right)\left|\begin{array}{l}{}^1\\ {}_0\end{array}\right.=\frac{11}{12}$