Cho $f(x)$ là hàm số liên tục có đạo hàm $f^{\prime}(x)$ trên $[0...

Đặt $I=\int_{0}^{1}{f}(x)dx=-\dfrac{1}{3}.$
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {u}'={f}'\left( x \right)dx \\
& v=x \\
\end{aligned} \right..$
$I=x.f\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{x.{f}'\left( x \right)}.dx\Leftrightarrow \dfrac{1}{3}=\int\limits_{0}^{1}{x.{f}'\left( x \right)}.dx\Rightarrow 2\int\limits_{0}^{1}{x.{f}'\left( x \right)}.dx=\dfrac{2}{3}.$
$\int_{0}^{1}{{{\left( {{f}^{\prime }}(x) \right)}^{2}}}dx-2\int_{0}^{1}{xf}(x)dx+\int\limits_{0}^{1}{{{x}^{2}}.dx}=\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{3}=0\Rightarrow \int_{0}^{1}{{{\left( {{f}^{\prime }}(x)-x \right)}^{2}}}dx=0\Rightarrow {{f}^{\prime }}(x)-x=0$
$\Rightarrow {{f}^{\prime }}(x)=x\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}+C.$
Theo bài ra $f(1)=0\Rightarrow C=\dfrac{-1}{2}\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}-\dfrac{1}{2}\Rightarrow \int\limits_{0}^{\dfrac{1}{2}}{\left( \dfrac{{{x}^{2}}}{2}-\dfrac{1}{2} \right)}=-\dfrac{11}{48}.$