Cho hàm số $f(x)$ có đạo hàm liên tục trên $\left[ 0;2 \right]$...

Đặt $\left\{ \begin{aligned}
& u=f(x) \\
& dv={{x}^{2}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=f'(x)dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right. $, khi đó $ \int\limits_{0}^{2}{{{x}^{2}}f(x)dx=}\dfrac{{{x}^{3}}}{3}.\left. f(x) \right|_{0}^{2}-\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}}f'(x)dx$
Suy ra $\dfrac{40}{21}=\dfrac{8}{3}f(2)-2\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}}f'(x)dx\Rightarrow \int\limits_{0}^{2}{{{x}^{3}}}f'(x)dx=\dfrac{16}{7}$
Ta chọn k sao cho: $\int\limits_{0}^{2}{{{\left[ f'(x)+k{{x}^{3}} \right]}^{2}}dx=}\int\limits_{0}^{2}{{{\left[ f'(x) \right]}^{2}}dx+2k\int\limits_{0}^{2}{f'(x){{x}^{3}}dx}}+{{k}^{2}}\int\limits_{0}^{2}{{{x}^{6}}dx}=0$
$=\dfrac{2}{7}+\dfrac{32}{7}k+\dfrac{128{{k}^{2}}}{7}=0\Rightarrow k=\dfrac{-1}{8}\Rightarrow \int\limits_{0}^{1}{{{\left[ f'(x)-\dfrac{1}{8}{{x}^{3}} \right]}^{2}}dx=0\Rightarrow f'(x)=\dfrac{{{x}^{3}}}{8}}\Rightarrow f(x)=\dfrac{{{x}^{4}}}{32}+C$
Do $f(2)=1\Rightarrow C=\dfrac{1}{2}\Rightarrow f(x)=\dfrac{{{x}^{4}}}{32}+\dfrac{1}{2}\Rightarrow \int\limits_{0}^{2}{f(x)dx=\dfrac{6}{5}}$. Chọn B.