Cho hàm số f9(x) có đạo hàm liên tục trên [0;2] thỏa mãn

Đáp án B

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = {x^2}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = \frac{{{x^3}}}{3}\end{array} \right.\), khi đó \(\int\limits_0^2 {{x^2}.f\left( x \right)dx} = \frac{{{x^3}}}{3}.f\left( x \right)\left| \begin{array}{l}^2\\_0\end{array} \right. - \int\limits_0^2 {\frac{{{x^3}}}{3}f'\left( x \right)dx} \)

Suy ra \(\frac{{40}}{{21}} = \frac{8}{3}f\left( 2 \right) - \int\limits_0^2 {\frac{{{x^3}}}{3}f'\left( x \right)dx} \Rightarrow \int\limits_0^2 {{x^3}f'\left( x \right)dx} = \frac{{16}}{7}\)

Ta chọn k sao cho: \(\int\limits_0^2 {{{\left[ {f'\left( x \right) + k{x^3}} \right]}^2}dx} = \int\limits_0^2 {{{\left[ {f'\left( x \right)} \right]}^2}dx + 2k\int\limits_0^2 {f'\left( x \right){x^3}dx} + {k^2}\int\limits_0^2 {{x^6}dx} = 0} \)

\( = \frac{2}{7} + \frac{{32}}{7}k + \frac{{128{k^2}}}{7} = 0 \Rightarrow k = \frac{{ - 1}}{8} \Rightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right) - \frac{1}{8}{x^3}} \right]}^2}dx = 0} \)

\( \Rightarrow f'\left( x \right) = \frac{{{x^3}}}{8} \Rightarrow f\left( x \right) = \frac{{{x^4}}}{{32}} + C\)

Do \(f\left( 2 \right) = 1 \Rightarrow C = \frac{1}{2} \Rightarrow f\left( x \right) = \frac{{{x^4}}}{{32}} + \frac{1}{2} \Rightarrow \int\limits_0^2 {f\left( x \right)dx} = \frac{6}{5}\)