Cho hàm số \(f\left( x \right)\) có đạo hàm trên \...

A

- Hướng dẫn giải

Phương pháp giải:

Tìm hàm số f(x) qua hàm số f’(x) dựa vào tạo hằng đẳng thức tích phân

Giải chi tiết:

Đặt \(\left\{ {\begin{array}{*{20}{l}}{u = f\left( x \right)}\\{{\rm{d}}v = x{\mkern 1mu} {\rm{d}}x}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{du = f'\left( x \right){\rm{d}}x}\\{v = \dfrac{{{x^2}}}{2}}\end{array}} \right..\)

Khi đó \(\int\limits_0^1 {x.f\left( x \right){\mkern 1mu} {\rm{d}}x}  = \left. {\dfrac{{{x^2}}}{2}.f\left( x \right)} \right|_0^1 - \int\limits_0^1 {\dfrac{{{x^2}}}{2}f'\left( x \right){\mkern 1mu} {\rm{d}}x}  = \dfrac{1}{2} - \int\limits_0^1 {\dfrac{{{x^2}}}{2}f'\left( x \right){\mkern 1mu} {\rm{d}}x} \)

\(\begin{array}{l} \Rightarrow \dfrac{1}{2}\int\limits_0^1 {{x^2}f'\left( x \right){\mkern 1mu} {\rm{d}}x}  = \dfrac{1}{2} - \int\limits_0^1 {x.f\left( x \right){\mkern 1mu} {\rm{d}}x} \\ \Rightarrow \dfrac{1}{2}\int\limits_0^1 {{x^2}f'\left( x \right){\mkern 1mu} {\rm{d}}x}  = \dfrac{1}{2} - \dfrac{4}{{15}}\\ \Rightarrow \int\limits_0^1 {{x^2}f'\left( x \right){\mkern 1mu} {\rm{d}}x}  = \dfrac{7}{{15}}\end{array}\)

Ta có

\(\begin{array}{l}\int\limits_0^1 {{{\left( {f'\left( x \right) - k{x^2}} \right)}^{{\kern 1pt} 2}}{\mkern 1mu} {\rm{d}}x}  = \int\limits_0^1 {{{\left( {f'\left( x \right)} \right)}^{{\kern 1pt} 2}}{\mkern 1mu} {\rm{d}}x}  - 2k\int\limits_0^1 {{x^2}f'\left( x \right){\mkern 1mu} {\rm{d}}x}  + {k^2}\int\limits_0^1 {{x^4}{\mkern 1mu} {\rm{d}}x} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{49}}{{45}} - 2.\dfrac{7}{{15}}k + \dfrac{1}{5}{k^2}.\end{array}\)

Suy ra \(\int\limits_0^1 {{{\left( {f'\left( x \right) - k{x^2}} \right)}^{{\kern 1pt} 2}}{\rm{d}}x}  = 0 \Leftrightarrow k = \dfrac{7}{3}\)\( \Rightarrow f'\left( x \right) = \dfrac{7}{3}{x^2} \Leftrightarrow f\left( x \right) = \dfrac{7}{9}{x^3} + C.\)

Do \(f\left( 1 \right) = 1 \Leftrightarrow C = \dfrac{2}{9} \Rightarrow f\left( x \right) = \dfrac{7}{9}{x^3} + \dfrac{2}{9}\).

Vậy \(\int\limits_0^1 {{{\left( {f\left( x \right)} \right)}^2}dx}  = \dfrac{2}{9}.\)

Chọn A