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

A

- Hướng dẫn giải

Phương pháp giải:

Sử dụng phương pháp tích phân từng phần, đặt \(\left\{ \begin{align} & u=f\left( x \right) \\& \text{d}v=3{{x}^{2}}\,\text{d}x \\\end{align} \right.\)

Giải chi tiết:

Đặt

\(\left\{ \begin{array}{l}
u = f\left( x \right)\\
{\rm{d}}v = 3{x^2}\,{\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u = f'\left( x \right)\,{\rm{d}}x\\
v = {x^3}
\end{array} \right.,\)

khi đó \(\int\limits_{0}^{1}{3{{x}^{2}}f\left( x \right)\,\text{d}x}=\left. {{x}^{3}}.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{{{x}^{3}}\,{f}'\left( x \right)\,\text{d}x}\)

Suy ra \(1=f\left( 1 \right)-\int\limits_{0}^{1}{{{x}^{3}}\,{f}'\left( x \right)\,\text{d}x}\Rightarrow \int\limits_{0}^{1}{{{x}^{3}}\,{f}'\left( x \right)\,\text{d}x}=-\,1\Leftrightarrow \int\limits_{0}^{1}{7{{x}^{3}}\,{f}'\left( x \right)\,\text{d}x}=-\,7.\)

Khi đó \(\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{\,2}}\,\text{d}x}+\int\limits_{0}^{1}{7{{x}^{3}}\,{f}'\left( x \right)\,\text{d}x}=0\Leftrightarrow \int\limits_{0}^{1}{{f}'\left( x \right)\left[ {f}'\left( x \right)+7{{x}^{3}} \right]\,\text{d}x}=0\)

Vậy \({f}'\left( x \right)+7{{x}^{3}}=0\Rightarrow f\left( x \right)=-\frac{7}{4}{{x}^{4}}+C\) mà \(f\left( 1 \right)=0\,\,\Rightarrow \,\,f\left( x \right)=\frac{7}{4}\left( 1-{{x}^{2}} \right)\Rightarrow \,\,\int\limits_{0}^{1}{f\left( x \right)\,\text{d}x}=\frac{7}{5}.\)

Chọn A.