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

A

- Hướng dẫn giải

Phương pháp giải:

Đối với tích \(\int\limits_{0}^{1}{{{x}^{3}}f\left( x \right)\,\text{d}x}\), sử dụng phương pháp tích phân từng phần.

Tìm k để \(\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k{{x}^{4}} \right]}^{\,2}}\,\text{d}x}=0\)

Giải chi tiết:

Ta có \(\int\limits_{0}^{1}{{{x}^{3}}f\left( x \right)\,\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\,\text{d}\left( \frac{{{x}^{4}}}{4} \right)}=\left. \frac{{{x}^{4}}.f\left( x \right)}{4} \right|_{0}^{1}-\frac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)\,\text{d}x}=\frac{f\left( 1 \right)}{4}-\frac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)\,\text{d}x}\)

Mà \(f\left( 1 \right)=1\) và \(\int\limits_{0}^{1}{{{x}^{3}}f\left( x \right)\,\text{d}x}=\frac{1}{2}\) suy ra \(\frac{1}{2}=\frac{1}{4}-\frac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)\,\text{d}x}\Leftrightarrow \int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)\,\text{d}x}=-\,1.\)

Xét \(\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k{{x}^{4}} \right]}^{\,2}}\,\text{d}x}=\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{\,2}}\,\text{d}x}+2k.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)\,\text{d}x}+{{k}^{2}}.\int\limits_{0}^{1}{{{x}^{8}}\,\text{d}x}=9-2k+\frac{{{k}^{2}}}{9}=0\Rightarrow k=9.\)

Khi đó \(\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+9{{x}^{4}} \right]}^{\,2}}\,\text{d}x}=0\Rightarrow {f}'\left( x \right)+9{{x}^{4}}=0\Leftrightarrow {f}'\left( x \right)=-\,9{{x}^{4}}\)\(\Rightarrow \,\,f\left( x \right)=\int{{f}'\left( x \right)\,\text{d}x}=-\frac{9{{x}^{5}}}{5}+C.\)

Mặt khác \(f\left( 1 \right)=1\Rightarrow C-\frac{9}{5}=1\Leftrightarrow C=\frac{14}{5}.\) Vậy \(f\left( x \right)=-\frac{9{{x}^{5}}}{5}+\frac{14}{5}\)\(\Rightarrow \,\,\int\limits_{0}^{1}{f\left( x \right)\,\text{d}x}=\frac{5}{2}.\)

Chọn A.