Cho hàm số \(f(x)\) có đạo hàm liên tục trên đoạn...

B

- Hướng dẫn giải

Phương pháp giải:

Sử dụng công thức tích phân từng phần: \(\int\limits_{a}^{b}{udv}=\left. uv \right|_{a}^{b}-\int\limits_{a}^{b}{vdu}\)

Giải chi tiết:

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

Mà  \(f(1)=\frac{3}{5}\,\,,\,\,\int\limits_{0}^{1}{{{x}^{3}}f(x)dx=\frac{37}{180}}\)  suy ra \(\frac{37}{180}=\frac{3}{20}-\frac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}f'(x)dx}\Leftrightarrow \int\limits_{0}^{1}{{{x}^{4}}f'(x)dx}=-\frac{2}{9}\)

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

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

Mà \(f(1)=\frac{3}{5}\Rightarrow -\frac{2}{5}{{.1}^{5}}+C=\frac{3}{5}\Rightarrow C=1\Rightarrow f(x)-1=-\frac{2}{5}{{x}^{5}}\)

\(\Rightarrow \int\limits_{0}^{1}{\left[ f(x)-1 \right]dx=}\int\limits_{0}^{1}{\left[ -\frac{2}{5}{{x}^{5}} \right]dx=-\left. \frac{1}{15}{{x}^{6}} \right|_{0}^{1}=-\frac{1}{15}}\)

Chọn: B