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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0; 1 \right]$ thỏa mãn $f\left( 1 \right)=0$, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7$ và $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x.\cos x.f\left( \sin x \right)\text{d}x}=\dfrac{1}{3}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{7}{5}$.
B. $4$.
C. $\dfrac{7}{4}$.
D. $1$.
+ Xét $I=\dfrac{1}{3}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x.cosxf\left( \sin x \right)\text{d}x}$ $=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x.f\left( \sin x \right)\text{ds}inx}$.
$\Rightarrow I=\int\limits_{0}^{1}{{{u}^{2}}.f\left( u \right)\text{du}}$, hay $\int\limits_{0}^{1}{{{x}^{2}}.f\left( x \right)\text{dx=}\dfrac{1}{3}}$.
+ Ta có $\dfrac{1}{3}=\int\limits_{0}^{1}{{{x}^{2}}.f\left( x \right)\text{dx=}}$ $\int\limits_{0}^{1}{f\left( x \right)\text{d}\left( \dfrac{{{x}^{3}}}{3} \right)}$ $\left. =\dfrac{{{x}^{3}}}{3}f\left( x \right) \right|_{0}^{1}-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}$.
$\Rightarrow \int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}=-1$.
+Ta có $\int\limits_{0}^{1}{{{x}^{6}}}dx=\dfrac{1}{7}$.
Do $1={{\left[ \int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}} \right]}^{2}}\le $ $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}.\int\limits_{0}^{1}{{{x}^{6}}dx}=1$.
Dấu bằng xảy ra khi ${f}'\left( x \right)=k{{x}^{3}}$.
Mặt khác $\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}=-1\Rightarrow k=-7$ $\Rightarrow {f}'\left( x \right)=-7{{x}^{3}}$.
$\Rightarrow f\left( x \right)=-\dfrac{7}{4}{{x}^{4}}+C$. Do $f\left( 1 \right)=0\Rightarrow C=\dfrac{7}{4}$.
Vậy $f\left( x \right)=-\dfrac{7}{4}{{x}^{4}}+\dfrac{7}{4}$, nên $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{7}{5}$.
A. $\dfrac{7}{5}$.
B. $4$.
C. $\dfrac{7}{4}$.
D. $1$.
+ Xét $I=\dfrac{1}{3}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x.cosxf\left( \sin x \right)\text{d}x}$ $=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x.f\left( \sin x \right)\text{ds}inx}$.
$\Rightarrow I=\int\limits_{0}^{1}{{{u}^{2}}.f\left( u \right)\text{du}}$, hay $\int\limits_{0}^{1}{{{x}^{2}}.f\left( x \right)\text{dx=}\dfrac{1}{3}}$.
+ Ta có $\dfrac{1}{3}=\int\limits_{0}^{1}{{{x}^{2}}.f\left( x \right)\text{dx=}}$ $\int\limits_{0}^{1}{f\left( x \right)\text{d}\left( \dfrac{{{x}^{3}}}{3} \right)}$ $\left. =\dfrac{{{x}^{3}}}{3}f\left( x \right) \right|_{0}^{1}-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}$.
$\Rightarrow \int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}=-1$.
+Ta có $\int\limits_{0}^{1}{{{x}^{6}}}dx=\dfrac{1}{7}$.
Do $1={{\left[ \int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}} \right]}^{2}}\le $ $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}.\int\limits_{0}^{1}{{{x}^{6}}dx}=1$.
Dấu bằng xảy ra khi ${f}'\left( x \right)=k{{x}^{3}}$.
Mặt khác $\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)\text{dx}}=-1\Rightarrow k=-7$ $\Rightarrow {f}'\left( x \right)=-7{{x}^{3}}$.
$\Rightarrow f\left( x \right)=-\dfrac{7}{4}{{x}^{4}}+C$. Do $f\left( 1 \right)=0\Rightarrow C=\dfrac{7}{4}$.
Vậy $f\left( x \right)=-\dfrac{7}{4}{{x}^{4}}+\dfrac{7}{4}$, nên $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{7}{5}$.
Đáp án A.