Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0 ; 1 \right]$, thỏa mãn ${{\left( {f}'\left( x \right) \right)}^{2}}+4f\left( x \right)=8{{x}^{2}}+4, \forall x\in \left[ 0 ; 1 \right]$ và $f\left( 1 \right)=2$. Tính $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
A. $\dfrac{1}{3}$.
B. $2$.
C. $\dfrac{4}{3}$.
D. $\dfrac{21}{4}$.
A. $\dfrac{1}{3}$.
B. $2$.
C. $\dfrac{4}{3}$.
D. $\dfrac{21}{4}$.
Có ${{\left( {f}'\left( x \right) \right)}^{2}}+4f\left( x \right)=8{{x}^{2}}+4\Rightarrow \int\limits_{0}^{1}{{{\left( {f}'\left( x \right) \right)}^{2}}\text{d}x}+4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( 8{{x}^{2}}+4 \right)\text{d}x=\dfrac{20}{3}}$. (1)
Ta có $\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=\left. xf\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=2-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}\Rightarrow -4\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=-8+4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$. (2)
$\int\limits_{0}^{1}{{{\left( 2x \right)}^{2}}\text{d}x}=\dfrac{4}{3}$. (3)
Cộng vế với vế của (1), (2), (3) ta được $\int\limits_{0}^{1}{{{\left( {f}'\left( x \right)-2x \right)}^{2}}\text{d}x}=0\Rightarrow {f}'\left( x \right)=2x\Rightarrow f\left( x \right)={{x}^{2}}+C$.
Có $f\left( 1 \right)=C+1=2\Rightarrow C=1\Rightarrow f\left( x \right)={{x}^{2}}+1$.
Do đó $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{2}}+1 \right)\text{d}x}=\dfrac{4}{3}$.
Ta có $\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=\left. xf\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=2-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}\Rightarrow -4\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=-8+4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$. (2)
$\int\limits_{0}^{1}{{{\left( 2x \right)}^{2}}\text{d}x}=\dfrac{4}{3}$. (3)
Cộng vế với vế của (1), (2), (3) ta được $\int\limits_{0}^{1}{{{\left( {f}'\left( x \right)-2x \right)}^{2}}\text{d}x}=0\Rightarrow {f}'\left( x \right)=2x\Rightarrow f\left( x \right)={{x}^{2}}+C$.
Có $f\left( 1 \right)=C+1=2\Rightarrow C=1\Rightarrow f\left( x \right)={{x}^{2}}+1$.
Do đó $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{2}}+1 \right)\text{d}x}=\dfrac{4}{3}$.
Đáp án C.