Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ -1;1 \right]$ và thỏa mãn $f\left( 1 \right)=0$,
$\left( {f}'{{(x)}^{2}}+4f(x) \right)=8{{\text{x}}^{2}}+16\text{x}-8$ với mọi x thuộc $\left[ -1;1 \right]$. Giá trị của $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $-\dfrac{5}{3}$
B. $\dfrac{2}{3}$
C. $\dfrac{1}{5}$
D. $-\dfrac{1}{3}$
$\left( {f}'{{(x)}^{2}}+4f(x) \right)=8{{\text{x}}^{2}}+16\text{x}-8$ với mọi x thuộc $\left[ -1;1 \right]$. Giá trị của $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $-\dfrac{5}{3}$
B. $\dfrac{2}{3}$
C. $\dfrac{1}{5}$
D. $-\dfrac{1}{3}$
Ta có:
${{\left( {f}'(x) \right)}^{2}}+4f\left( x \right)=8{{\text{x}}^{2}}+16\text{x}-8\Rightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}+2\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\int\limits_{-1}^{1}{\left( 8{{\text{x}}^{2}}+16\text{x}-8 \right)d\text{x}}$ (1).
Xét $I=\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}$, đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=2\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)d\text{x} \\
& v=2\text{x}+2 \\
\end{aligned} \right.$.
Do đó $I=\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\left. \left( 2\text{x}+2 \right)f\left( x \right) \right|_{-1}^{1}-\int\limits_{-1}^{1}{\left( 2\text{x}+2 \right){f}'\left( x \right)d\text{x}}=-\int\limits_{-1}^{1}{\left( 2x+2 \right){f}'\left( x \right)d\text{x}}$.
Từ (1) suy ra $\int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}+2\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\int\limits_{-1}^{1}{\left( 8{{\text{x}}^{2}}+16\text{x}-8 \right)d\text{x}}$
$\Leftrightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}-2\int\limits_{-1}^{1}{\left( 2\text{x}+2 \right){f}'\left( x \right)d\text{x}}+\int\limits_{-1}^{1}{{{\left( 2\text{x}+2 \right)}^{2}}d\text{x}}=\int\limits_{-1}^{1}{\left( 12{{\text{x}}^{2}}+24\text{x}-4 \right)d\text{x}}$
$\Leftrightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right)-\left( 2\text{x}+2 \right) \right]}^{2}}d\text{x}}=0\Rightarrow {f}'\left( x \right)=2\text{x}+2\Rightarrow f\left( x \right)={{x}^{2}}+2\text{x}+C$.
Vì $f\left( 1 \right)=0$ nên $C=-3$. Suy ra $\int\limits_{0}^{1}{f\left( x \right)d\text{x}}=\int\limits_{0}^{1}{\left( {{x}^{2}}+2\text{x}-3 \right)d\text{x}}=-\dfrac{5}{3}$.
${{\left( {f}'(x) \right)}^{2}}+4f\left( x \right)=8{{\text{x}}^{2}}+16\text{x}-8\Rightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}+2\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\int\limits_{-1}^{1}{\left( 8{{\text{x}}^{2}}+16\text{x}-8 \right)d\text{x}}$ (1).
Xét $I=\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}$, đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=2\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)d\text{x} \\
& v=2\text{x}+2 \\
\end{aligned} \right.$.
Do đó $I=\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\left. \left( 2\text{x}+2 \right)f\left( x \right) \right|_{-1}^{1}-\int\limits_{-1}^{1}{\left( 2\text{x}+2 \right){f}'\left( x \right)d\text{x}}=-\int\limits_{-1}^{1}{\left( 2x+2 \right){f}'\left( x \right)d\text{x}}$.
Từ (1) suy ra $\int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}+2\int\limits_{-1}^{1}{2f\left( x \right)d\text{x}}=\int\limits_{-1}^{1}{\left( 8{{\text{x}}^{2}}+16\text{x}-8 \right)d\text{x}}$
$\Leftrightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}d\text{x}}-2\int\limits_{-1}^{1}{\left( 2\text{x}+2 \right){f}'\left( x \right)d\text{x}}+\int\limits_{-1}^{1}{{{\left( 2\text{x}+2 \right)}^{2}}d\text{x}}=\int\limits_{-1}^{1}{\left( 12{{\text{x}}^{2}}+24\text{x}-4 \right)d\text{x}}$
$\Leftrightarrow \int\limits_{-1}^{1}{{{\left[ {f}'\left( x \right)-\left( 2\text{x}+2 \right) \right]}^{2}}d\text{x}}=0\Rightarrow {f}'\left( x \right)=2\text{x}+2\Rightarrow f\left( x \right)={{x}^{2}}+2\text{x}+C$.
Vì $f\left( 1 \right)=0$ nên $C=-3$. Suy ra $\int\limits_{0}^{1}{f\left( x \right)d\text{x}}=\int\limits_{0}^{1}{\left( {{x}^{2}}+2\text{x}-3 \right)d\text{x}}=-\dfrac{5}{3}$.
Đáp án A.