Cho hàm số $y=f(x)$ có đạo hàm liên tục trên $\mathbb{R}$ và thỏa mãn $2f(x)+f'(x)=2x+1$, $\forall x\in\mathbb{R}$ và $f(0)=1$. Giá trị của $\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x$ bằng
 | $1-\dfrac{1}{2\mathrm{e}^2}$ |
 | $1+\dfrac{1}{2\mathrm{e}^2}$ |
 | $\dfrac{1}{2\mathrm{e}^2}$ |
 | $-\dfrac{1}{2\mathrm{e}^2}$ |
Chọn phương án A.
$\begin{array}{lll}
&2f(x)+f'(x)&=2x+1\\
\Leftrightarrow&f'(x)-1&=-2\big(f(x)-x\big)\\
\Leftrightarrow&\dfrac{f'(x)-1}{f(x)-x}&=-2.
\end{array}$
$\begin{array}{lll}
&\displaystyle\int\dfrac{f'(x)-1}{f(x)-x}\mathrm{\,d}x&=\int(-2)\mathrm{\,d}x\\
\Leftrightarrow&\ln\big|f(x)-x\big|&=-2x+C\\
\Leftrightarrow&\big|f(x)-x\big|&=\mathrm{e}^{-2x+C}\\
\Leftrightarrow&\left[\begin{array}{l}
f(x)-x=\mathrm{e}^{-2x+C}\\
f(x)-x=-\mathrm{e}^{-2x+C}
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
f(x)=x+\mathrm{e}^{-2x+C}\\
f(x)=x-\mathrm{e}^{-2x+C}
\end{array}\right.\\
\Rightarrow&\left[\begin{array}{l}
f(0)=0+\mathrm{e}^{-2\cdot0+C}\\
f(0)=0-\mathrm{e}^{-2\cdot0+C}
\end{array}\right.\\
\Rightarrow&\left[\begin{array}{l}
1=\mathrm{e}^C\\
1=-\mathrm{e}^C\,\,\text{(vô lý)}
\end{array}\right.\\
\Rightarrow&C=0.
\end{array}$
Vậy $f(x)=x+\mathrm{e}^{-2x}$.
Suy ra $\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x=\int\limits_{0}^{1}\big(x+\mathrm{e}^{-2x}\big)\mathrm{\,d}x=1-\dfrac{1}{2\mathrm{e}^2}$.