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Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=2$ và $f'\left( x \right)=\dfrac{2{{e}^{x}}+1}{{{e}^{x}}},\forall x\in \mathbb{R}$, khi đó $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng:
A. $\dfrac{3e-1}{e}.$
B. $\dfrac{3\left( e+1 \right)}{e}.$
C. $\dfrac{3e+1}{e}.$
D. $\dfrac{3\left( e-1 \right)}{e}.$
A. $\dfrac{3e-1}{e}.$
B. $\dfrac{3\left( e+1 \right)}{e}.$
C. $\dfrac{3e+1}{e}.$
D. $\dfrac{3\left( e-1 \right)}{e}.$
Ta có: $f\left( x \right)=\int{f'\left( x \right)dx}=\int{\left( \dfrac{2{{e}^{x}}+1}{{{e}^{x}}} \right)dx}=2x-{{e}^{-x}}+C$.
Theo bài ra ta có: $f\left( 0 \right)=2\Leftrightarrow -1+C=2\Leftrightarrow C=3$. Suy ra $f\left( x \right)=2x-{{e}^{-x}}+3$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( 2x-{{e}^{-x}}+3 \right)dx}=\left( {{x}^{2}}+{{e}^{-x}}+3x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\left( 4+\dfrac{1}{e} \right)-1=\dfrac{3e+1}{e}$.
Theo bài ra ta có: $f\left( 0 \right)=2\Leftrightarrow -1+C=2\Leftrightarrow C=3$. Suy ra $f\left( x \right)=2x-{{e}^{-x}}+3$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( 2x-{{e}^{-x}}+3 \right)dx}=\left( {{x}^{2}}+{{e}^{-x}}+3x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\left( 4+\dfrac{1}{e} \right)-1=\dfrac{3e+1}{e}$.
Đáp án C.