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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${f}'\left( x \right)=\left( x+1 \right){{e}^{x}}$ và $f\left( 0 \right)=1$. Tính $f\left( 2 \right)$.
A. $f\left( 2 \right)=4{{\text{e}}^{2}}+1.$
B. $f\left( 2 \right)=2{{\text{e}}^{2}}+1.$
C. $f\left( 2 \right)=3{{\text{e}}^{2}}+1.$
D. $f\left( 2 \right)={{\text{e}}^{2}}+1.$
A. $f\left( 2 \right)=4{{\text{e}}^{2}}+1.$
B. $f\left( 2 \right)=2{{\text{e}}^{2}}+1.$
C. $f\left( 2 \right)=3{{\text{e}}^{2}}+1.$
D. $f\left( 2 \right)={{\text{e}}^{2}}+1.$
Hàm số ${f}'\left( x \right)=\left( x+1 \right){{e}^{x}}$ liên tục trên $\mathbb{R}$ nên liên tục trên đoạn $\left[ 0;2 \right]$. Do đó ta có
$\int\limits_{0}^{2}{{f}'\left( x \right)} dx=\left. f\left( x \right) \right|_{0}^{2}=f\left( 2 \right)-f\left( 0 \right).$
Theo đề bài cho ta có $\int\limits_{0}^{2}{{f}'\left( x \right)} dx=\int\limits_{0}^{2}{\left( x+1 \right){{e}^{x}}dx}$.
Đặt $\left\{ \begin{aligned}
& u=x+1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{2}{f'\left( x \right)} dx=\left. \left( x+1 \right){{e}^{x}} \right|_{0}^{2}-\int\limits_{0}^{2}{{{e}^{x}}dx}=\left. \left( x+1 \right){{e}^{x}} \right|_{0}^{2}-\left. {{e}^{x}} \right|_{0}^{2}=3{{e}^{2}}-{{e}^{0}}-\left( {{e}^{2}}-{{e}^{0}} \right)=2{{e}^{2}}.$
Suy ra $f\left( 2 \right)-f\left( 0 \right)=2{{e}^{2}}\Leftrightarrow f\left( 2 \right)-1=2{{e}^{2}}$
$\Leftrightarrow f\left( 2 \right)=2{{e}^{2}}+1.$
Vậy $f\left( 2 \right)=2{{e}^{2}}+1.$
$\int\limits_{0}^{2}{{f}'\left( x \right)} dx=\left. f\left( x \right) \right|_{0}^{2}=f\left( 2 \right)-f\left( 0 \right).$
Theo đề bài cho ta có $\int\limits_{0}^{2}{{f}'\left( x \right)} dx=\int\limits_{0}^{2}{\left( x+1 \right){{e}^{x}}dx}$.
Đặt $\left\{ \begin{aligned}
& u=x+1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{2}{f'\left( x \right)} dx=\left. \left( x+1 \right){{e}^{x}} \right|_{0}^{2}-\int\limits_{0}^{2}{{{e}^{x}}dx}=\left. \left( x+1 \right){{e}^{x}} \right|_{0}^{2}-\left. {{e}^{x}} \right|_{0}^{2}=3{{e}^{2}}-{{e}^{0}}-\left( {{e}^{2}}-{{e}^{0}} \right)=2{{e}^{2}}.$
Suy ra $f\left( 2 \right)-f\left( 0 \right)=2{{e}^{2}}\Leftrightarrow f\left( 2 \right)-1=2{{e}^{2}}$
$\Leftrightarrow f\left( 2 \right)=2{{e}^{2}}+1.$
Vậy $f\left( 2 \right)=2{{e}^{2}}+1.$
Đáp án B.