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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 0 \right)=-1$ và ${f}'\left( x \right)=x\left( 6+12x+{{e}^{-x}} \right),\forall x\in \mathbb{R}$. Khi đó $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$ bằng
A. $3{{e}^{-1}}$.
B. $4-3{{e}^{-1}}$.
C. $-3{{e}^{-1}}$.
D. $3e$.
Ta có: ${f}'\left( x \right)=x\left( 6+12x+{{e}^{-x}} \right),\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
$\int{{f}'\left( x \right)\text{d}x=\int{x\left( 6+12x+{{e}^{-x}} \right)\text{d}x}}=\int{\left( 6x+12{{x}^{2}} \right)\text{d}x+\int{x{{e}^{-x}}\text{d}x}}$
Mà $\int{\left( 6x+12{{x}^{2}} \right)\text{d}x=3{{x}^{2}}+4{{x}^{3}}}+C$
Xét $\int{x{{e}^{-x}}\text{d}x}$ : Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$
$\int{x{{e}^{-x}}\text{d}x=-x{{e}^{-x}}+\int{{{e}^{-x}}}}\text{d}x=-x{{e}^{-x}}-{{e}^{-x}}+C=-\left( x+1 \right){{e}^{-x}}+C$
Suy ra $f\left( x \right)=3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}}+C,\forall x\in \mathbb{R}$.
Mà $f\left( 0 \right)=-1\Rightarrow C=0$ nên $f\left( x \right)=3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}},\forall x\in \mathbb{R}$.
Ta có
$\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\int\limits_{0}^{1}{\left( 3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}} \right)}\text{d}x=\left. \left( {{x}^{3}}+{{x}^{4}} \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}=2-\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}$
Xét $\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}$ : Đặt $\left\{ \begin{aligned}
& u=x+1 \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$
$\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}=\left. -\left( x+1 \right){{e}^{-x}} \right|_{0}^{1}+\int\limits_{0}^{1}{{{e}^{-x}}\text{d}x}=-2{{e}^{-1}}+1-\left. {{e}^{-x}} \right|_{0}^{1}=-2{{e}^{-1}}+1-{{e}^{-1}}+1=2-3{{e}^{-1}}$
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=3{{e}^{-1}}$.
A. $3{{e}^{-1}}$.
B. $4-3{{e}^{-1}}$.
C. $-3{{e}^{-1}}$.
D. $3e$.
Ta có: ${f}'\left( x \right)=x\left( 6+12x+{{e}^{-x}} \right),\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
$\int{{f}'\left( x \right)\text{d}x=\int{x\left( 6+12x+{{e}^{-x}} \right)\text{d}x}}=\int{\left( 6x+12{{x}^{2}} \right)\text{d}x+\int{x{{e}^{-x}}\text{d}x}}$
Mà $\int{\left( 6x+12{{x}^{2}} \right)\text{d}x=3{{x}^{2}}+4{{x}^{3}}}+C$
Xét $\int{x{{e}^{-x}}\text{d}x}$ : Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$
$\int{x{{e}^{-x}}\text{d}x=-x{{e}^{-x}}+\int{{{e}^{-x}}}}\text{d}x=-x{{e}^{-x}}-{{e}^{-x}}+C=-\left( x+1 \right){{e}^{-x}}+C$
Suy ra $f\left( x \right)=3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}}+C,\forall x\in \mathbb{R}$.
Mà $f\left( 0 \right)=-1\Rightarrow C=0$ nên $f\left( x \right)=3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}},\forall x\in \mathbb{R}$.
Ta có
$\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\int\limits_{0}^{1}{\left( 3{{x}^{2}}+4{{x}^{3}}-\left( x+1 \right){{e}^{-x}} \right)}\text{d}x=\left. \left( {{x}^{3}}+{{x}^{4}} \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}=2-\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}$
Xét $\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}$ : Đặt $\left\{ \begin{aligned}
& u=x+1 \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$
$\int\limits_{0}^{1}{\left( x+1 \right){{e}^{-x}}\text{d}x}=\left. -\left( x+1 \right){{e}^{-x}} \right|_{0}^{1}+\int\limits_{0}^{1}{{{e}^{-x}}\text{d}x}=-2{{e}^{-1}}+1-\left. {{e}^{-x}} \right|_{0}^{1}=-2{{e}^{-1}}+1-{{e}^{-1}}+1=2-3{{e}^{-1}}$
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=3{{e}^{-1}}$.
Đáp án A.