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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là ${f}'(x)=\left( 1-x \right){{e}^{-x}},\forall x\in \mathbb{R}$ và $f\left( 2 \right)=\dfrac{2}{{{e}^{2}}}$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thỏa mãn $F\left( 0 \right)=3+\dfrac{2}{e}$, khi đó $F\left( 1 \right)$ bằng
A. $1$.
B. $2$.
C. $3$.
D. $4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}$ $=\int{\left( 1-x \right){{e}^{-x}}\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=1-x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=-\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$.
$\Rightarrow f\left( x \right)=-\left( 1-x \right){{e}^{-x}}-\int{{{e}^{-x}}\text{d}x=\left( x-1 \right){{e}^{-x}}+{{e}^{-x}}+C}=x{{e}^{-x}}+C$.
Do $f\left( 2 \right)=\dfrac{2}{{{e}^{2}}}$ $\Rightarrow \dfrac{2}{{{e}^{2}}}+C=\dfrac{2}{{{e}^{2}}}$ $\Leftrightarrow C=0$. Suy ra $f\left( x \right)=x{{e}^{-x}}$.
Ta lại có: $\left. F\left( x \right) \right|_{0}^{1}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( 1 \right)-F\left( 0 \right)=\int\limits_{0}^{1}{x{{e}^{-x}}\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$.
Ta có: $F\left( 1 \right)-F\left( 0 \right)=\left. \left( -x{{e}^{-x}} \right) \right|_{0}^{1}+\int\limits_{0}^{1}{{{e}^{-x}}\text{d}x}$ $\Leftrightarrow F\left( 1 \right)-\left( 3+\dfrac{2}{e} \right)=-{{e}^{-1}}+\left. \left( -{{e}^{-x}} \right) \right|_{0}^{1}$
$\Leftrightarrow F\left( 1 \right)-\left( 3+\dfrac{2}{e} \right)=-2{{e}^{-1}}+1$ $\Leftrightarrow F\left( 1 \right)=4$.
Vậy $F\left( 1 \right)=4$.
A. $1$.
B. $2$.
C. $3$.
D. $4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}$ $=\int{\left( 1-x \right){{e}^{-x}}\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=1-x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=-\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$.
$\Rightarrow f\left( x \right)=-\left( 1-x \right){{e}^{-x}}-\int{{{e}^{-x}}\text{d}x=\left( x-1 \right){{e}^{-x}}+{{e}^{-x}}+C}=x{{e}^{-x}}+C$.
Do $f\left( 2 \right)=\dfrac{2}{{{e}^{2}}}$ $\Rightarrow \dfrac{2}{{{e}^{2}}}+C=\dfrac{2}{{{e}^{2}}}$ $\Leftrightarrow C=0$. Suy ra $f\left( x \right)=x{{e}^{-x}}$.
Ta lại có: $\left. F\left( x \right) \right|_{0}^{1}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( 1 \right)-F\left( 0 \right)=\int\limits_{0}^{1}{x{{e}^{-x}}\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={{e}^{-x}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=-{{e}^{-x}} \\
\end{aligned} \right.$.
Ta có: $F\left( 1 \right)-F\left( 0 \right)=\left. \left( -x{{e}^{-x}} \right) \right|_{0}^{1}+\int\limits_{0}^{1}{{{e}^{-x}}\text{d}x}$ $\Leftrightarrow F\left( 1 \right)-\left( 3+\dfrac{2}{e} \right)=-{{e}^{-1}}+\left. \left( -{{e}^{-x}} \right) \right|_{0}^{1}$
$\Leftrightarrow F\left( 1 \right)-\left( 3+\dfrac{2}{e} \right)=-2{{e}^{-1}}+1$ $\Leftrightarrow F\left( 1 \right)=4$.
Vậy $F\left( 1 \right)=4$.
Đáp án D.