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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là ${f}'\left( x \right)=2{{e}^{x}}+x{{e}^{x}}, \forall x\in \mathbb{R}$ và $f\left( 0 \right)=1$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thoả mãn $F\left( 4 \right)=4{{e}^{4}}+3$, khi đó $F\left( 1 \right)$ bằng
A. $e$.
B. $e+2$.
C. $e+3$.
D. $e+4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}=\int{\left( 2{{e}^{x}}+x{{e}^{x}} \right)\text{d}x}=\int{\left( {{e}^{x}}+{{\left( x{{e}^{x}} \right)}^{\prime }} \right)\text{d}x}={{e}^{x}}+x{{e}^{x}}+C$.
Mà: $f\left( 0 \right)=1\Rightarrow 1+C=1\Rightarrow C=0$.
Do đó: $f\left( x \right)={{e}^{x}}+x{{e}^{x}}$.
Ta có: $F\left( x \right)=\int{f\left( x \right)\text{d}x}=\int{\left( {{e}^{x}}+x{{e}^{x}} \right)\text{d}x}=\int{{{\left( x{{e}^{x}} \right)}^{\prime }}\text{d}x}=x{{e}^{x}}+K$.
Mà: $F\left( 4 \right)=4{{e}^{4}}+3\Rightarrow 4{{e}^{4}}+K=4{{e}^{4}}+3\Rightarrow K=3$.
Do đó: $F\left( x \right)=x{{e}^{x}}+3$.
Vậy $F\left( 1 \right)=e+3$.
A. $e$.
B. $e+2$.
C. $e+3$.
D. $e+4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}=\int{\left( 2{{e}^{x}}+x{{e}^{x}} \right)\text{d}x}=\int{\left( {{e}^{x}}+{{\left( x{{e}^{x}} \right)}^{\prime }} \right)\text{d}x}={{e}^{x}}+x{{e}^{x}}+C$.
Mà: $f\left( 0 \right)=1\Rightarrow 1+C=1\Rightarrow C=0$.
Do đó: $f\left( x \right)={{e}^{x}}+x{{e}^{x}}$.
Ta có: $F\left( x \right)=\int{f\left( x \right)\text{d}x}=\int{\left( {{e}^{x}}+x{{e}^{x}} \right)\text{d}x}=\int{{{\left( x{{e}^{x}} \right)}^{\prime }}\text{d}x}=x{{e}^{x}}+K$.
Mà: $F\left( 4 \right)=4{{e}^{4}}+3\Rightarrow 4{{e}^{4}}+K=4{{e}^{4}}+3\Rightarrow K=3$.
Do đó: $F\left( x \right)=x{{e}^{x}}+3$.
Vậy $F\left( 1 \right)=e+3$.
Đáp án C.