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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là ${f}'(x)=-3{{x}^{2}}+6x-2,\forall x\in \mathbb{R}$ và $f\left( -1 \right)=6$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thỏa mãn $F\left( 1 \right)=\dfrac{3}{4}$, khi đó $F\left( 2 \right)$ bằng
A. $1$.
B. $2$.
C. $3$.
D. $4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}$ $=\int{\left( -3{{x}^{2}}+6x-2 \right)\text{d}x=-{{x}^{3}}+3{{x}^{2}}-2x+C}$.
Có $f\left( -1 \right)=6$ $\Leftrightarrow 6+C=6\Leftrightarrow C=0$. Suy ra $f\left( x \right)=-{{x}^{3}}+3{{x}^{2}}-2x$.
Ta lại có: $\left. F\left( x \right) \right|_{1}^{2}=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( 2 \right)-F\left( 1 \right)=\int\limits_{1}^{2}{\left( -{{x}^{3}}+3{{x}^{2}}-2x \right)\text{d}x}$
$\Leftrightarrow F\left( 2 \right)-\dfrac{3}{4}=\left. \left( -\dfrac{{{x}^{4}}}{4}+{{x}^{3}}-{{x}^{2}} \right) \right|_{1}^{2}$ $\Leftrightarrow F\left( 2 \right)-\dfrac{3}{4}=0-\left( -\dfrac{1}{4} \right)$ $\Leftrightarrow F\left( 2 \right)=1$.
Vậy $F\left( 2 \right)=1$.
A. $1$.
B. $2$.
C. $3$.
D. $4$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}$ $=\int{\left( -3{{x}^{2}}+6x-2 \right)\text{d}x=-{{x}^{3}}+3{{x}^{2}}-2x+C}$.
Có $f\left( -1 \right)=6$ $\Leftrightarrow 6+C=6\Leftrightarrow C=0$. Suy ra $f\left( x \right)=-{{x}^{3}}+3{{x}^{2}}-2x$.
Ta lại có: $\left. F\left( x \right) \right|_{1}^{2}=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( 2 \right)-F\left( 1 \right)=\int\limits_{1}^{2}{\left( -{{x}^{3}}+3{{x}^{2}}-2x \right)\text{d}x}$
$\Leftrightarrow F\left( 2 \right)-\dfrac{3}{4}=\left. \left( -\dfrac{{{x}^{4}}}{4}+{{x}^{3}}-{{x}^{2}} \right) \right|_{1}^{2}$ $\Leftrightarrow F\left( 2 \right)-\dfrac{3}{4}=0-\left( -\dfrac{1}{4} \right)$ $\Leftrightarrow F\left( 2 \right)=1$.
Vậy $F\left( 2 \right)=1$.
Đáp án A.
