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Câu hỏi: Cho hàm số $y={{x}^{3}}$ có một nguyên hàm là $F\left( x \right)$. Khẳng định nào sau đây là đúng?
A. $F\left( 2 \right)-F\left( 0 \right)=16\cdot $
B. $F\left( 2 \right)-F\left( 0 \right)=1\cdot $
C. $F\left( 2 \right)-F\left( 0 \right)=8\cdot $
D. $F\left( 2 \right)-F\left( 0 \right)=4\cdot $
A. $F\left( 2 \right)-F\left( 0 \right)=16\cdot $
B. $F\left( 2 \right)-F\left( 0 \right)=1\cdot $
C. $F\left( 2 \right)-F\left( 0 \right)=8\cdot $
D. $F\left( 2 \right)-F\left( 0 \right)=4\cdot $
$\begin{aligned}
& F\left( x \right)=\int{{{x}^{3}}dx=}\dfrac{{{x}^{4}}}{4}+C \\
& F\left( 2 \right)-F\left( 0 \right)=\left( \dfrac{{{2}^{4}}}{4}+C \right)-\left( \dfrac{{{0}^{4}}}{4}+C \right)=4\cdot \\
\end{aligned}$
& F\left( x \right)=\int{{{x}^{3}}dx=}\dfrac{{{x}^{4}}}{4}+C \\
& F\left( 2 \right)-F\left( 0 \right)=\left( \dfrac{{{2}^{4}}}{4}+C \right)-\left( \dfrac{{{0}^{4}}}{4}+C \right)=4\cdot \\
\end{aligned}$
Đáp án D.