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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$, thỏa mãn $f\left( x \right)=\left\{ \begin{matrix}
2x & \text{khi} x\ge 0 \\
{{x}^{3}}+x+1 & \text{khi} x<0 \\
\end{matrix} \right.$.
Tính $\int\limits_{0}^{2}{f\left( x-1 \right)\text{d}x}$.
A. $\dfrac{3}{2}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{5}{4}$.
D. $\dfrac{11}{4}$.
2x & \text{khi} x\ge 0 \\
{{x}^{3}}+x+1 & \text{khi} x<0 \\
\end{matrix} \right.$.
Tính $\int\limits_{0}^{2}{f\left( x-1 \right)\text{d}x}$.
A. $\dfrac{3}{2}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{5}{4}$.
D. $\dfrac{11}{4}$.
$\int\limits_{0}^{2}{f\left( x-1 \right)\text{d}x}=\int\limits_{-1}^{1}{f\left( x \right)\text{d}x}=\int\limits_{-1}^{0}{f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ $=\int\limits_{-1}^{0}{\left( {{x}^{3}}+x+1 \right)\text{d}x}+\int\limits_{0}^{1}{2x\text{d}x}$ $=\left. \left( \dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{2}}}{2}+x \right) \right|_{-1}^{0}+\left. {{x}^{2}} \right|_{0}^{1}=-\dfrac{1}{4}-\dfrac{1}{2}+1+1=\dfrac{5}{4}$.
Đáp án C.