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Câu hỏi: Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned}
& 3{{x}^{2}} \text{khi} 0\le x\le 1 \\
& 4-x \text{khi} 1\le x\le 2 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{2}{f\left( x \right)\text{d}x}$.
A. $\dfrac{7}{2}$.
B. $1$.
C. $\dfrac{5}{2}$.
D. $\dfrac{3}{2}$.
& 3{{x}^{2}} \text{khi} 0\le x\le 1 \\
& 4-x \text{khi} 1\le x\le 2 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{2}{f\left( x \right)\text{d}x}$.
A. $\dfrac{7}{2}$.
B. $1$.
C. $\dfrac{5}{2}$.
D. $\dfrac{3}{2}$.
Ta có
$\int\limits_{0}^{2}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{1}{\left( 3{{x}^{2}} \right)\text{d}x}+\int\limits_{1}^{2}{\left( 4-x \right)\text{d}x}$ $=\left. \dfrac{3{{x}^{3}}}{3} \right|_{1}^{2}+\left. \left( 4x-\dfrac{{{x}^{2}}}{2} \right) \right|_{1}^{2}=$ $=\dfrac{7}{2}$.
$\int\limits_{0}^{2}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{1}{\left( 3{{x}^{2}} \right)\text{d}x}+\int\limits_{1}^{2}{\left( 4-x \right)\text{d}x}$ $=\left. \dfrac{3{{x}^{3}}}{3} \right|_{1}^{2}+\left. \left( 4x-\dfrac{{{x}^{2}}}{2} \right) \right|_{1}^{2}=$ $=\dfrac{7}{2}$.
Đáp án A.