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Câu hỏi: Tích phân $\int_0^2 \min \left\{x^2, 3 x-2\right\} \mathrm{d} x$ bằng
A. $\dfrac{17}{6}$.
B. $\dfrac{-2}{3}$.
C. $\dfrac{11}{6}$.
D. $\dfrac{2}{3}$.
A. $\dfrac{17}{6}$.
B. $\dfrac{-2}{3}$.
C. $\dfrac{11}{6}$.
D. $\dfrac{2}{3}$.
Ta có $x^2=3 x-2 \Leftrightarrow x^2-3 x+2=0 \Leftrightarrow\left[\begin{array}{l}x=1 \\ x=2\end{array}\right.$.
Suy ra $x^2-3 x+2$ âm trên khoảng $(0,1)$ ; dương trên $(1,2)$.
Vậy $\min _{[0,1]}\left\{x^2, 3 x-2\right\}=3 x-2, \min _{[1,2]}\left\{x^2, 3 x-2\right\}=x^2$
Vậy $\int_0^2 \min \left\{x^2, 3 x-2\right\} \mathrm{d} x=\int_0^1(3 x-2) \mathrm{d} x+\int_1^2 x^2 \mathrm{~d} x=-\dfrac{1}{2}+\dfrac{7}{3}=\dfrac{11}{6}$.
Suy ra $x^2-3 x+2$ âm trên khoảng $(0,1)$ ; dương trên $(1,2)$.
Vậy $\min _{[0,1]}\left\{x^2, 3 x-2\right\}=3 x-2, \min _{[1,2]}\left\{x^2, 3 x-2\right\}=x^2$
Vậy $\int_0^2 \min \left\{x^2, 3 x-2\right\} \mathrm{d} x=\int_0^1(3 x-2) \mathrm{d} x+\int_1^2 x^2 \mathrm{~d} x=-\dfrac{1}{2}+\dfrac{7}{3}=\dfrac{11}{6}$.
Đáp án C.