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Câu hỏi: Tính tích phân $I=\int\limits_{0}^{2}{\max \left\{ {{x}^{2}};x \right\}dx}$.
A. $I=\dfrac{17}{6}$
B. $I=\dfrac{11}{6}$
C. $I=\dfrac{7}{6}$
D. $I=\dfrac{8}{3}$
A. $I=\dfrac{17}{6}$
B. $I=\dfrac{11}{6}$
C. $I=\dfrac{7}{6}$
D. $I=\dfrac{8}{3}$
Trên đoạn $\left[ 0;2 \right]$, xét: ${{x}^{2}}\ge x\Leftrightarrow x(x-1)\ge 0\Leftrightarrow \left[ \begin{aligned}
& x\le 0 \\
& x\ge 1 \\
\end{aligned} \right.\xrightarrow{x\in \left[ 0;2 \right]}x\in \left[ 1;2 \right]\cup \left\{ 0 \right\}$.
Nghĩa là: $\max \left\{ {{x}^{2}};x \right\}=\left\{ \begin{aligned}
& {{x}^{2}}\text{ khi x}\in \left[ 1;2 \right] \\
& x\text{ khi x}\in \left[ 0;1 \right] \\
\end{aligned} \right.$.
Suy ra: $I=\int\limits_{0}^{2}{\max \left\{ {{x}^{2}};x \right\}dx}=\int\limits_{0}^{1}{xdx}+\int\limits_{1}^{2}{{{x}^{2}}dx}=\dfrac{17}{6}$.
& x\le 0 \\
& x\ge 1 \\
\end{aligned} \right.\xrightarrow{x\in \left[ 0;2 \right]}x\in \left[ 1;2 \right]\cup \left\{ 0 \right\}$.
Nghĩa là: $\max \left\{ {{x}^{2}};x \right\}=\left\{ \begin{aligned}
& {{x}^{2}}\text{ khi x}\in \left[ 1;2 \right] \\
& x\text{ khi x}\in \left[ 0;1 \right] \\
\end{aligned} \right.$.
Suy ra: $I=\int\limits_{0}^{2}{\max \left\{ {{x}^{2}};x \right\}dx}=\int\limits_{0}^{1}{xdx}+\int\limits_{1}^{2}{{{x}^{2}}dx}=\dfrac{17}{6}$.
Đáp án A.