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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}-2x\text{ khi }x\le 1 \\
& -{{x}^{3}}\text{ khi }x>1 \\
\end{aligned} \right.. $ Tính $ I=\int\limits_{0}^{2}{f\left( x \right)dx}.$
A. $I=-\dfrac{53}{12}.$
B. $I=-\dfrac{4}{3}.$
C. $I=-4$
D. $I=-\dfrac{16}{3}.$
& {{x}^{2}}-2x\text{ khi }x\le 1 \\
& -{{x}^{3}}\text{ khi }x>1 \\
\end{aligned} \right.. $ Tính $ I=\int\limits_{0}^{2}{f\left( x \right)dx}.$
A. $I=-\dfrac{53}{12}.$
B. $I=-\dfrac{4}{3}.$
C. $I=-4$
D. $I=-\dfrac{16}{3}.$
$I=\int\limits_{0}^{2}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{1}^{2}{f\left( x \right)dx}$
$=\int\limits_{0}^{1}{\left( {{x}^{2}}-2x \right)dx}+\int\limits_{1}^{2}{\left( -{{x}^{3}} \right)dx}$
$=\left( \dfrac{{{x}^{3}}}{3}-{{x}^{2}} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\left( -\dfrac{{{x}^{4}}}{4} \right)\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.$
$=-\dfrac{53}{12}.$
$=\int\limits_{0}^{1}{\left( {{x}^{2}}-2x \right)dx}+\int\limits_{1}^{2}{\left( -{{x}^{3}} \right)dx}$
$=\left( \dfrac{{{x}^{3}}}{3}-{{x}^{2}} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\left( -\dfrac{{{x}^{4}}}{4} \right)\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.$
$=-\dfrac{53}{12}.$
Đáp án A.