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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& 2x-4\text{ khi }x\ge 1 \\
& 3{{x}^{2}}-7x+2\text{ khi }x<1 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-2\cos x \right)\sin xdx}$ bằng
A. $\dfrac{-3}{4}$.
B. $\dfrac{3}{4}$.
C. 3.
D. $\dfrac{-1}{2}$.
& 2x-4\text{ khi }x\ge 1 \\
& 3{{x}^{2}}-7x+2\text{ khi }x<1 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-2\cos x \right)\sin xdx}$ bằng
A. $\dfrac{-3}{4}$.
B. $\dfrac{3}{4}$.
C. 3.
D. $\dfrac{-1}{2}$.
Ta có
$\underset{x\to {{1}^{+}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }} \left( 2x-4 \right)=-2$ ; $\underset{x\to {{1}^{-}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} \left( 3{{x}^{2}}-7x+2 \right)=-2$ ; $f\left( 1 \right)=-2$
$\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} f\left( x \right)=f\left( 1 \right)$.
Suy ra hàm số đã cho liên tục tại x = 1.
Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-2\cos x \right)\sin xdx}$. Đặt $2-2\cos x=t\Rightarrow \sin xdx=\dfrac{1}{2}dt$.
Với $x=0\Rightarrow t=0$ ; $x=\dfrac{\pi }{2}\Rightarrow t=2$.
$\Rightarrow I=\int\limits_{0}^{2}{f\left( t \right)\dfrac{1}{2}dt}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{0}^{1}{\left( 3{{t}^{2}}-7t+2 \right)dt}+\dfrac{1}{2}\int\limits_{1}^{2}{\left( 2t-4 \right)dt}=\dfrac{-3}{4}$.
$\underset{x\to {{1}^{+}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }} \left( 2x-4 \right)=-2$ ; $\underset{x\to {{1}^{-}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} \left( 3{{x}^{2}}-7x+2 \right)=-2$ ; $f\left( 1 \right)=-2$
$\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} f\left( x \right)=f\left( 1 \right)$.
Suy ra hàm số đã cho liên tục tại x = 1.
Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2-2\cos x \right)\sin xdx}$. Đặt $2-2\cos x=t\Rightarrow \sin xdx=\dfrac{1}{2}dt$.
Với $x=0\Rightarrow t=0$ ; $x=\dfrac{\pi }{2}\Rightarrow t=2$.
$\Rightarrow I=\int\limits_{0}^{2}{f\left( t \right)\dfrac{1}{2}dt}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{0}^{1}{\left( 3{{t}^{2}}-7t+2 \right)dt}+\dfrac{1}{2}\int\limits_{1}^{2}{\left( 2t-4 \right)dt}=\dfrac{-3}{4}$.
Đáp án A.