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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{matrix}
{{x}^{2}}-1 khi x\ge 2 \\
{{x}^{2}}-2x+3 khi x<2 \\
\end{matrix} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos 2x+1 \right)\sin 2x\text{d}x}$ bằng
A. $\dfrac{43}{3}$.
B. $\dfrac{43}{12}$.
C. $\dfrac{14}{12}$.
D. $\dfrac{14}{3}$.
{{x}^{2}}-1 khi x\ge 2 \\
{{x}^{2}}-2x+3 khi x<2 \\
\end{matrix} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos 2x+1 \right)\sin 2x\text{d}x}$ bằng
A. $\dfrac{43}{3}$.
B. $\dfrac{43}{12}$.
C. $\dfrac{14}{12}$.
D. $\dfrac{14}{3}$.
Đặt $t=2\cos 2x+1\Leftrightarrow \text{d}t=-4\sin 2x\text{d}x$.
Đổi cận $x=0\Rightarrow t=3; x=\dfrac{\pi }{2}\Rightarrow t=-1$.
Tích phân trở thành:
$I=-\dfrac{1}{4}\int\limits_{3}^{-1}{f\left( t \right)\text{d}t}=\dfrac{1}{4}\int\limits_{-1}^{3}{f\left( t \right)\text{d}t}=\dfrac{1}{4}\left( \int\limits_{-1}^{2}{f\left( t \right)\text{d}t}+\int\limits_{2}^{3}{f\left( t \right)\text{d}t} \right)$
$=\dfrac{1}{4}\left( \int\limits_{-1}^{2}{\left( {{t}^{2}}-2t+3 \right)\text{d}t}+\int\limits_{2}^{3}{\left( {{t}^{2}}-1 \right)\text{d}t} \right)$
$=\dfrac{1}{4}\left( 9+\dfrac{16}{3} \right)=\dfrac{43}{12}$.
Đổi cận $x=0\Rightarrow t=3; x=\dfrac{\pi }{2}\Rightarrow t=-1$.
Tích phân trở thành:
$I=-\dfrac{1}{4}\int\limits_{3}^{-1}{f\left( t \right)\text{d}t}=\dfrac{1}{4}\int\limits_{-1}^{3}{f\left( t \right)\text{d}t}=\dfrac{1}{4}\left( \int\limits_{-1}^{2}{f\left( t \right)\text{d}t}+\int\limits_{2}^{3}{f\left( t \right)\text{d}t} \right)$
$=\dfrac{1}{4}\left( \int\limits_{-1}^{2}{\left( {{t}^{2}}-2t+3 \right)\text{d}t}+\int\limits_{2}^{3}{\left( {{t}^{2}}-1 \right)\text{d}t} \right)$
$=\dfrac{1}{4}\left( 9+\dfrac{16}{3} \right)=\dfrac{43}{12}$.
Đáp án B.