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Câu hỏi: Nếu ${\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 2f\left( x \right)-3\sin x \right]}\text{d}x=1}$ thì ${\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x}$ bằng
A. $\dfrac{1}{2}$.
B. $-1$.
C. $2$.
D. $\dfrac{3}{2}$.
A. $\dfrac{1}{2}$.
B. $-1$.
C. $2$.
D. $\dfrac{3}{2}$.
Ta có $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 2f\left( x \right)-3\sin x \right]}\text{d}x=1\Rightarrow 2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=3\int\limits_{0}^{\dfrac{\pi }{2}}{\text{sin}x\text{d}x}+1=-3\cos x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.+1=4$.
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x=2$.
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.+1=4$.
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x=2$.
Đáp án C.
