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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ và thỏa mãn $f(x)+f(-x)=2sinx$. Tích phân $\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f(x)dx}$ bằng
A. $-1$
B. $0$
C. $1$
D. $2$
A. $-1$
B. $0$
C. $1$
D. $2$
Xét $I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f(x)dx}$. Đặt $x=-t\Rightarrow I=\int\limits_{\dfrac{\pi }{2}}^{-\dfrac{\pi }{2}}{f(-t)d(-t)}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f(-t)dt}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f(-x)dx}$
$\Rightarrow I+I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\left[ f(x)+f(-x) \right]dx}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\sin xdx}=-2\cos x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& -\dfrac{\pi }{2} \\
\end{aligned} \right.=0\Rightarrow I=0$
$\Rightarrow I+I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\left[ f(x)+f(-x) \right]dx}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\sin xdx}=-2\cos x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& -\dfrac{\pi }{2} \\
\end{aligned} \right.=0\Rightarrow I=0$
Đáp án B.