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Câu hỏi: Nếu $\int\limits_{0}^{\dfrac{\pi }{3}}{\left[ \sin & x-3f\left( x \right) \right]}\text{d}x=6$ thì $\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x$ bằng
A. $\dfrac{13}{2}.$
B. $-\dfrac{11}{2}.$
C. $-\dfrac{13}{4}.$
D. $-\dfrac{11}{6}.$
$6=\int\limits_{0}^{\dfrac{\pi }{3}}{\left[ \sin & x-3f\left( x \right) \right]}\text{d}x=\int\limits_{0}^{\dfrac{\pi }{3}}{\sin & x\text{d}x-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}}\text{d}x=-\left. \text{co} & \text{s} x \right|_{0}^{\dfrac{\pi }{3}}-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=\dfrac{1}{2}-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x$
Suy ra $3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=\dfrac{1}{2}-6\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=-\dfrac{11}{6}$.
A. $\dfrac{13}{2}.$
B. $-\dfrac{11}{2}.$
C. $-\dfrac{13}{4}.$
D. $-\dfrac{11}{6}.$
Ta có$6=\int\limits_{0}^{\dfrac{\pi }{3}}{\left[ \sin & x-3f\left( x \right) \right]}\text{d}x=\int\limits_{0}^{\dfrac{\pi }{3}}{\sin & x\text{d}x-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}}\text{d}x=-\left. \text{co} & \text{s} x \right|_{0}^{\dfrac{\pi }{3}}-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=\dfrac{1}{2}-3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x$
Suy ra $3\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=\dfrac{1}{2}-6\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}\text{d}x=-\dfrac{11}{6}$.
Đáp án D.