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