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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của $f\left( x \right)=\sin 2x$ và $F\left( \dfrac{\pi }{4} \right)=1$. Tính $F\left( \dfrac{\pi }{6} \right)$.
A. $F\left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
B. $F\left( \dfrac{\pi }{6} \right)=\dfrac{5}{4}$.
C. $F\left( \dfrac{\pi }{6} \right)=\dfrac{3}{4}$.
D. $F\left( \dfrac{\pi }{6} \right)=0$.
A. $F\left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
B. $F\left( \dfrac{\pi }{6} \right)=\dfrac{5}{4}$.
C. $F\left( \dfrac{\pi }{6} \right)=\dfrac{3}{4}$.
D. $F\left( \dfrac{\pi }{6} \right)=0$.
Ta có: $F\left( x \right)=\int{f\left( x \right)\text{d}}x=\int{\sin 2x\text{d}}x=\dfrac{1}{2}\int{\sin 2x\text{d}}\left( 2x \right)=-\dfrac{1}{2}\text{cos}2x+C$.
$F\left( \dfrac{\pi }{4} \right)=1\Leftrightarrow -\dfrac{1}{2}\text{cos}\dfrac{\pi }{2}+C=1\Leftrightarrow C=1$.
Suy ra $F\left( x \right)=-\dfrac{1}{2}\text{cos}2x+1$.
Vậy $F\left( \dfrac{\pi }{6} \right)=\dfrac{3}{4}$.
$F\left( \dfrac{\pi }{4} \right)=1\Leftrightarrow -\dfrac{1}{2}\text{cos}\dfrac{\pi }{2}+C=1\Leftrightarrow C=1$.
Suy ra $F\left( x \right)=-\dfrac{1}{2}\text{cos}2x+1$.
Vậy $F\left( \dfrac{\pi }{6} \right)=\dfrac{3}{4}$.
Đáp án C.