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Câu hỏi: : Cho hàm số $y=\cos 4x$ có một nguyên hàm $F\left( x \right).$ Khẳng định nào sau đây đúng?
A. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=1.$
B. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=\dfrac{1}{4}.$
C. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=-1.$
D. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=\dfrac{-1}{4}.$
A. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=1.$
B. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=\dfrac{1}{4}.$
C. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=-1.$
D. $F\left( \dfrac{\pi }{8} \right)-F\left( 0 \right)=\dfrac{-1}{4}.$
Ta có
$\int\limits_{0}^{\dfrac{\pi }{8}}{\cos 4xdx}=\dfrac{1}{4}\left( \sin 4x \right)\left| \begin{aligned}
& \dfrac{\pi }{8} \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{4}\left[ \left( \sin 4.\dfrac{\pi }{8} \right)-\left( \sin 4.0 \right) \right]=\dfrac{1}{4}\left[ \left( \sin \dfrac{\pi }{2} \right)-\left( \sin 0 \right) \right]=\dfrac{1}{4}\left( 1-0 \right)=\dfrac{1}{4}.$
$\int\limits_{0}^{\dfrac{\pi }{8}}{\cos 4xdx}=\dfrac{1}{4}\left( \sin 4x \right)\left| \begin{aligned}
& \dfrac{\pi }{8} \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{4}\left[ \left( \sin 4.\dfrac{\pi }{8} \right)-\left( \sin 4.0 \right) \right]=\dfrac{1}{4}\left[ \left( \sin \dfrac{\pi }{2} \right)-\left( \sin 0 \right) \right]=\dfrac{1}{4}\left( 1-0 \right)=\dfrac{1}{4}.$
Đáp án B.