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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 0 \right)=\dfrac{1}{21}$ và ${f}'\left( x \right)=\sin 3x.{{\cos }^{2}}2x,\forall x\in \mathbb{R}$. Biết ${F\left( x \right)}$ là nguyên hàm của ${f\left( x \right)}$ thỏa mãn $F\left( 0 \right)=0$, khi đó $F\left( \dfrac{\pi }{2} \right)$ bằng
A. $\dfrac{137}{441}$.
B. $-\dfrac{137}{441}$.
C. $\dfrac{247}{441}$.
D. $\dfrac{167}{882}$.
A. $\dfrac{137}{441}$.
B. $-\dfrac{137}{441}$.
C. $\dfrac{247}{441}$.
D. $\dfrac{167}{882}$.
Ta có ${f}'\left( x \right)=\sin 3x.{{\cos }^{2}}2x,\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\sin 3x.{{\cos }^{2}}2x\text{d}x}=\int{\sin 3x.\dfrac{1+\cos 4x}{2}\text{d}x}=\int{\dfrac{\sin 3x}{2}\text{d}x+\int{\dfrac{\sin 3x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\sin 3x}\text{d}x+\dfrac{1}{4}\int{\left( \sin 7x-\sin x \right)\text{d}x=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x+C}$.
Suy ra $f\left( x \right)=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x+C,\forall x\in \mathbb{R}$. Mà $f\left( 0 \right)=\dfrac{1}{21}\Rightarrow C=0$.
Do đó $f\left( x \right)=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \dfrac{\pi }{2} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{18}\sin 3x-\dfrac{1}{196}\sin 7x+\dfrac{1}{4}\sin x \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{137}{441} \\
& \Rightarrow F\left( \dfrac{\pi }{2} \right)=F\left( 0 \right)+\dfrac{137}{441}=0+\dfrac{137}{441}=\dfrac{137}{441} \\
\end{aligned}$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\sin 3x.{{\cos }^{2}}2x\text{d}x}=\int{\sin 3x.\dfrac{1+\cos 4x}{2}\text{d}x}=\int{\dfrac{\sin 3x}{2}\text{d}x+\int{\dfrac{\sin 3x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\sin 3x}\text{d}x+\dfrac{1}{4}\int{\left( \sin 7x-\sin x \right)\text{d}x=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x+C}$.
Suy ra $f\left( x \right)=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x+C,\forall x\in \mathbb{R}$. Mà $f\left( 0 \right)=\dfrac{1}{21}\Rightarrow C=0$.
Do đó $f\left( x \right)=-\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \dfrac{\pi }{2} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\dfrac{1}{6}\cos 3x-\dfrac{1}{28}\cos 7x+\dfrac{1}{4}\cos x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{18}\sin 3x-\dfrac{1}{196}\sin 7x+\dfrac{1}{4}\sin x \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{137}{441} \\
& \Rightarrow F\left( \dfrac{\pi }{2} \right)=F\left( 0 \right)+\dfrac{137}{441}=0+\dfrac{137}{441}=\dfrac{137}{441} \\
\end{aligned}$.
Đáp án A.