Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=0$ và ${f}'\left( x \right)=\sin x.{{\sin }^{2}}2x,\forall x\in \mathbb{R}$. 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. $\dfrac{104}{225}$.
B. $-\dfrac{104}{225}$.
C. $\dfrac{121}{225}$.
D. $\dfrac{167}{225}$.
A. $\dfrac{104}{225}$.
B. $-\dfrac{104}{225}$.
C. $\dfrac{121}{225}$.
D. $\dfrac{167}{225}$.
Ta có ${f}'\left( x \right)=\sin x.{{\sin }^{2}}2x,\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\sin x.{{\sin }^{2}}2x\text{d}x}=\int{\sin x.\dfrac{1-\cos 4x}{2}\text{d}x}=\int{\dfrac{\sin x}{2}\text{d}x-\int{\dfrac{\sin x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\sin x}\text{d}x-\dfrac{1}{4}\int{\left( \sin 5x-\sin 3x \right)\text{d}x=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x+C}$.
Suy ra $f\left( x \right)=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x+C,\forall x\in \mathbb{R}$. Mà $f\left( \dfrac{\pi }{2} \right)=0\Rightarrow C=0$.
Do đó $f\left( x \right)=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \dfrac{\pi }{2} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\sin x+\dfrac{1}{100}\sin 5x-\dfrac{1}{36}\sin 3x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{104}{225} \\
& \Rightarrow F\left( \dfrac{\pi }{2} \right)=F\left( 0 \right)-\dfrac{104}{225}=0-\dfrac{104}{225}=-\dfrac{104}{225} \\
\end{aligned}$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\sin x.{{\sin }^{2}}2x\text{d}x}=\int{\sin x.\dfrac{1-\cos 4x}{2}\text{d}x}=\int{\dfrac{\sin x}{2}\text{d}x-\int{\dfrac{\sin x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\sin x}\text{d}x-\dfrac{1}{4}\int{\left( \sin 5x-\sin 3x \right)\text{d}x=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x+C}$.
Suy ra $f\left( x \right)=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x+C,\forall x\in \mathbb{R}$. Mà $f\left( \dfrac{\pi }{2} \right)=0\Rightarrow C=0$.
Do đó $f\left( x \right)=-\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \dfrac{\pi }{2} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\dfrac{1}{2}\cos x+\dfrac{1}{20}\cos 5x-\dfrac{1}{12}\cos 3x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\sin x+\dfrac{1}{100}\sin 5x-\dfrac{1}{36}\sin 3x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{104}{225} \\
& \Rightarrow F\left( \dfrac{\pi }{2} \right)=F\left( 0 \right)-\dfrac{104}{225}=0-\dfrac{104}{225}=-\dfrac{104}{225} \\
\end{aligned}$.
Đáp án B.
