Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{2} \right)=\dfrac{27}{8}$ và ${f}'\left( x \right)=12\sin 2x.{{\cos }^{2}}3x,\forall x\in \mathbb{R}$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thỏa mãn $F\left( 0 \right)=0$, khi đó $F\left( \pi \right)$ bằng
A. $0$.
B. $-\dfrac{87}{64}$.
C. $-\dfrac{21}{8}$.
D. $\dfrac{87}{64}$.
A. $0$.
B. $-\dfrac{87}{64}$.
C. $-\dfrac{21}{8}$.
D. $\dfrac{87}{64}$.
Ta có ${f}'\left( x \right)=12\sin 2x.{{\cos }^{2}}3x,\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{12\sin 2x.{{\cos }^{2}}3x\text{d}x}=\int{12.\sin 2x.\dfrac{1+\cos 6x}{2}\text{d}x}=\int{6.\sin 2x\text{d}x+\int{6\sin 2x.\cos 6x\text{d}x}}$
$=6\int{\sin 2x}\text{d}x+3\int{\left( \sin 8x-\sin 4x \right)\text{d}x=-3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x+C}$.
Suy ra $f\left( x \right)=-3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x+C$. Mà $f\left( \dfrac{\pi }{2} \right)=\dfrac{27}{8}\Rightarrow C=0$.
Do đó. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( -3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x \right)\text{d}x} \\
& =\left. \left( -\dfrac{3}{2}\sin 2x-\dfrac{3}{64}\sin 8x+\dfrac{3}{16}\sin 4x \right) \right|_{0}^{\pi }=0 \\
& \Rightarrow F\left( \pi \right)=F\left( 0 \right)+0=-\dfrac{21}{8}+0=-\dfrac{21}{8} \\
\end{aligned}$
Có $\int{{f}'\left( x \right)\text{d}x}=\int{12\sin 2x.{{\cos }^{2}}3x\text{d}x}=\int{12.\sin 2x.\dfrac{1+\cos 6x}{2}\text{d}x}=\int{6.\sin 2x\text{d}x+\int{6\sin 2x.\cos 6x\text{d}x}}$
$=6\int{\sin 2x}\text{d}x+3\int{\left( \sin 8x-\sin 4x \right)\text{d}x=-3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x+C}$.
Suy ra $f\left( x \right)=-3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x+C$. Mà $f\left( \dfrac{\pi }{2} \right)=\dfrac{27}{8}\Rightarrow C=0$.
Do đó. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( -3\cos 2x-\dfrac{3}{8}\cos 8x+\dfrac{3}{4}\cos 4x \right)\text{d}x} \\
& =\left. \left( -\dfrac{3}{2}\sin 2x-\dfrac{3}{64}\sin 8x+\dfrac{3}{16}\sin 4x \right) \right|_{0}^{\pi }=0 \\
& \Rightarrow F\left( \pi \right)=F\left( 0 \right)+0=-\dfrac{21}{8}+0=-\dfrac{21}{8} \\
\end{aligned}$
Đáp án C.