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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \dfrac{\pi }{4} \right)=-\dfrac{8}{3}$ và ${f}'\left( x \right)=16\cos 4x.{{\sin }^{2}}x,\forall x\in \mathbb{R}$. Biết ${F\left( x \right)}$ là nguyên hàm của ${f\left( x \right)}$ thỏa mãn $F\left( 0 \right)=\dfrac{31}{18}$, khi đó $F\left( \pi \right)$ bằng
A. $\dfrac{16}{3}$.
B. $\dfrac{64}{27}$.
C. $0$.
D. $\dfrac{31}{8}$.
A. $\dfrac{16}{3}$.
B. $\dfrac{64}{27}$.
C. $0$.
D. $\dfrac{31}{8}$.
Ta có ${f}'\left( x \right)=16\cos 4x.{{\sin }^{2}}x,\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{16\cos 4x.{{\sin }^{2}}x\text{d}x}=\int{16.\cos 4x.\dfrac{1-\cos 2x}{2}\text{d}x}=\int{8.\cos 4x\text{d}x-\int{8\cos 4x.\cos 2x\text{d}x}}$
$=8\int{\cos 4x}\text{d}x-8\int{\left( \cos 6x+\cos 2x \right)\text{d}x=2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+C}$.
Suy ra $f\left( x \right)=2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+C$. Mà $f\left( \dfrac{\pi }{4} \right)=-\dfrac{8}{3}\Rightarrow C=0$.
Do đó. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( 2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+ \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\cos 4x+\dfrac{2}{9}\cos 6x+2\cos 2x \right) \right|_{0}^{\pi }=0 \\
& F\left( \pi \right)=F\left( 0 \right)+0=\dfrac{31}{18} \\
\end{aligned}$
Có $\int{{f}'\left( x \right)\text{d}x}=\int{16\cos 4x.{{\sin }^{2}}x\text{d}x}=\int{16.\cos 4x.\dfrac{1-\cos 2x}{2}\text{d}x}=\int{8.\cos 4x\text{d}x-\int{8\cos 4x.\cos 2x\text{d}x}}$
$=8\int{\cos 4x}\text{d}x-8\int{\left( \cos 6x+\cos 2x \right)\text{d}x=2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+C}$.
Suy ra $f\left( x \right)=2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+C$. Mà $f\left( \dfrac{\pi }{4} \right)=-\dfrac{8}{3}\Rightarrow C=0$.
Do đó. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( 2\sin 4x-\dfrac{4}{3}\sin 6x-4\sin 2x+ \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\cos 4x+\dfrac{2}{9}\cos 6x+2\cos 2x \right) \right|_{0}^{\pi }=0 \\
& F\left( \pi \right)=F\left( 0 \right)+0=\dfrac{31}{18} \\
\end{aligned}$
Đáp án D.
