Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là $f'\left( x \right)=\sin x+x.\cos x,\forall x\in \mathbb{R}$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thỏa mãn $F\left( 0 \right)=F\left( \pi \right)=1$, khi đó giá trị của $F\left( 2\pi \right)$ bằng.
A. $1+2\pi $.
B. $1-4\pi $.
C. $1-2\pi $.
D. $4\pi $.
A. $1+2\pi $.
B. $1-4\pi $.
C. $1-2\pi $.
D. $4\pi $.
Ta có: $f'\left( x \right)=\sin x+x.\cos x,\forall x\in \mathbb{R}\Rightarrow f\left( x \right)=\int{\left( \sin x+x.\cos x \right)dx=-\cos x+\int{x.\cos xdx}}$
$f\left( x \right)=-\cos x+x.\sin x-\int{\sin xdx=-\cos x+x\sin x+\cos x+{{C}_{1}}=x\sin x+{{C}_{1}}}$.
$F\left( x \right)=\int{f\left( x \right)dx=\int{\left( x\sin x+{{C}_{1}} \right)}dx=-x\cos x+\int{\cos xdx+{{C}_{1}}x=-x\cos x+\sin x+{{C}_{1}}x+{{C}_{2}}}}$.
Vì $F\left( 0 \right)=F\left( \pi \right)=1$ nên $\left\{ \begin{aligned}
& {{C}_{2}}=1 \\
& \pi +{{C}_{1}}\pi +{{C}_{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=1 \\
& {{C}_{1}}=-1 \\
\end{aligned} \right.$.
Do đó $F\left( x \right)=-x\cos x+\sin x-x+1$.
Vậy $F\left( 2\pi \right)=-2\pi -2\pi +1=1-4\pi $.
$f\left( x \right)=-\cos x+x.\sin x-\int{\sin xdx=-\cos x+x\sin x+\cos x+{{C}_{1}}=x\sin x+{{C}_{1}}}$.
$F\left( x \right)=\int{f\left( x \right)dx=\int{\left( x\sin x+{{C}_{1}} \right)}dx=-x\cos x+\int{\cos xdx+{{C}_{1}}x=-x\cos x+\sin x+{{C}_{1}}x+{{C}_{2}}}}$.
Vì $F\left( 0 \right)=F\left( \pi \right)=1$ nên $\left\{ \begin{aligned}
& {{C}_{2}}=1 \\
& \pi +{{C}_{1}}\pi +{{C}_{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=1 \\
& {{C}_{1}}=-1 \\
\end{aligned} \right.$.
Do đó $F\left( x \right)=-x\cos x+\sin x-x+1$.
Vậy $F\left( 2\pi \right)=-2\pi -2\pi +1=1-4\pi $.
Đáp án B.