Cho hàm số y = f(x) liên tục trên R và thỏa mãn f(-x) + 2021f(x) = xsinx

Chọn C.

Từ giả thuyết: $f\left(-x\right)+2021f\left(x\right)+x\sin x,\forall x\in ℝ$

$\iff \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)dx+2021\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}x\sin xdx\left(*\right)$

Tính: $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)dx\stackrel{t=-x}{=}-\underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}f\left(t\right)dt=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=I.$

Tính: $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}x\sin xdx$. Đặt $\left\{\begin{array}{l}u=x\\ dv=\sin xdx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=dx\\ v=-\cos x\end{array}\right.$

$\Rightarrow \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}x\sin xdx=-x\cos x\left|\begin{array}{l}\frac{\pi }{2}\\ -\frac{\pi }{2}\end{array}\right.+\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos xdx=\sin x\left|\begin{array}{l}\frac{\pi }{2}\\ -\frac{\pi }{2}\end{array}\right.=2$

$\left(*\right)\iff I+2021.I=2\iff I=\frac{1}{1011}$.