Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và...

Ta có $\int\limits_{0}^{2}{xf'\left( 2x \right)\text{d}x}=\dfrac{1}{4}\int\limits_{0}^{4}{t{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\int\limits_{0}^{4}{x{f}'\left( x \right)\text{d}x}=\dfrac{1}{4}\left[ xf\left( x \right)|_{0}^{4}-\int\limits_{0}^{4}{f\left( x \right)\text{d}x} \right]$ $=\dfrac{1}{4}\left[ 4.f\left( 4 \right)-\int\limits_{0}^{4}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 4.2023-4 \right]=2022$.