Nếu \(\int\limits_{0}^{\frac{\pi }{2}}{\left[ 2020f\left( x \right)+\sin 2x \right]}\text{d}x=2021\) thì \(\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)}\text{d}x\) bằng

* Hướng dẫn giải

Ta có \(\int\limits_{0}^{\frac{\pi }{2}}{\left[ 2020f\left( x \right)+\sin 2x \right]}\text{d}x=2021\Leftrightarrow 2020\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)}\text{d}x+\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x}\text{d}x=2021\).

Khi đó ta có \(2020\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)}\text{d}x-\frac{1}{2}\left. \left( c\text{os}2x \right) \right|_{0}^{\frac{\pi }{2}}=2021\Leftrightarrow 2020\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)}\text{d}x+1=2021\).

Do đó \(\int\limits_{0}^{\frac{\pi }{2}}{f\left( x \right)}\text{d}x=1\).