If    $g\left(x\right)=f\left(x\right){\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)}^{2021}$  is decreasing function and $f\left(x\right)$ is continuous in R. Then $f\left(2022\right)$ is

 

$$\begin{gathered} Let G\left(x\right)={\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)}^{2022}\mathrm{......91} \\ G\text{'}\left(x\right)=2022f\left(x\right){\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)}^{2021}\Rightarrow G\text{'}\left(x\right)=2022g\left(x\right)\Rightarrow G\text{'}\left(0\right)=2022g\left(0\right) \\ also g\left(x\right) is decrea\sin g \Rightarrow g\left(x\right)\ge 0\forall x\in (-\infty ,0) \\ and g\left(x\right)\le 0\forall x\in [0,\infty )\therefore G\text{'}\left(x\right)\ge 0 and G\text{'}\left(x\right)\le 0 for x\ge 0 , G\left(g\right)=0 \end{gathered}$$

$\therefore G\left(x\right)\le 0\forall x\in R$ but  $G\left(x\right)\ge 0\forall x\in R$ (from 1)

$\therefore \Rightarrow G\left(x\right)=0\Rightarrow \underset{0}{\overset{x}{\int }}f\left(t\right)dt=0\forall x\in R\therefore f\left(x\right)=0\forall x\in R$