Let $f\left(x\right)$ be non-negative continuous function defined on R such that  $f\left(x\right)+f\left(x+\frac{1}{2}\right)=3;2\underset{0}{\overset{2023}{\int }}f\left(x\right)dx=$

$Given f\left(x\right)+f\left(x+\frac{1}{2}\right)=3;\to \left(1\right)$

$\mathrm{Re}placexbyx+\frac{1}{2}$

$f\left(x+\frac{1}{2}\right)+f\left(x+1\right)=3;\to \left(2\right)$

$\left(1\right)-\left(2\right):f\left(x\right)-f\left(x+1\right)=0$

$f\left(x\right)=f\left(x+1\right)$

$f\left(x\right)$ is a periodic function with period 1

$Let I=\underset{0}{\overset{2023}{\int }}f\left(x\right)dx=2023\underset{0}{\overset{1}{\int }}f\left(x\right)dx=2023\left[\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(x\right)dx+\underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx\right]$

$=2023\left[\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(x\right)dx+\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(x+\frac{1}{2}\right)dx\right]$

$=2023\underset{0}{\overset{\frac{1}{2}}{\int }}3dx$

$=2023\times 3\left[\frac{1}{2}-0\right]$

$I=\frac{6069}{2}$

$2I=6069$