If $f\left(x\right)$ and $g\left(x\right)$ are continuous functions on the interval $\left[0, 4\right]$ satisfying $f\left(x\right)=f\left(4-x\right)$ and $g\left(x\right)+g\left(4-x\right)=3$ and $\underset{0}{\overset{4}{\int }}f\left(x\right)dx=2$ then $\underset{0}{\overset{4}{\int }}f\left(x\right)g\left(x\right)dx=$

$\begin{array}{l}f\left(x\right)=f(4-x)\\ g\left(x\right)+g(4-x)=3\\ \Rightarrow g(4-x)=3-g\left(x\right)\\ \underset{0}{\overset{4}{\int }}f\left(x\right)dx=2\\ I=\underset{0}{\overset{4}{\int }}f\left(x\right)g\left(x\right)dx\\ =\underset{0}{\overset{4}{\int }}f(4-x)g(4-x)dx\\ =\underset{0}{\overset{4}{\int }}f\left(x\right)(3-g(x\left)\right)dx\\ =3\underset{0}{\overset{4}{\int }}f\left(x\right)dx-\underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx\\ I=3\left(2\right)-I\\ 2I=6\\ \Rightarrow I=3\end{array}$