If $f\left(a-x\right)=f\left(x\right)$  and $g\left(x\right)+g\left(a-x\right)=2,I_1={\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx,I_2={\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx}}$ , then $\frac{I_2}{I_1}=$

$I_2={\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx,x\to a-x}$

$I_2={\displaystyle \underset{0}{\overset{a}{\int }}f\left(a-x\right)g\left(a-x\right)dx={\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)g\left(a-x\right)dx}}$

$2I_2=2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)\left[g\left(x\right)+g\left(a-x\right)\right]dx}$

        $=2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx}\to I_2=I_1$