If $f\left(a+b-x\right)=f\left(x\right)$ , then${\displaystyle {\int }_a^{b}xf\left(x\right)dx=}$

$puta+b-x=t\Rightarrow -dx=dt$

$putx=a\Rightarrow a+b-a=t\Rightarrow t=b$

$x=b\Rightarrow a+b-b=t\Rightarrow t=a$

$I={\displaystyle {\int }_a^{b}xf\left(x\right)dx=}{\displaystyle {\int }_b^a\left(a+b-t\right)f\left(a+b-t\right)\left(-dt\right)}$

$={\displaystyle {\int }_a^b\left(a+b\right)f\left(t\right)dt}-{\displaystyle {\int }_a^{b}tf\left(t\right)\left(dt\right)}$

$=\left(a+b\right){\displaystyle {\int }_a^{b}f\left(t\right)dt}-{\displaystyle {\int }_a^{b}xf\left(x\right)dx}$

$=\left(a+b\right){\displaystyle {\int }_a^{b}f\left(t\right)dt-I}$

$2I=\left(a+b\right){\displaystyle {\int }_a^{b}f\left(t\right)dt}\Rightarrow I=\left(\frac{a+b}{2}\right){\displaystyle {\int }_a^{b}f\left(x\right)dx}$ .