Question

If $f\left(a+b-x\right)=f\left(x\right)$, then prove that ${\int }_a^{b}xf\left(x\right)dx=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$.

Answer 1

$$\begin{gathered} {\int }_a^{b}xf\left(x\right)dx \\ ={\int }_a^b\left(a+b-x\right)f\left(a+b-x\right)dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_a^b\left(a+b-x\right)f\left(x\right)dx\left[f\left(a+b-x\right)=f\left(x\right)\right] \\ \therefore {\int }_a^{b}xf\left(x\right)dx={\int }_a^b\left(a+b\right)f\left(x\right)dx-{\int }_a^{b}xf\left(x\right)dx \end{gathered}$$
$$\begin{gathered} \Rightarrow {\int }_a^{b}xf\left(x\right)dx+{\int }_a^{b}xf\left(x\right)dx=\left(a+b\right){\int }_a^{b}f\left(x\right)dx \\ \Rightarrow 2{\int }_a^{b}xf\left(x\right)dx=\left(a+b\right){\int }_a^{b}f\left(x\right)dx \\ \Rightarrow {\int }_a^{b}xf\left(x\right)dx=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx \end{gathered}$$