Question
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that, $\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$

Answer

$\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})(\text{a}+\text{b}-\text{x}){}\text{dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$ $\big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f}(\text{x})\big]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b})\text{f}(\text{x})\text{dx}-\int\limits^{\text{b}}_{\text{a}}\text{x}\text{f}(\text{x})\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$

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