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

Answer

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

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