MCQ
Choose the correct answer in Exercise : If $\text{f (a}+\text{b)}-\text{x}=\text{f (x)},$ then $\int^{\text{b}}_{\text{a}}\text{x f (x)}\ \text{dx}$ is equal to
  • A
    $\frac{\text{a}+\text{b}}{2}\int^{\text{a}}_{\text{b}}\text{f (b}-\text{x)}\text{dx}$
  • B
    $\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (b}+\text{x)}\text{dx}$
  • C
    $\frac{\text{b}-\text{a}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
  • $\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$

Answer

Correct option: D.
$\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
Let $\text{I}=\int^{\text{b}}\limits_{a}\text{x f (x)}\text{dx}$
$\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\text{f (a}+\text{b}-\text{x)}\text{dx}\ \ \ \ \ \Bigg[\int^{\text{b}}\limits_{\text{a}}\text{f (x)dx}=\int^{\text{b}}\limits_{\text{a}}\text{f (a}+\text{b}-\text{x)}\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}\ \ \ -\text{I}\ [$Using $(1)]$
$\Rightarrow\text{I}+\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow2\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\bigg(\frac{\text{a}+\text{b}}{2}\bigg)\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$

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