Question
$\text{Show that}\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\text,$if and are defined as $\text{f(x)}=\text{f(a}-\text{x)}$ and $\text{g(x)}+\text{g(a}-\text{x)}=4$

Answer

$\text{Here}\ \text{f}\text{(x)}=\text{f}\text{(a}-\text{x)}\ .....(\text{i})\text{and}\ \ \text{g}\text{(x)}+\text{g}\text{(a}-\text{x)}=4 \ ......\text{(ii)}$$\text{Let}\ \text{I}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}\ ....\text{(iii)}$
$\therefore\ \ \text{I}=\int^{\text{a}}\limits_{0}\text{(a}-\text{x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}\ \ \ [\text{from eq.(i)}]\ ...\text{(iv)}$
Adding eq. (iii) and (iv)
$21=\int^{\text{a}}\limits_{0}(\text{f}\text{(x)}\text{g}\text{(x)}+\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}(\text{g}\text{(x)}+\text{g}\text{(a}-\text{x)})\text{dx}$
$\Rightarrow\ \ \ \ 21\int^{\text{a}}_{0}\text{f}\text{(x)}(4)\text{dx}.......[\text{from eq. (ii)}]$
$\Rightarrow\ \ \ \ 21=4\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\ \Rightarrow\ \text{I}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}$ Hence proved.

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