If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ${\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

Solution:

Let,

$\begin{array}{l}\mathrm{I}={\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}----\left(\mathrm{i}\right)\\ \mathrm{I}={\int }_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{g}(\mathrm{a}-\mathrm{x})\mathrm{dx}\\ \left[\because {\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}\right]\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\{4-\mathrm{g}(\mathrm{x}\left)\right\}\mathrm{dx}---\left(\mathrm{ii}\right)\\ [\because \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}-\mathrm{x}\left)\text{ and }\mathrm{g}\right(\mathrm{x})+\mathrm{g}(\mathrm{a}-\mathrm{x})=4\text{ (given) }]\end{array}$

On adding Eqs. (i) and (ir), we get

$2\mathrm{I}={\int }_0^{\mathrm{a}} 4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{I}=2{\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$