Let \(f\) and \(g\) be continuous functions on \([0, \mathrm{a}]\) such that \(\mathrm{f}(x)=\mathrm{f}(\mathrm{a}-x)\) and \(g(x)+g(a-x)=4\), then \(\int_0^a f(x) g(x) d x\) is equal to

Given,
\(\begin{aligned}
& \mathrm{f}(x)=\mathrm{f}(\mathrm{a}-\mathrm{x}) \\
& \mathrm{g}(x)+\mathrm{g}(\mathrm{a}-x)=4 \\
& \text { Let } \mathrm{I}=\int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x \\
& =\int_0^a f(a-x) \cdot g(a-x) d x \\
& =\int_0^{\mathrm{a}} \mathrm{f}(x) \cdot[(4-\mathrm{g}(x)] \mathrm{d} x \\
& =4 \int_0^a \mathrm{f}(x) \mathrm{d} x-\int_0^a \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x \\
& \mathrm{I}=4 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x-\mathrm{I} \\
& 2 \mathrm{I}=4 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x \\
& \therefore \quad \mathrm{I}=2 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x
\end{aligned}\)