If \(f\) and \(g\) are continuous function in [0,1] satisfying \(f(x)=f(a-x)\) and \(g(x)+g(a-x)=a\), then \(\int_{0}^{a}f(x)g(x)dx=\)

\(\frac{a}{2}\int_{0}^{a}f(x)dx\)

\(I = \int\limits_{0}^{a} f(x) g(x) dx\)

\(I = \int\limits_{0}^{a} f(a − x) g(a − x) dx\)

\(I = \int\limits_{0}^{a} f(x) g(a −g(x)) dx\)

\(I = a\int\limits_{0}^{a}f(x) dx − \int\limits_{0}^{a}f(x)g(x)dx\)

\(I = a\int\limits_{0}^{a}f(x)dx − I\)

⇒ \(2I = a\int\limits_{0}^{a}f(x) dx\)

⇒\(I = \frac{a}{2}\int \limits_{0}^{a}f(x)dx\)