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(A) Let $I = \int_0^a f(x)g(x) dx$. Using the property $\int_0^a h(x) dx = \int_0^a h(a - x) dx$,we have: $I = \int_0^a f(a - x)g(a - x) dx$. Given $f

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$\int_0^a f(x) dx$

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(A) Let $I = \int_0^a f(x)g(x) dx$. Using the property $\int_0^a h(x) dx = \int_0^a h(a - x) dx$,we have: $I = \int_0^a f(a - x)g(a - x) dx$. Given $f(x) = f(a - x)$ and $g(a - x) = 2 - g(x)$,we substitute these into the integral: $I = \int_0^a f(x)[2 - g(x)] dx$. $I = 2\int_0^a f(x) dx - \int_0^a f(x)g(x) dx$. $I = 2\int_0^a f(x) dx - I$. $2I = 2\int_0^a f(x) dx$. $I = \int_0^a f(x) dx$.

If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 2$,then $\int_0^a f(x)g(x) dx = $

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