We are given:

• $f(x)=f(a-x)$ on $[0,a]$
• $g(x)+g(a-x)=4$

We need to find $I=\int_0^a f(x)g(x)\,dx.$

Use the standard substitution $x\mapsto a-x$ in the integral: $I=\int_0^a f(a-x)g(a-x)\,dx.$

Since $f(a-x)=f(x)$, this becomes $I=\int_0^a f(x)g(a-x)\,dx.$

Now add the two expressions for $I$: $I=\int_0^a f(x)g(x)\,dx,$ $I=\int_0^a f(x)g(a-x)\,dx.$

Therefore, $2I=\int_0^a f(x)\big(g(x)+g(a-x)\big)\,dx.$

Using $g(x)+g(a-x)=4$, $2I=\int_0^a f(x)\cdot 4\,dx=4\int_0^a f(x)\,dx.$

Hence, $I=2\int_0^a f(x)\,dx.$

So the correct option is: $\boxed{\text{D}}$

Verification with stored answer:

• Derived answer: D
• Stored correct answer: D
• They match.