1. We are given

   \quad \text{for all } x\in(0,2),$$ with $$f(0)=1,\qquad f(2)=e^2.$$

2. Define a new function $g(x)=f(x)+f(2-x).$ Differentiate it: $g'(x)=f'(x)-f'(2-x).$ Here, derivative of $f(2-x)$ is $-f'(2-x)$ by chain rule.

3. Using the given condition $f'(x)=f'(2-x)$, we get $g'(x)=f'(x)-f'(2-x)=0.$ Hence, $g(x)$ is constant on $[0,2]$.

4. Find this constant using $x=0$: $g(0)=f(0)+f(2)=1+e^2.$ Therefore, $f(x)+f(2-x)=1+e^2 \quad \text{for all } x\in[0,2].$

5. Integrate both sides from $0$ to $2$: $\int_0^2 \big(f(x)+f(2-x)\big)\,dx=\int_0^2 (1+e^2)\,dx.$

6. Now, $\int_0^2 f(2-x)\,dx=\int_0^2 f(x)\,dx$ by substitution $u=2-x$. So the left side becomes $2\int_0^2 f(x)\,dx.$

7. Hence, $2\int_0^2 f(x)\,dx = 2(1+e^2).$ Therefore, $\int_0^2 f(x)\,dx = 1+e^2.$

8. Checking options:

   • A: $1+e^2$ ✓
   • B: $2(1+e^2)$ ✗
   • C: $1-e^2$ ✗
   • D: $2(1-e^2)$ ✗

So the correct option is A.