Key Concepts and Formulas

1. Integration of Derivatives: The Fundamental Theorem of Calculus states that if $f'(x)$ is continuous on $[a, b]$, then $\int_a^b f'(x) dx = f(b) - f(a)$. More generally, $\int f'(x) dx = f(x) + C$.
2. Substitution Rule for Integration: For an integral of the form $\int g(h(x)) h'(x) dx$, we can substitute $u = h(x)$, so $du = h'(x) dx$, leading to $\int g(u) du$.
3. Property of Definite Integrals: For a function $g(x)$ integrable on $[0, 2a]$, $\int_0^{2a} g(x) dx = \int_0^a [g(x) + g(2a-x)] dx$. This property is useful for integrals with symmetric limits and integrands exhibiting symmetry.

Step-by-Step Solution

Step 1: Derive a Functional Equation for $f(x)$ We are given the condition $f'(x) = f'(2-x)$ for $x \in (0, 2)$. To find a relationship involving $f(x)$, we integrate both sides with respect to $x$. Integrating the left side: $\int f'(x) dx = f(x) + C_1$ Integrating the right side requires a substitution. Let $u = 2-x$. Then, $du = -dx$, so $dx = -du$. $\int f'(2-x) dx = \int f'(u) (-du) = -\int f'(u) du = -f(u) + C_2$ Substituting back $u = 2-x$: $-f(2-x) + C_2$ Equating the results from both sides: $f(x) + C_1 = -f(2-x) + C_2$ Rearranging the terms, we get: $f(x) + f(2-x) = C_2 - C_1$ Let $C = C_2 - C_1$. This gives us the functional equation: $f(x) + f(2-x) = C \quad (*)$ This equation holds for all $x \in [0, 2]$.

Step 2: Determine the Value of the Constant $C$ We are given $f(0) = 1$ and $f(2) = e^2$. We can use these values in the functional equation $(*)$ to find $C$. Substitute $x=0$ into $(*)$: $f(0) + f(2-0) = C$ $f(0) + f(2) = C$ Using the given values: $1 + e^2 = C$ So, the functional equation becomes: $f(x) + f(2-x) = 1 + e^2 \quad (**)$

Step 3: Evaluate the Definite Integral We need to find the value of $I = \int_0^2 f(x) dx$. This integral has the form $\int_0^{2a} g(x) dx$ with $a=1$ and $g(x)=f(x)$. We can use the property $\int_0^{2a} g(x) dx = \int_0^a [g(x) + g(2a-x)] dx$. Applying this property to our integral: $I = \int_0^2 f(x) dx = \int_0^1 [f(x) + f(2-x)] dx$ Now, substitute the functional equation $(**)$, $f(x) + f(2-x) = 1 + e^2$, into the integral: $I = \int_0^1 (1 + e^2) dx$ Since $(1 + e^2)$ is a constant, we can integrate it directly: $I = (1 + e^2) \int_0^1 dx$ $I = (1 + e^2) [x]_0^1$ $I = (1 + e^2) (1 - 0)$ $I = 1 + e^2$

Common Mistakes and Tips to Avoid

1. Incorrectly integrating $f'(2-x)$: Remember to apply the chain rule when integrating $f'(2-x)$. The integral is $-f(2-x) + C$, not $f(2-x) + C$.
2. Forgetting the Constant of Integration: While constants may cancel out, it's good practice to include them during intermediate steps when deriving functional relationships to ensure correctness.
3. Not recognizing the use of the integral property: The property $\int_0^{2a} g(x) dx = \int_0^a [g(x) + g(2a-x)] dx$ is a powerful tool for integrals with symmetric limits. If a relationship between $f(x)$ and $f(2a-x)$ can be found, this property is often the key to simplification.

Summary The problem requires us to first deduce a functional relationship between $f(x)$ and $f(2-x)$ by integrating the given derivative condition $f'(x) = f'(2-x)$. This leads to $f(x) + f(2-x) = C$. Using the boundary conditions $f(0)=1$ and $f(2)=e^2$, we determined the constant $C$ to be $1+e^2$. Finally, we applied a property of definite integrals, $\int_0^{2a} g(x) dx = \int_0^a [g(x) + g(2a-x)] dx$, to evaluate $\int_0^2 f(x) dx$, substituting the derived functional equation to find the integral's value.

The final answer is $\boxed{1 + e^2}$.