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Step-by-Step Solution
Step 1: Understand the given conditions
We have a differentiable function $f(x)$ on the interval $[0,2]$ such that:
$f'(x) = f'(2 - x)$ for all $x \in (0,2)$.
$f(0) = 1$ and $f(2) = e^2$.
We want to find the value of $\displaystyle \int_{0}^{2} f(x)\, dx$.
Step 2: Derive the relationship $f(x) + f(2 - x) = \text{constant}$
From $f'(x) = f'(2 - x)$, we consider:
\[
f'(x) - f'(2 - x) = 0.
\]
Integrate both sides with respect to $x$ from $0$ to any $x$ in the interval:
\[
\int_{0}^{x} \bigl[f'(t) - f'(2 - t)\bigr]\,dt = 0.
\]
This becomes:
\[
\bigl(f(x) - f(0)\bigr) - \bigl(f(2 - x) - f(2)\bigr) = 0.
\]
Rearranging yields:
\[
f(x) + f(2 - x) = f(0) + f(2).
\]
Since $f(0) = 1$ and $f(2) = e^2$, we obtain:
\[
f(x) + f(2 - x) = 1 + e^2.
\]
This holds for all $x$ in $[0, 2]$.
Step 3: Express the definite integral using the derived identity
We want to evaluate
\[
I = \int_{0}^{2} f(x)\, dx.
\]
Using the fact that $f(x) + f(2 - x) = 1 + e^2,$ we can split the integral at $x = 1$ and use a substitution for the second part:
Write
\[
I = \int_{0}^{1} f(x)\, dx + \int_{1}^{2} f(x)\, dx.
\]
In the second integral, substitute $t = 2 - x.$ Then when $x=1,$ $t=1$; and when $x=2,$ $t=0.$ Also, $dx = -dt.$ Thus,
\[
\int_{1}^{2} f(x)\, dx = \int_{1}^{0} f(2 - t)\, (-dt) = \int_{0}^{1} f(2 - t)\, dt.
\]
But $f(2 - t) = 1 + e^2 - f(t).$ So
\[
\int_{1}^{2} f(x)\, dx = \int_{0}^{1} \bigl[1 + e^2 - f(t)\bigr]\, dt.
\]
This expands to
\[
\int_{0}^{1} (1 + e^2)\, dt - \int_{0}^{1} f(t)\, dt = (1 + e^2)\cdot 1 - \int_{0}^{1} f(t)\, dt = 1 + e^2 - \int_{0}^{1} f(t)\, dt.
\]
Step 4: Combine the parts and simplify
Therefore,
\[
I = \int_{0}^{1} f(x)\, dx + \left[1 + e^2 - \int_{0}^{1} f(t)\, dt \right].
\]
Notice that
\[
\int_{0}^{1} f(x)\, dx - \int_{0}^{1} f(t)\, dt = 0
\]
(since they are the same integral). Hence,
\[
I = 1 + e^2.
\]
Final Answer
\[
\int_{0}^{2} f(x)\, dx = 1 + e^2.
\]
