To solve the problem, we need to evaluate the integral
\[
I = \frac{1}{5} \int_{2}^{3} x [f(x) + f(x+1)] \, dx
\]
given the property \( f(6 - x) = f(x) \) for all \( x \).
### Step 1: Rewrite the Integral
We can use the property of integrals that states:
\[
\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx
\]
In our case, we can set \( a = 2 \) and \( b = 3 \). Therefore, we can rewrite the integral as:
\[
I = \frac{1}{5} \int_{2}^{3} x [f(6 - x) + f(6 - (x - 1))] \, dx
\]
### Step 2: Substitute the Function
Using the property \( f(6 - x) = f(x) \), we can simplify the integral:
\[
I = \frac{1}{5} \int_{2}^{3} x [f(x) + f(6 - x + 1)] \, dx
\]
### Step 3: Change of Variable
Now let’s perform a change of variable in the second part of the integral. Let \( u = 6 - x \). Then, when \( x = 2 \), \( u = 4 \) and when \( x = 3 \), \( u = 3 \). The differential \( dx = -du \).
Thus, we can rewrite the integral as:
\[
I = \frac{1}{5} \int_{4}^{3} (6 - u) [f(6 - u) + f(6 - (u - 1))] (-du)
\]
Reversing the limits gives us:
\[
I = \frac{1}{5} \int_{3}^{4} (6 - u) [f(u) + f(u - 1)] \, du
\]
### Step 4: Add the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \frac{1}{5} \int_{2}^{3} x [f(x) + f(x + 1)] \, dx \)
2. \( I = \frac{1}{5} \int_{3}^{4} (6 - u) [f(u) + f(u - 1)] \, du \)
Adding these two equations gives:
\[
2I = \frac{1}{5} \left( \int_{2}^{3} x [f(x) + f(x + 1)] \, dx + \int_{3}^{4} (6 - u) [f(u) + f(u - 1)] \, du \right)
\]
### Step 5: Evaluate the Integral
Now, we can evaluate the integrals separately. However, we notice that the second integral can be transformed back into the first integral's form.
After simplification and using the property of the function, we find:
\[
I = \frac{1}{2} \int_{2}^{3} f(x) \, dx
\]
### Step 6: Final Calculation
Thus, we find that:
\[
I = \frac{1}{5} \cdot 2 \cdot \int_{2}^{3} f(x) \, dx
\]
This leads us to conclude that:
\[
I = \frac{1}{5} \cdot 2 \cdot \int_{2}^{3} f(x) \, dx = \frac{2}{5} \int_{2}^{3} f(x) \, dx
\]
### Conclusion
The final answer is:
\[
\frac{1}{5} \int_{2}^{3} x [f(x) + f(x + 1)] \, dx = \text{(value depending on } f \text{)}
\]
