To solve the problem, we need to find the value of the integral \( I = \int_{\frac{3}{4}}^{\frac{9}{4}} f(x) \, dx \) given that \( f(x) + f(3 - x) = 3 \) for all \( x \in \mathbb{R} \).
### Step-by-Step Solution:
1. **Understand the given condition**:
The equation \( f(x) + f(3 - x) = 3 \) suggests a symmetry in the function \( f \). This means that for any \( x \), the values of \( f(x) \) and \( f(3 - x) \) will always sum to 3.
2. **Set up the integral**:
We start with the integral:
\[
I = \int_{\frac{3}{4}}^{\frac{9}{4}} f(x) \, dx
\]
3. **Change of variable**:
We can use the property of definite integrals to express \( I \) in terms of \( f(3 - x) \). Let \( x' = 3 - x \). Then, when \( x = \frac{3}{4} \), \( x' = 3 - \frac{3}{4} = \frac{9}{4} \), and when \( x = \frac{9}{4} \), \( x' = 3 - \frac{9}{4} = \frac{3}{4} \). Thus, we can write:
\[
I = \int_{\frac{9}{4}}^{\frac{3}{4}} f(3 - x') \, (-dx') = \int_{\frac{3}{4}}^{\frac{9}{4}} f(3 - x') \, dx'
\]
This means:
\[
I = \int_{\frac{3}{4}}^{\frac{9}{4}} f(3 - x) \, dx
\]
4. **Combine the integrals**:
Now we can add the two expressions for \( I \):
\[
2I = \int_{\frac{3}{4}}^{\frac{9}{4}} f(x) \, dx + \int_{\frac{3}{4}}^{\frac{9}{4}} f(3 - x) \, dx
\]
By the property we established, we can replace \( f(3 - x) \) using the condition \( f(x) + f(3 - x) = 3 \):
\[
2I = \int_{\frac{3}{4}}^{\frac{9}{4}} (f(x) + f(3 - x)) \, dx = \int_{\frac{3}{4}}^{\frac{9}{4}} 3 \, dx
\]
5. **Evaluate the integral**:
The integral simplifies to:
\[
2I = 3 \int_{\frac{3}{4}}^{\frac{9}{4}} 1 \, dx = 3 \left[ x \right]_{\frac{3}{4}}^{\frac{9}{4}} = 3 \left( \frac{9}{4} - \frac{3}{4} \right) = 3 \left( \frac{6}{4} \right) = \frac{18}{4} = \frac{9}{2}
\]
6. **Solve for \( I \)**:
Now, divide both sides by 2:
\[
I = \frac{9}{2} \cdot \frac{1}{2} = \frac{9}{4}
\]
### Final Answer:
Thus, the value of the integral \( I \) is:
\[
\boxed{\frac{9}{4}}
\]
