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To solve the given problem, we start with the functional equation and the integral condition.

Given:
\[ f(x+2) - f(x) = x \]
\[ \int_{0}^{2} f(x) \, dx = 0 \]

We are to find:
\[ \int_{0}^{3} f(x) \, dx \]

First, let's solve the functional equation. We know that \( f(x+2) - f(x) = x \). To find \( f(x) \), let's assume a general form:
\[ f(x) = g(x) + h(x) \]
where \( g(x) \) is a particular solution to the non-homogeneous equation \( g(x+2) - g(x) = x \) and \( h(x) \) is a solution to the homogeneous equation \( h(x+2) - h(x) = 0 \).

### Step 1: Find the particular solution \( g(x) \).

Assume \( g(x) = ax + b \).

\[ g(x+2) = a(x+2) + b = ax + 2a + b \]
\[ g(x+2) - g(x) = ax + 2a + b - (ax + b) = 2a \]
To satisfy \( g(x+2) - g(x) = x \), we must have \( 2a = x \), but this is not possible since \( x \) is variable. Therefore, assume another form. Try \( g(x) = \frac{x^2}{2} \).

\[ g(x+2) = \frac{(x+2)^2}{2} = \frac{x^2 + 4x + 4}{2} = \frac{x^2}{2} + 2x + 2 \]
\[ g(x+2) - g(x) = \left(\frac{x^2}{2} + 2x + 2\right) - \frac{x^2}{2} = 2x + 2 \]

This still does not satisfy the equation. To find a simpler particular solution, use:
\[ g(x) = \frac{x(x+2)}{4} \]

Then:
\[ g(x+2) = \frac{(x+2)(x+4)}{4} = \frac{x^2 + 6x + 8}{4} \]
\[ g(x+2) - g(x) = \frac{x^2 + 6x + 8}{4} - \frac{x^2 + 2x}{4} = \frac{4x + 8}{4} = x + 2 \]

Thus, the correct particular solution to satisfy \( g(x+2) - g(x) = x \) is:
\[ g(x) = \frac{x(x+2)}{4} \]

### Step 2: Find the homogeneous solution \( h(x) \).

Given \( h(x+2) - h(x) = 0 \), we conclude \( h(x+2) = h(x) \), meaning \( h(x) \) is a periodic function with period 2.

### Step 3: Combine solutions and satisfy the integral condition.

Since \( \int_{0}^{2} f(x) \, dx = 0 \),
\[ f(x) = \frac{x(x+2)}{4} + h(x) \]
We calculate:
\[ \int_{0}^{2} f(x) \, dx = \int_{0}^{2} \left(\frac{x(x+2)}{4} + h(x)\right) dx \]
\[ = \int_{0}^{2} \frac{x^2 + 2x}{4} \, dx + \int_{0}^{2} h(x) \, dx \]
\[ = \frac{1}{4} \int_{0}^{2} (x^2 + 2x) \, dx + \int_{0}^{2} h(x) \, dx \]

Since \(\int_{0}^{2} f(x) \, dx = 0\),
\[ \int_{0}^{2} h(x) \, dx = 0 \]

### Step 4: Calculate \( \int_{0}^{3} f(x) \, dx \).

\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{3} \left(\frac{x(x+2)}{4} + h(x)\right) dx \]
Since \( h(x) \) is periodic with period 2,
\[ \int_{0}^{3} h(x) \, dx = \int_{0}^{2} h(x) \, dx + \int_{2}^{3} h(x) \, dx = 0 + \int_{0}^{1} h(x) \, dx \]

We thus only need to consider the integral of the particular solution:
\[ \int_{0}^{3} \frac{x(x+2)}{4} \, dx = \frac{1}{4} \int_{0}^{3} (x^2 + 2x) \, dx \]
\[ = \frac{1}{4} \left( \int_{0}^{3} x^2 \, dx + 2 \int_{0}^{3} x \, dx \right) \]
\[ = \frac{1}{4} \left( \frac{x^3}{3} \bigg|_{0}^{3} + 2 \cdot \frac{x^2}{2} \bigg|_{0}^{3} \right) \]
\[ = \frac{1}{4} \left( \frac{27}{3} + 2 \cdot \frac{9}{2} \right) \]
\[ = \frac{1}{4} \left( 9 + 9 \right) \]
\[ = \frac{1}{4} \cdot 18 = 4.5 \]

Thus, 
\[ \int_{0}^{3} f(x) \, dx = 4.5 \]

Would you like more details or have any questions?

Here are 5 questions you might ask next:
1. How do you solve a different type of functional equation?
2. Can you explain how to find homogeneous solutions?
3. How can I verify if my solution to a functional equation is correct?
4. Can you solve an integral involving periodic functions?
5. How does the assumption of \( h(x) \) being periodic help in solving the equation?

**Tip:** Always verify your solution by checking if it satisfies both the functional equation and any given integral conditions.

Explore how to solve a functional equation with integral conditions step-by-step. This problem involves functional equations, integral calculus, and integration by parts. Suitable for advanced high school or college-level students.

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