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.