1. The problem is to evaluate the definite integral $\int_2^2 3x \, dx$. 2. The formula for the definite integral of a function $f(x)$ from $a$ to $b$ is: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. 3. Here, $f(x) = 3x$. The antiderivative of $3x$ is: $$F(x) = \frac{3x^2}{2}$$ 4. Now, evaluate $F(x)$ at the upper and lower limits: $$F(2) = \frac{3 \times 2^2}{2} = \frac{3 \times 4}{2} = \frac{12}{2} = 6$$ $$F(2) = 6$$ 5. The definite integral is: $$\int_2^2 3x \, dx = F(2) - F(2) = 6 - 6 = 0$$ 6. Since the limits of integration are the same, the integral evaluates to zero. **Final answer:** $$\int_2^2 3x \, dx = 0$$