1. **Problem:** Calculate the definite integral $$\int_0^3 x \, dx$$ 2. **Formula:** The integral of $$x$$ is $$\frac{x^2}{2}$$. For definite integrals, evaluate the antiderivative at the upper and lower limits and subtract: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $$F'(x) = f(x)$$. 3. **Calculation:** $$F(x) = \frac{x^2}{2}$$ Evaluate at the limits: $$F(3) = \frac{3^2}{2} = \frac{9}{2}$$ $$F(0) = \frac{0^2}{2} = 0$$ 4. **Result:** $$\int_0^3 x \, dx = \frac{9}{2} - 0 = \frac{9}{2} = 4.5$$ This means the area under the curve $$y=x$$ from $$x=0$$ to $$x=3$$ is 4.5.