1. We are asked to evaluate the definite integral $$\int_0^3 (x^2 + 4) \, dx$$ which means finding the area under the curve of the function $f(x) = x^2 + 4$ from $x=0$ to $x=3$. 2. The formula for the definite integral of a sum is the sum of the integrals: $$\int_a^b (f(x) + g(x)) \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx$$ 3. We split the integral: $$\int_0^3 (x^2 + 4) \, dx = \int_0^3 x^2 \, dx + \int_0^3 4 \, dx$$ 4. Use the power rule for integration: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ and the integral of a constant: $$\int c \, dx = cx + C$$ 5. Calculate each integral: $$\int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9$$ $$\int_0^3 4 \, dx = \left[ 4x \right]_0^3 = 4 \times 3 - 4 \times 0 = 12 - 0 = 12$$ 6. Add the results: $$9 + 12 = 21$$ 7. Therefore, the value of the integral is $$\boxed{21}$$.