To calculate the area under the curve $y = x + 2$ from $x = 0$ to $x = 3$, we follow these steps:

1. Set up the integral: The total area under the curve is given by the definite integral: $A = \int_0^3 (x + 2) \, dx$

2. Integrate $x + 2$: The antiderivative of $x + 2$ is: $\int (x + 2) \, dx = \frac{x^2}{2} + 2x$

3. Evaluate the definite integral: Substitute the limits of integration into the antiderivative: $A = \left[ \frac{x^2}{2} + 2x \right]_0^3$ Evaluate at $x = 3$: $\frac{3^2}{2} + 2(3) = \frac{9}{2} + 6 = \frac{9}{2} + \frac{12}{2} = \frac{21}{2}$ Evaluate at $x = 0$: $\frac{0^2}{2} + 2(0) = 0$ Subtract the two values: $A = \frac{21}{2} - 0 = \frac{21}{2}$

4. Conclusion: The area under the curve $y = x + 2$ from $x = 0$ to $x = 3$ is: $\boxed{\frac{21}{2} \text{ or } 10.5 \text{ square units}}$

---

Do you want more details or have further questions?

Here are 5 related questions to expand your understanding:

1. How do you find the area between two curves instead of a single curve and the x-axis?
2. What happens if the curve dips below the x-axis—how does that affect the area calculation?
3. Can you find the area under a curve when it is defined piecewise?
4. How would you calculate the area for curves with vertical or horizontal asymptotes?
5. Can you use numerical integration to approximate areas when the function cannot be integrated analytically?

Tip: When integrating to find areas, always check whether the curve lies above or below the x-axis. If below, take the absolute value to avoid negative areas!