The question requires calculating the area of the region bounded by the following:

1. $y = f(x) = x + 1$
2. $x = 0$
3. $x = 3$
4. The x-axis, which is $y = 0$

Solution

To find the area of the region bounded by these curves, we integrate the function $f(x) = x + 1$ from $x = 0$ to $x = 3$.

1. Set up the integral: The area $A$ under the curve from $x = 0$ to $x = 3$ is given by: $A = \int_{0}^{3} (x + 1) \, dx$

2. Evaluate the integral: $A = \int_{0}^{3} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{0}^{3}$

3. Calculate the values:

   • At $x = 3$: $\frac{3^2}{2} + 3 = \frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}$
   • At $x = 0$: $\frac{0^2}{2} + 0 = 0$

   So, the area $A$ is: $A = \frac{15}{2} - 0 = \frac{15}{2} = 7.5$

Final Answer

The area of the region is $7.5$ square units.

Would you like a detailed explanation of each integration step or have any specific questions?

1. What is the importance of definite integrals in calculating bounded areas?
2. How would the answer change if the bounds were altered?
3. How can this method be applied to functions with curves?
4. What if the function was quadratic? How would integration change?
5. Can this problem be solved graphically for better understanding?

Tip: When calculating areas, always confirm the limits and whether the area is above or below the x-axis.