Find the area enclosed betweent the curve `y = x^(2)+3, y = 0, x = - 1, x = 2`.

Answer

To find the area enclosed between the curve \( y = x^2 + 3 \), the x-axis \( y = 0 \), and the vertical lines \( x = -1 \) and \( x = 2 \), we will follow these steps:

### Step 1: Identify the curves and boundaries
We have the following curves and lines:
- The curve: \( y = x^2 + 3 \)
- The x-axis: \( y = 0 \)
- The vertical lines: \( x = -1 \) and \( x = 2 \)

### Step 2: Determine the points of intersection
We need to find the points where the curve intersects the x-axis:
\[ x^2 + 3 = 0 \]
This equation has no real solutions since \( x^2 + 3 \) is always positive. Therefore, the curve does not intersect the x-axis.

### Step 3: Set up the integral for the area
The area \( A \) enclosed between the curve and the x-axis from \( x = -1 \) to \( x = 2 \) can be calculated using the definite integral:
\[ A = \int_{-1}^{2} (x^2 + 3) \, dx \]

### Step 4: Calculate the integral
Now we will compute the integral:
\[ A = \int_{-1}^{2} (x^2 + 3) \, dx = \int_{-1}^{2} x^2 \, dx + \int_{-1}^{2} 3 \, dx \]

Calculating each part separately:
1. **Integral of \( x^2 \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \( -1 \) to \( 2 \):
\[ \left[ \frac{x^3}{3} \right]_{-1}^{2} = \left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) = \left( \frac{8}{3} + \frac{1}{3} \right) = \frac{9}{3} = 3 \]

2. **Integral of \( 3 \)**:
\[ \int 3 \, dx = 3x \]
Evaluating from \( -1 \) to \( 2 \):
\[ \left[ 3x \right]_{-1}^{2} = (3 \cdot 2) - (3 \cdot -1) = 6 + 3 = 9 \]

### Step 5: Combine the results
Now, adding both parts together:
\[ A = 3 + 9 = 12 \]

### Final Answer
The area enclosed between the curve \( y = x^2 + 3 \), the x-axis, and the lines \( x = -1 \) and \( x = 2 \) is \( \boxed{12} \).