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

### Step 2: Set up the integral for the area
The area \( A \) 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 3: Calculate the integral
Now we will compute the integral:
\[ A = \int_{-1}^{2} (x^2 + 3) \, dx \]
We can split this into two separate integrals:
\[ A = \int_{-1}^{2} x^2 \, dx + \int_{-1}^{2} 3 \, dx \]

### Step 4: Evaluate the first integral
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \( -1 \) to \( 2 \):
\[ \left[ \frac{x^3}{3} \right]_{-1}^{2} = \left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) = \left( \frac{8}{3} - \left(-\frac{1}{3}\right) \right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3 \]

### Step 5: Evaluate the second integral
The integral of a constant \( 3 \) is:
\[ \int 3 \, dx = 3x \]
Evaluating from \( -1 \) to \( 2 \):
\[ \left[ 3x \right]_{-1}^{2} = 3(2) - 3(-1) = 6 + 3 = 9 \]

### Step 6: Combine the results
Now, we combine the results of the two integrals:
\[ A = 3 + 9 = 12 \]

### Final Answer
Thus, the area enclosed between the curve \( y = x^2 + 3 \), the x-axis, and the lines \( x = -1 \) and \( x = 2 \) is:
\[ \boxed{12} \text{ square units} \]