Find the area of the region enclosed by the given curves .
`y=x^(3)+3, y=0 , x=-1, x=2`

Answer

To find the area of the region enclosed by the curves \( y = x^3 + 3 \), \( y = 0 \), \( x = -1 \), and \( x = 2 \), we will follow these steps:

### Step 1: Identify the curves and the area of interest
The curves are:
1. \( y = x^3 + 3 \) (a cubic curve)
2. \( y = 0 \) (the x-axis)
3. Vertical lines at \( x = -1 \) and \( x = 2 \)

We need to find the area between the curve \( y = x^3 + 3 \) and the x-axis from \( x = -1 \) to \( x = 2 \).

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

### Step 3: Calculate the integral
We will evaluate the integral:
\[ A = \int_{-1}^{2} (x^3 + 3) \, dx \]
This can be split into two separate integrals:
\[ A = \int_{-1}^{2} x^3 \, dx + \int_{-1}^{2} 3 \, dx \]

### Step 4: Evaluate the first integral
The integral of \( x^3 \) is:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Now we evaluate it from \( -1 \) to \( 2 \):
\[ \left[ \frac{x^4}{4} \right]_{-1}^{2} = \frac{2^4}{4} - \frac{(-1)^4}{4} = \frac{16}{4} - \frac{1}{4} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \]

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

### Step 6: Combine the results
Now we can combine the results of the two integrals:
\[ A = \frac{15}{4} + 9 \]
To add these, convert \( 9 \) into a fraction:
\[ 9 = \frac{36}{4} \]
Thus,
\[ A = \frac{15}{4} + \frac{36}{4} = \frac{51}{4} \]

### Final Result
The area of the region enclosed by the curves is:
\[ \boxed{\frac{51}{4}} \text{ square units} \]