The area of the region
`A= {(x,y) in R xx R|0 le x le 3, 0 le y le 4, x^(2) + 3x}`is

Answer

To find the area of the region defined by \( A = \{(x,y) \in \mathbb{R}^2 | 0 \leq x \leq 3, 0 \leq y \leq 4, y = x^2 + 3x\} \), we will follow these steps:

### Step 1: Determine the intersection points
We need to find the intersection points of the parabola \( y = x^2 + 3x \) with the line \( y = 4 \).

Set the equations equal to each other:
\[ x^2 + 3x = 4 \]
Rearranging gives:
\[ x^2 + 3x - 4 = 0 \]

### Step 2: Factor the quadratic equation
To factor the quadratic equation \( x^2 + 3x - 4 = 0 \), we look for two numbers that multiply to \(-4\) and add to \(3\). The factors are \(4\) and \(-1\):
\[ (x + 4)(x - 1) = 0 \]

### Step 3: Solve for \( x \)
Setting each factor to zero gives:
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \quad (\text{not in the interval } [0, 3]) \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \quad (\text{valid}) \]

### Step 4: Find the area under the parabola from \( x = 0 \) to \( x = 1 \)
The area under the parabola from \( x = 0 \) to \( x = 1 \) can be calculated using integration:
\[ \text{Area}_1 = \int_0^1 (x^2 + 3x) \, dx \]

Calculating the integral:
\[ \int (x^2 + 3x) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} \]
Evaluating from \(0\) to \(1\):
\[ \text{Area}_1 = \left[ \frac{1^3}{3} + \frac{3(1^2)}{2} \right] - \left[ \frac{0^3}{3} + \frac{3(0^2)}{2} \right] = \frac{1}{3} + \frac{3}{2} \]
Finding a common denominator (6):
\[ \text{Area}_1 = \frac{2}{6} + \frac{9}{6} = \frac{11}{6} \]

### Step 5: Calculate the area of the rectangle from \( x = 1 \) to \( x = 3 \)
The area of the rectangle can be calculated as:
\[ \text{Area}_2 = \text{length} \times \text{height} = (3 - 1) \times 4 = 2 \times 4 = 8 \]

### Step 6: Total area
Now, we add the areas from both regions:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{11}{6} + 8 \]
Converting \(8\) to sixths:
\[ 8 = \frac{48}{6} \]
Thus,
\[ \text{Total Area} = \frac{11}{6} + \frac{48}{6} = \frac{59}{6} \]

### Final Answer
The area of the region \( A \) is:
\[ \frac{59}{6} \text{ square units} \]