The area (in sq. units) of the region `A={(x,y) inRRxxRR|0lexle3,0leyle4,ylex^(2)+3x}` is:

Options

A. `(53)/(6)`
B. 8
C. `(59)/(6)`
D. `(26)/(3)`

Answer

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

### Step 1: Identify the boundaries of the region
The region is bounded by:
- The vertical line \( x = 3 \)
- The horizontal line \( y = 4 \)
- The curve \( y = x^2 + 3x \)

### Step 2: Find the intersection points
We need to find where the curve intersects the line \( y = 4 \):
\[ 4 = x^2 + 3x \]
Rearranging gives:
\[ x^2 + 3x - 4 = 0 \]
Now, we can factor or use the quadratic formula:
\[ (x - 1)(x + 4) = 0 \]
Thus, the roots are \( x = 1 \) and \( x = -4 \). Since we are only interested in the region where \( 0 \leq x \leq 3 \), we take the intersection point \( (1, 4) \).

### Step 3: Determine the area under the curve from \( x = 0 \) to \( x = 1 \)
We will calculate the area under the curve from \( x = 0 \) to \( x = 1 \):
\[ \text{Area}_1 = \int_{0}^{1} (x^2 + 3x) \, dx \]
Calculating the integral:
\[ = \left[ \frac{x^3}{3} + \frac{3x^2}{2} \right]_{0}^{1} = \left( \frac{1^3}{3} + \frac{3 \cdot 1^2}{2} \right) - \left( \frac{0^3}{3} + \frac{3 \cdot 0^2}{2} \right) \]
\[ = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} \]

### Step 4: Determine the area of the rectangle from \( x = 1 \) to \( x = 3 \)
The area of the rectangle formed between \( x = 1 \) and \( x = 3 \) with height \( y = 4 \):
\[ \text{Area}_2 = \text{width} \times \text{height} = (3 - 1) \times 4 = 2 \times 4 = 8 \]

### Step 5: Total area
The total area \( A \) is the sum of the two areas calculated:
\[ A = \text{Area}_1 + \text{Area}_2 = \left( \frac{1}{3} + \frac{3}{2} \right) + 8 \]
Calculating \( \frac{1}{3} + \frac{3}{2} \):
\[ = \frac{2}{6} + \frac{9}{6} = \frac{11}{6} \]
So,
\[ A = \frac{11}{6} + 8 = \frac{11}{6} + \frac{48}{6} = \frac{59}{6} \]

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

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