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

### Step 1: Understand the boundaries
The region is bounded by:
1. The vertical line \( x = 0 \)
2. The vertical line \( x = 3 \)
3. The horizontal line \( y = 0 \)
4. The horizontal line \( y = 4 \)
5. The curve \( y = x^2 + 3x \)

### Step 2: Find the intersection points
We need to find where the curve \( y = x^2 + 3x \) intersects the line \( y = 4 \):
\[ 4 = x^2 + 3x \]
Rearranging gives us:
\[ x^2 + 3x - 4 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 3, c = -4 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]
Calculating the two possible values:
1. \( x = \frac{2}{2} = 1 \)
2. \( x = \frac{-8}{2} = -4 \) (not in our region since \( x \) must be non-negative)

Thus, the intersection point is \( (1, 4) \).

### Step 3: Set up the area calculation
The area can be split into two parts:
1. From \( x = 0 \) to \( x = 1 \): the area under the curve \( y = x^2 + 3x \)
2. From \( x = 1 \) to \( x = 3 \): the area under the line \( y = 4 \)

### Step 4: Calculate the area from \( x = 0 \) to \( x = 1 \)
The area under the curve from \( x = 0 \) to \( x = 1 \) is given by the integral:
\[ \text{Area}_1 = \int_0^1 (x^2 + 3x) \, dx \]
Calculating this integral:
\[ = \left[ \frac{x^3}{3} + \frac{3x^2}{2} \right]_0^1 = \left( \frac{1^3}{3} + \frac{3 \cdot 1^2}{2} \right) - \left( 0 \right) \]
\[ = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{4.5}{3} = \frac{5.5}{3} = \frac{11}{6} \]

### Step 5: Calculate the area from \( x = 1 \) to \( x = 3 \)
The area under the line \( y = 4 \) from \( x = 1 \) to \( x = 3 \):
\[ \text{Area}_2 = \int_1^3 4 \, dx = 4[x]_1^3 = 4(3 - 1) = 4 \cdot 2 = 8 \]

### Step 6: Total Area
Now, we can find the total area:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{11}{6} + 8 = \frac{11}{6} + \frac{48}{6} = \frac{59}{6} \]

Thus, the area of the region \( A \) is \( \frac{59}{6} \) square units.