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

Options

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

Answer

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

### Step 1: Identify the boundaries
We have the following boundaries from the inequalities:
1. \( x = 0 \) (the y-axis)
2. \( x = 3 \) (a vertical line)
3. \( y = 0 \) (the x-axis)
4. \( y = 4 \) (a horizontal line)
5. \( y = x^2 + 3x \) (a parabola)

### Step 2: Sketch the region
We will sketch the region defined by these inequalities on the coordinate plane. The parabola \( y = x^2 + 3x \) opens upwards and intersects the x-axis at points where \( y = 0 \).

### Step 3: Find the intersection points
To find the intersection points of the parabola with the line \( y = 4 \):
Set \( x^2 + 3x = 4 \):
\[ x^2 + 3x - 4 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]
Calculating the roots:
1. \( x = 1 \)
2. \( x = -4 \) (not in our region since \( x \) must be between 0 and 3)

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

### Step 4: Determine the area
The area can be divided into two parts:
1. Area under the parabola from \( x = 0 \) to \( x = 1 \).
2. Area of the rectangle from \( x = 1 \) to \( x = 3 \).

#### Area under the parabola (A1):
\[ A_1 = \int_0^1 (x^2 + 3x) \, dx \]
Calculating the integral:
\[ = \left[ \frac{x^3}{3} + \frac{3x^2}{2} \right]_0^1 = \left( \frac{1^3}{3} + \frac{3 \cdot 1^2}{2} \right) - \left( 0 \right) = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{4.5}{3} = \frac{5.5}{3} = \frac{11}{6} \]

#### Area of the rectangle (A2):
The rectangle extends from \( x = 1 \) to \( x = 3 \) with height 4.
\[ A_2 = \text{width} \times \text{height} = (3 - 1) \times 4 = 2 \times 4 = 8 \]

### Step 5: Total area
The total area \( A \) is the sum of \( A_1 \) and \( A_2 \):
\[ A = A_1 + A_2 = \frac{11}{6} + 8 = \frac{11}{6} + \frac{48}{6} = \frac{59}{6} \]

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

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