The area (in sqaure units) of the region `{(x,y):x ge 0, x + y le 3, x^(2) le 4y and y le 1 + sqrt(x)}` is

Options

A. `5/2`
B. `59/12`
C. `3/2`
D. `7/3`

Answer

To find the area of the region defined by the inequalities \( x \geq 0 \), \( x + y \leq 3 \), \( x^2 \leq 4y \), and \( y \leq 1 + \sqrt{x} \), we will follow these steps:

### Step 1: Understand the inequalities
We need to analyze the boundaries defined by the inequalities:
1. \( x \geq 0 \) - This indicates we are in the right half of the Cartesian plane.
2. \( x + y \leq 3 \) - This is a line with a y-intercept of 3 and an x-intercept of 3.
3. \( x^2 \leq 4y \) - This represents a parabola opening upwards.
4. \( y \leq 1 + \sqrt{x} \) - This is a curve that starts at (0, 1) and increases.

### Step 2: Find the points of intersection
To find the area, we need the points where these curves intersect. We will find the intersection points of the lines and curves:
1. Set \( y = 3 - x \) and \( y = \frac{x^2}{4} \):
\[ 3 - x = \frac{x^2}{4} \]
Rearranging gives:
\[ x^2 + 4x - 12 = 0 \]
Solving this quadratic equation gives:
\[ x = 2 \quad \text{and} \quad x = -6 \quad (\text{ignore } -6) \]
Plugging \( x = 2 \) back into \( y = 3 - x \):
\[ y = 1 \quad \Rightarrow \quad (2, 1) \]

2. Set \( y = 3 - x \) and \( y = 1 + \sqrt{x} \):
\[ 3 - x = 1 + \sqrt{x} \]
Rearranging gives:
\[ \sqrt{x} + x - 2 = 0 \]
Let \( \sqrt{x} = t \):
\[ t^2 + t - 2 = 0 \]
Solving gives:
\[ t = 1 \quad \Rightarrow \quad x = 1 \quad \Rightarrow \quad y = 2 \quad \Rightarrow \quad (1, 2) \]

3. Set \( y = 1 + \sqrt{x} \) and \( y = \frac{x^2}{4} \):
\[ 1 + \sqrt{x} = \frac{x^2}{4} \]
Rearranging gives:
\[ x^2 - 4\sqrt{x} + 4 = 0 \]
Let \( \sqrt{x} = t \):
\[ t^2 - 4t + 4 = 0 \quad \Rightarrow \quad (t - 2)^2 = 0 \quad \Rightarrow \quad t = 2 \quad \Rightarrow \quad x = 4 \quad \Rightarrow \quad y = 3 \]

### Step 3: Identify the area to integrate
The area is bounded by the curves from \( x = 0 \) to \( x = 2 \) and from \( x = 1 \) to \( x = 4 \).

### Step 4: Set up the integral
The area \( A \) can be calculated as:
\[ A = \int_0^1 (1 + \sqrt{x}) \, dx + \int_1^2 (3 - x) \, dx - \int_0^2 \left(\frac{x^2}{4}\right) \, dx \]

### Step 5: Calculate the integrals
1. For \( \int_0^1 (1 + \sqrt{x}) \, dx \):
\[ = \left[x + \frac{2}{3}x^{3/2}\right]_0^1 = 1 + \frac{2}{3} = \frac{5}{3} \]

2. For \( \int_1^2 (3 - x) \, dx \):
\[ = \left[3x - \frac{x^2}{2}\right]_1^2 = (6 - 2) - (3 - \frac{1}{2}) = 4 - 2.5 = 1.5 \]

3. For \( \int_0^2 \left(\frac{x^2}{4}\right) \, dx \):
\[ = \frac{1}{4} \left[\frac{x^3}{3}\right]_0^2 = \frac{1}{4} \cdot \frac{8}{3} = \frac{2}{3} \]

### Step 6: Combine the areas
Now, combine the areas:
\[ A = \frac{5}{3} + 1.5 - \frac{2}{3} = \frac{5}{3} + \frac{3}{2} - \frac{2}{3} \]
Converting to a common denominator (6):
\[ A = \frac{10}{6} + \frac{9}{6} - \frac{4}{6} = \frac{15}{6} = \frac{5}{2} \]

### Final Answer
The area of the region is \( \frac{5}{2} \) square units.