The area (in sq units) of the region `A={(x,y): x^(2) le y le x +2}` is

Options

A. `4`
B. `3`
C. `3/2`
D. `9/2`

Explanation

To find the area of the region defined by the inequalities \( A = \{(x,y): x^2 \leq y \leq x + 2\} \), we need to follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = x + 2 \) (a straight line)

### Step 2: Find the points of intersection
To find the area between these curves, we first need to find the points where they intersect. We set the equations equal to each other:
\[ x^2 = x + 2 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Now, we can factor this quadratic equation:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Calculate the corresponding y-values
Next, we find the y-values at these x-values:
- For \( x = -1 \):
\[ y = (-1)^2 = 1 \]
- For \( x = 2 \):
\[ y = 2 + 2 = 4 \]

So, the points of intersection are \( (-1, 1) \) and \( (2, 4) \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{-1}^{2} ((x + 2) - (x^2)) \, dx \]
This simplifies to:
\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \]

### Step 5: Evaluate the integral
Now we compute the integral:
\[ A = \int_{-1}^{2} (-x^2 + x + 2) \, dx \]
Calculating the antiderivative:
\[ = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} \]

### Step 6: Plug in the limits
Now we evaluate at the upper and lower limits:
1. At \( x = 2 \):
\[ -\frac{2^3}{3} + \frac{2^2}{2} + 2(2) = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3} \]
2. At \( x = -1 \):
\[ -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1}{3} - \frac{9}{6} = \frac{1 - 9}{6} = -\frac{8}{6} = -\frac{4}{3} \]

### Step 7: Calculate the area
Now, we subtract the two results:
\[ A = \left(\frac{10}{3}\right) - \left(-\frac{4}{3}\right) = \frac{10}{3} + \frac{4}{3} = \frac{14}{3} \]

### Final Result
The area of the region \( A \) is:
\[ \boxed{\frac{14}{3}} \]