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 will follow these steps:

### Step 1: Identify the curves
The inequalities define two curves:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = x + 2 \) (a straight line with a slope of 1 and y-intercept of 2)

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine where they intersect. We set the equations equal to each other:

\[ x^2 = x + 2 \]

Rearranging gives:

\[ x^2 - x - 2 = 0 \]

### Step 3: Solve the quadratic equation
We can factor the quadratic:

\[ (x - 2)(x + 1) = 0 \]

Thus, the solutions are:

\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) is given by the integral of the upper curve minus the lower curve:

\[ A = \int_{-1}^{2} ((x + 2) - (x^2)) \, dx \]

### Step 5: Simplify the integrand
The integrand simplifies to:

\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \]

### Step 6: Integrate
Now we compute the integral:

\[ A = \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

Calculating the integral:

\[ = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} \]

### Step 7: Evaluate the definite integral
Now we evaluate the integral at the bounds:

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 \( x = -1 \):

\[ -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1 - 4.5}{3} = -\frac{3.5}{3} = -\frac{7}{6} \]

### Step 8: Calculate the area
Now we subtract the two results:

\[ A = \left(\frac{10}{3}\right) - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} \]

To add these fractions, we need a common denominator, which is 6:

\[ A = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
Thus, the area of the region \( A \) is:

\[ \boxed{\frac{9}{2}} \text{ square units} \]