The area (in sq. units) of the region `{(x,y), x ge 0, y ge 0, y ge x-2, and y le sqrt(x)}` is

Options

A. `(10)/(3)`
B. `(13)/(3)`
C. `(5)/(3)`
D. `(8)/(3)`

Answer

To find the area of the region defined by the inequalities \( y \geq x - 2 \) and \( y \leq \sqrt{x} \) in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)), we will follow these steps:

### Step 1: Identify the curves and their intersection points
We have two curves:
1. \( y = \sqrt{x} \)
2. \( y = x - 2 \)

To find the intersection points, we set these equations equal to each other:
\[ \sqrt{x} = x - 2 \]

### Step 2: Solve for \( x \)
Squaring both sides:
\[ x = (x - 2)^2 \]
Expanding the right side:
\[ x = x^2 - 4x + 4 \]
Rearranging gives:
\[ 0 = x^2 - 5x + 4 \]
Factoring the quadratic:
\[ 0 = (x - 4)(x - 1) \]
Thus, the solutions are:
\[ x = 4 \quad \text{and} \quad x = 1 \]

### Step 3: Determine the corresponding \( y \) values
For \( x = 1 \):
\[ y = \sqrt{1} = 1 \]
For \( x = 4 \):
\[ y = \sqrt{4} = 2 \]
So, the intersection points are \( (1, 1) \) and \( (4, 2) \).

### Step 4: Set up the area integral
The area \( A \) can be computed by integrating the difference between the upper curve and the lower curve from \( x = 1 \) to \( x = 4 \):
\[ A = \int_{1}^{4} (\sqrt{x} - (x - 2)) \, dx \]
This simplifies to:
\[ A = \int_{1}^{4} (\sqrt{x} - x + 2) \, dx \]

### Step 5: Compute the integral
We can break this integral into three parts:
\[ A = \int_{1}^{4} \sqrt{x} \, dx - \int_{1}^{4} x \, dx + \int_{1}^{4} 2 \, dx \]

Calculating each integral:
1. For \( \int \sqrt{x} \, dx \):
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from 1 to 4:
\[ \left[ \frac{2}{3} (4)^{3/2} - \frac{2}{3} (1)^{3/2} \right] = \frac{2}{3} (8 - \frac{2}{3}) = \frac{16}{3} \]

2. For \( \int x \, dx \):
\[ \int x \, dx = \frac{1}{2} x^2 \]
Evaluating from 1 to 4:
\[ \left[ \frac{1}{2} (4^2) - \frac{1}{2} (1^2) \right] = \left[ \frac{1}{2} (16 - 1) \right] = \frac{15}{2} \]

3. For \( \int 2 \, dx \):
\[ \int 2 \, dx = 2x \]
Evaluating from 1 to 4:
\[ \left[ 2(4) - 2(1) \right] = 8 - 2 = 6 \]

### Step 6: Combine the results
Now we can combine the results from the three integrals:
\[ A = \frac{16}{3} - \frac{15}{2} + 6 \]
Finding a common denominator (which is 6):
\[ A = \frac{32}{6} - \frac{45}{6} + \frac{36}{6} = \frac{32 - 45 + 36}{6} = \frac{23}{6} \]

### Final Answer
The area of the region is:
\[ \boxed{\frac{23}{6}} \text{ square units} \]