The area (in sq units) of the region bounded by the curve `y = sqrtx` and the lines `y = 0, y = x - 2,` is

Options

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

Explanation

To find the area of the region bounded by the curve \( y = \sqrt{x} \) and the lines \( y = 0 \) and \( y = x - 2 \), we will follow these steps:

### Step 1: Identify the curves and points of intersection
We have the curve \( y = \sqrt{x} \) and the line \( y = x - 2 \). We also have the line \( y = 0 \) (the x-axis).

To find the points of intersection between \( y = \sqrt{x} \) and \( y = x - 2 \), we set them equal to each other:
\[ \sqrt{x} = x - 2 \]
Squaring both sides gives:
\[ x = (x - 2)^2 \]
Expanding the right side:
\[ x = x^2 - 4x + 4 \]
Rearranging gives:
\[ x^2 - 5x + 4 = 0 \]

### Step 2: Solve the quadratic equation
We can factor the quadratic:
\[ (x - 1)(x - 4) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = 4 \]

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

### Step 4: Set up the area integral
The area can be calculated by integrating the difference between the upper curve and the lower curve. The area \( A \) can be expressed as:
\[ A = \int_{0}^{2} \sqrt{x} \, dx + \int_{2}^{4} (\sqrt{x} - (x - 2)) \, dx \]

### Step 5: Calculate the first integral
Calculating the first integral from \( 0 \) to \( 2 \):
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from \( 0 \) to \( 2 \):
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2} = \frac{2}{3} (2^{3/2}) - 0 = \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3} \]

### Step 6: Calculate the second integral
Calculating the second integral from \( 2 \) to \( 4 \):
\[ \int (\sqrt{x} - (x - 2)) \, dx = \int \sqrt{x} \, dx - \int (x - 2) \, dx \]
Calculating \( \int (x - 2) \, dx \):
\[ \int (x - 2) \, dx = \frac{x^2}{2} - 2x \]
Now we evaluate:
\[ \int_{2}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{2}^{4} = \frac{2}{3} (4^{3/2} - 2^{3/2}) = \frac{2}{3} (8 - 2\sqrt{2}) = \frac{16}{3} - \frac{4\sqrt{2}}{3} \]
And for \( \int_{2}^{4} (x - 2) \, dx \):
\[ \left[ \frac{x^2}{2} - 2x \right]_{2}^{4} = \left( \frac{16}{2} - 8 \right) - \left( \frac{4}{2} - 4 \right) = (8 - 8) - (2 - 4) = 0 + 2 = 2 \]

### Step 7: Combine the areas
Now we combine the areas:
\[ A = \frac{4\sqrt{2}}{3} + \left( \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2 \right) \]
This simplifies to:
\[ A = \frac{4\sqrt{2}}{3} + \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2 = \frac{16}{3} - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3} \]

### Final Answer
Thus, the area of the region bounded by the curves is:
\[ \boxed{\frac{10}{3}} \text{ square units} \]