The area (in square units) bounded by the curves `y=sqrt(x),2y-x+3=0,` x-axis, and lying in the first quadrant is

Options

A. 9
B. 36
C. 18
D. `(27)/(4)`

Explanation

To find the area bounded by the curves \( y = \sqrt{x} \), \( 2y - x + 3 = 0 \), the x-axis, and lying in the first quadrant, we will follow these steps:

### Step 1: Find the points of intersection of the curves
We need to find the intersection points of the curves \( y = \sqrt{x} \) and \( 2y - x + 3 = 0 \).

1. Substitute \( y = \sqrt{x} \) into the line equation:
\[ 2\sqrt{x} - x + 3 = 0 \]
Rearranging gives:
\[ 2\sqrt{x} = x - 3 \]

2. Square both sides to eliminate the square root:
\[ 4x = (x - 3)^2 \]
Expanding the right side:
\[ 4x = x^2 - 6x + 9 \]
Rearranging gives:
\[ x^2 - 10x + 9 = 0 \]

3. Factor the quadratic:
\[ (x - 9)(x - 1) = 0 \]
Thus, \( x = 9 \) and \( x = 1 \).

### Step 2: Find the corresponding y-values
Substituting \( x = 1 \) and \( x = 9 \) back into \( y = \sqrt{x} \):
- For \( x = 1 \):
\[ y = \sqrt{1} = 1 \]
- For \( x = 9 \):
\[ y = \sqrt{9} = 3 \]
Thus, the points of intersection are \( (1, 1) \) and \( (9, 3) \).

### Step 3: Set up the area calculation
The area \( A \) bounded by the curves can be found using the integral:
\[ A = \int_{1}^{9} (\text{upper function} - \text{lower function}) \, dx \]
The upper function is \( y = \sqrt{x} \) and the lower function is derived from the line equation rearranged as:
\[ y = \frac{x - 3}{2} \]

### Step 4: Calculate the area
Now we set up the integral:
\[ A = \int_{1}^{9} \left( \sqrt{x} - \frac{x - 3}{2} \right) \, dx \]
Simplifying the integrand:
\[ A = \int_{1}^{9} \left( \sqrt{x} - \frac{x}{2} + \frac{3}{2} \right) \, dx \]

### Step 5: Evaluate the integral
1. Calculate the integral:
\[ A = \int_{1}^{9} \sqrt{x} \, dx - \int_{1}^{9} \frac{x}{2} \, dx + \int_{1}^{9} \frac{3}{2} \, dx \]

2. Evaluate each integral:
- \( \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \)
- \( \int \frac{x}{2} \, dx = \frac{1}{4} x^2 \)
- \( \int \frac{3}{2} \, dx = \frac{3}{2} x \)

3. Now evaluate from 1 to 9:
\[ A = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{9} - \left[ \frac{1}{4} x^2 \right]_{1}^{9} + \left[ \frac{3}{2} x \right]_{1}^{9} \]

4. Calculate each term:
- For \( \frac{2}{3} x^{3/2} \):
\[ \frac{2}{3} (9^{3/2}) - \frac{2}{3} (1^{3/2}) = \frac{2}{3} (27 - 2) = \frac{2}{3} \times 25 = \frac{50}{3} \]
- For \( \frac{1}{4} x^2 \):
\[ \frac{1}{4} (9^2) - \frac{1}{4} (1^2) = \frac{1}{4} (81 - 1) = \frac{1}{4} \times 80 = 20 \]
- For \( \frac{3}{2} x \):
\[ \frac{3}{2} (9) - \frac{3}{2} (1) = \frac{3}{2} (9 - 1) = \frac{3}{2} \times 8 = 12 \]

5. Combine the results:
\[ A = \frac{50}{3} - 20 + 12 = \frac{50}{3} - \frac{60}{3} + \frac{36}{3} = \frac{26}{3} \]

### Final Area Calculation
Thus, the area bounded by the curves in the first quadrant is:
\[ \boxed{\frac{26}{3}} \text{ square units} \]