Find the area bounded by the curve `y=sqrtx,x=2y+3` in the first quadrant and X-axis.

Explanation

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

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

1. Substitute \( y = \sqrt{x} \) into \( x = 2y + 3 \):
\[ x = 2\sqrt{x} + 3 \]
2. Rearranging gives:
\[ x - 2\sqrt{x} - 3 = 0 \]
3. Let \( \sqrt{x} = y \) (thus \( x = y^2 \)):
\[ y^2 - 2y - 3 = 0 \]
4. Factor the quadratic:
\[ (y - 3)(y + 1) = 0 \]
5. The solutions are \( y = 3 \) and \( y = -1 \). Since we are in the first quadrant, we take \( y = 3 \).

### Step 2: Find the corresponding x-coordinate
Substituting \( y = 3 \) back into \( x = 2y + 3 \):
\[ x = 2(3) + 3 = 6 + 3 = 9 \]
Thus, the intersection point is \( (9, 3) \).

### Step 3: Set up the integral for the area
The area \( A \) can be found by integrating the difference between the line and the curve from \( y = 0 \) to \( y = 3 \):
\[ A = \int_{0}^{3} \left( (2y + 3) - y^2 \right) \, dy \]

### Step 4: Calculate the integral
1. Simplify the integrand:
\[ A = \int_{0}^{3} (2y + 3 - y^2) \, dy \]
2. Integrate:
\[ A = \int_{0}^{3} (2y - y^2 + 3) \, dy = \left[ y^2 - \frac{y^3}{3} + 3y \right]_{0}^{3} \]
3. Evaluate the definite integral:
- At \( y = 3 \):
\[ A = \left[ 3^2 - \frac{3^3}{3} + 3(3) \right] = \left[ 9 - 9 + 9 \right] = 9 \]
- At \( y = 0 \):
\[ A = 0 \]
4. Thus, the area is:
\[ A = 9 - 0 = 9 \]

### Final Answer
The area bounded by the curve \( y = \sqrt{x} \), the line \( x = 2y + 3 \), the x-axis, and the y-axis in the first quadrant is \( 9 \) square units.

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