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 \( x = 2y + 3 \), and the x-axis in the first quadrant, we can follow these steps:

### Step 1: Identify the curves and their intersection points
The first curve is \( y = \sqrt{x} \) and the second curve is \( x = 2y + 3 \). To find the intersection points, we can substitute \( y \) from the first equation into the second equation.

From \( y = \sqrt{x} \), we have:
\[ x = y^2 \]
Substituting this into the second equation:
\[ y^2 = 2y + 3 \]
Rearranging gives us:
\[ y^2 - 2y - 3 = 0 \]

### Step 2: Solve the quadratic equation
We can factor the quadratic equation:
\[ (y - 3)(y + 1) = 0 \]
This gives us the solutions:
\[ y = 3 \quad \text{and} \quad y = -1 \]
Since we are only interested in the first quadrant, we take \( y = 3 \).

### Step 3: Find the corresponding \( x \) value
Substituting \( y = 3 \) back into \( x = y^2 \):
\[ x = 3^2 = 9 \]
Thus, the intersection point in the first quadrant is \( (9, 3) \).

### Step 4: Set up the integral for the area
To find the area bounded by the curves and the x-axis, we can integrate with respect to \( y \). The limits of integration will be from \( y = 0 \) to \( y = 3 \).

The area \( A \) can be expressed as:
\[ A = \int_{0}^{3} (x_{\text{right}} - x_{\text{left}}) \, dy \]
Here, \( x_{\text{right}} \) is given by the line \( x = 2y + 3 \) and \( x_{\text{left}} \) is given by the curve \( x = y^2 \).

Thus, the area becomes:
\[ A = \int_{0}^{3} ((2y + 3) - (y^2)) \, dy \]

### Step 5: Evaluate the integral
Now we compute the integral:
\[ A = \int_{0}^{3} (2y + 3 - y^2) \, dy \]
This simplifies to:
\[ A = \int_{0}^{3} (-y^2 + 2y + 3) \, dy \]

Calculating the integral:
\[ A = \left[ -\frac{y^3}{3} + y^2 + 3y \right]_{0}^{3} \]
Evaluating at the limits:
\[ A = \left( -\frac{3^3}{3} + 3^2 + 3 \cdot 3 \right) - \left( 0 \right) \]
\[ A = \left( -9 + 9 + 9 \right) = 9 \]

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

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