Find the area of the region bounded by `y^2=9x ,``x = 2, x = 4`and the x-axis in the first quadrant.

Options

A. `16 - 4sqrt2`
B. `15 - 4sqrt5`
C. `17 - 4sqrt3`
D. `18 - 4sqrt7`

Explanation

To find the area of the region bounded by the curve \( y^2 = 9x \), the lines \( x = 2 \), \( x = 4 \), and the x-axis in the first quadrant, we can follow these steps:

### Step 1: Understand the curve
The equation \( y^2 = 9x \) can be rewritten to express \( y \) in terms of \( x \):
\[ y = 3\sqrt{x} \]
This represents a parabola that opens to the right.

### Step 2: Identify the bounds
We are given the vertical lines \( x = 2 \) and \( x = 4 \). We will find the area under the curve \( y = 3\sqrt{x} \) between these two lines.

### Step 3: Set up the integral
The area \( A \) under the curve from \( x = 2 \) to \( x = 4 \) can be calculated using the definite integral:
\[ A = \int_{2}^{4} 3\sqrt{x} \, dx \]

### Step 4: Calculate the integral
To solve the integral, we first find the antiderivative of \( 3\sqrt{x} \):
\[ \int 3\sqrt{x} \, dx = 3 \cdot \frac{2}{3} x^{3/2} = 2x^{3/2} \]
Now we evaluate this from \( x = 2 \) to \( x = 4 \):
\[ A = \left[ 2x^{3/2} \right]_{2}^{4} \]

### Step 5: Evaluate the definite integral
Now we substitute the upper and lower limits into the antiderivative:
\[ A = 2(4^{3/2}) - 2(2^{3/2}) \]
Calculating \( 4^{3/2} \) and \( 2^{3/2} \):
\[ 4^{3/2} = (2^2)^{3/2} = 2^{3} = 8 \]
\[ 2^{3/2} = 2 \sqrt{2} \]
Now substituting these values back:
\[ A = 2(8) - 2(2\sqrt{2}) = 16 - 4\sqrt{2} \]

### Final Answer
Thus, the area of the region bounded by the curve, the lines \( x = 2 \), \( x = 4 \), and the x-axis in the first quadrant is:
\[ A = 16 - 4\sqrt{2} \text{ square units} \]