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 \( x = 2 \) and \( x = 4 \), and the x-axis in the first quadrant, we can follow these steps:

### Step 1: Understand the Curve
The equation \( y^2 = 9x \) represents a rightward-opening parabola. To express \( y \) in terms of \( x \):
\[ y = \sqrt{9x} = 3\sqrt{x} \]
Since we are only interested in the first quadrant, we will consider the positive root.

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

### Step 3: Calculate the Integral
We can factor out the constant:
\[ A = 3 \int_{2}^{4} \sqrt{x} \, dx \]
Now, we rewrite \( \sqrt{x} \) as \( x^{1/2} \):
\[ A = 3 \int_{2}^{4} x^{1/2} \, dx \]

### Step 4: Find the Antiderivative
The antiderivative of \( x^{1/2} \) is:
\[ \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Thus, we can evaluate the integral:
\[ A = 3 \left[ \frac{2}{3} x^{3/2} \right]_{2}^{4} \]

### Step 5: Evaluate the Definite Integral
Now we substitute the limits:
\[ A = 3 \left( \frac{2}{3} \left( 4^{3/2} - 2^{3/2} \right) \right) \]
Calculating \( 4^{3/2} \) and \( 2^{3/2} \):
\[ 4^{3/2} = (2^2)^{3/2} = 2^3 = 8 \]
\[ 2^{3/2} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2} \]
Substituting these values back:
\[ A = 3 \left( \frac{2}{3} (8 - 2\sqrt{2}) \right) \]
The \( 3 \) cancels out:
\[ A = 2(8 - 2\sqrt{2}) = 16 - 4\sqrt{2} \]

### Step 6: Final Area
Thus, the area of the region bounded by the given curves is:
\[ \text{Area} = 16 - 4\sqrt{2} \text{ square units} \]