The area of the region bounded by `y^(2)=4x, x=0, x=4` and the X-axis in the first quadrant is

Options

A. 16
B. `16/3`
C. `32`
D. `32/3`

Answer

To find the area of the region bounded by the curve \(y^2 = 4x\), the lines \(x = 0\), \(x = 4\), and the x-axis in the first quadrant, we can follow these steps:

### Step 1: Understand the given equation
The equation \(y^2 = 4x\) represents a parabola that opens to the right. To express \(y\) in terms of \(x\), we can take the square root:

\[ y = \sqrt{4x} = 2\sqrt{x} \]

### Step 2: Identify the boundaries
The boundaries of the area we want to find are:
- The parabola \(y = 2\sqrt{x}\)
- The vertical line \(x = 0\) (the y-axis)
- The vertical line \(x = 4\)
- The x-axis (where \(y = 0\))

### Step 3: Set up the integral
The area \(A\) can be found by integrating the function \(y = 2\sqrt{x}\) from \(x = 0\) to \(x = 4\):

\[ A = \int_{0}^{4} 2\sqrt{x} \, dx \]

### Step 4: Calculate the integral
To solve the integral, we first find the antiderivative of \(2\sqrt{x}\):

\[ \int 2\sqrt{x} \, dx = \int 2x^{1/2} \, dx = 2 \cdot \frac{x^{3/2}}{3/2} = \frac{4}{3} x^{3/2} \]

Now, we evaluate this from \(0\) to \(4\):

\[ A = \left[ \frac{4}{3} x^{3/2} \right]_{0}^{4} \]

Calculating at the upper limit \(x = 4\):

\[ A = \frac{4}{3} (4)^{3/2} = \frac{4}{3} (8) = \frac{32}{3} \]

Calculating at the lower limit \(x = 0\):

\[ A = \frac{4}{3} (0)^{3/2} = 0 \]

Thus, the area is:

\[ A = \frac{32}{3} - 0 = \frac{32}{3} \]

### Final Answer
The area of the region bounded by \(y^2 = 4x\), \(x = 0\), \(x = 4\), and the x-axis in the first quadrant is:

\[ \boxed{\frac{32}{3}} \]