The area bounded by the curve `y^(2) = 16x` and the line x = 4 is

Options

A. `128/3 sq. units`
B. `64/3 sq. units`
C. `32/3 sq, units`
D. `16/3 sq. units`

Answer

To find the area bounded by the curve \(y^2 = 16x\) and the line \(x = 4\), we can follow these steps:

### Step 1: Understand the curve and the line
The equation \(y^2 = 16x\) represents a parabola that opens to the right. The line \(x = 4\) is a vertical line that intersects the parabola.

### Step 2: Find the points of intersection
To find the points where the parabola intersects the line \(x = 4\), we substitute \(x = 4\) into the equation of the parabola:
\[ y^2 = 16(4) = 64 \]
Taking the square root, we find:
\[ y = \pm 8 \]
Thus, the points of intersection are \((4, 8)\) and \((4, -8)\).

### Step 3: Set up the integral for the area
The area between the curve and the line from \(x = 0\) to \(x = 4\) can be calculated using the integral of the function that represents the upper part of the curve minus the lower part. The upper part of the curve is \(y = 4\sqrt{x}\) and the lower part is \(y = -4\sqrt{x}\).

The area \(A\) can be expressed as:
\[ A = \int_{0}^{4} (4\sqrt{x} - (-4\sqrt{x})) \, dx = \int_{0}^{4} 8\sqrt{x} \, dx \]

### Step 4: Calculate the integral
Now we calculate the integral:
\[ A = \int_{0}^{4} 8\sqrt{x} \, dx = 8 \int_{0}^{4} x^{1/2} \, dx \]

Using the power rule for integration:
\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]
we have:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \]
Thus,
\[ A = 8 \cdot \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4} = \frac{16}{3} \left[ 4^{3/2} - 0^{3/2} \right] \]

### Step 5: Evaluate the limits
Now we evaluate \(4^{3/2}\):
\[ 4^{3/2} = (4^1)^{3/2} = 8 \]
So,
\[ A = \frac{16}{3} \cdot 8 = \frac{128}{3} \]

### Conclusion
The area bounded by the curve \(y^2 = 16x\) and the line \(x = 4\) is:
\[ \boxed{\frac{128}{3}} \text{ square units} \]