Using integration, find the area of the region bounded by the parabola `y^(2)=4x` and the line `x=4`.

Answer

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

### Step 1: Understand the curves
The equation of the parabola \( y^2 = 4x \) can be rewritten as:
\[ y = \pm 2\sqrt{x} \]
This indicates that the parabola opens to the right. The line \( x = 4 \) is a vertical line.

### Step 2: Determine the points of intersection
We need to find the points where the parabola intersects the line \( x = 4 \). Substituting \( x = 4 \) into the parabola's equation:
\[ y^2 = 4(4) = 16 \implies y = \pm 4 \]
Thus, the points of intersection are \( (4, 4) \) and \( (4, -4) \).

### Step 3: Set up the integral for the area
The area between the curves can be calculated by integrating the upper curve minus the lower curve. The upper curve is \( y = 2\sqrt{x} \) and the lower curve is \( y = -2\sqrt{x} \). The area \( A \) can be expressed as:
\[ A = \int_{0}^{4} (2\sqrt{x} - (-2\sqrt{x})) \, dx \]
This simplifies to:
\[ A = \int_{0}^{4} (2\sqrt{x} + 2\sqrt{x}) \, dx = \int_{0}^{4} 4\sqrt{x} \, dx \]

### Step 4: Evaluate the integral
Now we compute the integral:
\[ A = 4 \int_{0}^{4} x^{1/2} \, dx \]
Using the power rule for integration:
\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]
we find:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Now, substituting the limits from 0 to 4:
\[ A = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} \]
Calculating this gives:
\[ A = 4 \left( \frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2} \right) = 4 \left( \frac{2}{3} \cdot 8 \right) = 4 \cdot \frac{16}{3} = \frac{64}{3} \]

### Step 5: Conclusion
Thus, the area of the region bounded by the parabola and the line is:
\[ \boxed{\frac{64}{3}} \text{ square units} \]