Let `alpha` be the area of the larger region bounded by the curve `y^(2)=8x` and the lines `y=x` and `x=2,` which lies in the first quadrant.Then the value of `3 alpha` is equal to

Answer

To find the area \( \alpha \) of the larger region bounded by the curve \( y^2 = 8x \) and the lines \( y = x \) and \( x = 2 \) in the first quadrant, we can follow these steps:

### Step 1: Understand the curves and lines
The curve \( y^2 = 8x \) is a rightward-opening parabola. The line \( y = x \) is a diagonal line through the origin, and \( x = 2 \) is a vertical line.

### Step 2: Find points of intersection
To find the points where the curves intersect, we set \( y = x \) in the equation of the parabola:

\[ x^2 = 8x \]

Rearranging gives:

\[ x^2 - 8x = 0 \]

Factoring out \( x \):

\[ x(x - 8) = 0 \]

Thus, \( x = 0 \) or \( x = 8 \). Since we are only interested in the first quadrant, we will also consider the line \( x = 2 \).

### Step 3: Determine the area of interest
The area we need to calculate is bounded by:
- The parabola from \( x = 0 \) to \( x = 2 \)
- The line \( y = x \) from \( x = 0 \) to \( x = 2 \)

### Step 4: Set up the integral
The area \( \alpha \) can be calculated using the integral:

\[ \alpha = \int_{0}^{2} \left( \sqrt{8x} - x \right) \, dx \]

### Step 5: Solve the integral
First, we simplify the integral:

\[ \alpha = \int_{0}^{2} (2\sqrt{2x} - x) \, dx \]

Now we can integrate term by term:

1. For \( 2\sqrt{2x} \):

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

2. For \( -x \):

\[ \int -x \, dx = -\frac{x^2}{2} \]

Putting it all together, we have:

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

### Step 6: Evaluate the definite integral
Now we evaluate from \( 0 \) to \( 2 \):

1. At \( x = 2 \):

\[ \frac{4\sqrt{2}}{3} (2)^{3/2} - \frac{(2)^2}{2} = \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2} - 2 = \frac{8}{3} \cdot 2 - 2 = \frac{16}{3} - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3} \]

2. At \( x = 0 \):

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

Thus, the area \( \alpha \) is:

\[ \alpha = \frac{10}{3} \]

### Step 7: Calculate \( 3\alpha \)
Finally, we calculate \( 3\alpha \):

\[ 3\alpha = 3 \cdot \frac{10}{3} = 10 \]

### Final Answer
Thus, the value of \( 3\alpha \) is:

\[ \boxed{10} \]