Find the area of the region bounded by the curve ` y = sqrt ( 6x + 4)`, the X - axis and the lines ` x = 0, x = 2`.

Answer

To find the area of the region bounded by the curve \( y = \sqrt{6x + 4} \), the X-axis, and the lines \( x = 0 \) and \( x = 2 \), we will follow these steps:

### Step 1: Set up the integral
The area \( A \) can be found using the definite integral of the function from \( x = 0 \) to \( x = 2 \):
\[ A = \int_{0}^{2} \sqrt{6x + 4} \, dx \]

### Step 2: Substitute to simplify the integral
Let \( t = 6x + 4 \). Then, we differentiate to find \( dx \):
\[ dt = 6 \, dx \quad \Rightarrow \quad dx = \frac{dt}{6} \]

Next, we need to change the limits of integration:
- When \( x = 0 \):
\[ t = 6(0) + 4 = 4 \]
- When \( x = 2 \):
\[ t = 6(2) + 4 = 16 \]

Now we can rewrite the integral:
\[ A = \int_{4}^{16} \sqrt{t} \cdot \frac{dt}{6} = \frac{1}{6} \int_{4}^{16} \sqrt{t} \, dt \]

### Step 3: Integrate the function
The integral of \( \sqrt{t} \) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]

Now we evaluate the definite integral:
\[ A = \frac{1}{6} \left[ \frac{2}{3} t^{3/2} \right]_{4}^{16} \]

### Step 4: Calculate the definite integral
Substituting the limits into the integral:
\[ A = \frac{1}{6} \left( \frac{2}{3} (16)^{3/2} - \frac{2}{3} (4)^{3/2} \right) \]

Calculating \( (16)^{3/2} \) and \( (4)^{3/2} \):
\[ (16)^{3/2} = 64 \quad \text{and} \quad (4)^{3/2} = 8 \]

Thus, we have:
\[ A = \frac{1}{6} \left( \frac{2}{3} (64 - 8) \right) = \frac{1}{6} \left( \frac{2}{3} \cdot 56 \right) \]

### Step 5: Simplify the expression
Calculating further:
\[ A = \frac{1}{6} \cdot \frac{112}{3} = \frac{112}{18} = \frac{56}{9} \]

### Final Answer
The area of the region bounded by the curve, the X-axis, and the lines \( x = 0 \) and \( x = 2 \) is:
\[ \boxed{\frac{56}{9}} \text{ square units} \]