Find the area bounded by the curve ` y ^ 2 = 4ax`, X - axis and the lines ` x = 0 and x = a `.

Answer

To find the area bounded by the curve \( y^2 = 4ax \), the X-axis, and the lines \( x = 0 \) and \( x = a \), we can follow these steps:

### Step 1: Understand the curve
The equation \( y^2 = 4ax \) represents a rightward-opening parabola with its vertex at the origin (0, 0). The curve is symmetric about the X-axis.

### Step 2: Determine the bounds
We need to find the area between the curve and the X-axis from \( x = 0 \) to \( x = a \).

### Step 3: Express \( y \) in terms of \( x \)
From the equation \( y^2 = 4ax \), we can express \( y \) as:
\[ y = 2\sqrt{ax} \]
This gives us the upper half of the parabola. The lower half would be \( y = -2\sqrt{ax} \), but since we are interested in the area above the X-axis, we will use the positive root.

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

### Step 5: Simplify the integral
We can factor out the constant \( 2\sqrt{a} \):
\[ A = 2\sqrt{a} \int_{0}^{a} \sqrt{x} \, dx \]

### Step 6: Evaluate the integral
The integral \( \int \sqrt{x} \, dx \) is:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Now, we evaluate it from 0 to \( a \):
\[ \int_{0}^{a} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = \frac{2}{3} a^{3/2} \]

### Step 7: Substitute back into the area formula
Substituting this back into our area formula gives:
\[ A = 2\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{4}{3} a^2 \]

### Step 8: Final result
Thus, the area bounded by the curve, the X-axis, and the lines \( x = 0 \) and \( x = a \) is:
\[ A = \frac{4}{3} a^2 \text{ square units} \]