Find the area of the loop of the curve ` y^2 = x ( 1 - x ) ^ 2 `.

Answer

To find the area of the loop of the curve given by the equation \( y^2 = x(1 - x)^2 \), we will follow these steps:

### Step 1: Understand the Curve
The equation \( y^2 = x(1 - x)^2 \) describes a curve that has a loop. To find the area of the loop, we first need to identify the points where the loop intersects the x-axis.

### Step 2: Find Intersection Points
Set \( y = 0 \):
\[ 0 = x(1 - x)^2 \]
This gives us:
\[ x = 0 \quad \text{or} \quad (1 - x)^2 = 0 \Rightarrow x = 1 \]
Thus, the curve intersects the x-axis at \( x = 0 \) and \( x = 1 \).

### Step 3: Determine the Area of the Loop
The area enclosed by the loop can be calculated using the definite integral of the upper half of the curve from \( x = 0 \) to \( x = 1 \). The upper half of the curve is given by:
\[ y = \sqrt{x(1 - x)^2} = (1 - x)\sqrt{x} \]
The area \( A \) can be expressed as:
\[ A = 2 \int_0^1 (1 - x)\sqrt{x} \, dx \]
We multiply by 2 because the area below the x-axis (the lower half) is symmetric to the area above the x-axis.

### Step 4: Solve the Integral
Now we will compute the integral:
\[ A = 2 \int_0^1 (1 - x)\sqrt{x} \, dx \]
Expanding the integrand:
\[ A = 2 \int_0^1 (\sqrt{x} - x\sqrt{x}) \, dx = 2 \left( \int_0^1 x^{1/2} \, dx - \int_0^1 x^{3/2} \, dx \right) \]

### Step 5: Evaluate the Integrals
1. For the first integral:
\[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \]
2. For the second integral:
\[ \int_0^1 x^{3/2} \, dx = \left[ \frac{x^{5/2}}{5/2} \right]_0^1 = \frac{2}{5} \]

### Step 6: Combine the Results
Now substitute back into the area formula:
\[ A = 2 \left( \frac{2}{3} - \frac{2}{5} \right) \]
Finding a common denominator (15):
\[ A = 2 \left( \frac{10}{15} - \frac{6}{15} \right) = 2 \left( \frac{4}{15} \right) = \frac{8}{15} \]

### Final Answer
Thus, the area of the loop of the curve \( y^2 = x(1 - x)^2 \) is:
\[ \boxed{\frac{8}{15}} \]