Prove that the curve `y = x^(2)` and, `x=y^(2)` divide the square bounded by x = 0, y = 0, x = 1, y = 1 into three equal parts.

Answer

To prove that the curves \( y = x^2 \) and \( x = y^2 \) divide the square bounded by \( x = 0 \), \( y = 0 \), \( x = 1 \), and \( y = 1 \) into three equal parts, we will follow these steps:

### Step 1: Find Points of Intersection
We need to find the points of intersection of the curves \( y = x^2 \) and \( x = y^2 \).

1. Substitute \( y = x^2 \) into \( x = y^2 \):
\[ x = (x^2)^2 = x^4 \]
Rearranging gives:
\[ x^4 - x = 0 \implies x(x^3 - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

2. The points of intersection are:
- When \( x = 0 \), \( y = 0^2 = 0 \) → Point \( (0, 0) \)
- When \( x = 1 \), \( y = 1^2 = 1 \) → Point \( (1, 1) \)

### Step 2: Sketch the Curves and the Square
We can visualize the curves and the square defined by the points \( (0, 0) \), \( (1, 0) \), \( (1, 1) \), and \( (0, 1) \).

### Step 3: Calculate the Area of the Square
The area of the square is:
\[ \text{Area} = \text{side}^2 = 1^2 = 1 \]

### Step 4: Calculate Area \( A_3 \) (Below the Curve \( y = x^2 \))
To find the area \( A_3 \) under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \):
\[ A_3 = \int_0^1 x^2 \, dx \]
Calculating the integral:
\[ A_3 = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Step 5: Calculate Area \( A_2 \) (Between the Curves)
To find the area \( A_2 \) between the curves \( y = x^2 \) and \( x = y^2 \) (or \( y = \sqrt{x} \)):
\[ A_2 = \int_0^1 (\sqrt{x} - x^2) \, dx \]
Calculating the integral:
\[ A_2 = \int_0^1 x^{1/2} \, dx - \int_0^1 x^2 \, dx \]
Calculating each part:
1. \( \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \)
2. \( \int_0^1 x^2 \, dx = \frac{1}{3} \)

Thus,
\[ A_2 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Step 6: Calculate Area \( A_1 \) (Above the Curve \( x = y^2 \))
The area \( A_1 \) is the remaining area in the square:
\[ A_1 = \text{Area of Square} - (A_2 + A_3) \]
Substituting the values:
\[ A_1 = 1 - \left(\frac{1}{3} + \frac{1}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3} \]

### Conclusion
We have found:
- \( A_1 = \frac{1}{3} \)
- \( A_2 = \frac{1}{3} \)
- \( A_3 = \frac{1}{3} \)

Thus, the areas \( A_1 \), \( A_2 \), and \( A_3 \) are equal, proving that the curves divide the square into three equal parts.