If the line `x = alpha` divides the area of region `R = {(x,y) in R^(2) : x^(3) le y le x, 0 le x le 1}` into two equal parts, then

Options

A. `2alpha^(4)-4alpha^(2)+1=0`
B. `alpha^(4)+4alpha^(2)-1=0`
C. `1/2 lt alpha lt 1`
D. `0 lt alpha le 1/2`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the line \( x = \alpha \) divides the area of the region \( R = \{(x,y) \in \mathbb{R}^2 : x^3 \leq y \leq x, 0 \leq x \leq 1\} \) into two equal parts.

### Step-by-Step Solution:

1. **Identify the Functions and Region**:
The region \( R \) is bounded by the curves \( y = x^3 \) and \( y = x \) for \( 0 \leq x \leq 1 \).

2. **Calculate the Total Area of Region \( R \)**:
The area \( A \) of region \( R \) can be calculated using the integral:
\[ A = \int_0^1 (x - x^3) \, dx \]
Here, \( x \) is the upper function and \( x^3 \) is the lower function.

3. **Evaluate the Integral**:
\[ A = \int_0^1 (x - x^3) \, dx = \int_0^1 x \, dx - \int_0^1 x^3 \, dx \]
Calculate each integral:
- \( \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \)
- \( \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \)

Therefore,
\[ A = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

4. **Set Up the Equation for Area Divided by \( x = \alpha \)**:
To find \( \alpha \) such that the area to the left of \( x = \alpha \) is half of the total area, we set up the equation:
\[ \int_0^\alpha (x - x^3) \, dx = \frac{1}{2} A = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \]

5. **Evaluate the Left Side Integral**:
\[ \int_0^\alpha (x - x^3) \, dx = \int_0^\alpha x \, dx - \int_0^\alpha x^3 \, dx \]
Calculate each integral:
- \( \int_0^\alpha x \, dx = \left[ \frac{x^2}{2} \right]_0^\alpha = \frac{\alpha^2}{2} \)
- \( \int_0^\alpha x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^\alpha = \frac{\alpha^4}{4} \)

Therefore,
\[ \int_0^\alpha (x - x^3) \, dx = \frac{\alpha^2}{2} - \frac{\alpha^4}{4} \]

6. **Set the Equation**:
Set the integral equal to \( \frac{1}{8} \):
\[ \frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8} \]

7. **Multiply Through by 8 to Eliminate Fractions**:
\[ 4\alpha^2 - 2\alpha^4 = 1 \]
Rearranging gives:
\[ 2\alpha^4 - 4\alpha^2 + 1 = 0 \]

8. **Solve the Quadratic in \( \alpha^2 \)**:
Let \( u = \alpha^2 \), then:
\[ 2u^2 - 4u + 1 = 0 \]
Using the quadratic formula:
\[ u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2} \]

9. **Find \( \alpha \)**:
Since \( u = \alpha^2 \), we take the positive root:
\[ \alpha^2 = 1 - \frac{\sqrt{2}}{2} \quad \text{(since } \alpha \text{ must be in } [0, 1]) \]
Thus,
\[ \alpha = \sqrt{1 - \frac{\sqrt{2}}{2}} \]