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

Explanation

To solve the problem of finding 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, we can follow these steps:

### Step 1: Determine the Total Area of the Region \( R \)

The area of the region \( R \) can be calculated using the integral of the difference between the upper curve \( y = x \) and the lower curve \( y = x^3 \) from \( x = 0 \) to \( x = 1 \).

\[ \text{Area} = \int_0^1 (x - x^3) \, dx \]

### Step 2: Calculate the Integral

Now, we calculate the integral:

\[ \int_0^1 (x - x^3) \, dx = \int_0^1 x \, dx - \int_0^1 x^3 \, dx \]

Calculating each integral separately:

1. \(\int_0^1 x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2}\)
2. \(\int_0^1 x^3 \, dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1^4}{4} - 0 = \frac{1}{4}\)

Now, substituting back into the area calculation:

\[ \text{Area} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

### Step 3: Set Up the Equation for Half the Area

Since we want the line \( x = \alpha \) to divide the area into two equal parts, we need to find \( \alpha \) such that the area from \( x = 0 \) to \( x = \alpha \) is half of the total area:

\[ \int_0^\alpha (x - x^3) \, dx = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \]

### Step 4: Calculate the Integral from 0 to \( \alpha \)

Now we compute the integral from \( 0 \) to \( \alpha \):

\[ \int_0^\alpha (x - x^3) \, dx = \int_0^\alpha x \, dx - \int_0^\alpha x^3 \, dx \]

Calculating each integral:

1. \(\int_0^\alpha x \, dx = \left[\frac{x^2}{2}\right]_0^\alpha = \frac{\alpha^2}{2}\)
2. \(\int_0^\alpha x^3 \, dx = \left[\frac{x^4}{4}\right]_0^\alpha = \frac{\alpha^4}{4}\)

Putting it together:

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

### Step 5: Set Up the Equation

Now we set this equal to \( \frac{1}{8} \):

\[ \frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8} \]

### Step 6: Clear the Denominators

Multiply through by 8 to eliminate the fractions:

\[ 4\alpha^2 - 2\alpha^4 = 1 \]

Rearranging gives:

\[ 2\alpha^4 - 4\alpha^2 + 1 = 0 \]

### Step 7: Final Equation

This is the equation we need to solve for \( \alpha \):

\[ 2\alpha^4 - 4\alpha^2 + 1 = 0 \]