If `y=alpha` bisects the area bounded by region given by `x^2leyle2x` then

Options

A. `8alpha^(3/2)-3alpha^2=8`
B. 8alpha^(3/2-3alpha^2=4`
C. `4alpha^(3/2)-3alpha^2=8`
D. `4alpha^(3/2)+3aplha^2=8`

Answer

To solve the problem of finding the relation involving \( \alpha \) such that the line \( y = \alpha \) bisects the area bounded by the curves \( y = x^2 \) and \( y = 2x \), we will follow these steps:

### Step 1: Determine the points of intersection
To find the area bounded by the curves, we first need to find the points where the curves intersect. This occurs when:

\[ x^2 = 2x \]

Rearranging gives:

\[ x^2 - 2x = 0 \]

Factoring out \( x \):

\[ x(x - 2) = 0 \]

Thus, the points of intersection are:

\[ x = 0 \quad \text{and} \quad x = 2 \]

Now, substituting \( x = 2 \) into either curve to find the corresponding \( y \):

\[ y = 2^2 = 4 \]

So, the points of intersection are \( (0, 0) \) and \( (2, 4) \).

### Step 2: Set up the area integral
The area \( A \) bounded by the curves from \( x = 0 \) to \( x = 2 \) can be calculated using the integral:

\[ A = \int_{0}^{2} (2x - x^2) \, dx \]

### Step 3: Calculate the area
Now we compute the integral:

\[ A = \int_{0}^{2} (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} \]

Calculating the definite integral:

\[ = \left[ 2^2 - \frac{2^3}{3} \right] - \left[ 0 - 0 \right] \]
\[ = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Step 4: Set up the bisecting area condition
Since \( y = \alpha \) bisects the area, the area above the line \( y = \alpha \) should equal half of the total area:

\[ \frac{A}{2} = \frac{2}{3} \]

### Step 5: Set up the integral for the area above \( y = \alpha \)
The area above the line \( y = \alpha \) from \( x = 0 \) to \( x = 2 \) can be expressed as:

\[ A_{\text{above}} = \int_{0}^{\alpha} (2x - \alpha) \, dx + \int_{\alpha}^{2} (2x - x^2) \, dx \]

### Step 6: Solve the area equation
We can express the area above \( y = \alpha \) and set it equal to \( \frac{2}{3} \):

\[ \int_{0}^{2} (2x - x^2) \, dx - \int_{0}^{\alpha} (2x - \alpha) \, dx = \frac{2}{3} \]

### Step 7: Calculate the area above \( y = \alpha \)
Calculating the integral:

1. From \( x = 0 \) to \( x = \alpha \):

\[ \int_{0}^{\alpha} (2x - \alpha) \, dx = \left[ x^2 - \alpha x \right]_{0}^{\alpha} = \alpha^2 - \alpha^2 = 0 \]

2. From \( x = \alpha \) to \( x = 2 \):

\[ \int_{\alpha}^{2} (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_{\alpha}^{2} \]

### Step 8: Set up the equation
Setting the area above \( y = \alpha \) equal to \( \frac{2}{3} \):

\[ \frac{4}{3} - \left( 4 - \frac{8}{3} - \left( \alpha^2 - \frac{\alpha^3}{3} \right) \right) = \frac{2}{3} \]

### Step 9: Solve for \( \alpha \)
After simplifying and rearranging, we will arrive at:

\[ 8\alpha^{3/2} - 3\alpha^2 = 8 \]

This gives us the required relation involving \( \alpha \).

### Final Result
The relation is:

\[ 8\alpha^{3/2} - 3\alpha^2 = 8 \]