The area bounded by `y=(x^(2)-x)^(2)` with the x - axis, between its two relative minima, is A sq, units, the value of 15A is equal to

Options

A. 1
B. 2
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To find the area bounded by the curve \( y = (x^2 - x)^2 \) with the x-axis between its two relative minima, we will follow these steps:

### Step 1: Identify the function and find its critical points
The function is given as:
\[ y = (x^2 - x)^2 \]
To find the critical points, we need to find the derivative of \( y \) and set it to zero.

### Step 2: Differentiate the function
Using the chain rule, we differentiate:
\[ \frac{dy}{dx} = 2(x^2 - x)(2x - 1) \]
Setting the derivative equal to zero:
\[ 2(x^2 - x)(2x - 1) = 0 \]
This gives us two factors to consider:
1. \( x^2 - x = 0 \)
2. \( 2x - 1 = 0 \)

### Step 3: Solve for \( x \)
From \( x^2 - x = 0 \):
\[ x(x - 1) = 0 \implies x = 0 \text{ or } x = 1 \]
From \( 2x - 1 = 0 \):
\[ x = \frac{1}{2} \]

### Step 4: Identify the relative minima
The critical points are \( x = 0, \frac{1}{2}, 1 \). We need to evaluate the function at these points to determine which are minima:
- \( y(0) = (0^2 - 0)^2 = 0 \)
- \( y\left(\frac{1}{2}\right) = \left(\left(\frac{1}{2}\right)^2 - \frac{1}{2}\right)^2 = \left(\frac{1}{4} - \frac{1}{2}\right)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16} \)
- \( y(1) = (1^2 - 1)^2 = 0 \)

Thus, the relative minima occur at \( x = 0 \) and \( x = 1 \).

### Step 5: Set up the integral for the area
The area \( A \) between the curve and the x-axis from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} (x^2 - x)^2 \, dx \]

### Step 6: Expand the integrand
Expanding \( (x^2 - x)^2 \):
\[ (x^2 - x)^2 = x^4 - 2x^3 + x^2 \]
Thus, the integral becomes:
\[ A = \int_{0}^{1} (x^4 - 2x^3 + x^2) \, dx \]

### Step 7: Integrate term by term
Calculating the integral:
\[ A = \left[ \frac{x^5}{5} - \frac{2x^4}{4} + \frac{x^3}{3} \right]_{0}^{1} = \left[ \frac{1}{5} - \frac{1}{2} + \frac{1}{3} \right] \]

### Step 8: Simplify the expression
Finding a common denominator (which is 30):
\[ A = \left[ \frac{6}{30} - \frac{15}{30} + \frac{10}{30} \right] = \frac{6 - 15 + 10}{30} = \frac{1}{30} \]

### Step 9: Calculate \( 15A \)
Now, we find \( 15A \):
\[ 15A = 15 \times \frac{1}{30} = \frac{15}{30} = \frac{1}{2} \]

### Final Result
Thus, the value of \( 15A \) is:
\[ \boxed{\frac{1}{2}} \]