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: Find the relative minima
To find the relative minima, we first need to differentiate the function \( y = (x^2 - x)^2 \).

1. **Differentiate the function**:
\[ y = (x^2 - x)^2 \]
Using the chain rule:
\[ \frac{dy}{dx} = 2(x^2 - x)(2x - 1) \]

2. **Set the derivative to zero**:
\[ 2(x^2 - x)(2x - 1) = 0 \]
This gives us two factors to consider:
- \( x^2 - x = 0 \) which simplifies to \( x(x - 1) = 0 \) giving \( x = 0 \) and \( x = 1 \).
- \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \).

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

### Step 2: Calculate the area under the curve from \( x = 0 \) to \( x = 1 \)
The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{1} (x^2 - x)^2 \, dx \]

### Step 3: Expand and simplify the integrand
First, expand \( (x^2 - x)^2 \):
\[ (x^2 - x)^2 = x^4 - 2x^3 + x^2 \]

### Step 4: Integrate the function
Now we integrate:
\[ A = \int_{0}^{1} (x^4 - 2x^3 + x^2) \, dx \]
Calculating the integral:
\[ A = \left[ \frac{x^5}{5} - \frac{2x^4}{4} + \frac{x^3}{3} \right]_{0}^{1} \]
Evaluating at the limits:
\[ A = \left( \frac{1}{5} - \frac{1}{2} + \frac{1}{3} \right) - \left( 0 - 0 + 0 \right) \]
Finding a common denominator (which is 30):
\[ A = \left( \frac{6}{30} - \frac{15}{30} + \frac{10}{30} \right) = \frac{1}{30} \]

### Step 5: Calculate \( 15A \)
Now, we find \( 15A \):
\[ 15A = 15 \times \frac{1}{30} = \frac{15}{30} = \frac{1}{2} \]

### Final Answer
Thus, the value of \( 15A \) is \( \frac{1}{2} \).

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