The area bounded by the curve `y=x^(2)(x-1)^(2)` with the x - axis is k sq. units. Then the value of 60 k is equal to

Options

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

Answer

To find the area bounded by the curve \( y = x^2 (x - 1)^2 \) with the x-axis, we will follow these steps:

### Step 1: Expand the function
First, we need to expand the function \( y = x^2 (x - 1)^2 \).

\[ y = x^2 (x^2 - 2x + 1) = x^4 - 2x^3 + x^2 \]

### Step 2: Find the points of intersection with the x-axis
Next, we find the points where the curve intersects the x-axis by setting \( y = 0 \).

\[ x^2 (x - 1)^2 = 0 \]

This gives us:

\[ x^2 = 0 \quad \text{or} \quad (x - 1)^2 = 0 \]

Thus, the roots are:

\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 3: Set up the integral for the area
The area \( A \) bounded by the curve and the x-axis from \( x = 0 \) to \( x = 1 \) can be found using integration:

\[ A = \int_0^1 (x^4 - 2x^3 + x^2) \, dx \]

### Step 4: Calculate the integral
Now we will calculate the integral:

\[ A = \int_0^1 x^4 \, dx - 2 \int_0^1 x^3 \, dx + \int_0^1 x^2 \, dx \]

Calculating each integral separately:

1. \(\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}\)
2. \(\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}\)
3. \(\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}\)

Now substituting these results back into the area calculation:

\[ A = \frac{1}{5} - 2 \cdot \frac{1}{4} + \frac{1}{3} \]

### Step 5: Simplify the expression
Now we need a common denominator to combine these fractions. The least common multiple of 5, 4, and 3 is 60.

Rewriting each term with a denominator of 60:

\[ A = \frac{12}{60} - \frac{30}{60} + \frac{20}{60} = \frac{12 - 30 + 20}{60} = \frac{2}{60} = \frac{1}{30} \]

### Step 6: Find the value of \( k \)
Since the area \( A = k \), we have:

\[ k = \frac{1}{30} \]

### Step 7: Calculate \( 60k \)
Now we need to find \( 60k \):

\[ 60k = 60 \cdot \frac{1}{30} = 2 \]

### Final Answer
Thus, the value of \( 60k \) is:

\[ \boxed{2} \]