Find all maxima and minima of the function `y = x(x-1)^2` for `0<=x<=2` Also, determine the area bounded by the curve `y = x (x-1)^2`, the X-axis and the line x = 2 .

Answer

To find the maxima and minima of the function \( y = x(x-1)^2 \) for \( 0 \leq x \leq 2 \), and to determine the area bounded by the curve, the x-axis, and the line \( x = 2 \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = x(x-1)^2 \).

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

Using the product rule:

\[ f'(x) = (x-1)^2 + x \cdot 2(x-1) \cdot (1) \]

Simplifying this:

\[ f'(x) = (x-1)^2 + 2x(x-1) \]

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find critical points:

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

Factoring out \( (x-1) \):

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

This gives us:

\[ (x-1)(3x-1) = 0 \]

### Step 3: Solve for critical points
Setting each factor to zero:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( 3x - 1 = 0 \) → \( x = \frac{1}{3} \)

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical points and at the endpoints \( x = 0 \) and \( x = 2 \):

1. \( f(0) = 0(0-1)^2 = 0 \)
2. \( f(1) = 1(1-1)^2 = 0 \)
3. \( f\left(\frac{1}{3}\right) = \frac{1}{3}\left(\frac{1}{3}-1\right)^2 = \frac{1}{3} \cdot \left(-\frac{2}{3}\right)^2 = \frac{4}{27} \)
4. \( f(2) = 2(2-1)^2 = 2 \)

### Step 5: Identify maxima and minima
From the evaluations:

- \( f(0) = 0 \)
- \( f(1) = 0 \)
- \( f\left(\frac{1}{3}\right) = \frac{4}{27} \)
- \( f(2) = 2 \)

The minimum value is \( 0 \) (at \( x = 0 \) and \( x = 1 \)), and the maximum value is \( 2 \) (at \( x = 2 \)).

### Step 6: Determine the area bounded by the curve
To find the area bounded by the curve, the x-axis, and the line \( x = 2 \), we will integrate the function from \( 0 \) to \( 2 \):

\[ \text{Area} = \int_0^2 f(x) \, dx = \int_0^2 x(x-1)^2 \, dx \]

### Step 7: Simplify and integrate
First, expand the integrand:

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

Now integrate term by term:

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

Calculating at the limits:

\[ = \left( \frac{2^4}{4} - \frac{2 \cdot 2^3}{3} + \frac{2^2}{2} \right) - \left( 0 \right) \]

Calculating each term:

\[ = \left( \frac{16}{4} - \frac{16}{3} + 2 \right) \]
\[ = 4 - \frac{16}{3} + 2 = 6 - \frac{16}{3} = \frac{18}{3} - \frac{16}{3} = \frac{2}{3} \]

### Final Results
- The minimum value of \( y \) is \( 0 \) at \( x = 0 \) and \( x = 1 \).
- The maximum value of \( y \) is \( 2 \) at \( x = 2 \).
- The area bounded by the curve, the x-axis, and the line \( x = 2 \) is \( \frac{2}{3} \) square units.