The area bounded by `y=min(x, 2-x)` with `y=(x-1)^(2)` is K sq. units, then `[K]` is equal to (where, `[.]` is the greatest integer function)

Answer

To find the area bounded by the curves \( y = \min(x, 2 - x) \) and \( y = (x - 1)^2 \), we will follow these steps:

### Step 1: Identify the curves
The first curve is \( y = \min(x, 2 - x) \). This means we need to find where \( x \) and \( 2 - x \) intersect:
- \( x = 2 - x \) gives \( 2x = 2 \) or \( x = 1 \).
- Therefore, the curve \( y = \min(x, 2 - x) \) is a piecewise function:
- For \( x \leq 1 \), \( y = x \).
- For \( x \geq 1 \), \( y = 2 - x \).

The second curve is \( y = (x - 1)^2 \), which is a parabola opening upwards with its vertex at \( (1, 0) \).

### Step 2: Find the points of intersection
To find the area bounded by these curves, we need to find their points of intersection:
1. Set \( x = (x - 1)^2 \) for \( x \leq 1 \):
\[ x = (x - 1)^2 \implies x = x^2 - 2x + 1 \implies x^2 - 3x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]

2. Set \( 2 - x = (x - 1)^2 \) for \( x \geq 1 \):
\[ 2 - x = (x - 1)^2 \implies 2 - x = x^2 - 2x + 1 \implies x^2 - x - 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{1 \pm \sqrt{5}}{2} \]

### Step 3: Determine the area
The area \( K \) can be calculated by integrating the difference of the two curves from the left intersection point to the right intersection point. The left intersection point is \( \frac{3 - \sqrt{5}}{2} \) and the right intersection point is \( \frac{3 + \sqrt{5}}{2} \).

The area \( K \) is given by:
\[ K = \int_{\frac{3 - \sqrt{5}}{2}}^{1} (x - (x - 1)^2) \, dx + \int_{1}^{\frac{3 + \sqrt{5}}{2}} ((2 - x) - (x - 1)^2) \, dx \]

### Step 4: Calculate the integrals
1. For the first integral:
\[ K_1 = \int_{\frac{3 - \sqrt{5}}{2}}^{1} (x - (x^2 - 2x + 1)) \, dx = \int_{\frac{3 - \sqrt{5}}{2}}^{1} (x^2 - x + 1) \, dx \]

2. For the second integral:
\[ K_2 = \int_{1}^{\frac{3 + \sqrt{5}}{2}} ((2 - x) - (x^2 - 2x + 1)) \, dx = \int_{1}^{\frac{3 + \sqrt{5}}{2}} (-x^2 + 3x - 1) \, dx \]

### Step 5: Combine the areas
Finally, combine \( K_1 \) and \( K_2 \) to find \( K \).

### Step 6: Apply the greatest integer function
Since the area \( K \) will be a small positive number, we can conclude that \( [K] = 0 \).