The area (in sq. units) bounded by the curve `y={{:(x.":",x in ["0, 1"]),(2-x,":" x in ["1, 2"]):}` with the x - axis from x = 0 to x= 2 is

Options

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

Answer

To find the area bounded by the curves \( y = x \) for \( x \) in the interval [0, 1] and \( y = 2 - x \) for \( x \) in the interval [1, 2], along with the x-axis from \( x = 0 \) to \( x = 2 \), we can follow these steps:

### Step 1: Identify the curves and their intervals
The curves are:
- \( y = x \) for \( 0 \leq x \leq 1 \)
- \( y = 2 - x \) for \( 1 < x \leq 2 \)

### Step 2: Sketch the curves
Sketching the curves helps visualize the area we need to calculate. The line \( y = x \) starts from the origin (0,0) and goes up to the point (1,1). The line \( y = 2 - x \) starts from the point (1,1) and goes down to (2,0).

### Step 3: Set up the integrals
To find the area, we will integrate the upper curve minus the lower curve over the specified intervals:
1. From \( x = 0 \) to \( x = 1 \), the upper curve is \( y = x \) and the lower curve is the x-axis (y = 0).
2. From \( x = 1 \) to \( x = 2 \), the upper curve is \( y = 2 - x \) and the lower curve is the x-axis (y = 0).

Thus, the area \( A \) can be expressed as:
\[ A = \int_{0}^{1} (x - 0) \, dx + \int_{1}^{2} ((2 - x) - 0) \, dx \]

### Step 4: Calculate the first integral
Calculating the first integral:
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

### Step 5: Calculate the second integral
Calculating the second integral:
\[ \int_{1}^{2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} \]
Calculating the limits:
\[ = \left( 2(2) - \frac{(2)^2}{2} \right) - \left( 2(1) - \frac{(1)^2}{2} \right) \]
\[ = (4 - 2) - (2 - 0.5) = 2 - 1.5 = 0.5 \]

### Step 6: Add the areas from both integrals
Now, we add the areas from both integrals:
\[ A = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
The area bounded by the curves and the x-axis from \( x = 0 \) to \( x = 2 \) is \( 1 \) square unit.