The area (in sq. units) bounded by the curve `y={{:(x.":",x in ["0, 1"]),(2-x,":",xin["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 \( x \in [0, 1] \) and \( y = 2 - x \) for \( x \in [1, 2] \) 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 defined as:
- \( y = x \) for \( x \) in the interval \( [0, 1] \)
- \( y = 2 - x \) for \( x \) in the interval \( [1, 2] \)

### Step 2: Sketch the curves
We can sketch the two lines:
- The line \( y = x \) starts at the origin (0,0) and goes up to the point (1,1).
- The line \( y = 2 - x \) starts at the point (1,1) and goes down to the point (2,0).

### Step 3: Calculate the area under each curve
We will calculate the area under each curve separately and then sum them up.

#### Area under \( y = x \) from \( x = 0 \) to \( x = 1 \):
The area under the curve \( y = x \) can be calculated using the formula for the area of a triangle:
\[ \text{Area}_{1} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is 1 (from \( x = 0 \) to \( x = 1 \)) and the height is also 1 (the value of \( y \) at \( x = 1 \)):
\[ \text{Area}_{1} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]

#### Area under \( y = 2 - x \) from \( x = 1 \) to \( x = 2 \):
The area under the curve \( y = 2 - x \) is also a triangle:
\[ \text{Area}_{2} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is 1 (from \( x = 1 \) to \( x = 2 \)) and the height is 1 (the value of \( y \) at \( x = 1 \)):
\[ \text{Area}_{2} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]

### Step 4: Total Area
Now, we sum the areas under both curves:
\[ \text{Total Area} = \text{Area}_{1} + \text{Area}_{2} = \frac{1}{2} + \frac{1}{2} = 1 \]

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

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