Area bounded by the curves `y=|x-1|, y=0` and `|x|=2`

Options

A. 4
B. 5
C. 3
D. 6

Answer

To find the area bounded by the curves \( y = |x - 1| \), \( y = 0 \), and \( |x| = 2 \), we can follow these steps:

### Step 1: Understand the curves
1. The curve \( y = |x - 1| \) can be expressed as:
- \( y = x - 1 \) for \( x \geq 1 \)
- \( y = 1 - x \) for \( x < 1 \)

2. The line \( y = 0 \) is simply the x-axis.

3. The equation \( |x| = 2 \) represents two vertical lines:
- \( x = 2 \)
- \( x = -2 \)

### Step 2: Identify the points of intersection
The area we want to find is bounded between \( x = -2 \) and \( x = 2 \).

- For \( x = -2 \):
\[ y = |(-2) - 1| = | -3 | = 3 \]
- For \( x = 2 \):
\[ y = |2 - 1| = |1| = 1 \]

### Step 3: Set up the integral
We will split the integral into two parts:
1. From \( x = -2 \) to \( x = 1 \) (where \( y = 1 - x \))
2. From \( x = 1 \) to \( x = 2 \) (where \( y = x - 1 \))

The area \( A \) can be calculated as:
\[ A = \int_{-2}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx \]

### Step 4: Calculate the first integral
\[ \int_{-2}^{1} (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_{-2}^{1} \]
Calculating the limits:
- Upper limit at \( x = 1 \):
\[ 1 - \frac{1^2}{2} = 1 - \frac{1}{2} = \frac{1}{2} \]
- Lower limit at \( x = -2 \):
\[ -2 - \frac{(-2)^2}{2} = -2 - 2 = -4 \]

So,
\[ \int_{-2}^{1} (1 - x) \, dx = \frac{1}{2} - (-4) = \frac{1}{2} + 4 = \frac{1}{2} + \frac{8}{2} = \frac{9}{2} \]

### Step 5: Calculate the second integral
\[ \int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} \]
Calculating the limits:
- Upper limit at \( x = 2 \):
\[ \frac{2^2}{2} - 2 = 2 - 2 = 0 \]
- Lower limit at \( x = 1 \):
\[ \frac{1^2}{2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]

So,
\[ \int_{1}^{2} (x - 1) \, dx = 0 - (-\frac{1}{2}) = \frac{1}{2} \]

### Step 6: Combine the areas
Now, we add the two areas:
\[ A = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \]

### Final Answer
The area bounded by the curves is \( 5 \) square units.