The area bounded by y = [x] , y = -[x] , x = 4 on the right side of the y-axis is

Options

A. (a)12 sq. unit
B. (b)Equal to `18 underset(0)overset(1)(int) [2 sin pi x]` dx
C. (c)Equal to area of circle of radius `sqrt((12)/(pi))`
D. (d)18 sq. units

Answer

To find the area bounded by the curves \( y = [x] \), \( y = -[x] \), and the line \( x = 4 \) on the right side of the y-axis, we can follow these steps:

### Step 1: Understand the Functions
The function \( y = [x] \) represents the greatest integer function (or floor function), which gives the largest integer less than or equal to \( x \). Similarly, \( y = -[x] \) gives the negative of that integer.

### Step 2: Identify the Points of Intersection
We need to find the area between \( y = [x] \) and \( y = -[x] \) from \( x = 0 \) to \( x = 4 \). The greatest integer function will take the following values:
- For \( 0 \leq x < 1 \), \( [x] = 0 \)
- For \( 1 \leq x < 2 \), \( [x] = 1 \)
- For \( 2 \leq x < 3 \), \( [x] = 2 \)
- For \( 3 \leq x < 4 \), \( [x] = 3 \)

### Step 3: Sketch the Graph
Sketch the graphs of \( y = [x] \) and \( y = -[x] \):
- From \( x = 0 \) to \( x = 1 \), \( y = 0 \) and \( y = 0 \).
- From \( x = 1 \) to \( x = 2 \), \( y = 1 \) and \( y = -1 \).
- From \( x = 2 \) to \( x = 3 \), \( y = 2 \) and \( y = -2 \).
- From \( x = 3 \) to \( x = 4 \), \( y = 3 \) and \( y = -3 \).

### Step 4: Calculate the Area
The area between the curves can be calculated as the sum of the areas of rectangles formed by the intervals:
1. From \( x = 0 \) to \( x = 1 \): Area = \( 0 \)
2. From \( x = 1 \) to \( x = 2 \): Area = \( (1 - (-1)) \cdot (2 - 1) = 2 \)
3. From \( x = 2 \) to \( x = 3 \): Area = \( (2 - (-2)) \cdot (3 - 2) = 4 \)
4. From \( x = 3 \) to \( x = 4 \): Area = \( (3 - (-3)) \cdot (4 - 3) = 6 \)

Now, sum these areas:
\[ \text{Total Area} = 0 + 2 + 4 + 6 = 12 \text{ square units} \]

### Final Answer
The area bounded by the curves is \( 12 \) square units.

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