The area bounded by the curve `y=[x^2/64+2],y=x-1,y=x-1 and x=0` above the x-axis will be-(Where [] represents greatest integer function) (a) 2 (b) 3 (c) 4 (d) none of these

Options

A. 2
B. 3
C. 4
D. None of these

Answer

To find the area bounded by the curves \( y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor \), \( y = x - 1 \), and \( x = 0 \) above the x-axis, we will follow these steps:

### Step 1: Understand the functions
The first function is \( y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor \). The greatest integer function means we take the integer part of \( \frac{x^2}{64} + 2 \).

### Step 2: Determine the range of \( x \)
To find where the two curves intersect, we set \( y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor \) equal to \( y = x - 1 \).

### Step 3: Solve the equation
We need to solve:
\[ \frac{x^2}{64} + 2 = x - 1 \]
This simplifies to:
\[ \frac{x^2}{64} - x + 3 = 0 \]
Multiplying through by 64 to eliminate the fraction gives:
\[ x^2 - 64x + 192 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 1 \cdot 192}}{2 \cdot 1} \]
\[ x = \frac{64 \pm \sqrt{4096 - 768}}{2} \]
\[ x = \frac{64 \pm \sqrt{3328}}{2} \]
Calculating \( \sqrt{3328} \):
\[ \sqrt{3328} = 57.6 \quad (\text{approximately}) \]
Thus,
\[ x = \frac{64 \pm 57.6}{2} \]
Calculating the two roots:
1. \( x = \frac{121.6}{2} = 60.8 \)
2. \( x = \frac{6.4}{2} = 3.2 \)

### Step 4: Identify the area
The area we need to find is between \( x = 0 \) and \( x = 3.2 \) since \( y = x - 1 \) intersects \( y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor \) at \( x = 3.2 \).

### Step 5: Calculate the area
1. **Area under \( y = x - 1 \)** from \( x = 0 \) to \( x = 3.2 \):
\[ \text{Area} = \int_0^{3.2} (x - 1) \, dx \]
\[ = \left[ \frac{x^2}{2} - x \right]_0^{3.2} = \left( \frac{(3.2)^2}{2} - 3.2 \right) - (0 - 0) \]
\[ = \left( \frac{10.24}{2} - 3.2 \right) = 5.12 - 3.2 = 1.92 \]

2. **Area under \( y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor \)** from \( x = 0 \) to \( x = 3.2 \):
Since \( \frac{x^2}{64} + 2 \) is less than 3 for \( x < 8 \), we have:
\[ y = 2 \quad \text{for } 0 \leq x < 3.2 \]
Thus, the area is:
\[ \text{Area} = \int_0^{3.2} 2 \, dx = 2 \cdot 3.2 = 6.4 \]

### Step 6: Total area
The total area above the x-axis is:
\[ \text{Total Area} = \text{Area under } y = \left\lfloor \frac{x^2}{64} + 2 \right\rfloor - \text{Area under } y = x - 1 \]
\[ = 6.4 - 1.92 = 4.48 \]

### Step 7: Apply the greatest integer function
Finally, applying the greatest integer function:
\[ \left\lfloor 4.48 \right\rfloor = 4 \]

### Final Answer
Thus, the area bounded by the curves is \( 4 \).