If `f(x)={{:(sqrt({x}),"for",xcancelinZ),(1,"for",x in Z):} and g(x)={x}^(2)` where {.} denotes fractional part of x then area bounded by f(x) and g(x) for `x in 0,6 ` is

Options

A. `(2)/(3)`
B. 2
C. `(10)/(3)`
D. 6

Explanation

To find the area bounded by the functions \( f(x) \) and \( g(x) \) for \( x \) in the interval \( [0, 6] \), we will follow these steps:

### Step 1: Define the functions
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \sqrt{x} & \text{for } x \notin \mathbb{Z} \\ 1 & \text{for } x \in \mathbb{Z} \end{cases} \]
The function \( g(x) = x^2 \).

### Step 2: Identify the intervals
The interval \( [0, 6] \) contains the integers \( 0, 1, 2, 3, 4, 5, 6 \). We will analyze the area between \( f(x) \) and \( g(x) \) in the intervals:
- \( [0, 1) \)
- \( [1, 2) \)
- \( [2, 3) \)
- \( [3, 4) \)
- \( [4, 5) \)
- \( [5, 6) \)

### Step 3: Calculate the area in each interval
1. **Interval \( [0, 1) \)**:
\[ f(x) = \sqrt{x}, \quad g(x) = x^2 \]
The area \( A_1 \) is given by:
\[ A_1 = \int_0^1 (\sqrt{x} - x^2) \, dx \]

2. **Interval \( [1, 2) \)**:
\[ f(x) = 1, \quad g(x) = x^2 \]
The area \( A_2 \) is given by:
\[ A_2 = \int_1^2 (1 - x^2) \, dx \]

3. **Interval \( [2, 3) \)**:
\[ f(x) = 1, \quad g(x) = x^2 \]
The area \( A_3 \) is given by:
\[ A_3 = \int_2^3 (1 - x^2) \, dx \]

4. **Interval \( [3, 4) \)**:
\[ f(x) = 1, \quad g(x) = x^2 \]
The area \( A_4 \) is given by:
\[ A_4 = \int_3^4 (1 - x^2) \, dx \]

5. **Interval \( [4, 5) \)**:
\[ f(x) = 1, \quad g(x) = x^2 \]
The area \( A_5 \) is given by:
\[ A_5 = \int_4^5 (1 - x^2) \, dx \]

6. **Interval \( [5, 6) \)**:
\[ f(x) = 1, \quad g(x) = x^2 \]
The area \( A_6 \) is given by:
\[ A_6 = \int_5^6 (1 - x^2) \, dx \]

### Step 4: Calculate each integral
1. **For \( A_1 \)**:
\[ A_1 = \int_0^1 (\sqrt{x} - x^2) \, dx = \left[ \frac{2}{3} x^{3/2} - \frac{1}{3} x^3 \right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

2. **For \( A_2 \)**:
\[ A_2 = \int_1^2 (1 - x^2) \, dx = \left[ x - \frac{1}{3} x^3 \right]_1^2 = \left( 2 - \frac{8}{3} \right) - \left( 1 - \frac{1}{3} \right) = \left( \frac{6}{3} - \frac{8}{3} \right) - \left( \frac{3}{3} - \frac{1}{3} \right) = -\frac{2}{3} + \frac{2}{3} = 0 \]

3. **For \( A_3 \)**:
\[ A_3 = \int_2^3 (1 - x^2) \, dx = \left[ x - \frac{1}{3} x^3 \right]_2^3 = \left( 3 - \frac{27}{3} \right) - \left( 2 - \frac{8}{3} \right) = \left( 3 - 9 \right) - \left( 2 - \frac{8}{3} \right) = -6 + \frac{6}{3} = -6 + 2 = -4 \]

4. **For \( A_4 \)**:
\[ A_4 = \int_3^4 (1 - x^2) \, dx = \left[ x - \frac{1}{3} x^3 \right]_3^4 = \left( 4 - \frac{64}{3} \right) - \left( 3 - \frac{27}{3} \right) = \left( 4 - \frac{64}{3} \right) - \left( 3 - 9 \right) = -\frac{12}{3} + \frac{64}{3} = \frac{52}{3} \]

5. **For \( A_5 \)**:
\[ A_5 = \int_4^5 (1 - x^2) \, dx = \left[ x - \frac{1}{3} x^3 \right]_4^5 = \left( 5 - \frac{125}{3} \right) - \left( 4 - \frac{64}{3} \right) = \left( 5 - \frac{125}{3} \right) - \left( 4 - \frac{64}{3} \right) = \frac{15}{3} - \frac{125}{3} + \frac{64}{3} = -\frac{46}{3} \]

6. **For \( A_6 \)**:
\[ A_6 = \int_5^6 (1 - x^2) \, dx = \left[ x - \frac{1}{3} x^3 \right]_5^6 = \left( 6 - \frac{216}{3} \right) - \left( 5 - \frac{125}{3} \right) = \left( 6 - 72 \right) - \left( 5 - \frac{125}{3} \right) = -66 + \frac{125}{3} = -66 + 41.67 = -24.33 \]

### Step 5: Sum the areas
Now, we sum the areas from all intervals:
\[ \text{Total Area} = A_1 + A_2 + A_3 + A_4 + A_5 + A_6 = \frac{1}{3} + 0 - 4 + \frac{52}{3} - \frac{46}{3} - 24.33 \]

### Final Calculation
After calculating the total area, we find the area bounded by \( f(x) \) and \( g(x) \) for \( x \in [0, 6] \).