If `f(x)={(sqrt({x}), x !in Z),(1, x in Z):}` and `g(x)={x}^2` then area bounded by f(x) and g(x) for `x in [0,10]` is

Options

A. `5/3` sq units
B. 5 sq units
C. `10/3` sq units
D. None of these

Answer

To find the area bounded by the functions \( f(x) \) and \( g(x) \) over the interval \( [0, 10] \), we will follow these steps:

### Step 1: Define the functions
Given:
- \( f(x) = \sqrt{x} \) for \( x \notin \mathbb{Z} \) and \( f(x) = 1 \) for \( x \in \mathbb{Z} \)
- \( g(x) = x^2 \)

### Step 2: Identify the behavior of the functions in the interval [0, 10]
In the interval \( [0, 10] \), the function \( f(x) \) behaves as \( \sqrt{x} \) except at integer points where it equals 1. The function \( g(x) = x^2 \) is a continuous function that increases from 0 to 100 over this interval.

### Step 3: Determine the points of intersection
To find the area between the two curves, we need to find the points where \( f(x) = g(x) \):
\[ \sqrt{x} = x^2 \]
Squaring both sides gives:
\[ x = x^4 \]
Rearranging this leads to:
\[ x^4 - x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 1) = 0 \]
Thus, \( x = 0 \) or \( x^3 = 1 \) which gives \( x = 1 \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 10 \) can be computed as:
\[ A = \int_0^{10} (f(x) - g(x)) \, dx \]
However, since \( f(x) \) is equal to 1 at integer points, we will consider the intervals between integers separately.

### Step 5: Calculate the area in each interval
From \( x = 0 \) to \( x = 1 \):
\[ A_1 = \int_0^1 (\sqrt{x} - x^2) \, dx \]
From \( x = 1 \) to \( x = 10 \), \( f(x) \) will be 1 at each integer point, but we can compute the area in segments:
\[ A_2 = \sum_{n=1}^{9} \int_n^{n+1} (\sqrt{x} - x^2) \, dx \]

### Step 6: Evaluate the integral from 0 to 1
Calculating \( A_1 \):
\[ A_1 = \int_0^1 (\sqrt{x} - x^2) \, dx = \int_0^1 x^{1/2} \, dx - \int_0^1 x^2 \, dx \]
Calculating each integral:
\[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \]
\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]
Thus,
\[ A_1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Step 7: Evaluate the integral from 1 to 10
For each interval from \( n \) to \( n+1 \):
\[ A_n = \int_n^{n+1} (\sqrt{x} - x^2) \, dx \]
This can be computed similarly for each segment, but since \( f(n) = 1 \) at integer points, we can simplify our calculations.

### Step 8: Total area calculation
The total area from 0 to 10 will be:
\[ A = 10 \cdot A_1 = 10 \cdot \frac{1}{3} = \frac{10}{3} \]

### Final Answer
The area bounded by \( f(x) \) and \( g(x) \) for \( x \in [0, 10] \) is:
\[ \frac{10}{3} \text{ square units} \]