For any natural number n, the value of the integral
`int_(0)^(sqrt(n)) [x^(2)]dx` is

Options

A. `n sqrt(n)+overset(n) underset(r=1)sum sqrt(r )`
B. `nsqrt(n)-overset(n)underset(r=1)sumsqrt(r )`
C. `overset(n)underset(r=1)sum sqrt(r)-nsqrt(n)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{\sqrt{n}} \lfloor x^2 \rfloor \, dx \), we will break it down into intervals based on the behavior of the greatest integer function \( \lfloor x^2 \rfloor \).

### Step 1: Identify the intervals for \( x \)
The function \( \lfloor x^2 \rfloor \) changes its value at points where \( x^2 \) is an integer. Therefore, we need to find the points where \( x^2 = k \) for \( k = 0, 1, 2, \ldots, n \). The corresponding values of \( x \) will be \( 0, 1, \sqrt{2}, \sqrt{3}, \ldots, \sqrt{n} \).

### Step 2: Set up the integral
We can express the integral as a sum of integrals over the intervals:
\[ I = \int_{0}^{1} \lfloor x^2 \rfloor \, dx + \int_{1}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx + \int_{\sqrt{2}}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx + \ldots + \int_{\sqrt{n-1}}^{\sqrt{n}} \lfloor x^2 \rfloor \, dx \]

### Step 3: Evaluate each integral
1. For \( x \) in \( [0, 1] \):
\[ \lfloor x^2 \rfloor = 0 \quad \Rightarrow \quad \int_{0}^{1} 0 \, dx = 0 \]

2. For \( x \) in \( [1, \sqrt{2}] \):
\[ \lfloor x^2 \rfloor = 1 \quad \Rightarrow \quad \int_{1}^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1 \]

3. For \( x \) in \( [\sqrt{2}, \sqrt{3}] \):
\[ \lfloor x^2 \rfloor = 2 \quad \Rightarrow \quad \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2}) \]

4. For \( x \) in \( [\sqrt{3}, \sqrt{4}] \):
\[ \lfloor x^2 \rfloor = 3 \quad \Rightarrow \quad \int_{\sqrt{3}}^{\sqrt{4}} 3 \, dx = 3(\sqrt{4} - \sqrt{3}) = 3(2 - \sqrt{3}) \]

Continuing this process, for \( x \) in \( [\sqrt{k}, \sqrt{k+1}] \) where \( k = 1, 2, \ldots, n-1 \):
\[ \lfloor x^2 \rfloor = k \quad \Rightarrow \quad \int_{\sqrt{k}}^{\sqrt{k+1}} k \, dx = k(\sqrt{k+1} - \sqrt{k}) \]

### Step 4: Sum the integrals
Now, we can sum these integrals:
\[ I = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) + \ldots + (n-1)(\sqrt{n} - \sqrt{n-1}) \]

### Step 5: Final expression
After evaluating the sum, we can express \( I \) in terms of \( n \) and the summation of square roots:
\[ I = n \sqrt{n} - \sum_{r=1}^{n} \sqrt{r} \]

### Conclusion
Thus, the value of the integral \( I \) is:
\[ I = n \sqrt{n} - \sum_{r=1}^{n} \sqrt{r} \]