`int_(0)^(n^(2)) [sqrtx] dx`=

Options

A. `(n)/(6) (n+1) [4n+1]`
B. `(n)/(6) (n-1) (4n+1)`
C. `(n)/(6) (n-1) (4n-1)`
D. none

Answer

To solve the integral \( I = \int_{0}^{n^2} \sqrt{x} \, dx \), we will follow these steps:

### Step 1: Identify the integral
We start with the integral:
\[ I = \int_{0}^{n^2} \sqrt{x} \, dx \]

### Step 2: Use the power rule for integration
The integral of \( \sqrt{x} \) can be rewritten as \( x^{1/2} \). According to the power rule of integration:
\[ \int x^m \, dx = \frac{x^{m+1}}{m+1} + C \quad (m \neq -1) \]
For \( m = \frac{1}{2} \):
\[ \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

### Step 3: Evaluate the definite integral
Now we will evaluate the definite integral from \( 0 \) to \( n^2 \):
\[ I = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{n^2} \]
Substituting the limits:
\[ I = \frac{2}{3} (n^2)^{3/2} - \frac{2}{3} (0)^{3/2} \]
Calculating \( (n^2)^{3/2} \):
\[ (n^2)^{3/2} = n^3 \]
Thus,
\[ I = \frac{2}{3} n^3 - 0 = \frac{2}{3} n^3 \]

### Final Result
The value of the integral is:
\[ I = \frac{2}{3} n^3 \]