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 \( \int_{0}^{\sqrt{n}} x^2 \, dx \), we will follow these steps:

### Step 1: Set up the integral
We need to evaluate the definite integral from \( 0 \) to \( \sqrt{n} \):
\[ \int_{0}^{\sqrt{n}} x^2 \, dx \]

### Step 2: Find the antiderivative
The antiderivative of \( x^2 \) is:
\[ \frac{x^3}{3} \]

### Step 3: Evaluate the integral using the Fundamental Theorem of Calculus
Now we will evaluate the integral by substituting the limits \( 0 \) and \( \sqrt{n} \):
\[ \left[ \frac{x^3}{3} \right]_{0}^{\sqrt{n}} = \frac{(\sqrt{n})^3}{3} - \frac{(0)^3}{3} \]

### Step 4: Simplify the expression
Calculating \( (\sqrt{n})^3 \):
\[ (\sqrt{n})^3 = n^{3/2} \]
Thus, the integral evaluates to:
\[ \frac{n^{3/2}}{3} - 0 = \frac{n^{3/2}}{3} \]

### Final Result
The value of the integral \( \int_{0}^{\sqrt{n}} x^2 \, dx \) is:
\[ \frac{n^{3/2}}{3} \]

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