Evaluate `int_0^a[x^n]dx,` (where,[*] denotes the greatest integer function).

Answer

To evaluate the integral \( \int_0^a [x^n] \, dx \), where \( [x^n] \) denotes the greatest integer function (also known as the floor function), we can break down the problem into manageable steps.

### Step 1: Understand the Greatest Integer Function
The greatest integer function \( [x^n] \) gives the largest integer less than or equal to \( x^n \). For \( x \) in the interval \( [0, a] \), we need to determine the integer values that \( [x^n] \) can take.

### Step 2: Determine the Intervals
For \( x \) in the range \( [0, a] \):
- When \( x \) is in the interval \( [k^{1/n}, (k+1)^{1/n}) \) for \( k = 0, 1, 2, \ldots, [a^n] \), we have \( [x^n] = k \).

### Step 3: Set Up the Integral
We can express the integral as a sum of integrals over the intervals where \( [x^n] \) is constant:
\[ \int_0^a [x^n] \, dx = \sum_{k=0}^{[a^n]} k \int_{k^{1/n}}^{(k+1)^{1/n}} 1 \, dx \]

### Step 4: Evaluate Each Integral
The integral \( \int_{k^{1/n}}^{(k+1)^{1/n}} 1 \, dx \) is simply the length of the interval:
\[ \int_{k^{1/n}}^{(k+1)^{1/n}} 1 \, dx = (k+1)^{1/n} - k^{1/n} \]

### Step 5: Combine the Results
Thus, we can rewrite the integral as:
\[ \int_0^a [x^n] \, dx = \sum_{k=0}^{[a^n]} k \left( (k+1)^{1/n} - k^{1/n} \right) \]

### Step 6: Simplify the Expression
This sum can be evaluated by recognizing that it is a finite series. We can compute it explicitly for small values of \( n \) or \( a \), or use numerical methods for larger values.

### Final Result
The final expression for the integral is:
\[ \int_0^a [x^n] \, dx = \sum_{k=0}^{[a^n]} k \left( (k+1)^{1/n} - k^{1/n} \right) \]