Evaluate `int_(1)^(e^(6))[(logx)/3]dx,` where [.] denotes the greatest integer function.

Explanation

To evaluate the integral \( I = \int_{1}^{e^{6}} \left\lfloor \frac{\log x}{3} \right\rfloor dx \), we will break it down step by step.

### Step 1: Determine the intervals for the greatest integer function
The function \( \left\lfloor \frac{\log x}{3} \right\rfloor \) changes its value at specific points. We need to find the values of \( x \) where \( \frac{\log x}{3} \) is an integer.

1. Set \( \frac{\log x}{3} = n \), where \( n \) is an integer.
2. This implies \( \log x = 3n \) or \( x = e^{3n} \).

Now, we will find the integer values of \( n \) for the interval \( x \in [1, e^{6}] \):

- For \( n = 0 \): \( x = e^{0} = 1 \)
- For \( n = 1 \): \( x = e^{3} \)
- For \( n = 2 \): \( x = e^{6} \)

Thus, the critical points are \( x = 1, e^{3}, e^{6} \).

### Step 2: Evaluate the integral in segments
We can split the integral into two parts based on the critical points:

\[ I = \int_{1}^{e^{3}} \left\lfloor \frac{\log x}{3} \right\rfloor dx + \int_{e^{3}}^{e^{6}} \left\lfloor \frac{\log x}{3} \right\rfloor dx \]

### Step 3: Evaluate the first integral \( \int_{1}^{e^{3}} \left\lfloor \frac{\log x}{3} \right\rfloor dx \)
In the interval \( [1, e^{3}] \):
- For \( x \in [1, e^{3}] \), \( \frac{\log x}{3} \) ranges from \( 0 \) to \( 1 \).
- Thus, \( \left\lfloor \frac{\log x}{3} \right\rfloor = 0 \).

So, we have:

\[ \int_{1}^{e^{3}} 0 \, dx = 0 \]

### Step 4: Evaluate the second integral \( \int_{e^{3}}^{e^{6}} \left\lfloor \frac{\log x}{3} \right\rfloor dx \)
In the interval \( [e^{3}, e^{6}] \):
- For \( x \in [e^{3}, e^{6}] \), \( \frac{\log x}{3} \) ranges from \( 1 \) to \( 2 \).
- Thus, \( \left\lfloor \frac{\log x}{3} \right\rfloor = 1 \).

So, we have:

\[ \int_{e^{3}}^{e^{6}} 1 \, dx = \left[ x \right]_{e^{3}}^{e^{6}} = e^{6} - e^{3} \]

### Step 5: Combine the results
Now we combine the results from both integrals:

\[ I = 0 + (e^{6} - e^{3}) = e^{6} - e^{3} \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{e^{6} - e^{3}} \]