`int_(0)^(9)[sqrt(t)]dt`.

Answer

To solve the integral \( I = \int_{0}^{9} \lfloor \sqrt{t} \rfloor \, dt \), we will break it down into intervals based on the values of \( \lfloor \sqrt{t} \rfloor \).

### Step 1: Identify the intervals for \( \lfloor \sqrt{t} \rfloor \)

- For \( 0 \leq t < 1 \): \( \lfloor \sqrt{t} \rfloor = 0 \)
- For \( 1 \leq t < 4 \): \( \lfloor \sqrt{t} \rfloor = 1 \)
- For \( 4 \leq t < 9 \): \( \lfloor \sqrt{t} \rfloor = 2 \)
- For \( t = 9 \): \( \lfloor \sqrt{9} \rfloor = 3 \)

### Step 2: Split the integral into parts

We can express the integral as:

\[ I = \int_{0}^{1} \lfloor \sqrt{t} \rfloor \, dt + \int_{1}^{4} \lfloor \sqrt{t} \rfloor \, dt + \int_{4}^{9} \lfloor \sqrt{t} \rfloor \, dt \]

### Step 3: Evaluate each integral

1. **For \( \int_{0}^{1} \lfloor \sqrt{t} \rfloor \, dt \)**:
\[ \int_{0}^{1} 0 \, dt = 0 \]

2. **For \( \int_{1}^{4} \lfloor \sqrt{t} \rfloor \, dt \)**:
\[ \int_{1}^{4} 1 \, dt = 1 \cdot (4 - 1) = 3 \]

3. **For \( \int_{4}^{9} \lfloor \sqrt{t} \rfloor \, dt \)**:
\[ \int_{4}^{9} 2 \, dt = 2 \cdot (9 - 4) = 10 \]

### Step 4: Combine the results

Now, we combine the results from each part:

\[ I = 0 + 3 + 10 = 13 \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{13} \]