The value of the integral `int_(0)^(2)x[x]dx`

Options

A. `(7)/(2)`
B. `(3)/(2)`
C. `(5)/(2)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{2} x \lfloor x \rfloor \, dx \), we will break the integral into two parts based on the behavior of the greatest integer function \( \lfloor x \rfloor \).

### Step 1: Identify the intervals
The function \( \lfloor x \rfloor \) changes its value at integer points. Therefore, we will split the integral at \( x = 1 \):
- From \( 0 \) to \( 1 \), \( \lfloor x \rfloor = 0 \)
- From \( 1 \) to \( 2 \), \( \lfloor x \rfloor = 1 \)

### Step 2: Break the integral
We can now express the integral as:
\[ I = \int_{0}^{1} x \lfloor x \rfloor \, dx + \int_{1}^{2} x \lfloor x \rfloor \, dx \]

### Step 3: Evaluate the first integral
For the first integral:
\[ \int_{0}^{1} x \lfloor x \rfloor \, dx = \int_{0}^{1} x \cdot 0 \, dx = \int_{0}^{1} 0 \, dx = 0 \]

### Step 4: Evaluate the second integral
For the second integral:
\[ \int_{1}^{2} x \lfloor x \rfloor \, dx = \int_{1}^{2} x \cdot 1 \, dx = \int_{1}^{2} x \, dx \]
Now, we compute this integral:
\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} \]
Calculating the limits:
\[ = \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \left( \frac{4}{2} - \frac{1}{2} \right) = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

### Step 5: Combine the results
Now, we combine the results from both integrals:
\[ I = 0 + \frac{3}{2} = \frac{3}{2} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{2} x \lfloor x \rfloor \, dx \) is:
\[ \boxed{\frac{3}{2}} \]