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 \( I = \int_{0}^{2} x [x] \, dx \), where \([x]\) is the greatest integer function, we can break the integral into two parts based on the behavior of the greatest integer function.

### Step 1: Break the integral into two parts
The limits of integration are from 0 to 2. We can split this integral at \( x = 1 \):

\[ I = \int_{0}^{1} x [x] \, dx + \int_{1}^{2} x [x] \, dx \]

### Step 2: Evaluate the first integral from 0 to 1
For \( x \) in the interval \([0, 1)\), the value of \([x]\) is 0. Therefore, the first integral becomes:

\[ \int_{0}^{1} x [x] \, dx = \int_{0}^{1} x \cdot 0 \, dx = \int_{0}^{1} 0 \, dx = 0 \]

### Step 3: Evaluate the second integral from 1 to 2
For \( x \) in the interval \([1, 2)\), the value of \([x]\) is 1. Therefore, the second integral becomes:

\[ \int_{1}^{2} x [x] \, dx = \int_{1}^{2} x \cdot 1 \, dx = \int_{1}^{2} x \, dx \]

### Step 4: Calculate the integral \(\int_{1}^{2} x \, dx\)
The integral of \( x \) is given by:

\[ \int x \, dx = \frac{x^2}{2} \]

Now, we evaluate this from 1 to 2:

\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2} \]

### Step 5: Combine the results
Now, we can combine the results of both integrals:

\[ I = 0 + \frac{3}{2} = \frac{3}{2} \]

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

\[ \boxed{\frac{3}{2}} \]