The value of `int_(0)^(2)[x^(2)-x+1] dx` (where , [.] denotes the greatest integer function ) is equal to

Options

A. `(5+sqrt5)/(2)`
B. `(1+sqrt5)/(2)`
C. `(1-sqrt5)/(2)`
D. `(5-sqrt5)/(2)`

Answer

To solve the integral \( \int_{0}^{2} [x^2 - x + 1] \, dx \), where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = x^2 - x + 1 \).

### Step 2: Find the range of \( f(x) \) on the interval [0, 2]
We will evaluate \( f(x) \) at the endpoints and find its minimum value in the interval.

1. Calculate \( f(0) \):
\[ f(0) = 0^2 - 0 + 1 = 1 \]

2. Calculate \( f(2) \):
\[ f(2) = 2^2 - 2 + 1 = 3 \]

3. Find the critical points by taking the derivative and setting it to zero:
\[ f'(x) = 2x - 1 \]
Setting \( f'(x) = 0 \):
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

4. Calculate \( f\left(\frac{1}{2}\right) \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 3: Determine the greatest integer function values
Now we have:
- \( f(0) = 1 \)
- \( f\left(\frac{1}{2}\right) = \frac{3}{4} \)
- \( f(2) = 3 \)

The minimum value of \( f(x) \) in the interval [0, 2] is \( \frac{3}{4} \) and the maximum is \( 3 \).

### Step 4: Identify the intervals for the greatest integer function
- For \( x \in [0, \frac{1}{2}) \), \( f(x) \) ranges from \( 1 \) to \( \frac{3}{4} \) (but since \( f(x) \) is always greater than \( \frac{3}{4} \), the greatest integer is \( 0 \)).
- For \( x = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = \frac{3}{4} \) (greatest integer is \( 0 \)).
- For \( x \in (\frac{1}{2}, 2] \), \( f(x) \) ranges from \( \frac{3}{4} \) to \( 3 \) (greatest integer is \( 1 \) for \( x \in (\frac{1}{2}, 1) \) and \( 2 \) for \( x \in [1, 2] \)).

### Step 5: Set up the integral
We can break the integral into parts based on the intervals:
\[ \int_{0}^{2} [f(x)] \, dx = \int_{0}^{\frac{1}{2}} [f(x)] \, dx + \int_{\frac{1}{2}}^{1} [f(x)] \, dx + \int_{1}^{2} [f(x)] \, dx \]

### Step 6: Evaluate each integral
1. For \( x \in [0, \frac{1}{2}) \):
\[ \int_{0}^{\frac{1}{2}} [f(x)] \, dx = \int_{0}^{\frac{1}{2}} 0 \, dx = 0 \]

2. For \( x \in [\frac{1}{2}, 1) \):
\[ \int_{\frac{1}{2}}^{1} [f(x)] \, dx = \int_{\frac{1}{2}}^{1} 0 \, dx = 0 \]

3. For \( x \in [1, 2] \):
\[ \int_{1}^{2} [f(x)] \, dx = \int_{1}^{2} 1 \, dx = 1 \]

### Step 7: Combine the results
Now, we combine the results from each interval:
\[ \int_{0}^{2} [f(x)] \, dx = 0 + 0 + 1 = 1 \]

### Final Answer
Thus, the value of \( \int_{0}^{2} [x^2 - x + 1] \, dx \) is \( \boxed{1} \).