The value of the integral, `int_1^3 [x^2-2x–2]dx` , where [x] denotes the greatest integer less than or equal to x, is :

Options

A. `-sqrt2-sqrt3+1`
B. `-sqrt2-sqrt3-1`
C. -5
D. `-4`

Answer

To solve the integral \( \int_1^3 [x^2 - 2x - 2] \, dx \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we will follow these steps:

### Step 1: Simplify the expression inside the integral
The expression \(x^2 - 2x - 2\) can be rewritten by completing the square.

\[ x^2 - 2x - 2 = (x - 1)^2 - 3 \]

### Step 2: Rewrite the integral
Now we can rewrite the integral as:

\[ \int_1^3 [(x - 1)^2 - 3] \, dx = \int_1^3 [(x - 1)^2] \, dx - \int_1^3 3 \, dx \]

### Step 3: Calculate the integral of the constant
The integral of the constant \(3\) from \(1\) to \(3\) is straightforward:

\[ \int_1^3 3 \, dx = 3[x]_1^3 = 3(3 - 1) = 3 \cdot 2 = 6 \]

### Step 4: Calculate the integral of \((x - 1)^2\)
Next, we need to calculate the integral of \((x - 1)^2\):

\[ \int_1^3 (x - 1)^2 \, dx \]

Let’s perform the substitution \(u = x - 1\), then \(du = dx\). The limits change as follows:
- When \(x = 1\), \(u = 0\)
- When \(x = 3\), \(u = 2\)

Thus, the integral becomes:

\[ \int_0^2 u^2 \, du \]

Calculating this integral:

\[ \int_0^2 u^2 \, du = \left[ \frac{u^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

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

\[ \int_1^3 [(x - 1)^2 - 3] \, dx = \frac{8}{3} - 6 = \frac{8}{3} - \frac{18}{3} = -\frac{10}{3} \]

### Step 6: Apply the greatest integer function
Since the integral involves the greatest integer function, we need to evaluate the expression \([x^2 - 2x - 2]\) over the interval from \(1\) to \(3\).

1. At \(x = 1\):
\[ 1^2 - 2(1) - 2 = -3 \quad \Rightarrow \quad [ -3 ] = -3 \]

2. At \(x = 2\):
\[ 2^2 - 2(2) - 2 = -2 \quad \Rightarrow \quad [ -2 ] = -2 \]

3. At \(x = 3\):
\[ 3^2 - 2(3) - 2 = -1 \quad \Rightarrow \quad [ -1 ] = -1 \]

### Step 7: Evaluate the integral piecewise
Now we can evaluate the integral piecewise based on the intervals where the greatest integer function remains constant:

- From \(1\) to \(2\), \([x^2 - 2x - 2] = -3\)
- From \(2\) to \(3\), \([x^2 - 2x - 2] = -2\)

Thus, we can write:

\[ \int_1^2 -3 \, dx + \int_2^3 -2 \, dx \]

Calculating these integrals:

1. \(\int_1^2 -3 \, dx = -3[x]_1^2 = -3(2 - 1) = -3\)

2. \(\int_2^3 -2 \, dx = -2[x]_2^3 = -2(3 - 2) = -2\)

### Final Calculation
Adding these results together:

\[ -3 - 2 = -5 \]

### Conclusion
The value of the integral \( \int_1^3 [x^2 - 2x - 2] \, dx \) is:

\[ \boxed{-5} \]