if `I = int_0 ^1.7 [x^2]dx`, then `I` equal is

Options

A. `2.4 + sqrt(2)`
B. `2.4 - sqrt(2)`
C. `2.4 + (1)/(sqrt(2))`
D. `2.4-(1)/(sqrt(2))`

Answer

To solve the integral \( I = \int_0^{1.7} \text{gif}(x^2) \, dx \), where \(\text{gif}(x)\) is the greatest integer function, we will break the integral into segments based on the behavior of the function \(\text{gif}(x^2)\).

### Step 1: Identify the intervals for \(x\)
The function \(x^2\) will take different ranges as \(x\) varies from 0 to 1.7. We need to find the critical points where the value of \(\text{gif}(x^2)\) changes.

1. For \(0 \leq x < 1\):
- \(x^2\) ranges from \(0\) to \(1\).
- Thus, \(\text{gif}(x^2) = 0\).

2. For \(1 \leq x < \sqrt{2}\):
- \(x^2\) ranges from \(1\) to \(2\).
- Thus, \(\text{gif}(x^2) = 1\).

3. For \(\sqrt{2} \leq x \leq 1.7\):
- \(x^2\) ranges from \(2\) to \(2.89\).
- Thus, \(\text{gif}(x^2) = 2\).

### Step 2: Break the integral into segments
We can break the integral \(I\) into three parts based on the intervals identified:

\[ I = \int_0^1 \text{gif}(x^2) \, dx + \int_1^{\sqrt{2}} \text{gif}(x^2) \, dx + \int_{\sqrt{2}}^{1.7} \text{gif}(x^2) \, dx \]

### Step 3: Evaluate each integral
1. **First Integral**:
\[ \int_0^1 \text{gif}(x^2) \, dx = \int_0^1 0 \, dx = 0 \]

2. **Second Integral**:
\[ \int_1^{\sqrt{2}} \text{gif}(x^2) \, dx = \int_1^{\sqrt{2}} 1 \, dx = \left[ x \right]_1^{\sqrt{2}} = \sqrt{2} - 1 \]

3. **Third Integral**:
\[ \int_{\sqrt{2}}^{1.7} \text{gif}(x^2) \, dx = \int_{\sqrt{2}}^{1.7} 2 \, dx = 2 \left[ x \right]_{\sqrt{2}}^{1.7} = 2(1.7 - \sqrt{2}) \]

### Step 4: Combine the results
Now, we can combine the results of the three integrals:

\[ I = 0 + (\sqrt{2} - 1) + 2(1.7 - \sqrt{2}) \]

Simplifying this expression:

\[ I = \sqrt{2} - 1 + 3.4 - 2\sqrt{2} = 3.4 - \sqrt{2} - 1 = 2.4 - \sqrt{2} \]

### Final Result
Thus, the value of the integral \(I\) is:

\[ \boxed{2.4 - \sqrt{2}} \]