What is `int_0^sqrt2 [x^2]dx` equal to where [.] is the greatest integer function

Options

A. `sqrt2-1`
B. `1-sqrt2`
C. `2(sqrt2-1)`
D. `sqrt3-1`

Answer

To solve the integral \( \int_0^{\sqrt{2}} [x^2] \, dx \), where \([x^2]\) denotes the greatest integer function (also known as the floor function), we will break the integral into two parts based on the values of \(x^2\) within the limits of integration.

### Step 1: Identify the intervals
First, we need to determine the values of \(x^2\) as \(x\) varies from \(0\) to \(\sqrt{2}\):
- When \(x = 0\), \(x^2 = 0\).
- When \(x = 1\), \(x^2 = 1\).
- When \(x = \sqrt{2}\), \(x^2 = 2\).

This means we can break the integral into two parts:
1. From \(0\) to \(1\) where \(0 \leq x < 1\) and thus \([x^2] = 0\).
2. From \(1\) to \(\sqrt{2}\) where \(1 \leq x < \sqrt{2}\) and thus \([x^2] = 1\).

### Step 2: Set up the integral
We can express the integral as:
\[ \int_0^{\sqrt{2}} [x^2] \, dx = \int_0^1 [x^2] \, dx + \int_1^{\sqrt{2}} [x^2] \, dx \]

### Step 3: Evaluate the first integral
For the first part:
\[ \int_0^1 [x^2] \, dx = \int_0^1 0 \, dx = 0 \]

### Step 4: Evaluate the second integral
For the second part:
\[ \int_1^{\sqrt{2}} [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx \]
This integral evaluates to:
\[ \int_1^{\sqrt{2}} 1 \, dx = [x]_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

### Step 5: Combine the results
Now, combining both parts:
\[ \int_0^{\sqrt{2}} [x^2] \, dx = 0 + (\sqrt{2} - 1) = \sqrt{2} - 1 \]

### Final Answer
Thus, the value of the integral \( \int_0^{\sqrt{2}} [x^2] \, dx \) is:
\[ \sqrt{2} - 1 \]