The value of `int_(0)^(sqrt(2)) [ x^(2)] dx ` where `[*]` is the greatest integer function

Options

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

Explanation

To solve the integral \( I = \int_{0}^{\sqrt{2}} [x^2] \, dx \), 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 behavior of the greatest integer function.

### Step 1: Identify the intervals
The function \( x^2 \) ranges from \( 0 \) to \( 2 \) as \( x \) goes from \( 0 \) to \( \sqrt{2} \). We need to find the intervals where the value of \([x^2]\) remains constant.

1. For \( 0 \leq x < 1 \):
- \( x^2 \) ranges from \( 0 \) to \( 1 \).
- Therefore, \([x^2] = 0\).

2. For \( 1 \leq x < \sqrt{2} \):
- \( x^2 \) ranges from \( 1 \) to \( 2 \).
- Therefore, \([x^2] = 1\).

### Step 2: Break the integral
Now we can break the integral into two parts:
\[ I = \int_{0}^{1} [x^2] \, dx + \int_{1}^{\sqrt{2}} [x^2] \, dx \]

### Step 3: Evaluate the first integral
For the first integral from \( 0 \) to \( 1 \):
\[ \int_{0}^{1} [x^2] \, dx = \int_{0}^{1} 0 \, dx = 0 \]

### Step 4: Evaluate the second integral
For the second integral from \( 1 \) to \( \sqrt{2} \):
\[ \int_{1}^{\sqrt{2}} [x^2] \, dx = \int_{1}^{\sqrt{2}} 1 \, dx \]
This integral evaluates to:
\[ = \left[ x \right]_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

### Step 5: Combine the results
Now we combine the results of both integrals:
\[ I = 0 + (\sqrt{2} - 1) = \sqrt{2} - 1 \]

### Final Answer
Thus, the value of the integral is:
\[ I = \sqrt{2} - 1 \]

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