Let f(x) denote the fractional part of a real number x. Then the value of `int_(0)^(sqrt3)f(x^(2))dx` is

Options

A. `2sqrt3-sqrt2-1`
B. 0
C. `sqrt2-sqrt3+1`
D. `sqrt3-sqrt2+1`

Answer

To solve the integral \( \int_{0}^{\sqrt{3}} f(x^2) \, dx \), where \( f(x) \) denotes the fractional part of \( x \), we can follow these steps:

### Step 1: Understanding the Function \( f(x) \)
The fractional part function \( f(x) \) can be expressed as:
\[ f(x) = x - \lfloor x \rfloor \]
Thus, for \( f(x^2) \):
\[ f(x^2) = x^2 - \lfloor x^2 \rfloor \]

### Step 2: Setting Up the Integral
We can rewrite the integral:
\[ \int_{0}^{\sqrt{3}} f(x^2) \, dx = \int_{0}^{\sqrt{3}} \left( x^2 - \lfloor x^2 \rfloor \right) \, dx \]
This can be separated into two integrals:
\[ \int_{0}^{\sqrt{3}} x^2 \, dx - \int_{0}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx \]

### Step 3: Calculating the First Integral
The first integral \( \int_{0}^{\sqrt{3}} x^2 \, dx \) can be computed as follows:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \( 0 \) to \( \sqrt{3} \):
\[ \left[ \frac{x^3}{3} \right]_{0}^{\sqrt{3}} = \frac{(\sqrt{3})^3}{3} - \frac{0^3}{3} = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

### Step 4: Calculating the Second Integral
Next, we need to evaluate \( \int_{0}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx \). We will break this integral into segments based on the values of \( \lfloor x^2 \rfloor \):

1. **From \( 0 \) to \( 1 \)**: Here, \( 0 \leq x^2 < 1 \) so \( \lfloor x^2 \rfloor = 0 \).
\[ \int_{0}^{1} \lfloor x^2 \rfloor \, dx = 0 \]

2. **From \( 1 \) to \( \sqrt{2} \)**: Here, \( 1 \leq x^2 < 2 \) so \( \lfloor x^2 \rfloor = 1 \).
\[ \int_{1}^{\sqrt{2}} \lfloor x^2 \rfloor \, dx = \int_{1}^{\sqrt{2}} 1 \, dx = \left[ x \right]_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]

3. **From \( \sqrt{2} \) to \( \sqrt{3} \)**: Here, \( 2 \leq x^2 < 3 \) so \( \lfloor x^2 \rfloor = 2 \).
\[ \int_{\sqrt{2}}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2 \left[ x \right]_{\sqrt{2}}^{\sqrt{3}} = 2(\sqrt{3} - \sqrt{2}) \]

Now, we can combine these results:
\[ \int_{0}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} = 2\sqrt{3} - \sqrt{2} - 1 \]

### Step 5: Putting It All Together
Now we can substitute back into our original integral:
\[ \int_{0}^{\sqrt{3}} f(x^2) \, dx = \sqrt{3} - (2\sqrt{3} - \sqrt{2} - 1) \]
Simplifying this gives:
\[ \sqrt{3} - 2\sqrt{3} + \sqrt{2} + 1 = -\sqrt{3} + \sqrt{2} + 1 \]

### Final Answer
Thus, the value of the integral is:
\[ \sqrt{2} - \sqrt{3} + 1 \]