`int_(0)^(4) {sqrt(x)}` is equal to, where {x} denotes the fraction part of x.

Options

A. `(2)/(3)`
B. `(16)/(3)`
C. `(5)/(3)`
D. `(7)/(3)`

Explanation

To solve the integral \( \int_{0}^{4} \{ \sqrt{x} \} \, dx \), where \( \{x\} \) denotes the fractional part of \( x \), we can break down the integral into manageable parts based on the behavior of the function \( \sqrt{x} \) over the interval from 0 to 4.

### Step 1: Understand the fractional part function
The fractional part of a number \( x \) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).

### Step 2: Identify the intervals
We will break the integral into intervals where \( \sqrt{x} \) takes on different integer values:
- From \( 0 \) to \( 1 \): \( \sqrt{x} \) varies from \( 0 \) to \( 1 \).
- From \( 1 \) to \( 4 \): \( \sqrt{x} \) varies from \( 1 \) to \( 2 \).

### Step 3: Set up the integral
We can express the integral as:
\[ \int_{0}^{4} \{ \sqrt{x} \} \, dx = \int_{0}^{1} \{ \sqrt{x} \} \, dx + \int_{1}^{4} \{ \sqrt{x} \} \, dx \]

### Step 4: Evaluate the first integral \( \int_{0}^{1} \{ \sqrt{x} \} \, dx \)
In the interval \( [0, 1] \):
- \( \sqrt{x} \) is between \( 0 \) and \( 1 \).
- Thus, \( \{ \sqrt{x} \} = \sqrt{x} \).

So,
\[ \int_{0}^{1} \{ \sqrt{x} \} \, dx = \int_{0}^{1} \sqrt{x} \, dx \]
Calculating this integral:
\[ \int_{0}^{1} \sqrt{x} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} \]

### Step 5: Evaluate the second integral \( \int_{1}^{4} \{ \sqrt{x} \} \, dx \)
In the interval \( [1, 4] \):
- \( \sqrt{x} \) ranges from \( 1 \) to \( 2 \).
- Thus, \( \{ \sqrt{x} \} = \sqrt{x} - 1 \).

So,
\[ \int_{1}^{4} \{ \sqrt{x} \} \, dx = \int_{1}^{4} (\sqrt{x} - 1) \, dx \]
Calculating this integral:
\[ \int_{1}^{4} (\sqrt{x} - 1) \, dx = \int_{1}^{4} \sqrt{x} \, dx - \int_{1}^{4} 1 \, dx \]
Calculating \( \int_{1}^{4} \sqrt{x} \, dx \):
\[ \int_{1}^{4} \sqrt{x} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4} = \frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3} \]
Calculating \( \int_{1}^{4} 1 \, dx \):
\[ \int_{1}^{4} 1 \, dx = [x]_{1}^{4} = 4 - 1 = 3 \]
Thus,
\[ \int_{1}^{4} \{ \sqrt{x} \} \, dx = \frac{14}{3} - 3 = \frac{14}{3} - \frac{9}{3} = \frac{5}{3} \]

### Step 6: Combine the results
Now, we combine the results from both intervals:
\[ \int_{0}^{4} \{ \sqrt{x} \} \, dx = \int_{0}^{1} \{ \sqrt{x} \} \, dx + \int_{1}^{4} \{ \sqrt{x} \} \, dx = \frac{2}{3} + \frac{5}{3} = \frac{7}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{4} \{ \sqrt{x} \} \, dx = \frac{7}{3} \]