`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 follow these steps:

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

### Step 2: Identify the limits of integration
We need to evaluate the integral from \(0\) to \(4\). The function \(\sqrt{x}\) varies from \(0\) to \(2\) as \(x\) goes from \(0\) to \(4\).

### Step 3: Determine where \(\sqrt{x}\) crosses integer values
The integer values of \(\sqrt{x}\) in the interval \(0\) to \(4\) are:
- From \(0\) to \(1\) when \(x\) is in \([0, 1)\)
- From \(1\) to \(2\) when \(x\) is in \([1, 4)\)

### Step 4: Break the integral into parts
We can split the integral into two parts based on the intervals identified:
\[ \int_{0}^{4} \{\sqrt{x}\} \, dx = \int_{0}^{1} \{\sqrt{x}\} \, dx + \int_{1}^{4} \{\sqrt{x}\} \, dx \]

### Step 5: Evaluate the first integral \(\int_{0}^{1} \{\sqrt{x}\} \, dx\)
In the interval \([0, 1)\), \(\sqrt{x}\) varies from \(0\) to \(1\), thus:
\[ \{\sqrt{x}\} = \sqrt{x} - 0 = \sqrt{x} \]
So,
\[ \int_{0}^{1} \{\sqrt{x}\} \, dx = \int_{0}^{1} \sqrt{x} \, dx \]
Calculating this integral:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \Big|_{0}^{1} = \frac{2}{3}(1^{3/2} - 0^{3/2}) = \frac{2}{3} \]

### Step 6: Evaluate the second integral \(\int_{1}^{4} \{\sqrt{x}\} \, dx\)
In the interval \([1, 4]\), \(\sqrt{x}\) varies from \(1\) to \(2\), thus:
\[ \{\sqrt{x}\} = \sqrt{x} - 1 \]
So,
\[ \int_{1}^{4} \{\sqrt{x}\} \, dx = \int_{1}^{4} (\sqrt{x} - 1) \, dx \]
This can be split into two integrals:
\[ \int_{1}^{4} \sqrt{x} \, dx - \int_{1}^{4} 1 \, dx \]
Calculating these integrals:
1. For \(\int_{1}^{4} \sqrt{x} \, dx\):
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \Big|_{1}^{4} = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{2}{3} \cdot 7 = \frac{14}{3} \]
2. For \(\int_{1}^{4} 1 \, dx\):
\[ \int_{1}^{4} 1 \, dx = x \Big|_{1}^{4} = 4 - 1 = 3 \]

Combining these results:
\[ \int_{1}^{4} \{\sqrt{x}\} \, dx = \frac{14}{3} - 3 = \frac{14}{3} - \frac{9}{3} = \frac{5}{3} \]

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

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