The value of `int_(0)^(4) {x} dx` (where , {.} denotes fractional part of x) is equal to

Options

A. `(4)/(3)`
B. `(5)/(3)`
C. `(7)/(3)`
D. None of these

Answer

To solve the integral \( \int_{0}^{4} \{x\} \, dx \), where \( \{x\} \) denotes the fractional part of \( x \), we can use the property that the fractional part can be expressed as:

\[ \{x\} = x - \lfloor x \rfloor \]

Thus, we can rewrite the integral as:

\[ \int_{0}^{4} \{x\} \, dx = \int_{0}^{4} (x - \lfloor x \rfloor) \, dx \]

Next, we will split the integral into intervals where \( \lfloor x \rfloor \) is constant. The function \( \lfloor x \rfloor \) changes at integer values, so we will break the integral into the following intervals: \( [0, 1) \), \( [1, 2) \), \( [2, 3) \), and \( [3, 4) \).

### Step 1: Evaluate the integral from \( 0 \) to \( 1 \)

In the interval \( [0, 1) \):
- \( \lfloor x \rfloor = 0 \)
- Thus, \( \{x\} = x - 0 = x \)

So, we have:

\[ \int_{0}^{1} \{x\} \, dx = \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

### Step 2: Evaluate the integral from \( 1 \) to \( 2 \)

In the interval \( [1, 2) \):
- \( \lfloor x \rfloor = 1 \)
- Thus, \( \{x\} = x - 1 \)

So, we have:

\[ \int_{1}^{2} \{x\} \, dx = \int_{1}^{2} (x - 1) \, dx = \int_{1}^{2} x \, dx - \int_{1}^{2} 1 \, dx \]

Calculating each part:

\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

\[ \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]

Thus,

\[ \int_{1}^{2} \{x\} \, dx = \frac{3}{2} - 1 = \frac{1}{2} \]

### Step 3: Evaluate the integral from \( 2 \) to \( 3 \)

In the interval \( [2, 3) \):
- \( \lfloor x \rfloor = 2 \)
- Thus, \( \{x\} = x - 2 \)

So, we have:

\[ \int_{2}^{3} \{x\} \, dx = \int_{2}^{3} (x - 2) \, dx = \int_{2}^{3} x \, dx - \int_{2}^{3} 2 \, dx \]

Calculating each part:

\[ \int_{2}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{2}^{3} = \frac{3^2}{2} - \frac{2^2}{2} = \frac{9}{2} - \frac{4}{2} = \frac{5}{2} \]

\[ \int_{2}^{3} 2 \, dx = [2x]_{2}^{3} = 2(3) - 2(2) = 6 - 4 = 2 \]

Thus,

\[ \int_{2}^{3} \{x\} \, dx = \frac{5}{2} - 2 = \frac{1}{2} \]

### Step 4: Evaluate the integral from \( 3 \) to \( 4 \)

In the interval \( [3, 4) \):
- \( \lfloor x \rfloor = 3 \)
- Thus, \( \{x\} = x - 3 \)

So, we have:

\[ \int_{3}^{4} \{x\} \, dx = \int_{3}^{4} (x - 3) \, dx = \int_{3}^{4} x \, dx - \int_{3}^{4} 3 \, dx \]

Calculating each part:

\[ \int_{3}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{3}^{4} = \frac{4^2}{2} - \frac{3^2}{2} = \frac{16}{2} - \frac{9}{2} = \frac{7}{2} \]

\[ \int_{3}^{4} 3 \, dx = [3x]_{3}^{4} = 3(4) - 3(3) = 12 - 9 = 3 \]

Thus,

\[ \int_{3}^{4} \{x\} \, dx = \frac{7}{2} - 3 = \frac{1}{2} \]

### Step 5: Combine all parts

Now, we can combine all the parts:

\[ \int_{0}^{4} \{x\} \, dx = \int_{0}^{1} \{x\} \, dx + \int_{1}^{2} \{x\} \, dx + \int_{2}^{3} \{x\} \, dx + \int_{3}^{4} \{x\} \, dx \]

\[ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 \]

Thus, the value of the integral \( \int_{0}^{4} \{x\} \, dx \) is \( \boxed{2} \).