Let `y= {x}^([x])` then the value of `int_0^3 yd x` equals to (where {.} and [.] denote fractional part and integerpart function respectively)

Options

A. 1
B. `11/6`
C. 3
D. `5/6`

Answer

To solve the integral \( \int_0^3 y \, dx \) where \( y = \{x\}^{[x]} \) (with \(\{x\}\) denoting the fractional part of \(x\) and \([x]\) denoting the greatest integer function), we will break down the integral into parts based on the intervals defined by the integer values of \(x\).

### Step 1: Understand the functions involved

The fractional part function \(\{x\}\) is defined as:
\[ \{x\} = x - [x] \]
where \([x]\) is the greatest integer less than or equal to \(x\).

### Step 2: Identify the intervals

We will evaluate the integral from \(0\) to \(3\), which can be broken down into three intervals:
1. \(0 \leq x < 1\)
2. \(1 \leq x < 2\)
3. \(2 \leq x < 3\)

### Step 3: Evaluate \(y\) in each interval

1. **For \(0 \leq x < 1\)**:
- Here, \([x] = 0\) and \(\{x\} = x\).
- Thus, \(y = \{x\}^{[x]} = x^0 = 1\).

2. **For \(1 \leq x < 2\)**:
- Here, \([x] = 1\) and \(\{x\} = x - 1\).
- Thus, \(y = \{x\}^{[x]} = (x - 1)^1 = x - 1\).

3. **For \(2 \leq x < 3\)**:
- Here, \([x] = 2\) and \(\{x\} = x - 2\).
- Thus, \(y = \{x\}^{[x]} = (x - 2)^2\).

### Step 4: Set up the integral

Now we can set up the integral as follows:
\[ \int_0^3 y \, dx = \int_0^1 1 \, dx + \int_1^2 (x - 1) \, dx + \int_2^3 (x - 2)^2 \, dx \]

### Step 5: Calculate each integral

1. **Calculate \( \int_0^1 1 \, dx \)**:
\[ \int_0^1 1 \, dx = [x]_0^1 = 1 - 0 = 1 \]

2. **Calculate \( \int_1^2 (x - 1) \, dx \)**:
\[ \int_1^2 (x - 1) \, dx = \left[\frac{(x - 1)^2}{2}\right]_1^2 = \left[\frac{(2 - 1)^2}{2} - \frac{(1 - 1)^2}{2}\right] = \frac{1}{2} - 0 = \frac{1}{2} \]

3. **Calculate \( \int_2^3 (x - 2)^2 \, dx \)**:
\[ \int_2^3 (x - 2)^2 \, dx = \left[\frac{(x - 2)^3}{3}\right]_2^3 = \left[\frac{(3 - 2)^3}{3} - \frac{(2 - 2)^3}{3}\right] = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 6: Combine the results

Now, we combine all the results:
\[ \int_0^3 y \, dx = 1 + \frac{1}{2} + \frac{1}{3} \]

To combine these fractions, we find a common denominator (which is 6):
\[ 1 = \frac{6}{6}, \quad \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \]
Thus,
\[ \int_0^3 y \, dx = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6} \]

### Final Answer
The value of \( \int_0^3 y \, dx \) is \( \frac{11}{6} \).

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