Area bounded by x-axis and the curve `f(x) = e^`[x]`.e^|x|.e^`{x}`` between the lines x=-1 and x=2,where [x] represents greatest integer function (a) `(e+1)/(2)` (b) `(e^(2)+1)/(2)` (c) `(e^(3)+1)/(2)` (d) `(e^(4)+1)/(2)`

Options

A. `(e+1)/(2)`
B. `(e^(2)+1)/(2)`
C. `(e^(3)+1)/(2)`
D. `(e^(4)+1)/(2)`

Answer

To find the area bounded by the x-axis and the curve \( f(x) = e^{[x]} \cdot e^{|x|} \cdot e^{\{x\}} \) between the lines \( x = -1 \) and \( x = 2 \), we will follow these steps:

### Step 1: Understand the function
The function can be expressed as:
\[ f(x) = e^{[x]} \cdot e^{|x|} \cdot e^{\{x\}} \]
where \([x]\) is the greatest integer function and \(\{x\} = x - [x]\) is the fractional part of \(x\).

### Step 2: Analyze the intervals
We need to analyze the function in the intervals:
- For \( -1 \leq x < 0 \)
- For \( 0 \leq x < 1 \)
- For \( 1 \leq x < 2 \)

### Step 3: Calculate \( f(x) \) in each interval
1. **For \( -1 \leq x < 0 \)**:
- Here, \([x] = -1\) and \(|x| = -x\), thus:
\[ f(x) = e^{-1} \cdot e^{-x} \cdot e^{x} = e^{-1} \]

2. **For \( 0 \leq x < 1 \)**:
- Here, \([x] = 0\) and \(|x| = x\), thus:
\[ f(x) = e^{0} \cdot e^{x} \cdot e^{x} = e^{2x} \]

3. **For \( 1 \leq x < 2 \)**:
- Here, \([x] = 1\) and \(|x| = x\), thus:
\[ f(x) = e^{1} \cdot e^{x} \cdot e^{x-1} = e^{2x} \]

### Step 4: Set up the integral for the area
The area \( A \) can be calculated as:
\[ A = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \]

### Step 5: Calculate each integral
1. **Integral from \(-1\) to \(0\)**:
\[ \int_{-1}^{0} e^{-1} \, dx = e^{-1} \cdot (0 - (-1)) = e^{-1} \]

2. **Integral from \(0\) to \(1\)**:
\[ \int_{0}^{1} e^{2x} \, dx = \left[ \frac{e^{2x}}{2} \right]_{0}^{1} = \frac{e^{2}}{2} - \frac{1}{2} = \frac{e^{2} - 1}{2} \]

3. **Integral from \(1\) to \(2\)**:
\[ \int_{1}^{2} e^{2x} \, dx = \left[ \frac{e^{2x}}{2} \right]_{1}^{2} = \frac{e^{4}}{2} - \frac{e^{2}}{2} = \frac{e^{4} - e^{2}}{2} \]

### Step 6: Combine the areas
Now, adding all parts together:
\[ A = e^{-1} + \frac{e^{2} - 1}{2} + \frac{e^{4} - e^{2}}{2} \]
\[ = e^{-1} + \frac{e^{2} - 1 + e^{4} - e^{2}}{2} \]
\[ = e^{-1} + \frac{e^{4} - 1}{2} \]

### Step 7: Simplify the expression
To combine the terms, we can express \( e^{-1} \) as \( \frac{1}{e} \):
\[ A = \frac{1}{e} + \frac{e^{4} - 1}{2} \]
To combine these fractions, we can find a common denominator:
\[ A = \frac{2 + e^{4} - 1}{2e} = \frac{e^{4} + 1}{2e} \]

### Final Answer
Thus, the area bounded by the x-axis and the curve is:
\[ \frac{e^{4} + 1}{2} \]

### Conclusion
The correct answer is:
(d) \( \frac{e^{4} + 1}{2} \)