The value of the integral `int_(-2)^(2)(1+2sin x)e^(|x|)dx` is equal to

Options

A. 0
B. `e^(2)-1`
C. `2(e^(2)-1)`
D. 1

Answer

To solve the integral \( \int_{-2}^{2} (1 + 2 \sin x) e^{|x|} \, dx \), we can follow these steps:

### Step 1: Analyze the function
The function \( f(x) = (1 + 2 \sin x) e^{|x|} \) is composed of two parts: \( e^{|x|} \) and \( (1 + 2 \sin x) \).

### Step 2: Determine the nature of the function
- The function \( e^{|x|} \) is an even function because \( e^{|-x|} = e^{|x|} \).
- The function \( 2 \sin x \) is an odd function because \( 2 \sin(-x) = -2 \sin x \).
- Therefore, \( f(-x) = (1 - 2 \sin x) e^{|x|} \).

### Step 3: Check if \( f(x) \) is odd or even
We can express \( f(-x) \):
\[ f(-x) = (1 - 2 \sin x) e^{|x|} = e^{|x|} - 2 \sin x \cdot e^{|x|} \]
This shows that \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), indicating that \( f(x) \) is neither even nor odd.

### Step 4: Split the integral
Given the properties of even and odd functions, we can split the integral:
\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{0} f(x) \, dx + \int_{0}^{2} f(x) \, dx \]

### Step 5: Evaluate the integral from 0 to 2
For \( x \geq 0 \), \( |x| = x \):
\[ \int_{0}^{2} (1 + 2 \sin x) e^{x} \, dx \]

### Step 6: Evaluate the integral from -2 to 0
For \( x < 0 \), \( |x| = -x \):
\[ \int_{-2}^{0} (1 + 2 \sin x) e^{-x} \, dx \]

### Step 7: Combine the results
Using the symmetry of the integrals, we can evaluate:
\[ \int_{-2}^{0} (1 + 2 \sin x) e^{-x} \, dx = \int_{0}^{2} (1 + 2 \sin x) e^{x} \, dx \]

### Step 8: Final calculation
The total integral becomes:
\[ 2 \int_{0}^{2} (1 + 2 \sin x) e^{x} \, dx \]

### Step 9: Solve the integral
Now we can compute:
\[ \int (1 + 2 \sin x) e^{x} \, dx = e^{x} (1 + 2 \sin x) - \int 2 \cos x e^{x} \, dx \]
Using integration by parts for \( \int 2 \cos x e^{x} \, dx \).

### Step 10: Evaluate limits
Finally, we evaluate the definite integral from 0 to 2 and compute the result.