If `f(x)= {(e^(cos x)sin x,"for " |x| le 2),(2,"otherwise"):}` then `int_(-2)^(3) f(x) dx`=

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the integral \( \int_{-2}^{3} f(x) \, dx \) where

\[ f(x) = \begin{cases} e^{\cos x} \sin x & \text{for } |x| \leq 2 \\ 2 & \text{otherwise} \end{cases} \]

we will break the integral into two parts based on the definition of the function \( f(x) \).

### Step 1: Break the Integral
We can split the integral at the points where the definition of \( f(x) \) changes:

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

### Step 2: Evaluate the First Integral
For the first integral, where \( |x| \leq 2 \):

\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{2} e^{\cos x} \sin x \, dx \]

### Step 3: Check Symmetry
To evaluate \( \int_{-2}^{2} e^{\cos x} \sin x \, dx \), we can check if the function is odd or even.

We find \( f(-x) \):

\[ f(-x) = e^{\cos(-x)} \sin(-x) = e^{\cos x} (-\sin x) = -f(x) \]

Since \( f(-x) = -f(x) \), the function \( f(x) \) is odd. Therefore, the integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-2}^{2} e^{\cos x} \sin x \, dx = 0 \]

### Step 4: Evaluate the Second Integral
Now we evaluate the second integral where \( x \) is outside the interval \([-2, 2]\):

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

### Step 5: Calculate the Integral
Calculating this integral:

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

### Step 6: Combine Results
Now combine the results from both integrals:

\[ \int_{-2}^{3} f(x) \, dx = \int_{-2}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx = 0 + 2 = 2 \]

### Final Answer
Thus, the value of the integral \( \int_{-2}^{3} f(x) \, dx \) is:

\[ \boxed{2} \]

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