If `f(x)={e^(cosxsinx ,for|x|lt=2)2,ot h e r w i s e ,t h e nint_(-2)^3f(x)dx=` 0 (b) 1 (c) 2 (d) 3

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: from \(-2\) to \(2\) and from \(2\) to \(3\).

### Step 1: Break the Integral
We can express the integral as:

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

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

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

### Step 3: Check if \( f(x) \) is Odd
To evaluate this integral, we need to check if \( f(x) \) is an odd function. We compute \( f(-x) \):

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

Since \( f(-x) = -f(x) \), \( f(x) \) is an odd function.

### Step 4: Evaluate the Integral from \(-2\) to \(2\)
For an odd function, the integral over a symmetric interval around zero is zero:

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

### Step 5: Evaluate the Second Integral
Now, we evaluate the second integral from \(2\) to \(3\):

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

This integral simplifies to:

\[ = 2 \int_{2}^{3} 1 \, dx = 2 [x]_{2}^{3} = 2(3 - 2) = 2 \]

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

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

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

\[ \boxed{2} \]