If `f(x) = sqrt(cos^(2)x)+sqrt(sin^(2)x)` then `int_(-3//2)^(3//2)[f(x)]dx=`
Where [.] G.I.F.

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem, we need to evaluate the integral of the function \( f(x) = \sqrt{\cos^2 x} + \sqrt{\sin^2 x} \) over the interval \([-3/2, 3/2]\) and then apply the greatest integer function (GIF) to the result.

### Step 1: Simplify the function \( f(x) \)

The function can be simplified as follows:

\[ f(x) = \sqrt{\cos^2 x} + \sqrt{\sin^2 x} \]

Since \( \sqrt{\cos^2 x} = |\cos x| \) and \( \sqrt{\sin^2 x} = |\sin x| \), we can rewrite \( f(x) \) as:

\[ f(x) = |\cos x| + |\sin x| \]

### Step 2: Analyze the function \( f(x) \)

The function \( f(x) = |\cos x| + |\sin x| \) varies depending on the values of \( \cos x \) and \( \sin x \).

1. The maximum value of \( |\cos x| + |\sin x| \) occurs when both \( \cos x \) and \( \sin x \) are at their maximum values. This occurs at \( x = \frac{\pi}{4} + n\frac{\pi}{2} \) for integers \( n \), where \( |\cos x| + |\sin x| = \sqrt{2} \).
2. The minimum value occurs when either \( \cos x \) or \( \sin x \) is zero, yielding a minimum of \( 1 \).

### Step 3: Determine the range of \( f(x) \)

From the analysis, we find that:

\[ 1 \leq f(x) \leq \sqrt{2} \quad \text{(approximately 1.41)} \]

### Step 4: Apply the Greatest Integer Function (GIF)

Since \( f(x) \) ranges from \( 1 \) to \( \sqrt{2} \), the greatest integer function \( \lfloor f(x) \rfloor \) will yield:

\[ \lfloor f(x) \rfloor = 1 \quad \text{for all } x \text{ in } [-3/2, 3/2] \]

### Step 5: Evaluate the integral

Now we can evaluate the integral:

\[ \int_{-3/2}^{3/2} \lfloor f(x) \rfloor \, dx = \int_{-3/2}^{3/2} 1 \, dx \]

The integral of \( 1 \) over the interval \([-3/2, 3/2]\) is simply the length of the interval:

\[ = 3 - (-3) = 3 \]

### Final Answer

Thus, the value of the integral is:

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