Difference between the greatest and least values opf the function `f (x) = int _(0)^(x) (cos ^(2) t + cos t +2) ` dt in the interval `[0, 2pi]` is `K pi, `then K is equal to:

Options

A. 1
B. 3
C. 5
D. None of these

Answer

To find the difference between the greatest and least values of the function

\[ f(x) = \int_0^x (\cos^2 t + \cos t + 2) \, dt \]

in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Differentiate the function
Using the Fundamental Theorem of Calculus, we can differentiate \(f(x)\):

\[ f'(x) = \cos^2 x + \cos x + 2 \]

### Step 2: Find critical points
To find the critical points, we set the derivative equal to zero:

\[ \cos^2 x + \cos x + 2 = 0 \]

This is a quadratic equation in terms of \(\cos x\). Let \(y = \cos x\):

\[ y^2 + y + 2 = 0 \]

### Step 3: Analyze the quadratic equation
The discriminant of this quadratic equation is:

\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 \]

Since the discriminant is negative, there are no real solutions for \(y\). Therefore, \(f'(x) \neq 0\) for any \(x\) in the interval \([0, 2\pi]\). This means \(f(x)\) is either strictly increasing or strictly decreasing in this interval.

### Step 4: Evaluate the function at the endpoints
Since there are no critical points, we will evaluate \(f(x)\) at the endpoints of the interval:

1. **At \(x = 0\)**:

\[ f(0) = \int_0^0 (\cos^2 t + \cos t + 2) \, dt = 0 \]

2. **At \(x = 2\pi\)**:

\[ f(2\pi) = \int_0^{2\pi} (\cos^2 t + \cos t + 2) \, dt \]

To evaluate this integral, we can break it down:

\[ \int_0^{2\pi} \cos^2 t \, dt + \int_0^{2\pi} \cos t \, dt + \int_0^{2\pi} 2 \, dt \]

### Step 5: Calculate the integrals
1. **Calculate \(\int_0^{2\pi} \cos^2 t \, dt\)**:

Using the identity \(\cos^2 t = \frac{1 + \cos 2t}{2}\):

\[ \int_0^{2\pi} \cos^2 t \, dt = \int_0^{2\pi} \frac{1 + \cos 2t}{2} \, dt = \frac{1}{2} \left[ t + \frac{\sin 2t}{2} \right]_0^{2\pi} = \frac{1}{2} \left[ 2\pi + 0 - 0 \right] = \pi \]

2. **Calculate \(\int_0^{2\pi} \cos t \, dt\)**:

\[ \int_0^{2\pi} \cos t \, dt = [\sin t]_0^{2\pi} = \sin(2\pi) - \sin(0) = 0 \]

3. **Calculate \(\int_0^{2\pi} 2 \, dt\)**:

\[ \int_0^{2\pi} 2 \, dt = 2 \cdot [t]_0^{2\pi} = 2 \cdot (2\pi - 0) = 4\pi \]

### Step 6: Combine the results
Now, we can combine these results to find \(f(2\pi)\):

\[ f(2\pi) = \pi + 0 + 4\pi = 5\pi \]

### Step 7: Find the difference between maximum and minimum values
The maximum value of \(f(x)\) in the interval \([0, 2\pi]\) is \(f(2\pi) = 5\pi\) and the minimum value is \(f(0) = 0\).

Thus, the difference is:

\[ f(2\pi) - f(0) = 5\pi - 0 = 5\pi \]

### Step 8: Express the difference in terms of \(K\)
We are given that this difference is \(K\pi\). Therefore, we have:

\[ K\pi = 5\pi \implies K = 5 \]

### Conclusion
The value of \(K\) is:

\[ \boxed{5} \]