The integral `I=int_(0)^(100pi)[tan^(-1)x]dx` (where, `[.]` represents the greatest integer function) has the vlaue `K(pi)+ tan(p)` then value of `K + p` is equal to

Options

A. 101
B. 99
C. `100pi`
D. `99pi`

Answer

To solve the integral \( I = \int_{0}^{100\pi} \lfloor \tan^{-1} x \rfloor \, dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the behavior of \( \tan^{-1} x \)
The function \( \tan^{-1} x \) is defined for \( x \geq 0 \) and approaches \( \frac{\pi}{2} \) as \( x \) approaches infinity. Specifically, we have:
- \( \tan^{-1}(0) = 0 \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)
- \( \tan^{-1}(+\infty) = \frac{\pi}{2} \)

### Step 2: Determine the intervals
We need to analyze the function \( \lfloor \tan^{-1} x \rfloor \) over the interval \( [0, 100\pi] \).

1. For \( 0 \leq x < \tan(1) \) (approximately \( 1.5574 \)), \( \tan^{-1} x \) ranges from \( 0 \) to \( \tan^{-1}(1) \), so \( \lfloor \tan^{-1} x \rfloor = 0 \).
2. For \( \tan(1) \leq x < \tan\left(\frac{\pi}{2}\right) \) (which is not applicable since \( \tan^{-1} x \) never actually reaches \( \frac{\pi}{2} \)), \( \lfloor \tan^{-1} x \rfloor = 1 \) for \( x \) values greater than \( \tan(1) \).

### Step 3: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{\tan(1)} \lfloor \tan^{-1} x \rfloor \, dx + \int_{\tan(1)}^{100\pi} \lfloor \tan^{-1} x \rfloor \, dx \]

### Step 4: Evaluate the first integral
For the first part:
\[ \int_{0}^{\tan(1)} \lfloor \tan^{-1} x \rfloor \, dx = \int_{0}^{\tan(1)} 0 \, dx = 0 \]

### Step 5: Evaluate the second integral
For the second part:
\[ \int_{\tan(1)}^{100\pi} \lfloor \tan^{-1} x \rfloor \, dx = \int_{\tan(1)}^{100\pi} 1 \, dx = [x]_{\tan(1)}^{100\pi} = 100\pi - \tan(1) \]

### Step 6: Combine results
Thus, we have:
\[ I = 0 + (100\pi - \tan(1)) = 100\pi - \tan(1) \]

### Step 7: Relate to the given form
The problem states that \( I = K\pi + \tan(p) \). We can equate:
\[ 100\pi - \tan(1) = K\pi + \tan(p) \]
From this, we can see:
- \( K = 100 \)
- \( \tan(p) = -\tan(1) \) implies \( p = -1 \) (since \( \tan(-\theta) = -\tan(\theta) \))

### Step 8: Find \( K + p \)
Now, we compute:
\[ K + p = 100 - 1 = 99 \]

### Final Answer
Thus, the value of \( K + p \) is \( \boxed{99} \).