The value of `int_(0)^(10pi)[tan^(-1)x]dx` (where, [.] denotes the greatest integer functionof x) is equal to

Options

A. `tan 1`
B. `10 pi`
C. `10 pi -tan 1`
D. None of these

Answer

To solve the integral \( \int_{0}^{10\pi} \lfloor \tan^{-1} x \rfloor \, dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the function \( \tan^{-1} x \)
The function \( \tan^{-1} x \) (or arctan) is a continuous function that approaches \( \frac{\pi}{2} \) as \( x \) approaches infinity. We need to determine the values of \( \tan^{-1} x \) at specific points to understand how it behaves over the interval \( [0, 10\pi] \).

### Step 2: Calculate \( \tan^{-1} x \) at key points
- At \( x = 0 \):
\[ \tan^{-1}(0) = 0 \]
- At \( x = 1 \):
\[ \tan^{-1}(1) = \frac{\pi}{4} \approx 0.785 \]
- At \( x = 10 \):
\[ \tan^{-1}(10) \approx 1.5608 \]
- As \( x \) approaches infinity:
\[ \tan^{-1}(x) \to \frac{\pi}{2} \approx 1.5708 \]

### Step 3: Determine the intervals for \( \lfloor \tan^{-1} x \rfloor \)
From the calculations:
- For \( 0 \leq x < 1 \), \( \tan^{-1} x \) is in the interval \( [0, \frac{\pi}{4}) \) so \( \lfloor \tan^{-1} x \rfloor = 0 \).
- For \( 1 \leq x < 10 \), \( \tan^{-1} x \) is in the interval \( [\frac{\pi}{4}, \tan^{-1}(10)) \) so \( \lfloor \tan^{-1} x \rfloor = 1 \).
- For \( x \geq 10 \), \( \tan^{-1} x \) approaches \( \frac{\pi}{2} \) but stays less than \( 2 \), so \( \lfloor \tan^{-1} x \rfloor = 1 \) until \( x = 10\pi \).

### Step 4: Set up the integral
We can split the integral into two parts based on the intervals identified:
\[ \int_{0}^{10\pi} \lfloor \tan^{-1} x \rfloor \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{10\pi} 1 \, dx \]

### Step 5: Evaluate the integrals
- The first integral:
\[ \int_{0}^{1} 0 \, dx = 0 \]
- The second integral:
\[ \int_{1}^{10\pi} 1 \, dx = [x]_{1}^{10\pi} = 10\pi - 1 \]

### Step 6: Combine the results
Thus, the value of the original integral is:
\[ \int_{0}^{10\pi} \lfloor \tan^{-1} x \rfloor \, dx = 0 + (10\pi - 1) = 10\pi - 1 \]

### Final Answer
The value of \( \int_{0}^{10\pi} \lfloor \tan^{-1} x \rfloor \, dx \) is \( 10\pi - 1 \).

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