The value of `int_(0)^(100)[ tan ^(-1)x] d x ` is equal to (where `[.]` denotes the greatest integer function)

Options

A. `tan1-100`
B. `pi//2- tan 1`
C. `100-tan 1`
D. None of these

Answer

To solve the integral \( \int_{0}^{100} \tan^{-1}(x) \, dx \), we will follow these steps:

### Step 1: Understand the function and limits
The function \( \tan^{-1}(x) \) is defined for all \( x \) and approaches \( \frac{\pi}{2} \) as \( x \) approaches infinity. We need to evaluate the integral from \( 0 \) to \( 100 \).

### Step 2: Find the antiderivative of \( \tan^{-1}(x) \)
The antiderivative of \( \tan^{-1}(x) \) can be found using integration by parts. Let:
- \( u = \tan^{-1}(x) \) so that \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = dx \) so that \( v = x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \int \frac{x}{1+x^2} \, dx \]

### Step 3: Evaluate the remaining integral
The remaining integral \( \int \frac{x}{1+x^2} \, dx \) can be solved by substitution:
Let \( w = 1 + x^2 \), then \( dw = 2x \, dx \) or \( \frac{1}{2} dw = x \, dx \).
Thus,
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C \]

### Step 4: Combine results
Now substituting back, we have:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

### Step 5: Evaluate the definite integral
Now we evaluate the definite integral from \( 0 \) to \( 100 \):
\[ \int_{0}^{100} \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) \right]_{0}^{100} \]

Calculating at the limits:
1. At \( x = 100 \):
\[ 100 \tan^{-1}(100) - \frac{1}{2} \ln(1 + 100^2) = 100 \cdot \frac{\pi}{2} - \frac{1}{2} \ln(10001) \]
\[ = 50\pi - \frac{1}{2} \ln(10001) \]

2. At \( x = 0 \):
\[ 0 \cdot \tan^{-1}(0) - \frac{1}{2} \ln(1 + 0^2) = 0 - 0 = 0 \]

Thus,
\[ \int_{0}^{100} \tan^{-1}(x) \, dx = 50\pi - \frac{1}{2} \ln(10001) \]

### Step 6: Apply the greatest integer function
Now, we need to find the greatest integer value of \( 50\pi - \frac{1}{2} \ln(10001) \).

Using \( \pi \approx 3.14 \):
\[ 50\pi \approx 50 \cdot 3.14 = 157 \]
And calculating \( \ln(10001) \):
\[ \ln(10001) \approx 9.210 \quad \text{(using a calculator)} \]
Thus,
\[ \frac{1}{2} \ln(10001) \approx \frac{9.210}{2} \approx 4.605 \]
So,
\[ 50\pi - \frac{1}{2} \ln(10001) \approx 157 - 4.605 \approx 152.395 \]

Finally, applying the greatest integer function:
\[ \lfloor 152.395 \rfloor = 152 \]

### Final Answer
The value of \( \int_{0}^{100} \tan^{-1}(x) \, dx \) is equal to \( \lfloor 50\pi - \frac{1}{2} \ln(10001) \rfloor = 152 \).

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