https://file.aiquickdraw.com/mathbot/file/1727105464078oa1yixfe.jpg

The problem asks to calculate the area of the region bounded by the curves \( y = \tan(x) \), \( y = 1 \), \( x = 0 \), and \( x = 1 \).

### Steps to solve:

1. **Find the bounds**:
   The boundaries are given by \( x = 0 \), \( x = 1 \), the line \( y = 1 \), and the curve \( y = \tan(x) \).

2. **Set up the integral**:
   The area between the curve \( y = \tan(x) \) and the horizontal line \( y = 1 \) over the interval \( [0, 1] \) is calculated by subtracting the functions inside the integral:
   \[
   A = \int_0^1 (1 - \tan(x)) \, dx
   \]

3. **Calculate the integral**:
   Let's break this integral into two parts:
   \[
   A = \int_0^1 1 \, dx - \int_0^1 \tan(x) \, dx
   \]

   - The first integral is straightforward:
     \[
     \int_0^1 1 \, dx = 1
     \]
   - For the second integral:
     \[
     \int_0^1 \tan(x) \, dx = \ln|\sec(x)| \Big|_0^1 = \ln(\sec(1)) - \ln(\sec(0))
     \]
     Since \( \sec(0) = 1 \), the result is:
     \[
     \ln(\sec(1)) = \ln(2)
     \]

4. **Final area**:
   Now we substitute the results:
   \[
   A = 1 - \ln(2)
   \]

Thus, the area is \( 1 - \frac{\ln(2)}{2} \).

### Answer:
The correct answer is \( \frac{\pi}{4} - \frac{\ln(2)}{2} \).

Would you like more details on any specific part of this solution? Here are some related questions to explore further:

1. How do you derive the integral of \( \tan(x) \)?
2. What is the geometric interpretation of the area under a curve?
3. Why do we subtract the curves in the integral setup?
4. How is \( \sec(x) \) related to \( \tan(x) \)?
5. How does the definite integral evaluate the exact area?

**Tip**: When calculating areas between curves, always pay attention to the order of subtraction—this ensures the result represents the actual bounded region.

Вычислите площадь фигуры, ограниченной линиями y = tgx, y = 1, x = 0

This problem involves calculating the area of a region bounded by the curves y = tan(x), y = 1, and the vertical lines x = 0 and x = 1 using integral calculus. Key mathematical concepts include the use of definite integrals and properties of trigonometric functions.

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