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resuelve esas tres graficas con su procedimiento , el tema es area entre la grafica de dos funciones 

Let's solve the three graphs based on the visual information provided. The problem is about finding the area enclosed between two functions. I will describe each graph and the corresponding calculations:

### 1st Graph:
**Functions:**
- \( y = x \)
- \( y = 3x \)
- \( y = 4 - x \)

#### Steps to solve:
- **Find the points of intersection:**
   1. Intersection of \( y = x \) and \( y = 4 - x \):
      \[
      x = 4 - x \implies 2x = 4 \implies x = 2
      \]
      So, the intersection is at \( (2, 2) \).
      
   2. Intersection of \( y = 3x \) and \( y = 4 - x \):
      \[
      3x = 4 - x \implies 4x = 4 \implies x = 1
      \]
      So, the intersection is at \( (1, 3) \).
      
   3. Intersection of \( y = x \) and \( y = 3x \):
      \[
      x = 3x \implies 2x = 0 \implies x = 0
      \]
      So, the intersection is at \( (0, 0) \).

- **Set up the integrals for the area:**
  The area can be split into two regions:
  1. From \( x = 0 \) to \( x = 1 \): The top function is \( y = 3x \) and the bottom is \( y = x \).
     \[
     A_1 = \int_0^1 (3x - x) \, dx = \int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1
     \]
  
  2. From \( x = 1 \) to \( x = 2 \): The top function is \( y = 4 - x \) and the bottom is \( y = x \).
     \[
     A_2 = \int_1^2 ((4 - x) - x) \, dx = \int_1^2 (4 - 2x) \, dx = \left[ 4x - x^2 \right]_1^2 = (8 - 4) - (4 - 1) = 4 - 3 = 1
     \]

- **Total area:**
  \[
  A = A_1 + A_2 = 1 + 1 = 2
  \]

### 2nd Graph:
**Functions:**
- \( y = x \)
- \( y = 3x \)
- \( y = 4 - x \)

The same equations and procedure apply as in the first graph, meaning the enclosed area is also:

\[
A = 2
\]

### 3rd Graph:
**Functions:**
- \( y = 3x \)
- \( y = 4 - x \)

#### Steps to solve:
- **Find the points of intersection:**
   1. Intersection of \( y = 3x \) and \( y = 4 - x \):
      \[
      3x = 4 - x \implies 4x = 4 \implies x = 1
      \]
      So, the intersection is at \( (1, 3) \).
      
   2. The other bound is at \( x = 4 \), where the function \( y = 4 - x \) intersects the x-axis.

- **Set up the integral for the area:**
  The area is the region between \( y = 3x \) and \( y = 4 - x \) from \( x = 1 \) to \( x = 4 \).
  
  \[
  A = \int_1^4 ((4 - x) - 3x) \, dx = \int_1^4 (4 - 4x) \, dx = \left[ 4x - 2x^2 \right]_1^4
  \]

  Now, evaluate the integral:
  \[
  = (16 - 32) - (4 - 2) = -16 - 2 = -18
  \]

Since this result is negative, we take the absolute value (as area cannot be negative):
\[
A = 18
\]

### Summary of Areas:
- **1st graph area:** 2
- **2nd graph area:** 2
- **3rd graph area:** 18

Would you like further details or explanations on any part of the process? 

Here are 5 related questions:
1. How would the area change if one of the functions were modified?
2. What if we considered only a specific portion of the graph, e.g., from \( x = 0 \) to \( x = 1 \)?
3. How can these integrals be computed using a calculator or software?
4. What is the geometric meaning of these areas in a real-world context?
5. How does the method change if one of the curves becomes a more complex function, like a parabola?

**Tip:** When finding areas between curves, always identify which function is on top and which is on the bottom to set up the correct integral.

Resuelve esas tres gráficas con su procedimiento, el tema es área entre la gráfica de dos funciones.

Solve for the area between two functions as presented in three graphs. Learn to find intersection points and apply integral calculus to determine the enclosed area. Suitable for advanced high school or early university students studying calculus.

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