Compute the area bounded by the lines `x +2y=2,y-x=1" and " 2x+y=7`.

Explanation

To find the area bounded by the lines \(x + 2y = 2\), \(y - x = 1\), and \(2x + y = 7\), we will follow these steps:

### Step 1: Find the intersection points of the lines
We need to find the points where these lines intersect, as these points will form the vertices of the bounded area.

1. **Intersection of \(x + 2y = 2\) and \(y - x = 1\)**:
- From \(y - x = 1\), we can express \(y\) in terms of \(x\):
\[ y = x + 1 \]
- Substitute \(y\) in the first equation:
\[ x + 2(x + 1) = 2 \implies x + 2x + 2 = 2 \implies 3x + 2 = 2 \implies 3x = 0 \implies x = 0 \]
- Substitute \(x = 0\) back to find \(y\):
\[ y = 0 + 1 = 1 \]
- So, the intersection point is \((0, 1)\).

2. **Intersection of \(y - x = 1\) and \(2x + y = 7\)**:
- Again, using \(y = x + 1\):
\[ 2x + (x + 1) = 7 \implies 2x + x + 1 = 7 \implies 3x + 1 = 7 \implies 3x = 6 \implies x = 2 \]
- Substitute \(x = 2\) back to find \(y\):
\[ y = 2 + 1 = 3 \]
- So, the intersection point is \((2, 3)\).

3. **Intersection of \(x + 2y = 2\) and \(2x + y = 7\)**:
- From \(x + 2y = 2\), express \(y\):
\[ 2y = 2 - x \implies y = 1 - \frac{x}{2} \]
- Substitute \(y\) in the second equation:
\[ 2x + (1 - \frac{x}{2}) = 7 \implies 2x + 1 - \frac{x}{2} = 7 \implies \frac{4x - x}{2} = 6 \implies \frac{3x}{2} = 6 \implies 3x = 12 \implies x = 4 \]
- Substitute \(x = 4\) back to find \(y\):
\[ y = 1 - \frac{4}{2} = 1 - 2 = -1 \]
- So, the intersection point is \((4, -1)\).

### Step 2: Identify the vertices of the bounded area
The vertices of the bounded area formed by the intersection points are:
1. \(A(0, 1)\)
2. \(B(2, 3)\)
3. \(C(4, -1)\)

### Step 3: Set up the integral to find the area
The area can be calculated by integrating the difference between the upper and lower functions over the range of \(x\) values from \(0\) to \(4\).

1. From \(x = 0\) to \(x = 2\):
- The upper curve is \(y = 1 + x\) (from \(y - x = 1\)).
- The lower curve is \(y = 1 - \frac{x}{2}\) (from \(x + 2y = 2\)).
- The area \(A_1\) is given by:
\[ A_1 = \int_0^2 \left((1 + x) - \left(1 - \frac{x}{2}\right)\right) dx = \int_0^2 \left(1 + x - 1 + \frac{x}{2}\right) dx = \int_0^2 \left(\frac{3x}{2}\right) dx \]

2. From \(x = 2\) to \(x = 4\):
- The upper curve is \(y = 7 - 2x\) (from \(2x + y = 7\)).
- The lower curve is \(y = 1 - \frac{x}{2}\).
- The area \(A_2\) is given by:
\[ A_2 = \int_2^4 \left((7 - 2x) - \left(1 - \frac{x}{2}\right)\right) dx = \int_2^4 \left(6 - \frac{3x}{2}\right) dx \]

### Step 4: Calculate the integrals
1. Calculate \(A_1\):
\[ A_1 = \int_0^2 \frac{3x}{2} dx = \frac{3}{2} \cdot \left[\frac{x^2}{2}\right]_0^2 = \frac{3}{2} \cdot \left[\frac{2^2}{2} - 0\right] = \frac{3}{2} \cdot 2 = 3 \]

2. Calculate \(A_2\):
\[ A_2 = \int_2^4 \left(6 - \frac{3x}{2}\right) dx = \left[6x - \frac{3x^2}{4}\right]_2^4 \]
- Evaluating at the limits:
\[ = \left[6(4) - \frac{3(4^2)}{4}\right] - \left[6(2) - \frac{3(2^2)}{4}\right] \]
\[ = \left[24 - 12\right] - \left[12 - 3\right] = 12 - 9 = 3 \]

### Step 5: Total Area
The total area \(A\) is:
\[ A = A_1 + A_2 = 3 + 3 = 6 \]

### Final Answer
The area bounded by the lines \(x + 2y = 2\), \(y - x = 1\), and \(2x + y = 7\) is \(6\) square units.