Using the method of integration, find the area of the triangular region whose vertices are `(2,-2), (4,3) and (1,2).`

Answer

To find the area of the triangular region with vertices at \( A(2, -2) \), \( B(4, 3) \), and \( C(1, 2) \) using integration, we can follow these steps:

### Step 1: Plot the Points
First, we plot the points \( A(2, -2) \), \( B(4, 3) \), and \( C(1, 2) \) on the Cartesian plane to visualize the triangle.

### Step 2: Find the Equations of the Lines
Next, we need to find the equations of the lines that form the sides of the triangle.

**Line AB:**
The slope \( m \) of line \( AB \) can be calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{4 - 2} = \frac{5}{2} \]
Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y + 2 = \frac{5}{2}(x - 2) \]
Simplifying this gives:
\[ y = \frac{5}{2}x - 7 \]

**Line BC:**
The slope \( m \) of line \( BC \):
\[ m = \frac{2 - 3}{1 - 4} = \frac{-1}{-3} = \frac{1}{3} \]
Using point-slope form:
\[ y - 3 = \frac{1}{3}(x - 4) \]
Simplifying this gives:
\[ y = \frac{1}{3}x + \frac{5}{3} \]

**Line CA:**
The slope \( m \) of line \( CA \):
\[ m = \frac{-2 - 2}{2 - 1} = \frac{-4}{1} = -4 \]
Using point-slope form:
\[ y - 2 = -4(x - 1) \]
Simplifying this gives:
\[ y = -4x + 6 \]

### Step 3: Set Up the Integral
To find the area of the triangle, we will integrate the difference between the upper and lower functions. The area \( A \) can be computed as:
\[ A = \int_{x_1}^{x_2} (f_{upper}(x) - f_{lower}(x)) \, dx \]
Where \( f_{upper}(x) \) and \( f_{lower}(x) \) are the equations of the lines.

We will integrate from \( x = 1 \) to \( x = 4 \).

### Step 4: Identify Upper and Lower Functions
- From \( x = 1 \) to \( x = 2 \): \( f_{upper}(x) = -4x + 6 \) and \( f_{lower}(x) = \frac{1}{3}x + \frac{5}{3} \)
- From \( x = 2 \) to \( x = 4 \): \( f_{upper}(x) = \frac{5}{2}x - 7 \) and \( f_{lower}(x) = -4x + 6 \)

### Step 5: Calculate the Area
Now we can calculate the area by integrating:

1. **From \( x = 1 \) to \( x = 2 \)**:
\[ A_1 = \int_{1}^{2} \left((-4x + 6) - \left(\frac{1}{3}x + \frac{5}{3}\right)\right) \, dx \]

2. **From \( x = 2 \) to \( x = 4 \)**:
\[ A_2 = \int_{2}^{4} \left(\left(\frac{5}{2}x - 7\right) - (-4x + 6)\right) \, dx \]

### Step 6: Solve the Integrals
Calculating \( A_1 \):
\[ A_1 = \int_{1}^{2} \left(-4x + 6 - \frac{1}{3}x - \frac{5}{3}\right) \, dx = \int_{1}^{2} \left(-\frac{13}{3}x + \frac{13}{3}\right) \, dx \]
Calculating this integral gives:
\[ A_1 = \left[-\frac{13}{6}x^2 + \frac{13}{3}x\right]_{1}^{2} \]

Calculating \( A_2 \):
\[ A_2 = \int_{2}^{4} \left(\frac{5}{2}x - 7 + 4x - 6\right) \, dx = \int_{2}^{4} \left(\frac{13}{2}x - 13\right) \, dx \]
Calculating this integral gives:
\[ A_2 = \left[\frac{13}{4}x^2 - 13x\right]_{2}^{4} \]

### Step 7: Combine Areas
Finally, sum the areas \( A_1 + A_2 \) to get the total area of the triangle.