Using integration find the area of region bounded by the triangle whose vertices are `( 1, 0), (1, 3) a n d (3, 2)`.

Explanation

To find the area of the triangle with vertices at \( (1, 0) \), \( (1, 3) \), and \( (3, 2) \) using integration, we can follow these steps:

### Step 1: Identify the vertices and plot the triangle
The vertices of the triangle are:
- \( A(1, 0) \)
- \( B(1, 3) \)
- \( C(3, 2) \)

Plot these points on the coordinate plane and connect them to form the triangle.

### Step 2: Find the equations of the lines forming the triangle
1. **Line AB** (between points \( A(1, 0) \) and \( B(1, 3) \)):
- This line is vertical, so the equation is \( x = 1 \).

2. **Line BC** (between points \( B(1, 3) \) and \( C(3, 2) \)):
- Using the two-point form of the line equation:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
- Let \( (x_1, y_1) = (1, 3) \) and \( (x_2, y_2) = (3, 2) \):
\[ y - 3 = \frac{2 - 3}{3 - 1}(x - 1) \implies y - 3 = -\frac{1}{2}(x - 1) \]
\[ y - 3 = -\frac{1}{2}x + \frac{1}{2} \implies y = -\frac{1}{2}x + \frac{7}{2} \]

3. **Line AC** (between points \( A(1, 0) \) and \( C(3, 2) \)):
- Again using the two-point form:
\[ y - 0 = \frac{2 - 0}{3 - 1}(x - 1) \implies y = x - 1 \]

### Step 3: Set up the integral for the area
The area \( A \) of the triangle can be calculated using the integral of the upper curve minus the lower curve from \( x = 1 \) to \( x = 3 \):
\[ A = \int_{1}^{3} \left( \text{Upper curve} - \text{Lower curve} \right) \, dx \]
Here, the upper curve is the line BC \( y = -\frac{1}{2}x + \frac{7}{2} \) and the lower curve is the line AC \( y = x - 1 \).

Thus, the area can be expressed as:
\[ A = \int_{1}^{3} \left( \left(-\frac{1}{2}x + \frac{7}{2}\right) - (x - 1) \right) \, dx \]

### Step 4: Simplify the integrand
\[ A = \int_{1}^{3} \left(-\frac{1}{2}x + \frac{7}{2} - x + 1\right) \, dx \]
\[ = \int_{1}^{3} \left(-\frac{3}{2}x + \frac{9}{2}\right) \, dx \]

### Step 5: Compute the integral
Now, calculate the integral:
\[ A = \int_{1}^{3} \left(-\frac{3}{2}x + \frac{9}{2}\right) \, dx \]
\[ = \left[-\frac{3}{4}x^2 + \frac{9}{2}x\right]_{1}^{3} \]
Calculating the limits:
\[ = \left[-\frac{3}{4}(3^2) + \frac{9}{2}(3)\right] - \left[-\frac{3}{4}(1^2) + \frac{9}{2}(1)\right] \]
\[ = \left[-\frac{27}{4} + \frac{27}{2}\right] - \left[-\frac{3}{4} + \frac{9}{2}\right] \]
\[ = \left[-\frac{27}{4} + \frac{54}{4}\right] - \left[-\frac{3}{4} + \frac{18}{4}\right] \]
\[ = \left[\frac{27}{4}\right] - \left[\frac{15}{4}\right] = \frac{12}{4} = 3 \]

### Final Result
Thus, the area of the triangle is \( 3 \) square units.