Using integration, find the area of the triangle whose vertices are `(-1, 0) (1, 3)` and (3, 2).

Answer

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

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

### Step 2: Find the equations of the lines forming the triangle
We need to find the equations of the lines connecting the points \(A\), \(B\), and \(C\).

#### Equation of line AB:
Using the two points \(A(-1, 0)\) and \(B(1, 3)\):
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting the values:
\[ y - 0 = \frac{3 - 0}{1 + 1}(x + 1) \]
\[ y = \frac{3}{2}(x + 1) = \frac{3}{2}x + \frac{3}{2} \]

#### Equation of line BC:
Using points \(B(1, 3)\) and \(C(3, 2)\):
\[ y - 3 = \frac{2 - 3}{3 - 1}(x - 1) \]
\[ y - 3 = -\frac{1}{2}(x - 1) \]
\[ y = -\frac{1}{2}x + \frac{7}{2} \]

#### Equation of line AC:
Using points \(A(-1, 0)\) and \(C(3, 2)\):
\[ y - 0 = \frac{2 - 0}{3 + 1}(x + 1) \]
\[ y = \frac{1}{2}(x + 1) = \frac{1}{2}x + \frac{1}{2} \]

### Step 3: Set up the integrals for the area
The area of triangle \(ABC\) can be computed using the formula:
\[ \text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) \, dx \]
where \(f(x)\) is the upper function and \(g(x)\) is the lower function.

1. **Area under line AB from \(x = -1\) to \(x = 1\)**:
\[ \text{Area}_{AB} = \int_{-1}^{1} \left(\frac{3}{2}x + \frac{3}{2}\right) \, dx \]

2. **Area under line AC from \(x = 1\) to \(x = 3\)**:
\[ \text{Area}_{AC} = \int_{1}^{3} \left(-\frac{1}{2}x + \frac{7}{2}\right) \, dx \]

3. **Area under line BC from \(x = -1\) to \(x = 3\)**:
\[ \text{Area}_{BC} = \int_{-1}^{3} \left(-\frac{1}{2}x + \frac{7}{2}\right) \, dx \]

### Step 4: Calculate the integrals
1. **Calculate Area under AB**:
\[ \int_{-1}^{1} \left(\frac{3}{2}x + \frac{3}{2}\right) \, dx = \left[\frac{3}{4}x^2 + \frac{3}{2}x\right]_{-1}^{1} = \left(\frac{3}{4}(1)^2 + \frac{3}{2}(1)\right) - \left(\frac{3}{4}(-1)^2 + \frac{3}{2}(-1)\right) \]
\[ = \left(\frac{3}{4} + \frac{3}{2}\right) - \left(\frac{3}{4} - \frac{3}{2}\right) = \frac{3}{4} + \frac{6}{4} - \frac{3}{4} + \frac{6}{4} = 6/4 = \frac{3}{2} \]

2. **Calculate Area under AC**:
\[ \int_{1}^{3} \left(-\frac{1}{2}x + \frac{7}{2}\right) \, dx = \left[-\frac{1}{4}x^2 + \frac{7}{2}x\right]_{1}^{3} = \left[-\frac{1}{4}(3)^2 + \frac{7}{2}(3)\right] - \left[-\frac{1}{4}(1)^2 + \frac{7}{2}(1)\right] \]
\[ = \left[-\frac{9}{4} + \frac{21}{2}\right] - \left[-\frac{1}{4} + \frac{7}{2}\right] = \left[-\frac{9}{4} + \frac{42}{4}\right] - \left[-\frac{1}{4} + \frac{14}{4}\right] \]
\[ = \frac{33}{4} - \frac{13}{4} = \frac{20}{4} = 5 \]

3. **Calculate Area under BC**:
\[ \int_{-1}^{3} \left(-\frac{1}{2}x + \frac{7}{2}\right) \, dx = \left[-\frac{1}{4}x^2 + \frac{7}{2}x\right]_{-1}^{3} = \left[-\frac{1}{4}(3)^2 + \frac{7}{2}(3)\right] - \left[-\frac{1}{4}(-1)^2 + \frac{7}{2}(-1)\right] \]
\[ = \left[-\frac{9}{4} + \frac{21}{2}\right] - \left[-\frac{1}{4} - \frac{7}{2}\right] = \left[-\frac{9}{4} + \frac{42}{4}\right] - \left[-\frac{1}{4} - \frac{14}{4}\right] \]
\[ = \frac{33}{4} + \frac{15}{4} = \frac{48}{4} = 12 \]

### Step 5: Calculate the total area of triangle ABC
Now, we can find the area of triangle \(ABC\):
\[ \text{Area}_{ABC} = \text{Area}_{AB} + \text{Area}_{AC} - \text{Area}_{BC} = \frac{3}{2} + 5 - 12 \]
\[ = \frac{3}{2} + \frac{10}{2} - \frac{24}{2} = \frac{13}{2} - \frac{24}{2} = -\frac{11}{2} \]
Since the area cannot be negative, we take the absolute value:
\[ \text{Area}_{ABC} = \frac{11}{2} \text{ square units} \]

### Final Answer
The area of the triangle is \( \frac{11}{2} \) square units.