Problem Statement
Using integration, find the area bounded by the triangle with vertices $(-1, 0), (1, 3)$ and $(3, 2)$.
Verified Solution & Marking Scheme
Line Equations
Side $AB$: $y = \frac{3}{2}(x + 1)$. Side $BC$: $y = -\frac{1}{2}x + \frac{7}{2}$. Side $AC$: $y = \frac{1}{2}(x + 1)$.
Integrate Sides
$\int_{-1}^{1} \frac{3}{2}(x + 1) dx = 3$. $\int_{1}^{3} (-\frac{1}{2}x + \frac{7}{2}) dx = 5$. $\int_{-1}^{3} \frac{1}{2}(x + 1) dx = 4$.
Total Area
$\text{Area} = 3 + 5 - 4 = 4 \text{ square units}$