Using integration find the area of region bounded by the triangle whose vertices are $(1, 0)$, $(2, 2)$ and $(3, 1)$.

$\frac{3}{2}$

The correct answer is Option (2) → $\frac{3}{2}$

Let $A(1, 0)$, $B(2, 2)$ and $C(3, 1)$ be the vertices of a triangle $ABC$ (in Fig).

$\text{Area of } \Delta ABC = \text{Area of } \Delta ABD + \text{Area of trapezium } BDEC - \text{Area of } \Delta AEC$

Now equation of the sides $AB, BC$ and $CA$ are given by

$y = 2(x - 1), \quad y = 4 - x, \quad y = \frac{1}{2}(x - 1), \text{ respectively.}$

Hence, $\text{area of } \Delta ABC = \int\limits_{1}^{2} 2(x - 1) \, dx + \int\limits_{2}^{3} (4 - x) \, dx - \int_{1}^{3} \frac{x - 1}{2} \, dx$

$= 2 \left[ \frac{x^2}{2} - x \right]_{1}^{2} + \left[ 4x - \frac{x^2}{2} \right]_{2}^{3} - \frac{1}{2} \left[ \frac{x^2}{2} - x \right]_{1}^{3}$

$= 2 \left[ \left( \frac{2^2}{2} - 2 \right) - \left( \frac{1^2}{2} - 1 \right) \right] + \left[ \left( 4 \times 3 - \frac{3^2}{2} \right) - \left( 4 \times 2 - \frac{2^2}{2} \right) \right] - \frac{1}{2} \left[ \left( \frac{3^2}{2} - 3 \right) - \left( \frac{1^2}{2} - 1 \right) \right]$

$= \frac{3}{2}$