APPLICATION OF INTEGRALS — Maths STD 12 Science — Question

Consider the points A(2, 1), B(3, 4) and C(5, 2)

We need to find area of shaded triangle ABC

Equation of AB is 

$\text{y}-1=\Big(\frac{4-3}{3-2}\Big)(\text{x}-2)$

$\Rightarrow\text{x}-3\text{y}-5=0\ ...(\text{i})$

Equation of BC is 

 $\text{y}-4=\Big(\frac{2-4}{5-3}\Big)(\text{x}-3)$

$\Rightarrow\text{x}=\text{y}-7=0\ ...(\text{ii})$

Equation of CA is

 $\text{y}-2=\Big(\frac{2-1}{5-2}\Big)(\text{x}-5)$

$\Rightarrow\text{x}-3\text{y}+2=0\ ...(\text{iii})$

Area of $\triangle\text{ABC}= \text{Area of}\ \triangle\text{ABC}+ \text{Area of }\triangle\text{ABC}$ in the, 

Consided point P(x₁, y₂) on AB and Q(x₁, y₁) on AD

Thus, the area of appoximating with length = |y₂ - y₁| from x = 2, to x = 3

$\therefore$ Area of $$$\triangle\text{ABC}=\int\limits_{2}^{3}|\text{y}_{2}-\text{y}_{1}|\text{dx} $

$\Rightarrow\text{A}=\int\limits_{2}^{3}(\text{y}_{2}-\text{y}_{1})\text{dx} $

$\Rightarrow\text{A}=\int\limits_{2}^{3}(3\text{x}-5)-\Big(\frac{\text{x}-1}{3}\Big)\text{dx}$

$\Rightarrow\text{A}=\int\limits_{2}^{3}\frac{(9\text{x}-15-\text{x}-1)}{3}\text{dx}$

$\Rightarrow\text{A}=\int\limits_{2}^{3}\frac{(8\text{x}-16)}{3}\text{dx}$

$\Rightarrow\text{A}=\frac{1}{3}\Big[8\text{x}^{2}-16\text{x}\Big]\text{dx}$

$\Rightarrow\text{A}=\frac{1}{3}(68-64)$

$\Rightarrow\text{A}=\frac{4}{3}\ \text{sq.}\ \text{units}$