Using integration finds the area of the region bounded by the triangle whose vertices are
Question:

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

Solution:

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)

Equation of line segment AB is

$y-0=\frac{3-0}{1+1}(x+1)$

$y=\frac{3}{2}(x+1)$

$\therefore$ Area $($ ALBA $)=\int_{-1}^{1} \frac{3}{2}(x+1) d x=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]=3$ units

Equation of line segment BC is

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

$y=\frac{1}{2}(-x+7)$

$\therefore$ Area $($ BLMCB $)=\int_{1}^{3} \frac{1}{2}(-x+7) d x=\frac{1}{2}\left[-\frac{x^{2}}{2}+7 x\right]_{1}^{3}=\frac{1}{2}\left[-\frac{9}{2}+21+\frac{1}{2}-7\right]=5$ units

Equation of line segment AC is

$y-0=\frac{2-0}{3+1}(x+1)$

$y=\frac{1}{2}(x+1)$

$\therefore$ Area $($ AMCA $)=\frac{1}{2} \int_{-1}^{3}(x+1) d x=\frac{1}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{3}=\frac{1}{2}\left[\frac{9}{2}+3-\frac{1}{2}+1\right]=4$ units

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 4) = 4 units

 

 

 

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