Class 12

Math

Calculus

Application of Integrals

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

It can be observed in the following figure that,

$Area(ΔACB)=Area(ALBA)+Area(BLMCB)−Area(AMCA)$ ......... (1)

Equating of line segment $AB$ is

$y−0=1+13−0 (x+1)$

$y=23 (x+1)$

$∴Area(ALBA)=∫_{−1}23 (x+1)dx=23 [2x_{2} +x]_{−1}=23 [21 +1−21 +1]=3$sq. units

Equating of line segment $BC$ is

$y−3=3−12−3 (x−1)$

$y=21 (−x+7)$

Equating of line segment $BC$ is

$y−3=3−12−3 (x−1)$

$y=21 (−x+7)$

$∴Area(BLMCB)=∫_{1}21 (−x+7)dx=21 [−2x_{2} +7x]_{1}=21 [−29 +21+21 −7]=5$sq. units

Equation of line segment $AC$ is

$y−0=3+12−0 (x+1)$

$y=21 (x+1)$

Equation of line segment $AC$ is

$y−0=3+12−0 (x+1)$

$y=21 (x+1)$

$∴Area(AMCA)=21 ∫_{−1}(x+1)dx=21 [2x_{2} +x]_{−1}=21 [29 +3−21 +1]=4$sq.units

Therefore, from equation (1), we obtain

$Area(ΔABC)=(3+5−4)=4$ sq. units

$Area(ΔABC)=(3+5−4)=4$ sq. units