Question

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

Answer 1

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)

∴Area(ALBA)=

$\int_{-1}^{1} \frac{3}{2}(x+1) \,dx$

=$\frac{3}{2}$[$\frac{1}{2}$+1-$\frac{1}{2}$+1]= 3units

Equation of the segment BC is

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

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

∴Area(BLMCB)=$\int_{1}^{3} \frac{1}{2}(-x+7) \,dx$=$\frac{1}{2}$[$\frac{-x^2}{2}$+7x]31=$\frac{1}{2}$[$\frac{-9}{2}$+21+$\frac{1}{2}$-7]=5units

Equation of line segment AC is

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

∴Area(AMCA)=$\frac{1}{2}\int_{-1}^{3} (x+1) \,dx$=$\frac{1}{2}$[$\frac{-x^2}{2}$+x]3-1=$\frac{1}{2}$[$\frac{9}{2}$+3-$\frac{1}{2}$+1]=4units

Therefore, from equation(1),we obtain

Area(ΔABC)=(3+5-4)=4units.