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.