Using integration, find the area of the region bounded by the triangle whose vertices are (-1,2),(1,5), and (3,4).

Let A=(-1,2), B=(1,5), C=(3,4)
We have to find the area of ΔABC.
Find equation of line AB → y-2 = (5-2)/(1-(-1))(x-(-1))
y-2 = 3/2(x+1)
2y-4 = 3x+3
3x-2y+7 = 0 --- (1)
y = (3x+7)/2
Equation of line BC → y-5 = (5-4)/(1-3)(x-1)
y-5 = -1/2(x-1)
2y-10 = -x+1
x+2y-11 = 0 --- (2)
y = (11-x)/2
Equation of line AC → y-2 = (2-4)/(-1-3)(x-3)
y-2 = 1/2(x-3)
2y-4 = x-3
x-2y+1 = 0 --- (3)
y = (x+1)/2
So, the required area = ∫-11 ((3x+7)/2)dx + ∫31 ((11-x)/2)dx - ∫3-1 ((x+1)/2)dx
= 1/2[3x²/2 + 7x]-11 + 1/2[11x - x²/2]31 - 1/2[x²/2 + x]3-1
= 1/2[(3/2 + 7) - (3/2 - 7)] + 1/2[(33-9/2) - (11-1/2)] - 1/2[(9/2+3) - (1/2-1)]
= 1/2[14] + 1/2[22.5 - 10.5] - 1/2[7.5 - (-0.5)]
= 1/2[14 + 12 - 8]
= 1/2[18] = 9
Therefore, the area of the triangle is 4 sq. units.