Using integration find the area of the triangle ABC whose vertices are A(3,0), B(4,6) and C(6,2)

Hello student,

i)Area of an enclosed region bounder by the curve y = f(x), x-axis and the boundaries,x = a to b is given by A = ∫f(x) dx in [x = a to b]
ii) Hence, here the area of the triangle ABC is enclosed by the lines AB, BC & CA; its area byintegrationis given by Area under AB + Area under BC - Area under AC
iii) Using two point form equation of AB, Bc & CA are respectively:
y = 6(x -3); y-6 = -2(x-4) and y = (2x - 6)/3
iv) Area under AB = ∫(6x - 18) dx in [3 to 4] = ([6x²/2 - 18x] in [3 to 4]
Evaluating for the limits, A1 = 3
Similarlyarea under BC = (14x - x²) in [4 to 6] = 8
Area under AC = (x²/3 - 2x) in [3 to 6] =3
Hence net area = 3+ 8 - 3 = 8 sq units.