Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

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 by integration is given by Area under AB + Area under BC - Area under AC
iii) Using two point form equation of AB, Bc & CA are respectively:
y = (5x - 18)/2; y = (12 - x) and y = (3x - 8)/4
iv) Area under AB = ∫(5x - 18)/2 dx in [4 to 6] = (1/2[5x²/2 - 18x] in [4 to 6]
Evaluating for the limits, A1 = (1/2)[(90 - 108) - (40 - 72)] = 7
Similarly area under BC = (12x - x²/2) in [6 to 8] = 10
Area under AC = (1/4)(3x²/2 - 8x) in [4 to 8] = 10
Hence net area = 7 + 10 - 10 = 7 sq units.

Thanks and Regards

Shaik Aasif

askIITians faculty