Question

Find the area enclosed between the parabola 4y = 3x2 and the line 3x - 2y + 12 = 0.

Answer 1

The given curve is 4y = 3x2

⇒ x2 - 2x - 8= 0 ⇒ (x + 2)(x - 4) = 0 ⇒ x = -2, x = 4

∴ The point of intersection are A(-2, 3) and B (4,12)

= (12 - 16 + 24) - (3 + 2 - 12) A

= 20 + 7 = 27 sq. units