Question

Find the area of the region included between the parabola y = `(3x^2)/4` and the line 3x – 2y + 12 = 0.

Answer 1

Solving the equations of the given curves y = `(3x^2)/4` and 3x – 2y + 12 = 0

We get 3x² – 6x – 24 = 0

⇒ (x – 4)(x + 2) = 0

⇒ x = 4, x = –2

Which give y = 12, y = 3

From Fig.8.6, the required area = area of ABC

= `int_(-2)^4 ((12 + 3x)/2)"d"x - int_(-2)^4 (3x^2)/4  "d"x`

= `(6x +  (3x^2)/4)_-2^4 - |(3x^3)/12|_-2^4`

= 27 sq.units