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