Q.302. Area between a parabola and a line.

Question 302.
Find the area cut off form the parabola 4y=3x^2 by the line 2y=3x +12.

Answer 302.
Solving 4y=3x^2 by the line 2y=3x +12,
3x^2 - 6x - 24 = 0
=> x^2 - 2x - 8 = 0
=> (x + 2)(x - 4) = 0
=> x = -2 or 4

=> required area
= ∫ [(1/2) (3x + 12) - (3/4) x^2] dx ... [x = - 2 to 4]
= (1/2) [3x^2/2 + 12x] - (3/4) (x^3/3) ... [x = - 2 to 4]
= [(3/4) x^2 + 6x - (1/4)x^3] ... [x = - 2 to 4]
= (1/4) (3x^2 + 24x - x^3) ... [x = - 2 to 4]
= (1/4) [(48 + 96 - 64) - (12 - 48 + 8)]
= (1/4) (80 + 28)
= 27 sq. units.

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