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Question
Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x - 2y + 12 = 0.
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Solution 1
The given quations are
`y = (3x^2)/4` ....1
and, 3x – 2y + 12 = 0 ...(2)
Solving equation no. (1) & (2)
y = `(3x + 12)/2`
Solution 2
The equations of the curves are
4y = 3x2 .....(1)
3x – 2y + 12 = 0 .....(2)
The curve (1) represents a parabola having vertex at the origin, axis along the positive direction of y-axis and opens upwards.
The curve (2) represents a straight line. This straight line meets the x-axis at (−4, 0) and y-axis at (0, 6).
Soving (1) and (2), we get
`3x - 2((3x^2)/4) + 12 = 0`
⇒ 6x - 3x2 + 24 = 0
⇒ x2 - 2x−8=0
⇒ (x+2)(x−4) = 0
⇒ x = -2, 4
When x = − 2, y = 3
When x = 4, y = 12
So, the point of intersection of the given curves is (− 2, 3) and (4, 12).
The area bounded by the given curves is shown below.
Required area = Area of the shaded region
= `int_(-2)^4 (y"Line" - y"Parabola") dx`
= `int_(-2)^4 ((3x + 12)/2 - 3/4 x^2) dx`
`= (3/4 x^2 + 6x - x^3/4)^4`
`= (3/4 xx 16 + 6 xx 4 - 64/4) - (3/4 xx 4 +6 xx (-2) + 3/4)`
= 20 - (- 7)
= 27 square units
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