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Find the area enclosed between the parab...

Find the area enclosed between the parabola `4y=3x^2` and the straight line `3x-2y+12=0`

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Solving ` y = (3x^(2))/(4)` and `3x - 2y + 12= 0` , we get `x = -2 , 4`
Area = `int_(-2)^(4) [((3x + 12)/(2)) - (3x^(2))/(4)]` dx
`= (3)/(4) [x^(2)]_(-2)^(4) + 6[x]_(-2)^(4) - (3)/(4) [(x^(3))/(3)]_(-2)^(4)`
=27 sq. units
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