Find the area enclosed by the parabola `4y=3x^2` and the line `2y=3x+12.`

Explanation

`4y=3x^2` and `3x-2y+12=0`

or`4((3x+12)/2)=3x^2`

`=3x^2-6x-24=0`

`=x^2-2x-8=0`

`=(x-4)(x+2)=0`

x-coordinates of points of intersection are`x=2,x=4`

Area(A)=`int_-2^4[1/2(3x+12)-3/4x^2]dx`

`=[1/2(3x+12)^2/6-3/4.x^3/3]_-2^4`

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