Find the area enclosed by the parabola 4y= 3x2and the line 2y= 3x+ 12

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The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as The points of intersection of the given curves are A (–2, 3) and (4, 12). We draw AC and BD perpendicular to x-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area...
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The area enclosed between the parabola, 4y= 3x2, and the line, 2y= 3x+ 12, is represented by the shaded area OBAO as The points of intersection of the given curves are A (–2, 3) and (4, 12). We draw AC and BD perpendicular tox-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO) read less
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Nothing is impossible.Hard Work beats everything.

Let, f(x):2Y=3X+12 , Y=(3X+12)/2 implies equation 1. ,g(x):4Y=3X^2 , Then,Y=(3X^2)/4 implies equation 2. By comparing eq.1 and eq.2,we get (3X^2)/4=(3X+12)/2 Then do cross multiplication,i .e. 2×(3X^2)=4×(3X+12) 6X^2=12X+48 6X^2-12X-48=0 6(X^2-2X-8)=0 X^2-2X-8=0 implies equation 3. X^2-4X+2X-8=0 X(X-4)+2(X-4)=0 (X-4)(X+2)=0 X=...
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Let, f(x):2Y=3X+12 , Y=(3X+12)/2 implies equation 1. ,g(x):4Y=3X^2 , Then,Y=(3X^2)/4 implies equation 2. By comparing eq.1 and eq.2,we get (3X^2)/4=(3X+12)/2 Then do cross multiplication,i .e. 2×(3X^2)=4×(3X+12) 6X^2=12X+48 6X^2-12X-48=0 6(X^2-2X-8)=0 X^2-2X-8=0 implies equation 3. X^2-4X+2X-8=0 X(X-4)+2(X-4)=0 (X-4)(X+2)=0 X= 4 or -2.implies eq.4. Putting this value in eq.1&eq.2,we have Y=3 or 12. Implies eq.5. Area covered by parabola and line will be= \int [f(x)-g(x)]dx from limit a to b =\int [(3X+12)/2]dx - \int [(3X^2)/4]dx =[(3X^2)/4+(6X)-(3X^3)/12] from limit -2 to 4On applying limit,we getArea=[1/2(24+48-6+24)-1/4(64+8)]Area=[1/2(90)-1/4(72)]Area=45-18= 27 sq.units read less
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27 sq. units.
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