Question

Find the area enclosed by the parabola 4y=3x2 and the line 2y=3x+12

Answer 1

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 OACO)

=

\int_{-2}^{1} \frac12(3x+12) \,dx$$$$-\int_{-2}^{1} \frac{3x^2}{4} \,dx

=$\frac12$[$\frac{3x^2}{2}$+12x]4-2-$\frac34$[x3/3]4-2

=$\frac12$[24+48-6+24]-$\frac14$[64+8]

=$\frac12$[90]-$\frac14$72]

=45-18

=27units.