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.