Find the area enclosed by the parabola


Find the area enclosed by the parabola $4 y=3 x^{2}$ and the line $2 y=3 x+12$


The area enclosed between the parabola, $4 y=3 x^{2}$, and the line, $2 y=3 x+12$, 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)

$=\int_{-2}^{1} \frac{1}{2}(3 x+12) d x-\int_{-2}^{1} \frac{3 x^{2}}{4} d x$

$=\frac{1}{2}\left[\frac{3 x^{2}}{2}+12 x\right]_{-2}^{4}-\frac{3}{4}\left[\frac{x^{3}}{3}\right]_{-2}^{4}$




$=27$ units

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