#### Question

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

#### Solution

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)

Is there an error in this question or solution?

Solution Find the Area Enclosed by the Parabola 4y = 3x2 and the Line 2y = 3x + 12 Concept: Area Under Simple Curves.