Find the area between the curves $y=16x^2$ and $y=9$.

Points of intersection: $16x^2=9⟹x^2=9/16⟹x=\pm 3/4$.Area $={\int }_{-3/4}^{3/4}(9-16x^2)dx$ Since the function is even:Area $=2{\int }_0^{3/4}(9-16x^2)dx$ $
$=2[9x-\frac{16x^3}{3}{]}_0^{3/4}=2[9(3/4)-\frac{16}{3}(27/64)]$ $
$$=2[27/4-9/4]=2[18/4]=9\text{ sq. units.}$$