Solution
= Solution
The intersections satisfy $6x-x^2=2x+3$, or $(x-1)(x-3)=0$, so the bounded region has $1\leq x\leq3$. Over that interval the <parabola> is above the line, and their height difference is $(x-1)(3-x)$. Thus its <area> is
$$
\boxed{A=\int_1^3(4x-x^2-3)\,dx
=\left[2x^2-\frac{x^3}{3}-3x\right]_1^3=\frac43}.
$$
Using upper minus lower ensures a positive geometric <area>.