Area between graph and $x$-axis uses $x$-limits, integrand $y$. Integrate, then sub upper minus lower: $F(b)-F(a)$. E.g. $\int_0^3[(x-2)^2+3]dx=12$ units².

Reflex: With the x-axis: y dx and x-limits; F(b) − F(a).

Below the axis, area integrates negative (e.g. −9 → area 9). Integrating straight across a root cancels to 0 instead. Reflex: Find the roots first; one integral per sign-region; add absolute values.

Rearrange the equation to make $x$ the subject, then integrate using $y$-limits. Ex: $y^2=3x-2 \to x=\frac{y^2+2}{3}$; $\int_0^4 = \frac{88}{9}$ units². Reflex: y-axis → x dy → rearrange to x = f(y) → y-limits.

Limits = the intersections, so solve the two equations simultaneously first. Reflex: Intersect → identify top graph → integrate the difference.