Area Between Two Curves & Intersection Points

To find the points of intersection of $y = f(x)$ and $y = g(x)$:

• Set them equal: solve $f(x) = g(x)$. Each solution $x$ gives one intersection; substitute back for the matching $y$.
• The smallest and largest solutions become the limits $a$ and $b$ of the area integral.
• With a modulus present, split by sign or square carefully: e.g. $x^2 = 2|x|$ gives $|x|(|x| - 2) = 0$, so $|x| = 0$ or $2$ — the points $(0,0)$, $(2, 4)$, $(-2, 4)$, i.e. 3 intersections.

For two curves with $f(x) \ge g(x)$ on $[a, b]$ (so $f$ is the upper curve):

• Decide top vs bottom by testing one point between the crossings (or by sketching). On $[0, 1]$, $y = 2x - x^2$ lies above $y = 0$; $y = x$ lies above $y = x^3$.
• The result is the same whether the region is above or below the axis — only the relative position of the two curves matters.

For a sideways parabola $y^2 = 2x$ and a line $y = x$:

• Intersections: substitute $y = x$ into $y^2 = 2x$: $x^2 = 2x \Rightarrow x = 0, 2.$
• The parabola's upper branch $y = \sqrt{2x}$ lies above the line $y = x$ on $(0, 2)$:

• For two lines $y = x$ and $y = mx$ (with $m < 0$) up to $x = c$, the strip height is $(x - mx)$, giving area $(1 - m)\tfrac{c^2}{2}$ — a tidy way to solve for an unknown slope.

For an awkward region, decompose into known areas:

• Subtract: the region inside a quarter-circle but above $y = \sin x$ is (quarter-circle area) − (area under $\sin x$). For $x^2 + y^2 = \pi^2$ in the first quadrant minus $y = \sin x$ on $[0, \pi]$: $A = \tfrac{\pi^3}{4} - \int_0^{\pi}\sin x\,dx = \tfrac{\pi^3}{4} - 2.$
• Use sectors: a region cut by a line through the origin and an arc can equal a circular sector, area $\tfrac{1}{2}r^2\theta$.
• Horizontal strips: when two curves are easier as $x$ in terms of $y$ (two sideways parabolas), integrate $\int (x_{\text{right}} - x_{\text{left}})\,dy$ instead.