Area in Between Two Curves: Top Minus Bottom, Intersections, Splitting

Scale the engine room down to a picture any calculus student can draw. The line $y = 2x$ climbs steadily; the parabola $y = x^2$ starts slower, then catches up. They meet at $(0, 0)$ and $(2, 4)$, trapping one region — the area in between two curves.

The picture shows the whole method for the area in between two curves. Shade the region, slice it into vertical strips, and let each strip run from the bottom curve up to the top curve. The area in between two curves is the sum of those strips — the entire agenda, and every area in between two curves problem starts here.

Start with numbers. At $x = 1.2$ in the figure, the top curve $y = 2x$ gives $2.4$; the bottom curve $y = x^2$ gives $1.44$. Strip height: $2.4 - 1.44 = 0.96$ — the raw material of the area in between two curves.

In words: area = (top − bottom) × strip width, summed. As $dx$ shrinks to zero, the sum becomes an integral. With $f(x)$ on top, $g(x)$ below, and crossings at $a$ and $b$, the area in between two curves is

$A = \int_a^b \left[f(x) - g(x)\right]dx$

Todhunter's 1889 text states it nearly verbatim — “Let $y = \phi(x)$ be the equation to the upper curve, and $y = \psi(x)$ the equation to the lower curve” — giving $A = \int \left[\phi(x) - \psi(x)\right]dx$ (Article 138). Granville's 1911 calculus says it with a strip: area $(y' - y'')\,dx$, top ordinate (the y-value) minus bottom ordinate. Every method for the area in between two curves is that one subtraction.

Check the mini-example: $\int_0^2 (2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}$ — the shaded area in between two curves above really is $\frac{4}{3}$.

Every AP Calculus question on the area in between two curves runs on four steps, in the order Granville's worked examples use.

1. Find the intersections. Solve $f(x) = g(x)$; the solutions are the limits of the area in between two curves. Granville: “to determine the limits of integration we solve the equations simultaneously to find where the curves intersect.”
2. Decide which curve is on top on each piece. Test one x-value between crossings, or sketch the area in between two curves.
3. Integrate top minus bottom — the heart of every area in between two curves computation: $\int \left[f(x) - g(x)\right]dx$ between consecutive crossings.
4. Add the pieces. If the curves trade places, the area in between two curves arrives piecewise: compute each piece, add the positive results.

Follow the checklist and finding the area between curves is pure discipline — that is how to find area between two curves without wasted work. Skip step 1 and no integration will save the area in between two curves.

Why subtract? A strip between the curves is a thin rectangle — height (top − bottom), width $dx$; Granville writes its area as $(y' - y'')\,dx$. Integrating adds the strips into the area in between two curves. The subtraction also cancels everything below the bottom curve, so the x-axis never appears.

When to split? Areas between two curves that cross mid-interval demand it: on $[0, 1]$ one curve is on top, on $[1, 2]$ the other. Integrate each piece with its own top and bottom, then add: the area bounded by two curves stays positive piece by piece. Crossings are the top source of lost points on area in between two curves problems, and calculating area between curves piecewise is the fix.

Sideways strips: given $x = f(y)$ and $x = g(y)$ (parabolas like $y^2 = 4x$), slice horizontally: $A = \int \left[f(y) - g(y)\right]dy$ — Todhunter's Article 139 companion. Same rule, rotated 90 degrees, still the area in between two curves.

Problem. Find the area of the region enclosed by $y = 2x$ and $y = x^2$ — the most common area in between two curves pairing in textbooks.

Step 1 — intersect. $2x = x^2$ becomes $x(x - 2) = 0$, so $x = 0$ or $x = 2$: crossings at $(0, 0)$ and $(2, 4)$, integration from 0 to 2.

Step 2 — top and bottom. Test $x = 1$: line 2, parabola 1. The line stays on top — the typical area in between two curves picture.

Step 3 — integrate. $A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.

Answer. The area in between two curves is $\frac{4}{3} \approx 1.333$ square units. (Granville's 1911 set, Ex. 18 p. 369, builds this pair into a three-curve problem.)

Problem. Find the area of the region bounded by $y = x^2$, $y = x$, and $y = 2x$ — a three-curve area in between two curves problem. (Granville 1911, Ex. 18 p. 369; printed answer $\frac{7}{6}$.)

Find every crossing. $y = x$ meets $y = x^2$ at $x = 0, 1$; $y = 2x$ meets $y = x^2$ at $x = 0, 2$; $y = x$ meets $y = 2x$ only at the origin. The region runs from $x = 0$ to $x = 2$ with $y = 2x$ on top throughout.

Watch the bottom. For $0 \le x \le 1$ the bottom boundary is $y = x$; for $1 \le x \le 2$ it is $y = x^2$. At $x = 1$ the bottom switches — so split the area in between two curves into two integrals:

$A = \int_0^1 (2x - x)\,dx + \int_1^2 (2x - x^2)\,dx = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}$

Answer. The area in between two curves totals $\frac{7}{6}$ square units, matching the 1911 printed answer.

Problem. Find the area enclosed by the parabolas $y^2 = 4x$ and $x^2 = 4y$ — a classic area in between two curves problem still asked today. (Murray, An Elementary Course in the Integral Calculus, 1898, Ch. IV Ex. 3.)

Intersect. From $y^2 = 4x$ take the upper branch $y = 2\sqrt{x}$. Setting $2\sqrt{x} = \frac{x^2}{4}$ gives $x = 0$ and $x = 4$: the parabolas meet at $(0, 0)$ and $(4, 4)$, framing the area in between two curves.

Top minus bottom. The right-opening branch $y = 2\sqrt{x}$ is on top, the upward branch $y = \frac{x^2}{4}$ below — so the area in between two curves is

$A = \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right)dx = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$

Answer. The area in between two curves is $\frac{16}{3}$ square units.

1. Integrating before finding intersections. The limits are the crossings — solve $f(x) = g(x)$ first. Wrong limits are the most common way the area in between two curves goes wrong.

2. Subtracting bottom minus top. A negative result means the curves are inverted; Murray's 1898 text warns that areas below the axis carry a negative sign. Top minus bottom keeps the area in between two curves positive on every piece.

3. Assuming one curve stays on top. Curves that cross demand a split. In Example 2 the bottom switched from $y = x$ to $y = x^2$ at $x = 1$; integrating straight across it in one step returns a wrong area in between two curves.

4. Using x-limits on a y-integral. Horizontal strips pair with y-limits (Todhunter's Article 139 companion); mixing variables silently ruins the area when calculating area between curves.

5. Missing a third boundary curve. The area in between two curves can be stitched from three curves, as in Example 2. Audit the top and bottom of every piece before integrating.