Area Between Curves That Cross… — AP® Calculus AB Notes

When top and bottom trade places

So far the region between two curves had one curve on top the whole way. Now look at $f(x) = x^3 - 3x$ and $g(x) = x$ in the figure. They cross three times, at $x = -2$ , $0$ and $2$ . Between $-2$ and $0$ the cubic is on top. Between $0$ and $2$ the line is on top. Every crossing swaps the roles.

The cubic f(x) = x³ − 3x and the line g(x) = x cross at (−2, −2), (0, 0) and (2, 2). From −2 to 0 the cubic is on top (pink piece, rectangle height f − g); from 0 to 2 the line is on top (blue piece, rectangle height g − f). The side panel adds the pieces, 4 + 4 = 8, and warns that one integral of f − g from −2 to 2 is 0 because the pieces cancel. — **Figure 1.** The cubic and the line enclose two pieces. Each has area 4, but a single integral of $f - g$ over $[-2, 2]$ is 0. LevlPrep original diagram.

Try one integral anyway:

\int_{-2}^{2} \left(f(x) - g(x)\right)\,dx = \int_{-2}^{2} \left(x^3 - 4x\right)\,dx = \left[\frac{x^4}{4} - 2x^2\right]_{-2}^{2} = (4 - 8) - (4 - 8) = 0.

Zero, for a region you can clearly see. On the right piece, $f - g$ is negative, so its rectangles have negative heights and cancel the left piece. A definite integral of $f - g$ is a net amount: area where $f$ is on top minus area where $g$ is on top. Area itself is never negative, so you have to stop the cancelling.

Splitting the integral

The fix uses a property you already know: integrals over touching intervals add, $\int_a^c$ $= \int_a^b$ $+ \int_b^c$ . Cut the interval at every crossing. On each piece one curve stays on top, so top minus bottom is positive there.

Find every crossing in the interval: solve $f(x) = g(x)$ . Keep only the solutions inside the interval.
Test each piece with one $x$ to see which curve is on top.
Integrate top minus bottom on each piece and add the results.

Doing this by hand is the same as integrating the absolute value of the difference:

\text{Area} = \int_a^b \left|f(x) - g(x)\right|\,dx.

On a calculator, that single setup is the quickest route, because the calculator handles the swaps for you.

Worked example. Find the area between $f(x) = x^3 - 3x$ and $g(x) = x$ .

Crossings. $x^3 - 3x = x$ gives $x^3 - 4x$ $= x(x - 2)(x + 2)$ $= 0$ , so $x = -2, 0, 2$ .

Tests. At $x = -1$ : $f = 2$ and $g = -1$ , so $f$ is on top. At $x = 1$ : $f = -2$ and $g = 1$ , so $g$ is on top.

Pieces.

\int_{-2}^{0} \left(x^3 - 4x\right)\,dx = \left[\frac{x^4}{4} - 2x^2\right]_{-2}^{0} = 0 - (4 - 8) = 4,

\int_{0}^{2} \left(4x - x^3\right)\,dx = \left[2x^2 - \frac{x^4}{4}\right]_{0}^{2} = 8 - 4 = 4.

Area $= 4 + 4 = 8$ .

Net amount or area?

Expression	What it measures	For the example
$\int_{-2}^{2} (f - g)\,dx$	net: $f$ -on-top area minus $g$ -on-top area	$0$
$\left\|\int_{-2}^{2} (f - g)\,dx\right\|$	size of the net amount (pieces already canceled)	$0$
$\int_{-2}^{2} \|f - g\|\,dx$	area	$8$
$\int_{-2}^{0} (f - g)\,dx$ $+ \int_{0}^{2} (g - f)\,dx$	area, split by hand	$8$

The absolute value must go inside the integral. Outside, it comes too late.

Worked example. Find the area between $f(x) = x^2 - 2$ and $g(x) = x$ for $0 \le x \le 3$ .

Crossings. $x^2 - 2 = x$ gives $(x - 2)(x + 1) = 0$ . Only $x = 2$ is inside $[0, 3]$ ; the ends $0$ and $3$ are given by the problem.

Tests. At $x = 1$ : $g = 1 > f = -1$ . At $x = 2.5$ : $f = 4.25 > g = 2.5$ .

Pieces. $\int_0^2 \left(x - x^2 + 2\right)\,dx$ $= 2 - \frac{8}{3}$ $+ 4$ $= \frac{10}{3}$ and $\int_2^3 \left(x^2 - 2 - x\right)\,dx$ $= \frac{11}{6}$ .

Area $= \frac{10}{3} + \frac{11}{6} = \frac{31}{6}$ . One integral of $g - f$ over $[0, 3]$ would give the net value $\frac{3}{2}$ instead.

Traps to avoid

A crossing outside the interval does not make a piece. The interval's ends are limits, not crossings.
Do not assume the crossings are whole numbers. Factor, or use a calculator's solver and store the values.
The same idea works for horizontal strips: split where the right edge and the left edge trade places.
In context, the difference between net and total matters. Between two velocity graphs, $\int (v_1 - v_2)\,dt$ is how far ahead runner 1 ends up; $\int |v_1 - v_2|\,dt$ is the total of the gaps.