One integral per piece AB & BC

The integrand $f - g$ is a signed quantity. Where $f$ is on top it is positive; where $g$ is on top it is negative. On a region with a single piece that never matters, because the sign is constant and can be fixed by ordering the subtraction once.

Once the curves cross inside the interval, the integrand changes sign there, and a single integral adds a positive contribution to a negative one. That is subtraction of areas rather than addition of them.

Take $y = x^{3}$ and $y = x$. They meet where $x^{3} = x$, that is $x(x^{2} - 1) = 0$, at $x = -1$, $x = 0$ and $x = 1$: three points. On $(-1, 0)$ the cubic is above (at $x = -\frac{1}{2}$ it gives $-\frac{1}{8}$ against $-\frac{1}{2}$); on $(0, 1)$ the line is above. Each piece has area $\frac{1}{4}$, so the region's area is $\frac{1}{2}$.

Now the single integral:

$$\int_{-1}^{1} (x - x^{3})\,dx = 0.$$

The two pieces are congruent and oppositely signed, so they annihilate. A student who sets this up correctly except for the split gets $0$, which is not close to $\frac{1}{2}$ and is not a rounding problem. The setup is what is wrong.

Nothing in 8.4 is discarded. One step is added, and it is the second one.

1. Solve $f(x) = g(x)$ completely. Move everything to one side and factor. A cubic equation can have three roots and the third is the one that gets dropped.
2. Split the interval at every interior root. Three roots make two subintervals; four make three.
3. Order the curves on each piece separately. Test a point strictly inside each one. The ordering alternates in the usual case but is not guaranteed to.
4. Integrate each piece with its own order and add the results. Every term is then non-negative.

Written as one expression, this is

$$A = \int_a^b |f(x) - g(x)|\,dx,$$

and the absolute value bars are an instruction to carry out steps 1 through 4. They are not something to evaluate directly.

Step 1 is where most of these problems are lost, which is why CA8 sits alongside CA7 in this topic. Finding two roots of a cubic and stopping produces a setup with no split in it, and from there the cancellation is automatic.

It is tempting to compute one integral, notice the answer is negative or implausible, and take its absolute value. That works on a one-piece region and fails as soon as there are two, because

$$\left|\int_a^b (f - g)\,dx\right| \neq \int_a^b |f - g|\,dx$$

in general. The left side lets the pieces cancel first and then removes the sign of whatever survived; the right side removes the sign of each piece before adding.

The symmetric example makes the two sides differ by everything: the left is $0$ and the right is $\frac{1}{2}$. An asymmetric example shows the same failure without the tidy zero. The curve $y = x^{3} - x^{2} - 2x$ meets the $x$-axis at $x = -1$, $0$ and $2$, enclosing a piece of area $\frac{5}{12}$ above the axis and a piece of area $\frac{8}{3}$ below it, so the total area is $\frac{37}{12}$. The single integral across $[-1, 2]$ gives $\frac{5}{12} - \frac{8}{3} = -\frac{9}{4}$, and $\frac{9}{4}$ is not $\frac{37}{12}$.

Two numbers, neither of them the area, both produced by correct arithmetic on an incorrect setup. The moment to decide about splitting is before integrating, not after.

The count of intersections is a fact about the equation $f(x) = g(x)$, and it is usually visible before any solving.

1. Degree bounds it. A cubic set equal to a line gives a cubic equation, so up to three crossings. A quartic against a quadratic gives up to four.
2. Trigonometric pairs repeat. $\sin x$ and $\cos x$ cross once per half period, so an interval of length $2\pi$ holds two crossings and a longer one holds more.
3. A repeated root is a touch, not a crossing. If $f - g$ has a double root, the curves meet without the ordering changing, and no split is needed there. $y = x^{2}$ and $y = x^{3}$ meet at $x = 0$ with a double root and at $x = 1$ simply, so only $x = 1$ separates regions.

That last case is worth a moment, because splitting at a touch point is harmless while failing to split at a crossing is fatal. When in doubt, split: an unnecessary split adds two integrals that sum to the same total, while a missing one produces cancellation.