Topic 8.6 Finding the Area Between Curves That Intersect at More Than Two Points: split the area at each crossing where the top and bottom curves swap.

What this topic is asking

The College Board (Topic 8.6) handles areas where the two curves cross more than twice, so the upper and lower curves swap along the interval. You must split the integral at each crossing and integrate top minus bottom on each piece, keeping every contribution positive.

The method

A worked split area

Why a single signed integral fails

If you integrate $f - g$ over the whole interval without splitting, the regions where $f$ is on top contribute positively and those where $g$ is on top contribute negatively, so they partly cancel. For $y = x^3$ and $y = x$ on $[-1, 1]$, the single integral $\int_{-1}^1 (x^3 - x)\,dx = 0$ by symmetry, which is plainly not the area. The total area is $\frac{1}{2}$, obtained only by splitting at $x = 0$ and adding magnitudes. This cancellation is the entire reason the topic exists: "area" means the positive total, so each piece must be made positive before summing.

Finding all the crossings

The reliability of the method rests on finding every intersection inside the interval, not just the endpoints. A missed crossing leaves a subinterval where the top curve is misidentified, flipping the sign of that contribution. Solve $f(x) = g(x)$ completely (factor fully, or use the calculator on the calculator-active part), list the crossings in order, and treat each consecutive pair as a separate piece. On the no-calculator section the crossings usually factor nicely; on the calculator section you find them numerically. Either way, splitting at all of them and adding positive pieces gives the correct total area.

Using the absolute value on the calculator

On the calculator-active part you can compute the total area directly as $\int_a^b |f(x) - g(x)|\,dx$, letting the calculator handle the sign changes numerically without your splitting the interval by hand. This is fast and reliable when the crossings are messy, but you should still set up the absolute-value integral explicitly so the grader sees the correct expression, since the setup carries the marks. On the no-calculator part you cannot enter an absolute value into an antiderivative, so there you must find the crossings, split, and add magnitudes by hand. Knowing which approach the section allows, numerical absolute value versus hand-splitting, keeps you efficient.

Why this matters for area versus net signed area

This topic clarifies a distinction that runs through Unit 8: total area (always positive) versus net signed area (the plain integral, which can cancel). The definite integral $\int_a^b (f - g)\,dx$ computes net signed area and is the right tool when the question genuinely wants a net quantity, such as net displacement. But when the question asks for the area of a region, every piece must count positively, which requires splitting at sign changes or integrating the absolute difference. Reading the question to decide which is wanted, and applying the splitting only when total area is required, prevents both under-counting area and over-correcting a net-change question.