Top minus bottom, and test which is which AB & BC

Slice the region with a thin vertical strip at position $x$. Its height is the distance between the two boundaries there, and its width is $dx$, so its area is

$$\big(y_{\text{top}} - y_{\text{bottom}}\big)\,dx.$$

Adding the strips from the left edge of the region to the right edge gives

$$A = \int_a^b \big(f(x) - g(x)\big)\,dx, \qquad f \text{ on top}.$$

Two things follow immediately, and they are the whole topic. The limits $a$ and $b$ are $x$-values, because the strip moves in the $x$ direction. And the integrand is a distance, so it must be non-negative across the interval, which is what "$f$ on top" is doing in that formula.

Nothing here assumes either curve is above the $x$-axis. A region sitting entirely below the axis has exactly the same integrand, because the vertical distance between two curves does not care where the axis is. Subtracting handles the axis automatically, which is why this generalises Unit 6's area-under-a-curve rather than replacing it.

Unless the problem states the interval, the limits come from solving $f(x) = g(x)$, and the word that matters is every.

1. Set the two expressions equal and solve completely. A quadratic has two roots, and stopping at the first is the most common way to lose an interval.
2. Discard nothing without a reason. A negative root is still a root; the region does not have to live in the first quadrant.
3. Do not import numbers from the prompt. A problem mentioning $y = 4$ or the interval $[0, 3]$ has not thereby given you a limit of integration.

For $y = x + 2$ and $y = x^{2}$, solving $x^{2} = x + 2$ gives $x^{2} - x - 2 = 0$, so $(x - 2)(x + 1) = 0$ and the region runs from $x = -1$ to $x = 2$. Taking $0$ as the left limit, because it is where the parabola bottoms out or because the picture looks like it starts there, replaces the true area $\frac{9}{2}$ with $\frac{10}{3}$.

The same trap wears a different costume when a curve is bounded by a horizontal line. The region enclosed by $y = x^{2}$ and $y = 4$ runs from $x = -2$ to $x = 2$: two intersections, symmetric about the axis, and integrating only the right half gives exactly half the area.

Which curve is on top is a fact about an interval, and the reliable way to establish it takes one substitution.

1. Pick a value strictly between two consecutive intersections. Not an endpoint, where the curves agree.
2. Evaluate both functions there. The larger output is the top curve on that whole subinterval.
3. Subtract in that order. Once, for the whole subinterval.

Between $x = -1$ and $x = 2$, testing $x = 0$ gives $0 + 2 = 2$ for the line and $0$ for the parabola, so the line is on top and the integrand is $(x + 2) - x^{2}$.

What does not work is reasoning from the shape of the formulas. "The parabola grows faster, so it is on top" is true for large $x$ and false on the interval that matters. Take $y = x^{2}$ and $y = 2x$ on $[0, 2]$: at $x = 1$ the line gives $2$ and the parabola gives $1$, so the line is above throughout, even though the parabola overtakes it forever after $x = 2$. The heuristic is not merely unreliable, it is systematically wrong on bounded regions, because the region is precisely the interval where the ordering is reversed from its eventual one.

An area is never negative, so a negative answer is a diagnosis rather than a result.

If $\int_a^b (f - g)\,dx$ comes out negative and $f$ and $g$ do not cross inside $(a, b)$, then $g$ was on top and the subtraction ran the wrong way. The repair is to swap the order, not to attach an absolute value to the final number and move on: the two happen to agree here only because there is a single subinterval.

That distinction becomes real in 8.6. Once the curves cross inside the interval, $\left|\int (f - g)\right|$ and $\int |f - g|$ part company, and only the second is the area. Building the habit now, on regions with one piece, is what makes the multi-crossing case routine later.

One more consequence worth stating. Because the integrand is a distance, $\int_a^b (f - g)\,dx$ and $\int_a^b (g - f)\,dx$ differ only in sign, so a negative result carries the right magnitude. The number is not wrong; the report is. Writing $-\frac{9}{2}$ as an area is the error, and it is the one CA5 names.