In Evaluating Integrals with FTC Part 2 you found $\int_a^b f(x)\,dx$, the signed area between a curve and the $x$-axis. Now the region sits between two curves, $y = f(x)$ on top and $y = g(x)$ underneath. The $x$-axis was just the special curve $g(x) = 0$.
Why it works
Cut the region into thin vertical strips of width $\Delta x$. A strip at $x$ runs from the bottom curve up to the top curve, so its height is $f(x) - g(x)$ and its area is about $[f(x) - g(x)]\,\Delta x$. Adding the strips gives a Riemann sum, and as $\Delta x \to 0$ it becomes an integral:
$$\begin{aligned} A &= \lim_{n\to\infty}\sum_{i=1}^{n} \left[f(x_i) - g(x_i)\right]\Delta x \\ &= \int_a^b \left[f(x) - g(x)\right]dx. \end{aligned}$$
Only the difference matters. If both curves sit below the axis, or one above and one below, the strip height is still top minus bottom, which is positive.
Key idea
If $f(x) \ge g(x)$ on $[a, b]$, the area between them is
$$A = \int_a^b \left[f(x) - g(x)\right]dx$$ (top minus bottom).
If the curves cross, split the interval at each crossing so that every piece is top minus bottom. In one line: $A = \int_a^b \left|f(x) - g(x)\right|dx$.
- Find where the curves meet: solve $f(x) = g(x)$. These are often the limits $a$ and $b$.
- Pick a test point in each piece to see which curve is on top.
- Integrate top minus bottom on each piece and add the results.
Worked example: A region enclosed by two curves
Find the area enclosed by $y = x + 2$ and $y = x^{2}$.
- Meeting points: $x^{2} = x + 2$, so $x^{2} - x - 2 = (x - 2)(x + 1) = 0$, giving $x = -1$ and $x = 2$.
- Test $x = 0$: the line gives 2 and the parabola gives 0, so the line is on top.
- $$\begin{aligned} A &= \int_{-1}^{2} (x + 2 - x^{2})\,dx \\ &= \left[\frac{x^{2}}{2} + 2x - \frac{x^{3}}{3}\right]_{-1}^{2}. \end{aligned}$$
- At $2$: $2 + 4 - \frac{8}{3} = \frac{10}{3}$. At $-1$: $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$.
- $A = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.
Answer: $\frac{9}{2}$ square units
Worked example: Curves that cross
Find the total area between $y = x^{2}$ and $y = 2x$ from $x = 0$ to $x = 3$.
- They cross where $x^{2} = 2x$: at $x = 0$ and $x = 2$. So split at $x = 2$.
- On $[0, 2]$ the line is on top: $$\begin{aligned} &\int_0^2 (2x - x^{2})\,dx \\ &= \left[x^{2} - \frac{x^{3}}{3}\right]_0^2 \\ &= 4 - \frac{8}{3} \\ &= \frac{4}{3}. \end{aligned}$$
- On $[2, 3]$ the parabola is on top: $$\begin{aligned} &\int_2^3 (x^{2} - 2x)\,dx \\ &= \left[\frac{x^{3}}{3} - x^{2}\right]_2^3 \\ &= 0 - \left(-\frac{4}{3}\right) \\ &= \frac{4}{3}. \end{aligned}$$
- Total: $\frac{4}{3} + \frac{4}{3} = \frac{8}{3}$.
Answer: $\frac{8}{3}$ square units
Common mistake
Writing one integral across a crossing point. Here $\int_0^3 (x^{2} - 2x)\,dx = 0$: the two pieces have equal size and opposite signs, so they cancel and the "area" vanishes. An area can never be zero or negative when there is a region there. Quick check: before integrating, make sure the integrand is $\ge 0$ on every piece; if it changes sign, split.
Integrating with respect to y
Some regions are easier to slice sideways. If the region lies between a left curve $x = g(y)$ and a right curve $x = f(y)$ for $c \le y \le d$, use horizontal strips of height $\Delta y$ and length right minus left: $$A = \int_c^d \left[f(y) - g(y)\right]dy.$$
Worked example: Right minus left
Find the area enclosed by $x = y^{2}$ and $x = y + 2$.
- They meet where $y^{2} = y + 2$, at $y = -1$ and $y = 2$.
- At $y = 0$ the line gives $x = 2$ and the parabola $x = 0$, so the line is on the right.
- $A = \int_{-1}^{2} (y + 2 - y^{2})\,dy = \frac{9}{2}$, the same arithmetic as the first example.
- That is no accident: this region is the first one reflected across the line $y = x$. Slicing in $x$ here would need two integrals, because the bottom boundary changes at $x = 1$.
Answer: $\frac{9}{2}$ square units