The area between two curves $y = f(x)$ (top) and $y = g(x)$ (bottom) from $x = a$ to $x = b$ is:
$$A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx \quad \text{where } f(x) \ge g(x) \text{ on } [a,b]$$
If the curves switch which is on top, split the integral at the crossing point(s) and handle each piece separately.
Find the area enclosed by $y = x^2$ and $y = x + 2$.
Step 1 — Find intersection points: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$
Intersections at $x = -1$ and $x = 2$.
Step 2 — Determine which is on top: At $x = 0$: $f(0) = 2$ (line) and $g(0) = 0$ (parabola). So $y = x+2$ is on top on $[-1, 2]$.
Step 3 — Integrate:
$$A = \int_{-1}^{2} \bigl[(x+2) - x^2\bigr]\,dx = \int_{-1}^{2}(x + 2 - x^2)\,dx$$
$$= \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}$$
$$= \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \mathbf{\frac{27}{6} = \frac{9}{2}}$$
Interactive: Area between $y = x + 2$ and $y = x^2$ — the shaded region has area 9/2
Figure 7.1 — Area Between $y = x^2$ and $y = x + 2$
Find the area enclosed by $y = 4 - x^2$ and $y = x^2 - 4$.
Find the total area between $f(x) = \sin x$ and $g(x) = 0$ on $[0, 2\pi]$.
On $[0,\pi]$: $\sin x \ge 0$, so area = $\int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 2$
On $[\pi, 2\pi]$: $\sin x \le 0$, so area = $\int_\pi^{2\pi}|\sin x|\,dx = \int_\pi^{2\pi}(-\sin x)\,dx = [\cos x]_\pi^{2\pi} = 2$
Total area = 2 + 2 = 4
Note: $\int_0^{2\pi}\sin x\,dx = 0$ gives net signed area (not total area). Always split at zeros when computing total area.
When a region is rotated about the $x$-axis (or another axis), it sweeps out a solid. The Disk Method computes the volume by integrating cross-sectional areas.
If $f(x) \ge 0$ on $[a, b]$, rotating $y = f(x)$ about the $x$-axis gives a solid with volume:
$$V = \pi \int_a^b [f(x)]^2\,dx$$
Each cross-section is a disk (circle) with radius $r = f(x)$ and area $\pi r^2 = \pi[f(x)]^2$.
Find the volume of the solid formed by rotating $y = \sqrt{x}$ about the $x$-axis, from $x=0$ to $x=4$.
$$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = \mathbf{8\pi}$$
Area under $y = \sqrt{x}$ from 0 to 4 — rotate this region about the $x$-axis to get a solid of volume $8\pi$
Figure 7.2 — Region for Disk Method: $y = \sqrt{x}$ on $[0,4]$
When the region between two curves is rotated, each cross-section is an annulus (washer) — a disk with a hole. The area of a washer is $\pi(R^2 - r^2)$, where $R$ is the outer radius and $r$ is the inner radius.
If $f(x) \ge g(x) \ge 0$ on $[a, b]$, rotating about the $x$-axis gives:
$$V = \pi \int_a^b \bigl([f(x)]^2 - [g(x)]^2\bigr)\,dx$$
$R = f(x)$ = outer radius; $r = g(x)$ = inner radius (the hole).
Rotate the region between $y = x$ and $y = x^2$ about the $x$-axis (region on $[0,1]$).
On $[0,1]$: $x \ge x^2$, so $f(x) = x$ (outer) and $g(x) = x^2$ (inner).
$$V = \pi\int_0^1 (x^2 - x^4)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{2}{15} = \mathbf{\frac{2\pi}{15}}$$
Find the volume of the solid formed by rotating the region between $y = \sqrt{x}$ and $y = x$ about the $x$-axis.
Region between $y = \sqrt{x}$ and $y = x$ — rotating this region about the $x$-axis gives the washer method
Figure 7.3 — Washer Method: Region Between $y = \sqrt{x}$ and $y = x$
Integration naturally models accumulation: any rate function, integrated, gives the total accumulated quantity.
Given velocity $v(t)$:
If $v(t) > 0$: moving right/up. If $v(t) < 0$: moving left/down.
A particle moves with $v(t) = t^2 - 3t + 2$ m/s on $[0, 3]$.
Zeros: $t^2-3t+2=(t-1)(t-2)=0$ at $t=1, 2$.
Displacement:
$$\int_0^3 (t^2-3t+2)\,dt = \left[\frac{t^3}{3} - \frac{3t^2}{2} + 2t\right]_0^3 = 9 - \frac{27}{2} + 6 = \frac{3}{2}$$
Total distance: Check sign on each interval:
Total distance = $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \mathbf{\frac{11}{6}}$ m
AP Exam Tip: A very common Free Response question asks for (a) position at a given time, (b) total distance, (c) times when the particle changes direction. Always find zeros of $v(t)$ first, then check signs of $v$ in each subinterval. Remember: the particle changes direction only when $v$ changes sign (not just equals zero).
Find the area of the region enclosed by $y = 3 - x^2$ and $y = -1$.
Find the area of the region bounded by $y = x^3$, $y = x$, for $x \ge 0$.
Find the volume when $y = x^2$ on $[0, 3]$ is rotated about the $x$-axis.
Rotate the region between $y = x^2 + 1$ and $y = 1$ from $x=0$ to $x=2$ about the $x$-axis. Find the volume (Washer Method).
A particle has $v(t) = 6 - 2t$ m/s. Find (a) displacement on $[0, 5]$, (b) total distance on $[0, 5]$.
Water flows into a tank at rate $R(t) = 4 + 2t$ gallons per minute. How much water enters the tank from $t=1$ to $t=4$ minutes?
(AP Free Response) A particle moves with $v(t) = t^2 - 5t + 4$ on $[0,4]$. (a) Find all times when the particle changes direction. (b) Find total distance on $[0,4]$. (c) Find displacement on $[0,4]$.
Find the area bounded by $y = e^x$, $y = 1$, $x = 0$, and $x = 2$. Then find the volume when this region is rotated about the $x$-axis (Washer Method).
$A = \displaystyle\int_a^b [f(x) - g(x)]\,dx$ where $f(x) \geq g(x)$ on $[a,b]$. Find intersection points to determine limits.
Rotating $y=f(x)$ around the $x$-axis: $V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$. Each cross-section is a disk of radius $f(x)$.
Rotating region between two curves: $V = \pi\displaystyle\int_a^b \left([R(x)]^2 - [r(x)]^2\right)\,dx$ where $R$ is outer, $r$ is inner radius.
Rotating around the $y$-axis: $V = 2\pi\displaystyle\int_a^b x\,f(x)\,dx$. Useful when disk/washer requires solving for $x$.
$A = \displaystyle\int_a^b |f(x) - g(x)|\,dx$
$L = \displaystyle\int_a^b \sqrt{1 + [f'(x)]^2}\,dx$
$W = \displaystyle\int_a^b F(x)\,dx$ — total work equals the integral of force over displacement.
$f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$