Notes
Area between curves
- If $f \geq g$ on $[a, b]$, area $\displaystyle = \int_a^b (f - g)\,dx$. Find limits by setting the curves equal when none are given.
- If curves cross, split the interval at the crossing so the top function stays on top. Area is never negative.
Average value
- $\displaystyle f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx$.
Accumulation and net change
- $\displaystyle \int_a^b f'(x)\,dx = f(b) - f(a)$. The integral of a rate gives total change.
- Final amount $=$ starting amount $+$ net change.
Motion
- Displacement $\displaystyle = \int v\,dt$. Total distance $\displaystyle = \int |v|\,dt$, found by splitting where $v$ changes sign.
- Position $\displaystyle = s(a) + \int_a^b v\,dt$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
They meet at $x = 0$ and $x = 6$. $\displaystyle \int_0^6 (6x - x^2)\,dx = 108 - 72 = 36$.
$\displaystyle \frac{1}{2}\int_0^2 x^3\,dx = \frac{1}{2}(4) = 2$.
$\displaystyle \int_0^2 3t^2\,dt = [t^3]_0^2 = 8$.
$\displaystyle \int_0^4 (6 - 2t)\,dt = [6t - t^2]_0^4 = 24 - 16 = 8$.
$v < 0$ on $(0, 2)$ with distance 2, and $v > 0$ on $(2, 4)$ with distance 2. The displacement is 0 but the distance is 4.
$\displaystyle 400 + \int_0^6 50\,dt = 400 + 300 = 700$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A particle moves on a line with velocity $v(t) = t^2 - 4t + 3 = (t - 1)(t - 3)$ meters per second for $0 \leq t \leq 4$.
(a) Find the displacement from $t = 0$ to $t = 4$.
Show the answer and scoring guide
$\displaystyle \int_0^4 (t^2 - 4t + 3)\,dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4 = \frac{64}{3} - 32 + 12 = \frac{4}{3}$ meters.
Check off the points you earned:
(b) Find the total distance traveled.
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$v > 0$ on $(0, 1)$, $v < 0$ on $(1, 3)$, $v > 0$ on $(3, 4)$. The integrals are $\displaystyle \frac{4}{3}$, $\displaystyle -\frac{4}{3}$ and $\displaystyle \frac{4}{3}$. Total distance $\displaystyle = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ meters.
Check off the points you earned:
(c) Find the average velocity over $[0, 4]$.
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Average velocity $\displaystyle = \frac{1}{4}\int_0^4 v\,dt = \frac{1}{4}\cdot\frac{4}{3} = \frac{1}{3}$ m/s.
Check off the points you earned: