Using Accumulation Functions and Definite Integrals in Applied Contexts · 在应用背景下使用累积函数和定积分
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| definite integral/ˈdefɪnət ˈɪntɪɡrəl/ | 定积分 | dìng jī fēn |
Start here, add the change, land there
- Many applied problems give a rate and a starting amount, and ask for a later total.
- The key relationship: final value = initial value + accumulated change.
- The accumulated change is the definite integral of the rate over the interval.
- This one idea powers tanks, populations, accounts, and more.
从这里出发,加上变化,到达那里
- 许多应用题给出一个速率和一个初始量,问之后的总量。
- 关键关系:最终值 = 初始值 + 累积变化。
- 累积变化是速率在区间上的定积分。
- 这一个想法驱动水箱、人口、账户等等。
The accumulation formula
- If $R(t)$ is the rate and $Q$ the quantity, then:
-
$$Q(b)=Q(a)+\int_a^b R(t)\,dt$$
- Start with $Q(a)$, add the net accumulated change $\displaystyle\int_a^b R(t)\,dt$, arrive at $Q(b)$.
- It's the FTC in applied clothing: the integral of a rate is the net change of its quantity.
累积公式
- 若 $R(t)$ 是速率、$Q$ 是量,则:
-
$$Q(b)=Q(a)+\int_a^b R(t)\,dt$$
- 从 $Q(a)$ 开始,加上净累积变化 $\displaystyle\int_a^b R(t)\,dt$,到达 $Q(b)$。
- 这是穿了应用外衣的 FTC:速率的积分是其量的净变化。
Accumulated inflow is the area · 流入的累积量即为面积
y = 4t (inflow rate) · y = 4t（流入速率）
The area under the inflow rate $R(t)$ is the water added; the total is the initial amount plus that area. · 流入速率 $R(t)$ 下方的面积是加入的水量；总量等于初始量加上该面积。
The applied accumulation relationship is: final value =... · 应用的累积关系式为：最终值 = ...
Final = initial + accumulated change. · 最终值 = 初始值 + 累积变化量。
To find the final amount, add the accumulated change to the ____ value. · 要求最终量，需将累积变化量加到 ____ 值上。
Final = initial + integral of the rate. · 最终值 = 初始值 + 速率的积分。
Reading the context and units
- A definite integral 定积分 of a rate has units of (rate units)$\times$(time) = the quantity's units.
- $\int$ of litres/min over minutes gives litres; $\int$ of people/year over years gives people.
- The sign matters: a negative rate (draining, cooling) subtracts from the total.
- Always interpret the integral in the problem's own words and units.
读懂情境与单位
- 速率的定积分单位是*(速率单位)$\times$(时间)* = 量的单位。
- 升/分对分钟积分给出升;人/年对年积分给出人。
- 符号很重要:负速率(排水、冷却)从总量中减去。
- 永远用题目自己的措辞和单位来解读积分。
A tank has $50$ L; water flows in at $R(t)=4t$ for $0\le t\le 3$ ($\int_0^3 4t\,dt=18$). Find the total at $t=3$. · 一个水箱里有 $50$ L 水；水流以 $R(t)=4t$ 的速率流入持续了 $0\le t\le 3$ （$\int_0^3 4t\,dt=18$ ）。求 $t=3$ 时的总量。
$50+18=68$ L.
If $R$ is in litres/min and $t$ in minutes, $\int R\,dt$ has units of... · 如果 $R$ 的单位是升/分钟，$t$ 的单位是分钟，那么 $\int R\,dt$ 的单位是...
(litres/min)×(min) = litres. · (升/分钟)×(分钟) = 升。
Net vs. gross change
- The integral $\int_a^b R\,dt$ is the net change — inflow minus outflow.
- If a tank is filled and drained, the net integral accounts for both automatically.
- To find total added or total removed separately, integrate only the positive or negative parts.
- But "final = initial + net accumulated change" always uses the plain signed integral.
净变化 vs 总变化
- 积分 $\int_a^b R\,dt$ 是净变化——流入减流出。
- 若水箱既注水又排水,净积分自动兼顾两者。
- 要分别求总加入或总移除,只积分正部或负部。
- 但"最终 = 初始 + 净累积变化"总是用普通的带符号积分。
The definite integral of the rate gives the change, not the final total by itself. · 速率的定积分给出的是变化量，而非最终的总量。
You must add the initial value. · 你必须加上初始值。
For an outflow (draining) rate, the integral over the interval is... · 对于流出（排水）速率，其在区间上的积分是...
An outflow rate is negative → subtracts. · 流出速率为负 → 表示减少。
Don't report the integral alone as the final amount — it's only the change. You must add the initial value: final $=$ initial $+\int_a^b R\,dt$. And keep the sign of the rate: an outflow rate is negative, so its integral correctly subtracts from the starting total.
别把单独的积分当作最终量——它只是变化。你必须加上初始值:最终 $=$ 初始 $+\int_a^b R\,dt$。并保持速率的符号:流出速率为负,所以它的积分正确地从起始总量中减去。
A tank holds $50$ L at $t=0$. Water flows in at $R(t)=4t$ L/min for $0\le t\le 3$. How much at $t=3$?
- Accumulated change: $\displaystyle\int_0^3 4t\,dt=\Big[2t^2\Big]_0^3=18$ L.
- Final $=$ initial $+$ change $=50+18=68$ L.
- (The integral $18$ is the added water; $68$ is the total.)
一个水箱在 $t=0$ 装 $50$ L。水以 $R(t)=4t$ L/min 流入,$0\le t\le 3$。$t=3$ 时有多少?
- 累积变化:$\displaystyle\int_0^3 4t\,dt=\Big[2t^2\Big]_0^3=18$ L。
- 最终 $=$ 初始 $+$ 变化 $=50+18=68$ L。
- (积分 $18$ 是加入的水;$68$ 是总量。)
In applied contexts, final value = initial value + accumulated change, where the change is the definite integral of the rate: $Q(b)=Q(a)+\int_a^b R(t)\,dt$. The integral carries the quantity's units and its sign (inflow adds, outflow subtracts). Never forget to add the initial value.
在应用情境中,最终值 = 初始值 + 累积变化,其中变化是速率的定积分:$Q(b)=Q(a)+\int_a^b R(t)\,dt$。积分带着量的单位和符号(流入相加,流出相减)。永远别忘了加初始值。