Exploring Accumulations of Change · 探索变化量的累积
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| net change/net tʃeɪndʒ/ | 净变化 | jìng biàn huà |
| accumulation/əˌkjuːmjʊˈleɪʃn/ | 累积 | lěi jī |
| area/ˈeərɪə/ | 面积 | miàn jī |
| signed area/saɪnd ˈeərɪə/ | 带符号面积 | dài fú hào miàn jī |
Adding up change to get a total
- The derivative took things apart into rates. Integration puts them back together into totals.
- If you know the rate something changes, you can find the net change 净变化 it accumulates.
- Drive at a known speed for a while → total distance is speed accumulated over time.
- This "adding up a rate" is the second big idea of calculus: accumulation 累积.
把变化加起来得到总量
- 导数把事物拆成变化率。积分把它们重新拼成总量。
- 若你知道某物变化的速率,就能求出它累积的净变化。
- 以已知速度开一段时间 → 总距离是速度在时间上的累积。
- 这个"把速率加起来"是微积分的第二个大想法:累积。
Integration (accumulation) is essentially the reverse of... · 积分（累积）本质上是...的逆运算
Accumulating a rate undoes differentiating a total. · 累积一个速率抵消了对总量的微分。
The total change accumulated from a rate over an interval is called the ____ change. · 在某个区间内由速率累积的总变化称为 ____ 变化。
Net change = signed accumulation. · 净变化 = 有向累积量。
Area under a rate graph = accumulated amount
- Plot a rate of change against time. The area 面积 under that graph is the total accumulated change.
- Constant rate $60\,\tfrac{\text{km}}{\text{h}}$ for $2$ h → area of a rectangle $=60\times2=120$ km.
- A changing rate makes a curvy region, but the idea is the same: area = accumulated quantity.
- Height is the rate; width is the interval; area is the total.
速率图下的面积 = 累积量
- 把变化率对时间作图。那条图下的面积就是累积的总变化。
- 恒定速率 $60\,\tfrac{\text{km}}{\text{h}}$ 持续 $2$ 小时 → 矩形面积 $=60\times2=120$ km。
- 变化的速率形成弯曲的区域,但想法相同:面积 = 累积量。
- 高度是速率;宽度是区间;面积是总量。
Area under a rate is a total · 速率下的面积即为总量
y = 3 (constant rate) · y = 3（恒定速率）
The shaded area under a rate curve is the accumulated amount — widen the interval and the total grows. · 速率曲线下的阴影面积即为累积量——扩大区间，总量随之增长。
A tap runs at $3$ L/min for $4$ min. The area under the rate graph (accumulated volume) is... · 水龙头以 $3$ L/min 的速度运行 $4$ 分钟。速率图下的面积（累积体积）是……
Rectangle area $=3\times4=12$ L. · 矩形面积 $=3\times4=12$ L。
A rate rises linearly from $0$ to · 到 $6$ over $4$ min (a triangle). The accumulated amount (area) is... · 速率从 $0$ 线性上升到 $6$，历时 $4$ 分钟（呈三角形）。累积量（面积）是……
Triangle area $=\tfrac12\times4\times6=12$. · 三角形面积 $=\tfrac12\times4\times6=12$。
Signed area: rates can be negative
- When the rate is positive, the quantity grows (area counts as positive).
- When the rate is negative (below the axis), the quantity shrinks (area counts as negative).
- The net change is signed area 带符号面积: positive parts minus negative parts.
- So an object moving backward subtracts from the accumulated displacement.
带符号面积:速率可以为负
- 当速率为正,量增长(面积记为正)。
- 当速率为负(在轴下方),量减少(面积记为负)。
- 净变化是带符号面积:正的部分减去负的部分。
- 所以向后运动的物体从累积位移中减去。
Where a rate graph dips below the axis, that area subtracts from the net accumulated change. · 当速率图低于x轴时，该面积从净累积变化中减去。
Below-axis area is negative (signed area). · x轴下方的面积为负（有向面积）。
On a velocity-time graph, a portion below the axis means the object... · 在速度-时间图上，x轴下方的一段表示物体...
Negative velocity = backward motion = negative signed area. · 负速度 = 向后运动 = 负有向面积。
From rate back to amount
- If $f$ is a rate and you accumulate from $a$ to $b$, you get the net change of the original quantity.
- This is exactly the reverse of differentiating — undoing the rate to recover the total.
- The tool that computes this signed area precisely is the definite integral (coming up).
- For now: area under a rate graph = the amount that accumulated.
从速率回到量
- 若 $f$ 是一个速率,你从 $a$ 累积到 $b$,就得到原量的净变化。
- 这恰好是求导的逆过程——撤销速率以还原总量。
- 精确计算这个带符号面积的工具是定积分(即将学到)。
- 现在:速率图下的面积 = 累积的量。
Accumulated change is signed area, not just "area." A velocity graph that dips below the axis means the object moved backward, which subtracts from the net displacement. Don't add all the area as positive — regions below the axis count as negative.
累积变化是带符号面积,而不只是"面积"。速度图跌到轴下方意味着物体向后运动,这会从净位移中减去。别把所有面积都当正的加起来——轴下方的区域记为负。
A tap fills a tank at a rate of $3\,\tfrac{\text{L}}{\text{min}}$ for $4$ minutes, then $0$ after.
- The rate graph is a rectangle: height $3$, width $4$.
- Accumulated volume = area $=3\times4=12$ L.
- If the rate later went negative (draining), that area would subtract from the total.
一个水龙头以 $3\,\tfrac{\text{L}}{\text{min}}$ 的速率注水 $4$ 分钟,之后为 $0$。
- 速率图是一个矩形:高 $3$,宽 $4$。
- 累积体积 = 面积 $=3\times4=12$ L。
- 若之后速率变负(排水),那部分面积会从总量中减去。
Accumulation turns a rate back into a total: the signed area under a rate-of-change graph over $[a,b]$ is the net change of the quantity. Positive rate → area adds; negative rate → area subtracts. The definite integral makes this precise.
累积把速率还原成总量:变化率图在 $[a,b]$ 上的带符号面积就是量的净变化。正速率 → 面积相加;负速率 → 面积相减。定积分把这精确化。