Accumulation & Integration
AP Calculus AB Topic 6 10:18 English narration · English + 中文 subtitles burned in
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Transcript
Differentiation found rates of change.
微分求出了变化率。
Integration runs it in reverse: given a rate, it finds the total accumulated change.
积分把这个想法反过来：给定一个变化率，它求出累积的总变化。
Here is the key picture.
看这个关键图景。
Water flows into a tank at a changing rate.
水以变化的速率流入一个水箱。
The area under a rate graph is the total amount accumulated.
速率图象下方的面积，就是累积的总量。
And the units multiply: a rate in litres per minute, times minutes, gives litres.
而单位相乘：以每分钟多少升计的速率，乘以分钟，就得到升。
Area becomes an amount.
面积变成了一个总量。
Keep that tank picture.
记住这个水箱图。
If the rate is positive, water is flowing in, and the accumulated change is positive.
如果变化率为正，水在流入，累积变化为正。
If the rate is negative, water is flowing out, and the accumulation is negative.
如果变化率为负，水在流出，累积变化为负。
Area below the axis counts as negative.
横轴下方的面积算作负的。
For simple shapes — triangles and rectangles — you can find the accumulation with plain geometry, no fancy formula needed.
对于简单图形——三角形和矩形——你用纯几何就能求出累积量，不需要复杂公式。
Always check the units of the answer: a rate in vehicles per hour times hours gives vehicles.
永远检查答案的单位：以车每小时为单位的变化率乘以小时，得到的是车数。
Today we build the integral — from stacking rectangles, to the Fundamental Theorem that ties integration and differentiation together.
今天我们来建立积分——从堆叠矩形，到把积分与微分联系在一起的基本定理。
Let's begin.
让我们开始吧。
Exact area is often hard.
精确面积常常很难求。
So we approximate.
所以我们用近似。
Split the interval into thin strips, and fill each strip with a rectangle whose height comes from the function.
把区间分成一条条窄带， 用矩形填满每一条，矩形的高度来自函数值。
Add the rectangle areas. That sum is a Riemann sum.
把矩形面积加起来，这个和就是黎曼和。
Watch the strips get thinner: the staircase of tops gets closer and closer to the true curve.
看窄带变细：顶端的阶梯越来越贴近真正的曲线。
When exact area is hard, we approximate it.
当精确面积不好求时，我们就去近似它。
Split the interval into strips, and add up rectangles — each with height from the function, and width delta x.
把区间分成一条条窄带，然后把矩形加起来—— 每个矩形的高来自函数值，宽是增量。
This is a Riemann sum.
这就是黎曼和。
Take the heights from the left edge, or the right edge.
高可以取自左端，也可以取自右端。
For an increasing function, the left sum underestimates the area, and the right sum overestimates it.
对一个递增函数，左和低估面积，右和高估面积。
With the right endpoints here, the sum adds to forty-four — an overestimate.
这里用右端点，和加起来是四十四——一个高估值。
Each rectangle has area equal to the function value times its width.
每个矩形的面积等于函数值乘以它的宽度。
Add them to estimate the integral.
把它们加起来，就估计了积分。
Subintervals may be uniform — equal width — or nonuniform.
子区间可以是等宽的，也可以是不等宽的。
When the widths differ, read each width from the table and use that width for that strip.
当宽度不同时，从表格里读出每一条的宽度， 并用那条宽度去算。
Do not force equal spacing when the data is uneven.
数据不均匀时，不要硬凑成等间距。
How tall is each strip?
每条窄带该有多高？
There are four standard estimates. Height from the left endpoint. Or the right.
有四种标准估计。
Or the midpoint of the strip. Or a trapezoidal sum, which averages the two endpoint heights and is usually the most accurate. Then exams ask whether your estimate is over or under.
用左端点的高度，或右端点，或窄带的中点， 或用梯形法——取两个端点高度的平均，通常最准确。
For an increasing function, left is under and right is over.
接着考试会问：你的估计是偏大还是偏小？
A trapezoidal sum overestimates when the graph is concave up, and underestimates when it is concave down.
对递增函数，左和偏小，右和偏大。 梯形和在图像凹向上时偏大，凹向下时偏小。
Here is the trapezoidal idea in motion.
这是梯形法的动态图景。
Instead of a flat rectangle top, connect the two endpoint heights with a straight line.
不用水平的矩形顶，而是用直线连接两个端点的高度。
Each piece is a trapezoid.
每一块都是梯形。
The formula averages the left and right heights, then multiplies by the width.
公式是把左右高度取平均，再乘以宽度。
When the curve bows up, the straight chord sits above it, so the trap sum is too big.
当曲线向上拱起时，那条弦在曲线上方，所以梯形和偏大。
When the curve bows down, the chord is below, so the trap sum is too small.
当曲线向下弯时，弦在曲线下方，所以梯形和偏小。
Now do one, because the exam gives you a table far more often than a formula.
现在真做一道，因为考试给你表格的次数，远远多于给你公式。
Here is the table: f of zero is three, f of two is five, f of four is eight, f of six is nine.
表格是这样的：零处的函数值是三，二处是五，四处是八，六处是九。
Estimate the integral from zero to six with three equal strips, so delta x is two.
用三个等宽的窄条估计从零到六的积分，所以 delta x 是二。
Take the right endpoint of each strip.
取每一条的右端点。
The heights are f of two, f of four and f of six — five, eight and nine — and each strip is two wide.
高度是二处、四处和六处的函数值——五、八、九——每条宽二。
Two times five plus eight plus nine is forty-four.
二乘以五加八加九，等于四十四。
Now the left sum, using the left endpoint of each strip: three, five and eight.
再算左和，用每一条的左端点：三、五、八。
Two times three plus five plus eight is thirty-two.
二乘以三加五加八，等于三十二。
Notice what you now have for free.
注意你顺带得到了什么。
This f is increasing, so every right rectangle pokes above the curve and every left rectangle falls short — the right sum overestimates, the left sum underestimates, and the true integral is trapped between thirty-two and forty-four.
这个函数是递增的，所以每个右矩形都高出曲线， 每个左矩形都不够高——右和偏大，左和偏小， 真实的积分就被夹在三十二和四十四之间。
Exams ask for that reasoning by name, so say increasing, then say which way it errs.
考试会点名要这段推理，所以先说"递增"，再说偏哪一边。
Now shrink the strips.
现在把窄带变细。
As the strips grow narrower and more numerous, the staircase of rectangles smooths into the exact area.
当窄带越来越细、越来越多时，矩形的阶梯就平滑成精确的面积。
That exact value is the definite integral, written with the tall S sign, from a to b.
那个精确值就是定积分，用高高的求和记号写出，从甲到乙。
So a definite integral is nothing more than the limit of a Riemann sum — you should be able to translate one into the other.
所以定积分不过就是黎曼和的极限—— 你应该能够把两者互相转换。
Here is the deep connection.
这里有一个深刻的联系。
Let the upper limit move, and the integral becomes a new function — an accumulation function — that sweeps out signed area as it goes.
让积分的上限移动，积分就变成一个新的函数——累积函数—— 它一边前进，一边扫出带符号的面积。
The Fundamental Theorem of Calculus says something beautiful: if you differentiate this accumulation, you get back the original function.
微积分基本定理说了一件很美的事： 如果你对这个累积求导，就会得到原来的函数。
Integration and differentiation are inverse operations.
积分和微分是互逆的运算。
Let's write it.
我们来把它写出来。
If g of x is the integral of f from a to x — an accumulation function — then the derivative of g is just f again.
如果 函数 是从起点到自变量的积分——一个累积函数——那么它的导数就又是原来的被积函数。
Differentiation undoes the integration.
微分抵消了积分。
This is the engine behind the common "let g equal the integral" questions.
这正是常见的"设累积等于积分"这类题目的引擎。
And because the derivative of g is f, everything from the last unit applies: g is increasing wherever the rate is positive, and g has a peak where the rate crosses zero.
而因为累积的导数是变化率， 上一单元的一切都适用：只要变化率为正，累积就递增；而在变化率穿过零点的地方，累积就有一个峰值。
Read the whole behaviour of the accumulation straight off the graph of the rate.
累积函数的全部性态，都可以直接从变化率的图像上读出来。
The accumulation is increasing where the rate is positive, and decreasing where the rate is negative.
变化率为正的地方，累积递增；变化率为负的地方，累积递减。
It has a local extremum where the rate crosses zero with a sign change.
变化率带着变号穿过零的地方，累积有局部极值。
It is concave up where the rate is increasing, and concave down where the rate is decreasing.
变化率递增的地方，累积凹向上；变化率递减的地方，累积凹向下。
Inflection points of the accumulation sit where the rate has a local extremum.
变化率有局部极值的地方，累积有拐点。
To get a value, add signed area from the lower limit up to that input.
要求某个值，就把从下限到那个输入的带符号面积加起来。
A few properties make integrals easy to handle.
有几条性质让积分很好处理。
Over a zero-width interval, the integral is zero.
在宽度为零的区间上，积分是零。
If you swap the two limits, the integral flips sign.
如果你交换上下限，积分变号。
Constants pull out, and sums split term by term.
常数可以提出来，和可以逐项分开。
And the most useful one: you can split the interval at any interior point, and add the two pieces.
而最有用的一条：你可以在任意内部点把区间分开，再把两块相加。
This lets you build a total from areas read off a graph.
这让你能够从图象上读出的面积，拼出一个总量。
A definite integral is a signed area.
定积分是有符号的面积。
Above the axis the contribution is positive; below the axis it is negative.
横轴上方的贡献为正，下方为负。
So net accumulation can cancel.
所以净累积可以互相抵消。
If a question asks for total area, or total distance rather than displacement, split the integral at every sign change and add the absolute values of the pieces.
如果题目要的是总面积，或者总路程而不是位移，就在每个变号处把积分拆开， 再把各块的绝对值相加。
The picture makes the sign story clear: green regions add, and regions under the axis subtract.
图把符号说清楚了：上方区域相加，下方区域相减。
Part two of the Fundamental Theorem tells us how to get an exact value.
基本定理的第二部分告诉我们如何求出精确值。
Find any antiderivative — a function whose derivative is f — then subtract its values at the two limits.
找到任意一个原函数——一个导数为被积函数的函数—— 然后把它在上下限的值相减。
This is also the net change idea: the integral of a rate over an interval is the total change in the quantity.
这也是净变化的思想：一个变化率在区间上的积分，就是那个量的总变化。
For the integral of two x plus one, from one to three, an antiderivative is x squared plus x, and the answer is ten.
对二倍自变量加一从一到三的积分，一个原函数是平方加自变量，答案是十。
The indefinite integral is a whole family of antiderivatives, so we always add a constant of integration, plus C.
不定积分是一整族原函数，所以我们总要加上一个积分常数。
The workhorse is the power rule for integration: raise the power by one, and divide by the new power.
主力是积分的幂法则：把幂次加一，再除以新的幂次。
To get the rest, reverse each derivative you already know: the integral of one over x is the natural log, the integral of cosine is sine, and so on.
要得到其余的， 就把你已经知道的每个导数反过来：自变量分之一的积分是自然对数，余弦的积分是正弦，如此等等。
Finally, u-substitution — which reverses the chain rule.
最后是换元积分法——它把链式法则反过来。
Choose the inside function to be u, so its derivative times dx becomes du.
选内部的函数作为换元变量， 于是它的导数乘以微分就变成新的微分。
Then rewrite the whole integral in terms of u.
然后把整个积分用新变量改写。
Look for a function and its derivative both present.
要寻找一个函数和它的导数同时出现。
For two x times cosine of x squared, let u be x squared.
对二倍自变量乘以自变量平方的余弦，设换元为自变量的平方。
Its derivative, two x d x, is already there — so the integral becomes sine of x squared, plus C.
它的导数已经在那里了——于是积分就变成自变量平方的正弦，再加常数。
Two algebra moves unlock more integrals.
两个代数步骤能解开更多积分。
First, if a rational integrand is top heavy — the degree on top is at least the degree on the bottom — use polynomial long division.
第一，如果有理被积函数是头重的——分子次数不低于分母次数—— 就用多项式长除法。
It rewrites the fraction as a polynomial plus a proper fraction, and both pieces are then easy to integrate.
它把分式改写成多项式加真分式，两部分就都好积了。
Second, if the bottom is a quadratic that will not factor, try completing the square.
第二，如果分母是无法因式分解的二次式，就试着配方。
That turns the denominator into something squared plus a constant squared, and that shape has an arctangent antiderivative — or sometimes a logarithm.
配方把分母变成某个量的平方加一个常数的平方，这个形状的原函数是反正切——有时是对数。
Recognising the shape saves time.
认出形状就能省时间。
Selecting techniques for antidifferentiation is itself a skill.
把积分与方法配对，本身就是一项技能。
Build a short checklist.
建一张简短清单。
Does it match a basic antiderivative?
它符合基本原函数吗？
Then write it.
那就直接写。
Is there a chain-rule pattern — an inside function whose derivative is also present?
有没有链式法则模式——一个内层函数，而它的导数也在？
Then use u-substitution.
那就用换元。
Is it a rational function that is top heavy, or a quadratic bottom that will not factor?
它是头重的有理函数，或者分母是无法分解的二次式吗？
Then use algebra to reshape: long division, completing the square, or splitting a fraction into standard forms.
那就用代数改形：长除法、配方，或把分式拆成标准形式。
Name the structure first.
先给结构命名。
That prevents wasted effort.
这样就不会白费力气。
Before you go, three marks to keep.
结束之前，三个要守住的分。
First, the Fundamental Theorem links the two halves of calculus: the integral of a rate is the net change, and the derivative of an accumulation is the rate.
第一，基本定理把微积分的两半联系起来： 变化率的积分是净变化，累积的导数是变化率。
Second, for an indefinite integral, never forget the plus C.
第二，对不定积分，永远别忘加常数。
Third, when you use u-substitution on a definite integral, change the limits too, or convert back before you substitute.
第三，当你对定积分用换元法时，也要把上下限一起换，或者先换回去再代入。
This topic is yours.
这个专题就是你的了。