Exploring Accumulations of Change
AP Calculus BC Topic 6 10:02 English narration · English + 中文 subtitles burned in
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Water flows into a tank.
水正流进一个水箱。
The meter shows litres per minute.
仪表显示的是每分钟多少升。
That is a rate of change.
这是一个变化率。
Now, how much water arrived in the first ten minutes?
那么，前十分钟一共流进了多少水？
Draw the rate against time, and the answer is the area under it.
把这个变化率对时间画出来，答案就是它下面的面积。
That is the whole idea of this unit.
这就是本单元的全部核心。
A derivative measures a rate. An integral measures an accumulation, a total built up from a rate.
导数衡量变化率，积分衡量累积量，也就是由变化率累积出来的总量。
And the units prove it. Litres per minute, times minutes, gives litres.
单位就能证明这一点：升每分钟，乘以分钟，得到升。
Where a derivative measures a rate, an integral measures a total.
导数衡量变化率，积分衡量总量。
Let's begin.
让我们开始吧。
Exact area is often hard, so we approximate.
精确的面积常常很难求，所以我们用近似。
Split the interval into pieces and fill it with thin rectangles.
把区间分成若干小段，用很薄的矩形把它填满。
That is a Riemann sum.
这就是黎曼和。
But how tall is each rectangle?
但每个矩形该有多高？
There are four standard estimates.
有四种标准的估计方法。
The height at the left endpoint of each piece.
用每一小段左端点处的高度。
Or the right endpoint.
或者右端点。
Or the midpoint.
或者中点。
Or a trapezoidal sum, which averages the two endpoint heights and is usually the most accurate.
或者用梯形法，取两个端点高度的平均，通常最准确。
Then exams ask whether your estimate is over or under.
接着考试会问：你的估计是偏大还是偏小？
For an increasing function, a left sum underestimates and a right sum overestimates.
对递增函数，左和偏小，右和偏大。
A trapezoidal sum overestimates when the graph is concave up, and underestimates when it is concave down.
梯形和在图像凹向上时偏大，凹向下时偏小。
Write that sum compactly with summation notation. Add up the height times the width, over every rectangle.
用求和记号把那个和式紧凑地写出来：对每一个矩形，把高度乘以宽度再相加。
Now let the number of rectangles grow without bound, so each one gets thinner.
现在让矩形的个数无限增多，每一个都变得越来越薄。
The sum stops being an approximation and becomes exact.
这个和就不再是近似，而变成精确值。
That limit is the definite integral.
这个极限就是定积分。
The notation shows the history.
看这个记号，你能看出它的来历。
The integral sign is a stretched letter S, for sum.
积分号是拉长的字母 S，代表求和。
The function is the height.
函数是高度。
And the d x is what the width became.
而 dx 就是宽度变成的东西。
So a definite integral is the limit of a Riemann sum.
所以定积分就是黎曼和的极限，而你应该能把两者互相转换。
Now the theorem that ties the two halves of calculus together.
现在讲那条把微积分两半连接起来的定理。
Fix the left endpoint and let the right endpoint move.
固定左端点，让右端点移动。
What you get is an accumulation function.
你得到的就是一个累积函数。
Its input is where you stopped, its output is the area gathered so far.
它的输入是你停在哪里，输出是到目前为止累积的面积。
Here is the first form of the Fundamental Theorem of Calculus, the FTC. Differentiate that accumulation function, and you get the original function back.
这是微积分基本定理的第一种形式：对这个累积函数求导，就得到原来的函数。
They undo each other.
积分和求导互为逆运算。
And this lets you read its behaviour straight off the graph of the rate.
而这让你可以直接从变化率的图像上读出它的性态。
Rate positive means the accumulation increases.
变化率为正的地方，累积函数递增。
Where the rate is negative, the accumulation decreases. A sign change in the rate means the accumulation has extrema.
变化率带着变号穿过零的地方，累积函数有极值。
Rate increasing means it is concave up.
变化率递增的地方，累积函数凹向上。
One refinement, and it is asked constantly.
有一个补充，而且它被考得非常频繁。
So far the top limit was just x.
到目前为止上限就是 x。
But often the top limit is itself a function of x, like x squared.
但上限常常本身就是 x 的一个函数， 比如 x 平方。
Then you differentiate as before, evaluating the integrand at that top limit, and then you multiply by the derivative of that top limit.
这时你像之前那样求导，把被积函数在这个上限处取值，然后再乘以这个上限的导数。
It is the chain rule, exactly as in unit three, with the accumulation as the outer function.
这就是链式法则，和第三单元里完全一样，只不过外层函数变成了累积函数。
Work one.
来做一个。
The derivative, with respect to x, of the integral from one to x squared of sine t d t.
对 x 求导， 被求导的是从一到 x 平方的 sin t d t 的积分。
Evaluate sine at the top limit, giving sine of x squared.
把 sin 在上限处取值，得到 sin x 平方。
Then multiply by the derivative of x squared, which is two x.
然后乘以 x 平方的导数，也就是 二 x。
Miss that second factor and you lose the mark, in exactly the same way you would in unit three.
漏掉第二个因子，你就会丢分，和第三单元里的情况一模一样。
A few properties save a lot of work.
有几条性质能省下大量功夫。
Split an interval at any inner point and add the two pieces.
你可以在任意内点把区间拆开，再把两部分相加。
Swap the two limits and the integral changes sign.
如果交换上下限，积分变号。
Equal limits give zero.
如果上下限相等，积分为零。
Pull constants out in front, and integrate a sum term by term.
你可以把常数提到前面，也可以逐项积分。
And one to be careful with. A definite integral is a signed area.
还有一条要小心：定积分是有符号的面积。
Area below the axis counts as negative.
横轴下方的面积算作负的。
So if a question asks for total distance, or total area, split at every sign change first.
所以如果题目问的是总路程或总面积， 要先在每一个变号处把积分拆开。
Here is the second form, and it is the one you compute with.
这是第二种形式，也是你真正用来计算的那一个。
The integral of a derivative over an interval equals the function at the top limit, minus the function at the bottom limit.
一个导数在区间上的积分， 等于函数在上限处的值，减去函数在下限处的值。
Read it in words and it is the net change theorem. Definite integrals add over adjacent intervals. Total accumulated change equals final value minus starting value.
用文字读出来，就是净变化定理： 一个量累积的总变化，等于它的最终值减去初始值。
Obvious, and extremely useful.
这既显然，又极其有用。
So use the evaluation form of the FTC: to evaluate a definite integral, find any antiderivative, evaluate at the top, evaluate at the bottom, and subtract.
所以要计算一个定积分，就找出任意一个原函数，在上限处求值，在下限处求值，然后相减。
The constant cancels, which is why you do not need it here.
积分常数会被抵消掉，这就是这里不需要它的原因。
To do that you need antiderivatives, so run every derivative rule backwards.
要做到这一点你需要原函数，所以把每一条求导法则倒过来用。
The power rule in reverse says: raise the exponent by one, then divide by the new exponent.
幂法则反过来说：指数加一，然后除以新的指数。
That works for every power except minus one, where you get the natural logarithm instead.
除了指数为负一的情形之外，这对每个幂都成立； 负一时得到的是自然对数。
The exponential is its own antiderivative.
指数函数是它自己的原函数。
Sine and cosine swap, with a sign to watch.
正弦和余弦互换，符号要小心。
And an indefinite integral always carries the constant of integration, so the antiderivatives form a whole stack of parallel curves.
而不定积分总要带上积分常数。 很多不同的函数有相同的导数， 所以原函数构成的是一整族互相平行的曲线。
The first real technique is substitution, and it reverses the chain rule.
第一个真正的技巧是换元积分，它是链式法则的逆运算。
Look for the pattern. Something is inside, and the derivative of that inside is also there, multiplying.
找出这个模式： 有一个东西在里面，而这个内层的导数也在，作为一个乘数出现。
Name the inside u.
把内层记作 u。
Then the derivative of the inside, times d x, becomes d u, and the integral collapses into a much simpler one.
于是内层的导数乘以 dx 就变成 du，整个积分就塌缩成一个简单得多的积分。 有两个提醒。
Two warnings. Convert d x properly, never just swap the letter.
要正确地转换 dx，绝不能只是把字母换掉。
And for a definite integral, either change the limits, or convert back at the end.
而对于定积分，要么把上下限换成新变量的值，要么最后再换回去。
Do not mix the two.
不要把两者混着用。
Now the BC-only techniques begin.
现在开始讲 BC 独有的技巧。
Two of them are really algebra, not calculus.
其中有两个其实是代数，不是微积分。
First, if a rational integrand is top heavy, meaning the degree on top is at least the degree on the bottom, use long division.
第一，如果一个有理被积函数是"头重脚轻"的，也就是分子的次数不低于分母的次数，就用长除法。
It rewrites the fraction as a polynomial plus a proper fraction, and both pieces are then easy.
它把这个分式改写成一个多项式加上一个真分式，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，如果分母是一个无法因式分解的二次式，就试着配方。
That turns it into something squared plus a constant squared, and that shape has an arctangent antiderivative.
配方把它变成"某个东西的平方加上一个常数的平方"，而这个形状的原函数是反正切。
Recognising those two shapes saves a lot of wasted time.
认出这两种形状能省下大量白费的时间。
Integration by parts reverses the product rule.
分部积分是乘积法则的逆运算。
The integral of u d v equals u v, minus the integral of v d u.
公式说：u dv 的积分等于 uv 减去 v du 的积分。
The whole skill is choosing which factor is u.
全部的技巧在于选哪个因子作为 u。
Use the word LIATE as a guide.
用 LIATE 这个词作为指导： 对数、反三角、代数式、三角、指数。
Logarithms, inverse trig, algebraic, trig, exponential.
在这个顺序里排在前面的应该作为 u， 因为对它求导会让它变简单。
Whichever comes first in that list should be u, because it gets simpler when you differentiate it.
做一道。
Work one.
取 x 乘以 e 的 x 次方。
Take x times e to the x.
令 u 等于 x，它的导数是 1；令 dv 等于 e 的 x 次方 dx，那么 v 等于 e 的 x 次方。
Let u be x, and let d v be e to the x d x, so v is e to the x.
套用公式。
Apply the formula, and you get x times e to the x, minus the integral of e to the x.
得到 x 乘以 e 的 x 次方，减去 e 的 x 次方的积分， 也就是 x 乘以 e 的 x 次方减去 e 的 x 次方，再加一个常数。
Notice that second integral is easier than the first.
注意第二个积分比第一个更容易。
That is the whole point.
这正是这个方法的意义。
Sometimes you apply it twice.
有时候你需要用它两次。
Partial fractions handles a rational function whose bottom factors.
部分分式用来处理分母可以因式分解的有理函数。
The idea is to run the addition of fractions backwards.
思路是把分数的加法倒过来做。
Split it into simpler pieces, one over each factor, with an unknown number on top of each.
把它拆成更简单的几块，每个因式各占一块，每块的分子放一个未知数。
Then solve for the two numbers, either by clearing denominators and comparing coefficients, or by substituting the values that make each factor vanish.
然后解出这两个数，可以通过去分母比较系数，也可以代入让各因式为零的那些值。
Once it is split, each piece integrates to a logarithm.
一旦拆开，每一块的积分都是一个对数。
Learn this one properly, because you will need it in the next unit for the logistic equation.
这一条要学扎实， 因为下一个单元讲逻辑斯蒂方程时你会用到它。
One more kind of integral.
还有一类积分。
An improper integral is one with an infinite limit or an infinite discontinuity — where something runs to infinity. Either a limit of integration is infinite, or the function itself blows up somewhere inside.
反常积分是指某个东西跑到无穷的情形： 要么积分限是无穷大，要么被积函数本身在区间内某处趋于无穷。
Watch what happens as the region stretches out forever.
看看当这块区域无限伸展出去时会发生什么。
Sometimes the area settles on a finite number, and sometimes it grows without stopping.
有时候面积会稳定在一个有限的数上，有时候它会毫无止境地增大。
Never just write infinity as a limit of integration.
千万不要直接把无穷大写成积分限，而要把它改写成一个极限。
Rewrite it as a limit instead.
把无穷的那个端点换成一个字母，正常地积分，然后取这个字母趋于无穷时的极限。
Replace the infinite endpoint with a letter, integrate normally, then take the limit as that letter goes to infinity.
如果极限是一个有限的数，积分就收敛到它。
A finite limit means the integral converges to it.
如果不是，积分就发散。
Otherwise it diverges.
比较两个例子。
Compare two.
从 1 到无穷，x 平方分之一的积分，得到 1 减去 b 分之一，趋于 1。
One over x squared, from one to infinity, gives one minus one over b, which tends to one, so it converges.
所以这个积分收敛到 1。
But one over x, over the same interval, gives the natural logarithm of b, which grows without bound, so it diverges.
但在同一个区间上，x 分之一的积分得到 b 的自然对数， 它无限增大，所以发散。
Look how close those two functions are, and how completely different the answers are.
看看这两个函数有多接近，而答案又有多么完全不同。
The BC exam expects you to recognize which method fits, and that is a tested skill in itself.
BC 考试要求你能认出哪种方法适用，而这本身就是一个会被考的技能。
So build a checklist.
所以建一张清单。
Does it match a basic rule?
它符合某条基本法则吗？
Then just write the antiderivative.
那就直接写出原函数。
Is there a chain rule pattern, an inside with its derivative nearby?
有没有链式法则的模式， 也就是有一个内层，而它的导数就在旁边？
Then substitute.
那就换元。
Is it a product of two unlike things? Then integrate by parts.
它是两个不同类东西的乘积吗，比如一个代数式乘以一个指数？
Is it a rational function?
那就用分部积分。
Then check the degrees first.
它是有理函数吗？ 那就先看次数。
Top heavy means long division.
头重脚轻就用长除法。
Factorable bottom means partial fractions.
分母可因式分解就用部分分式。
An unfactorable quadratic bottom means complete the square.
分母是不可分解的二次式就配方。 动笔前先读结构，这才是省时间的地方。
Three marks students lose.
学生最常丢分的三个地方。
First, on an indefinite integral, never forget the constant.
第一，算不定积分时，绝不要忘记那个常数。
Second, when you substitute in a definite integral, change the limits too, or convert back before you evaluate.
第二，在定积分里换元时，上下限也要跟着换，或者在求值前换回去。
Third, remember a definite integral is a signed area.
第三，记住定积分是有符号的面积。
If the question wants total area, split it at every sign change.
如果题目要的是总面积，就要在每个变号处把它拆开。