积分与变化的累积

AP 微积分 AB · 第 6 主题

训练

讲义词汇表

6.1

探究变化的累积

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
CHA-4 Definite integrals allow us to solve problems involving the accumulation of change over an interval.	CHA-4.A Interpret the meaning of areas associated with the graph of a rate of change in context.	CHA-4.A.1 The area of the region between the graph of a rate of change function and the $x$ axis gives the accumulation of change. CHA-4.A.2 In some cases, accumulation of change can be evaluated by using geometry. CHA-4.A.3 If a rate of change is positive (negative) over an interval, then the accumulated change is positive (negative). CHA-4.A.4 The unit for the area of a region defined by rate of change is the unit for the rate of change multiplied by the unit for the independent variable.

来源:美国大学理事会 AP 课程与考试说明

求导找到率。积分反向运行这个思想:给定一个变化率(rate of change),它找到累积变化(accumulated change)。关键的图画:一个率函数的图象和 $x$ 轴之间的面积(area)给出总累积。

若率在一个区间上是正的,累积变化是正的;若是负的,是负的。坐标轴下方的面积算作负的。
简单的区域(三角形、矩形)能用几何(geometry)求出。
单位: 面积的单位是率的单位乘输入的单位。一个以车辆每小时计的率乘小时给出车辆。

词汇表训练

英文	中文	拼音
accumulated change	累积变化	lěi jī biàn huà
geometry	几何	jǐ hé

练习卷

6.2

用黎曼和近似面积

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
LIM-5 Definite integrals can be approximated using geometric and numerical methods.	LIM-5.A Approximate a definite integral using geometric and numerical methods.	LIM-5.A.1 Definite integrals can be approximated for functions that are represented graphically, numerically, analytically, and verbally. LIM-5.A.2 Definite integrals can be approximated using a left Riemann sum, a right Riemann sum, a midpoint Riemann sum, or a trapezoidal sum; approximations can be computed using either uniform or nonuniform partitions. LIM-5.A.3 Definite integrals can be approximated using numerical methods, with or without technology. LIM-5.A.4 Depending on the behavior of a function, it may be possible to determine whether an approximation for a definite integral is an underestimate or overestimate for the value of the definite integral.

来源:美国大学理事会 AP 课程与考试说明

定积分与黎曼和

梯形和近似面积

当精确的面积很难时,用一个黎曼和(Riemann sum)近似它——把区间分成子区间并加起矩形(或梯形)的面积。四个标准的估计:

一个黎曼和:宽度为 dx 的矩形近似一条曲线下的面积 — 每个矩形有面积 $f(x)\,\Delta x$;把它们相加估计面积,而随着条变窄和趋近定积分。

左黎曼和——高度来自每个子区间的左端点。
右黎曼和——高度来自右端点。
中点黎曼和——高度来自中点(midpoint)。
梯形法(trapezoidal sum)——对两个端点高度取平均(一个梯形)。

子区间可能是均匀的(相等宽度)或非均匀的——从表读宽度。

高估还是低估? 从函数的行为判断:对于一个递增函数,一个左和低估而一个右和高估;一个梯形和在函数上凹时高估而在下凹时低估。考试部分要求你陈述哪个以及为什么。

Worked example. 一张表给出 $f(0)=3$、$f(2)=5$、$f(4)=8$、$f(6)=9$。用三个相等子区间($\Delta x=2$)的一个右黎曼和估计 $\int_0^6 f(x)\,dx$。用每个条的右端点:

$$2\big(f(2)+f(4)+f(6)\big)=2(5+8+9)=44.$$

因为 $f$ 递增,这个右和是一个高估;左和 $2(3+5+8)=32$ 会是一个低估。

探索

Approximate area with rectangles

y = ax³ + bx² + cx + d

A Riemann sum approximates the area under a curve with rectangles. Add more, thinner rectangles and the estimate converges to the exact definite integral.

词汇表训练

英文	中文	拼音
Riemann sum	黎曼和	lí màn hé
midpoint	中点	zhōng diǎn
Trapezoidal sum	梯形法	tī xíng fǎ

练习卷

6.3

黎曼和、求和记号与定积分记号

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
LIM-5 Definite integrals can be approximated using geometric and numerical methods.	LIM-5.B Interpret the limiting case of the Riemann sum as a definite integral.	LIM-5.B.1 The limit of an approximating Riemann sum can be interpreted as a definite integral. LIM-5.B.2 A Riemann sum, which requires a partition of an interval $I$, is the sum of products, each of which is the value of the function at a point in a subinterval multiplied by the length of that subinterval of the partition.
	LIM-5.C Represent the limiting case of the Riemann sum as a definite integral.	LIM-5.C.1 The definite integral of a continuous function $f$ over the interval $[a, b]$, denoted by $\int_{a}^{b} f(x)\,dx$, is the limit of Riemann sums as the widths of the subintervals approach 0. That is, $\int_{a}^{b} f(x)\,dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(x_i^{})\Delta x_i$, where $n$ is the number of subintervals, $\Delta x_i$ is the width of the $i$th subinterval, and $x_i^{}$ is a value in the $i$th subinterval. LIM-5.C.2 A definite integral can be translated into the limit of a related Riemann sum, and the limit of a Riemann sum can be written as a definite integral.

来源:美国大学理事会 AP 课程与考试说明

随着子区间宽度收缩到零,黎曼和趋近一个精确的值——定积分(definite integral):

$$\int_a^b f(x)\,dx = \lim_{\max \Delta x_i \to 0}\sum_{i=1}^{n} f(x_i^{*})\,\Delta x_i.$$

这里 $\Delta x_i$ 是第 $i$ 个子区间的宽度而 $x_i^{*}$ 是它里面的一个点。所以一个定积分就是一个黎曼和的极限,而你应当能够把每个转换成另一个。

词汇表训练

英文	中文	拼音
definite integral	定积分	dìng jī fēn

练习卷

6.4

微积分基本定理与累积函数

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-5 The Fundamental Theorem of Calculus connects differentiation and integration.	FUN-5.A Represent accumulation functions using definite integrals.	FUN-5.A.1 The definite integral can be used to define new functions. Illustrative examples for FUN-5.A.1: $f(x) = \int_{0}^{x} e^{-t^2}\,dt$. FUN-5.A.2 If $f$ is a continuous function on an interval containing $a$, then $\dfrac{d}{dx}\left( \int_{a}^{x} f(t)\,dt \right) = f(x)$, where $x$ is in the interval.

来源:美国大学理事会 AP 课程与考试说明

微积分基本定理

一个有一个变的上限的定积分定义一个新的累积函数(accumulation function)。微积分基本定理(Fundamental Theorem of Calculus)(第一部分)说求导撤销这个累积:若 $f$ 连续,那么

$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x).$$

所以若 $g(x)=\int_a^x f(t)\,dt$,那么 $g'(x)=f(x)$ 而 $g''(x)=f'(x)$。这是那些很常见的"设 $g(x)=\int_a^x f(t)\,dt$"问题背后的引擎。

累积函数加上有符号的面积;FTC 说它的导数是 f

探索

Accumulate area as an integral

y = ax³ + bx² + cx + d

An accumulation function $\int_a^x f(t)\,dt$ builds up signed area as $x$ moves. The Fundamental Theorem says its derivative is just $f(x)$.

词汇表训练

英文	中文	拼音
accumulation function	累积函数	lěi jī hán shù
Fundamental Theorem of Calculus	微积分基本定理	wēi jī fēn jī běn dìng lǐ

练习卷

6.5

解释涉及面积的累积函数的性态

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-5 The Fundamental Theorem of Calculus connects differentiation and integration.	FUN-5.A Represent accumulation functions using definite integrals.	FUN-5.A.3 Graphical, numerical, analytical, and verbal representations of a function $f$ provide information about the function $g$ defined as $g(x) = \int_{a}^{x} f(t)\,dt$.

来源:美国大学理事会 AP 课程与考试说明

因为 $g'(x)=f(x)$,第 5 单元的一切都用 $f$ 的图象应用于一个累积函数:

$g$ 在 $f>0$ 的地方递增而在 $f<0$ 的地方递减;
$g$ 在 $f$ 穿越零(带一个符号变化)的地方有一个局部极值;
$g$ 在 $f$ 递增的地方上凹;$g$ 的拐点出现在 $f$ 有一个局部极值的地方。

要得到 $g$ 的一个值,计算有符号的面积:$g(x)=\int_a^x f(t)\,dt$,加坐标轴上方的面积并减坐标轴下方的面积。

练习卷

6.6

应用定积分的性质

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-6 Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.	FUN-6.A Calculate a definite integral using areas and properties of definite integrals.	FUN-6.A.1 In some cases, a definite integral can be evaluated by using geometry and the connection between the definite integral and area. FUN-6.A.2 Properties of definite integrals include the integral of a constant times a function, the integral of the sum of two functions, reversal of limits of integration, and the integral of a function over adjacent intervals. FUN-6.A.3 The definition of the definite integral may be extended to functions with removable or jump discontinuities.

来源:美国大学理事会 AP 课程与考试说明

这些性质简化计算并不断出现:

$$\int_a^a f = 0,\qquad \int_b^a f = -\int_a^b f,\qquad \int_a^b \big(f\pm g\big) = \int_a^b f \pm \int_a^b g,$$

$$\int_a^b k\,f = k\int_a^b f,\qquad \int_a^c f + \int_c^b f = \int_a^b f.$$

最后一个(在一个内部点 $c$ 拆分)让你从一个图象读出的片段构建一个总积分。

练习卷

6.7

微积分基本定理与定积分

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-6 Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.	FUN-6.B Evaluate definite integrals analytically using the Fundamental Theorem of Calculus.	FUN-6.B.1 An antiderivative of a function $f$ is a function $g$ whose derivative is $f$. FUN-6.B.2 If a function $f$ is continuous on an interval containing $a$, the function defined by $F(x) = \int_{a}^{x} f(t)\,dt$ is an antiderivative of $f$ for $x$ in the interval. FUN-6.B.3 If $f$ is continuous on the interval $[a, b]$ and $F$ is an antiderivative of $f$, then $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$.

来源:美国大学理事会 AP 课程与考试说明

基本定理的第二部分用一个原函数(antiderivative)求一个定积分。若 $F'=f$,那么

$$\int_a^b f(x)\,dx = F(b)-F(a).$$

所以:找到任何一个原函数 $F$,然后减去它在两个限处的值。这是大多数精确积分被计算的方式。它也给出净变化的视角:$\int_a^b g'(t)\,dt = g(b)-g(a)$,所以一个起始值加累积变化给出一个较后的值,例如 $g(5)=g(0)+\int_0^5 g'(t)\,dt$。

Worked example. 求 $\int_1^3 (2x+1)\,dx$。一个原函数是 $F(x)=x^2+x$,所以

$$\int_1^3(2x+1)\,dx=F(3)-F(1)=(9+3)-(1+1)=12-2=10.$$

一个定积分是曲线和 x 轴之间有符号的面积

词汇表训练

英文	中文	拼音
antiderivative	原函数	yuán hán shù

练习卷

6.8

求原函数与不定积分：基本法则与记号

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-6 Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.	FUN-6.C Determine antiderivatives of functions and indefinite integrals, using knowledge of derivatives.	FUN-6.C.1 $\int f(x)\,dx$ is an indefinite integral of the function $f$ and can be expressed as $\int f(x)\,dx = F(x) + C$, where $F'(x) = f(x)$ and $C$ is any constant. FUN-6.C.2 Differentiation rules provide the foundation for finding antiderivatives. FUN-6.C.3 Many functions do not have closed-form antiderivatives.

来源:美国大学理事会 AP 课程与考试说明

一个不定积分(indefinite integral)是所有原函数的族,带一个积分常数(constant of integration)写出:

$$\int f(x)\,dx = F(x)+C.$$

反转每个求导规则以构建基本的原函数:

$$\int x^n\,dx = \frac{x^{n+1}}{n+1}+C\ (n\neq -1),\quad \int \frac{1}{x}\,dx = \ln|x|+C,\quad \int e^x\,dx = e^x+C,$$

$$\int \cos x\,dx = \sin x + C,\quad \int \sin x\,dx = -\cos x + C.$$

词汇表训练

英文	中文	拼音
indefinite integral	不定积分	bù dìng jī fēn
constant of integration	积分常数	jī fēn cháng shù
u-substitution	换元积分法	huàn yuán jī fēn fǎ

练习卷

6.9

用换元法积分

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-6 Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.	FUN-6.D For integrands requiring substitution or rearrangements into equivalent forms: (a) Determine indefinite integrals. (b) Evaluate definite integrals.	FUN-6.D.1 Substitution of variables is a technique for finding antiderivatives. FUN-6.D.2 For a definite integral, substitution of variables requires corresponding changes to the limits of integration.

来源:美国大学理事会 AP 课程与考试说明

$u$-换元($u$-substitution,换元积分法)反转链式法则。选择一个里面的函数 $u=g(x)$,所以 $du=g'(x)\,dx$,并把积分完全以 $u$ 重写:

$$\int f\big(g(x)\big)g'(x)\,dx = \int f(u)\,du.$$

寻找一个函数和它的导数都存在。对于一个定积分,要么把限改成 $u$ 值,要么在代入原来的限之前转换回 $x$。

Worked example. 求 $\int 2x\cos(x^2)\,dx$。里面的函数是 $u=x^2$,它的导数 $2x\,dx=du$ 存在,所以

$$\int 2x\cos(x^2)\,dx=\int \cos u\,du=\sin u + C=\sin(x^2)+C.$$

认出 $2x$ 恰好是 $\dfrac{du}{dx}$ 是全部的诀窍。

练习卷

6.10

用长除法与配方法积分

大纲

Enduring Understanding	Learning Objective	Essential Knowledge
FUN-6 Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.	FUN-6.D For integrands requiring substitution or rearrangements into equivalent forms: (a) Determine indefinite integrals. (b) Evaluate definite integrals.	FUN-6.D.3 Techniques for finding antiderivatives include rearrangements into equivalent forms, such as long division and completing the square.

来源:美国大学理事会 AP 课程与考试说明

两个代数设置的做法让更多的积分符合基本形式:当一个有理函数的上部次数 $\ge$ 底部次数时的多项式长除法(polynomial long division),以及把一个二次分母变成一个积分到反正切或对数的形式的配方法(completing the square)。

词汇表训练

英文	中文	拼音
polynomial long division	多项式长除法	duō xiàng shì zhǎng chú fǎ
completing the square	配方法	pèi fāng fǎ

练习卷

6.14

选择求原函数的方法

大纲

This topic is intended to focus on the skill of selecting an appropriate procedure for antidifferentiation. Students should be given opportunities to practice when and how to apply all learning objectives relating to antidifferentiation.

来源:美国大学理事会 AP 课程与考试说明

一个技能主题:把积分匹配到一个方法。先尝试一个基本的原函数;寻找一个 $u$-换元(一个里面的函数,它的导数也存在);用代数(除法、配方法、拆分一个分数)把被积函数重塑成一个标准形式。先命名结构防止浪费的努力。

练习卷

6.14

考试技巧

积分是反求导;用幂法则 $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}+C$ 并不要忘记 $+C$。
基本定理联系这两者:$\int_a^b f'(x)\,dx=f(b)-f(a)$,而 $\tfrac{d}{dx}\int_a^x f(t)\,dt=f(x)$。
用黎曼和或从一张值表的梯形法则近似一个定积分。
一个定积分是一个有符号的面积(坐标轴下方算负的);在符号变化处拆分以求总面积。
用 u-换元并记得相应地改变限(或反代)。

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