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积分的应用

AP 微积分 BC · 第 8 主题

训练
讲义 词汇表
8.1

求函数在区间上的平均值

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-4
Definite integrals allow us to solve problems involving the accumulation of change over an interval.

CHA-4.B
Determine the average value of a function using definite integrals.

  • CHA-4.B.1 The average value of a continuous function $f$ over an interval $[a, b]$ is $\dfrac{1}{b-a}\int_a^b f(x)\,dx$.

来源:美国大学理事会 AP 课程与考试说明

$f$$[a,b]$ 上的平均值(average value)是积分除以宽度:

$$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$
它是一个矩形要有与 $f$ 下方区域相同面积所需的常数高度。不要把它与平均变化率(它用导数)搞混。

Worked example. $f(x)=x^2$$[0,3]$ 上的平均值是 $\dfrac{1}{3}\displaystyle\int_0^3 x^2\,dx=\dfrac13\left[\dfrac{x^3}{3}\right]_0^3=\dfrac13(9)=3$

探索

The average value of a function

y = ax³ + bx² + cx + d

The average value of $f$ on $[a,b]$ is its integral divided by the width — the constant height whose rectangle has the same area as under the curve.

词汇表 训练
英文 中文 拼音
average value 平均值 píng jūn zhí
8.2

用积分联系函数的位置、速度与加速度

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-4
Definite integrals allow us to solve problems involving the accumulation of change over an interval.

CHA-4.C
Determine values for positions and rates of change using definite integrals in problems involving rectilinear motion.

  • CHA-4.C.1 For a particle in rectilinear motion over an interval of time, the definite integral of velocity represents the particle's displacement over the interval of time, and the definite integral of speed represents the particle's total distance traveled over the interval of time.

来源:美国大学理事会 AP 课程与考试说明

对于直线运动,积分反转求导:

$$v(t)=\int a(t)\,dt,\qquad s(t)=\int v(t)\,dt.$$
两个关键的区别:位移(displacement)是 $\int_a^b v\,dt$(位置的净变化),而总路程(total distance)是 $\int_a^b |v|\,dt$(在 $v$ 改变符号的地方拆分)。速率是 $|v|$

词汇表 训练
英文 中文 拼音
displacement 位移 wèi yí
total distance 总路程 zǒng lù chéng
8.3

在实际情境中运用累积函数与定积分

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-4
Definite integrals allow us to solve problems involving the accumulation of change over an interval.

CHA-4.D
Interpret the meaning of a definite integral in accumulation problems.

  • CHA-4.D.1 A function defined as an integral represents an accumulation of a rate of change.
  • CHA-4.D.2 The definite integral of the rate of change of a quantity over an interval gives the net change of that quantity over that interval.

CHA-4.E
Determine net change using definite integrals in applied contexts.

  • CHA-4.E.1 The definite integral can be used to express information about accumulation and net change in many applied contexts.

来源:美国大学理事会 AP 课程与考试说明

当一个率被给出(流量、每天的销售)时,定积分给出累积总量,而 $\int_a^b R(t)\,dt$ 携带 $R$ 乘时间的单位。一个常见的设置:初始数量 $+\int(\text{rate in}-\text{rate out})\,dt$ 给出一个较后时刻的数量。总是在上下文里、带单位解释答案。

8.4

求以 x 为自变量表示的曲线之间的面积

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.A
Calculate areas in the plane using the definite integral.

  • CHA-5.A.1 Areas of regions in the plane can be calculated with definite integrals.

来源:美国大学理事会 AP 课程与考试说明

$y=f(x)$(顶部)和 $y=g(x)$(底部)从 $a$$b$ 之间的面积是

$$\int_a^b\big(f(x)-g(x)\big)\,dx.$$
为限找到交点,并总是减顶部减底部

两条曲线之间的面积是顶部减底部的积分
两条曲线之间的面积是顶部减底部的积分

Worked example.$y=x$$y=x^2$ 之间(在 $x=0,1$ 相交,$y=x$ 在顶部),面积是 $\displaystyle\int_0^1 (x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac16$

8.5

求以 y 为自变量表示的曲线之间的面积

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.A
Calculate areas in the plane using the definite integral.

  • CHA-5.A.2 Areas of regions in the plane can be calculated using functions of either $x$ or $y$.

来源:美国大学理事会 AP 课程与考试说明

当曲线更容易描述为 $x=f(y)$ 时,改为对 $y$ 积分,用右减左:

$$\int_c^d\big(f_{\text{right}}(y)-g_{\text{left}}(y)\big)\,dy.$$
选择对 $y$ 积分能避免把区域拆分成几块。

8.6

求相交多于两点的曲线之间的面积

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.A
Calculate areas in the plane using the definite integral.

  • CHA-5.A.3 Areas of certain regions in the plane may be calculated using a sum of two or more definite integrals or by evaluating a definite integral of the absolute value of the difference of two functions.

来源:美国大学理事会 AP 课程与考试说明

若两条曲线相交几次,顶部和底部交换。在每个交点处拆分区域并用正确的顶部减底部积分每一块(或用 $\int|f-g|$),然后把这些块相加。

8.7

用横截面求体积:正方形与矩形

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.B
Calculate volumes of solids with known cross sections using definite integrals.

  • CHA-5.B.1 Volumes of solids with square and rectangular cross sections can be found using definite integrals and the area formulas for these shapes.

来源:美国大学理事会 AP 课程与考试说明

若一个立体的、垂直于 $x$ 轴的横截面(cross sections)是正方形或矩形,积分它们的面积。以边长等于两条曲线之间的距离,一个正方形横截面给出

$$V=\int_a^b \big(f(x)-g(x)\big)^2\,dx.$$
方法总是"积分横截面面积"。

词汇表 训练
英文 中文 拼音
cross sections 横截面 héng jié miàn
8.8

用横截面求体积:三角形与半圆

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.B
Calculate volumes of solids with known cross sections using definite integrals.

  • CHA-5.B.2 Volumes of solids with triangular cross sections can be found using definite integrals and the area formulas for these shapes.
  • CHA-5.B.3 Volumes of solids with semicircular and other geometrically defined cross sections can be found using definite integrals and the area formulas for these shapes.
    • Illustrative examples for CHA-5.B.3:
      • The volume of a funnel whose cross sections are circles can be found using the area formula for a circle and definite integrals (see 2016 AB Exam FRQ #5(b)).
      • The volume of a solid whose cross sectional area is defined using a function can be found using the known area function and a definite integral (see 2009 AB Exam FRQ #4(c)).

来源:美国大学理事会 AP 课程与考试说明

相同的思想,不同的面积公式:对于等边三角形横截面用 $A=\tfrac{\sqrt3}{4}s^2$,而对于半圆的用 $A=\tfrac{\pi}{8}s^2$(以 $s$ 为曲线之间的距离)。代入面积公式并积分。

8.9

圆盘法求体积:绕 x 轴或 y 轴旋转

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.C
Calculate volumes of solids of revolution using definite integrals.

  • CHA-5.C.1 Volumes of solids of revolution around the $x$- or $y$-axis may be found by using definite integrals with the disc method.

来源:美国大学理事会 AP 课程与考试说明

旋转体:圆盘法

把一个区域绕一个轴旋转制造一个横截面是圆盘的立体。圆盘法(disc method)积分 $\pi(\text{radius})^2$:

$$V=\pi\int_a^b \big(R(x)\big)^2\,dx,$$
其中半径 $R$ 是从曲线到轴的距离。绕 $y$ 轴旋转时用 $dy$

圆盘法:把 y=f(x) 绕轴旋转扫出半径 f(x) 的圆盘
圆盘法:把 y=f(x) 绕轴旋转扫出半径 f(x) 的圆盘

Worked example.$y=\sqrt{x}$$0$$4$ 下方的区域绕 $x$ 轴旋转给出半径 $\sqrt{x}$ 的圆盘:$V=\pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=8\pi$

把一个区域绕一个轴旋转扫出一个旋转体
把一个区域绕一个轴旋转扫出一个旋转体
词汇表 训练
英文 中文 拼音
disc method 圆盘法 yuán pán fǎ
8.10

圆盘法求体积:绕其他轴旋转

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.C
Calculate volumes of solids of revolution using definite integrals.

  • CHA-5.C.2 Volumes of solids of revolution around any horizontal or vertical line in the plane may be found by using definite integrals with the disc method.

来源:美国大学理事会 AP 课程与考试说明

当旋转轴是一条像 $y=k$(不是一个轴)的水平或垂直线时,半径调整:$R=|f(x)-k|$。把半径设置为从曲线到那条线的距离,然后像之前一样积分 $\pi R^2$

8.11

垫圈法求体积:绕 x 轴或 y 轴旋转

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.C
Calculate volumes of solids of revolution using definite integrals.

  • CHA-5.C.3 Volumes of solids of revolution around the $x$- or $y$-axis whose cross sections are ring shaped may be found using definite integrals with the washer method.

来源:美国大学理事会 AP 课程与考试说明

垫圈法求体积

若区域不触及轴,旋转留下一个洞,所以横截面是垫圈(washers)(环)。垫圈法(washer method)减去内圆盘:

$$V=\pi\int_a^b\Big(R_{\text{outer}}^2-R_{\text{inner}}^2\Big)\,dx.$$
把外半径和内半径辨别为从每条曲线到轴的距离。

词汇表 训练
英文 中文 拼音
washer method 垫圈法 diàn juàn fǎ
8.12

垫圈法求体积:绕其他轴旋转

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-5
Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

CHA-5.C
Calculate volumes of solids of revolution using definite integrals.

  • CHA-5.C.4 Volumes of solids of revolution around any horizontal or vertical line whose cross sections are ring shaped may be found using definite integrals with the washer method.

来源:美国大学理事会 AP 课程与考试说明

与圆盘一样,绕一条线 $y=k$$x=k$ 旋转移动两个半径——每个变成从它的曲线到那条线的距离。草绘区域和轴、标注 $R_{\text{outer}}$$R_{\text{inner}}$,然后积分平方之差。

8.13

光滑平面曲线的弧长与所经过的距离

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

CHA-6
Definite integrals allow us to solve problems involving the accumulation of change in length over an interval.

CHA-6.A
Determine the length of a curve in the plane defined by a function, using a definite integral. BC ONLY

  • CHA-6.A.1 The length of a planar curve defined by a function can be calculated using a definite integral. BC ONLY

来源:美国大学理事会 AP 课程与考试说明

$y=f(x)$$a$$b$弧长(arc length)是

$$L=\int_a^b\sqrt{1+\big(f'(x)\big)^2}\,dx.$$
这个 BC 独有的公式来自把微小的斜边 $\sqrt{dx^2+dy^2}$ 相加。相同的思想给出一个质点沿一条弯曲路径行进的距离。

Worked example.$y=\tfrac{2}{3}x^{3/2}$$x=0$$x=3$ 的弧长。这里 $f'(x)=x^{1/2}$,所以 $1+(f')^2=1+x$

$$L=\int_0^3\sqrt{1+x}\,dx=\left[\tfrac{2}{3}(1+x)^{3/2}\right]_0^3=\tfrac{2}{3}(8-1)=\tfrac{14}{3}.$$

词汇表 训练
英文 中文 拼音
arc length 弧长 hú zhǎng
8.13

考试技巧

  • 曲线之间的面积是 $\int(\text{top}-\text{bottom})\,dx$ ——为限找到交点并保持顶部减底部。
  • 对于一个旋转体,加起面积 $\pi r^2$(或 $\pi(R^2-r^2)$)的圆盘/垫圈横截面。
  • $f$$[a,b]$ 上的平均值$\tfrac{1}{b-a}\int_a^b f\,dx$
  • 累积变化是一个率的 $\int$:总 = 初始值 $+\int_a^b(\text{rate})\,dt$
  • 积分意味着"加起无穷多个微小的片段"——把被积函数设置为一个薄切片。

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