Unit 6: Integration and
Accumulation of Change
单元 6：积分与
变化的累积
Connect differentiation and integration through the Fundamental Theorem of Calculus. Master Riemann sums, antiderivatives, and integration techniques.
通过微积分基本定理把微分与积分联系起来。掌握黎曼和、原函数与各种积分技巧。
Exploring Accumulations of Change
探索变化的累积
rate of change）函数的图像与 $x$ 轴之间的面积，代表变化的累积总量（total accumulation of change）。
Suppose $r(t)$ is a rate of change — for example, gallons per minute flowing into a tank. Then the area under the curve $r(t)$ from $t = a$ to $t = b$ gives the total quantity that accumulated over that time interval. In some cases, you can compute this area using basic geometry (triangles, rectangles, trapezoids) before learning any formal integration techniques.
设 $r(t)$ 是一个变化率——比如流入水箱的加仑/分钟。那么曲线 $r(t)$ 从 $t = a$ 到 $t = b$ 之间的曲线下面积（area under curve），就给出在该时间区间内累积的总量。有时候，无需任何正式的积分技巧，仅用基本几何（三角形、矩形、梯形）就能算出这块面积。
gallons/min and time in min → area has units gallons/min × min = gallons.
Example: Rate in cars/hour and time in hours → area has units cars.
gallons/min，时间单位是 min → 面积的单位是 gallons/min × min = gallons。
例：变化率单位是 cars/hour，时间单位是 hours → 面积的单位是 cars。
Worked Example — Accumulation from a Rate Graph例题——由变化率图求累积
A tank fills at rate $r(t)$ gallons/hour. The graph of $r(t)$ forms a triangle with base 4 hours and height 6 gal/hr.
Total water added $=$ area under the rate curve:
$$\text{Area} = \tfrac{1}{2} \times 4 \times 6 = 12 \text{ gallons}$$水箱以变化率 $r(t)$ 加仑/小时进水。$r(t)$ 的图像是一个底为 4 小时、高为 6 加仑/小时的三角形。
累积的总水量 $=$ 变化率曲线下面积：
$$\text{Area} = \tfrac{1}{2} \times 4 \times 6 = 12 \text{ gallons}$$Approximating Areas with Riemann Sums
用黎曼和近似面积
When the region under a curve isn't a simple geometric shape, we approximate the area by slicing it into thin rectangles. The total area of those rectangles — called a Riemann sum — estimates the true integral. The more rectangles we use, the better the estimate becomes.
当曲线下区域不是简单几何形状时，我们把它切成很多细长矩形来近似面积。这些矩形面积的总和——叫做黎曼和（Riemann sum）——就是积分（integral）的估计值。矩形越多，估计越准。
Drag the slider above and watch the rectangles multiply. Notice how the gaps and overlaps between the rectangles and the curve shrink toward zero as $n$ grows. That's the key geometric intuition: the Riemann sum converges to the exact area under the curve.
拖动上方的滑块，看着矩形越来越多。注意当 $n$ 增大时，矩形与曲线之间的缝隙和重叠是怎样逐步趋近于零的。这就是核心的几何直觉：黎曼和收敛（convergence）到曲线下精确面积。
| Method | Height Chosen At | Over/Under? |
|---|---|---|
| Left Riemann Sum (LRS) | Left endpoint of each subinterval | Depends on increasing / decreasing |
| Right Riemann Sum (RRS) | Right endpoint of each subinterval | Depends on increasing / decreasing |
| Midpoint Riemann Sum (MRS) | Midpoint of each subinterval | Usually the most accurate of the three |
| Trapezoidal Sum | Average of left & right (trapezoid, not rectangle) | Depends on concavity |
| 方法 | 高度取在 | 高估/低估？ |
|---|---|---|
左黎曼和（LRS） | 每个子区间的左端点 | 取决于递增/递减 |
右黎曼和（RRS） | 每个子区间的右端点 | 取决于递增/递减 |
中点黎曼和（MRS） | 每个子区间的中点 | 三者中通常最准 |
梯形和（trapezoidal sum） | 左右端点的平均（梯形而非矩形） | 取决于凹凸性 |
Worked Example — Left Riemann Sum from a Table例题——由表格求左黎曼和
Given table:
| $t$ | 0 | 1 | 3 | 6 |
| $f(t)$ | 2 | 5 | 4 | 7 |
Left Riemann Sum (3 nonuniform subintervals): $\Delta t_1 = 1,\; \Delta t_2 = 2,\; \Delta t_3 = 3$.
$$\begin{aligned} \text{LRS} &= f(0)\,\Delta t_1 + f(1)\,\Delta t_2 + f(3)\,\Delta t_3 \\[4pt] &= 2(1) + 5(2) + 4(3) \\[4pt] &= 2 + 10 + 12 = 24 \end{aligned}$$给定表格：
| $t$ | 0 | 1 | 3 | 6 |
| $f(t)$ | 2 | 5 | 4 | 7 |
左黎曼和（3 个非等距子区间）：$\Delta t_1 = 1,\; \Delta t_2 = 2,\; \Delta t_3 = 3$。
$$\begin{aligned} \text{LRS} &= f(0)\,\Delta t_1 + f(1)\,\Delta t_2 + f(3)\,\Delta t_3 \\[4pt] &= 2(1) + 5(2) + 4(3) \\[4pt] &= 2 + 10 + 12 = 24 \end{aligned}$$Riemann Sums, Summation Notation & Definite Integral Notation
黎曼和、求和记号与定积分记号
The interactive visualizer above showed you the key idea geometrically. Now we formalize it: the definite integral is defined as the limit of Riemann sums as the number of rectangles approaches infinity and their widths approach zero.
上方的交互可视化已从几何上展示了关键思路。现在把它正式化：定积分（definite integral）被定义为黎曼和的极限——当矩形数趋于无穷、宽度趋于零时的极限值。
This is the foundational definition. The integral symbol $\int$ is an elongated "S" for Sum. The $dx$ is the infinitesimal remnant of $\Delta x$. Every time you write $\int_a^b f(x)\,dx$, you are writing a limit of a sum of products — infinitely many, infinitely thin rectangles whose total area equals the exact area under the curve.
这是最根本的定义。积分号 $\int$ 是拉长了的字母 “S”，表示 Sum（求和）。$dx$ 是 $\Delta x$ 在取极限后留下的“无穷小”痕迹。每写一个 $\int_a^b f(x)\,dx$，你就是在写一个乘积之和的极限——无穷多个、无穷薄的矩形，其总面积等于曲线下精确面积。
limit）给出精确值。这个极限就是定积分。这与导数（derivative）被定义为差商的极限是类比的——积分被定义为求和的极限。
For equal subintervals of width $\Delta x = \frac{b-a}{n}$, the definition simplifies:
当子区间为等宽 $\Delta x = \frac{b-a}{n}$ 时，定义可简化为：
integrand），即我们要算其面积的函数
$dx$ —— 微分，标明积分变量，并提醒这是无穷小宽度 $\Delta x$
合起来，$f(x)\,dx$ 表示每个无穷小矩形的“高 × 宽”。
Worked Example — Converting Sum to Integral例题——把求和改写为积分
Express as a definite integral:
$$\lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + \frac{5i}{n}\right)^{\!2} \cdot \frac{5}{n}$$Identify the pieces: $\Delta x = \frac{5}{n}$, so $b - a = 5$. The starting value is $a = 3$, giving $b = 8$, and $f(x) = x^2$.
$$= \int_3^8 x^2\,dx$$把下式表示为定积分：
$$\lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + \frac{5i}{n}\right)^{\!2} \cdot \frac{5}{n}$$识别各部分：$\Delta x = \frac{5}{n}$，所以 $b - a = 5$。起点 $a = 3$，故 $b = 8$，且 $f(x) = x^2$。
$$= \int_3^8 x^2\,dx$$The Fundamental Theorem of Calculus & Accumulation Functions
微积分基本定理与累积函数
Fundamental Theorem of Calculus, FTC）是本单元最重要的结论：微分和积分是互为逆运算的过程。
where $f$ is continuous on an interval containing $a$
其中 $f$ 在包含 $a$ 的区间上连续
This tells us: if you accumulate a continuous function from a fixed starting point $a$ up to a variable endpoint $x$, then the rate at which that accumulation changes is just the original function evaluated at $x$.
这告诉我们：把一个连续函数从固定起点 $a$ 累积到一个可变的端点 $x$，那么这个累积量变化的速率恰好就是原函数在 $x$ 处的值。
Worked Example — FTC Part 1 with Chain Rule例题——FTC 第一部分配合链式法则
Find $F'(x)$ if $\displaystyle F(x) = \int_1^{x^2} \cos(t^3)\,dt$.
Upper limit $g(x) = x^2 \;\Longrightarrow\; g'(x) = 2x$. Integrand: $f(t) = \cos(t^3)$.
$$\begin{aligned} F'(x) &= f\bigl(g(x)\bigr)\cdot g'(x) \\[4pt] &= \cos\bigl((x^2)^3\bigr)\cdot 2x \\[4pt] &= 2x\cos(x^6) \end{aligned}$$若 $\displaystyle F(x) = \int_1^{x^2} \cos(t^3)\,dt$，求 $F'(x)$。
上限 $g(x) = x^2 \;\Longrightarrow\; g'(x) = 2x$。被积函数：$f(t) = \cos(t^3)$。
$$\begin{aligned} F'(x) &= f\bigl(g(x)\bigr)\cdot g'(x) \\[4pt] &= \cos\bigl((x^2)^3\bigr)\cdot 2x \\[4pt] &= 2x\cos(x^6) \end{aligned}$$Worked Example — FTC Part 1 with Chain Rule on the Lower Limit例题——FTC 第一部分作用在下限的链式法则
Find $H'(x)$ if $\displaystyle H(x) = \int_{x^2}^{3} \cos t\,dt$.
Step 1 — Reverse the bounds. Use the property $\int_a^b = -\int_b^a$ so the variable bound is on top:
$$ H(x) \;=\; -\int_{3}^{x^2} \cos t\,dt $$Step 2 — Apply FTC Part 1 with chain rule on the upper limit. Now $g(x)=x^2$, $g'(x)=2x$, integrand $f(t)=\cos t$:
$$\begin{aligned} H'(x) &= -\,f\!\bigl(g(x)\bigr)\cdot g'(x) \\[4pt] &= -\cos(x^2)\cdot 2x \\[4pt] &= -\,2x\cos(x^2) \end{aligned}$$Sanity check. The minus sign comes from reversing the bounds — not from anything in the integrand. If you skip Step 1 and apply the chain-rule formula directly to a lower-limit integral, you will get the wrong sign.
若 $\displaystyle H(x) = \int_{x^2}^{3} \cos t\,dt$，求 $H'(x)$。
第一步——交换上下限。用性质 $\int_a^b = -\int_b^a$，把变量限放到上方：
$$ H(x) \;=\; -\int_{3}^{x^2} \cos t\,dt $$第二步——对上限用 FTC 第一部分 + 链式法则。此时 $g(x)=x^2$，$g'(x)=2x$，被积函数 $f(t)=\cos t$：
$$\begin{aligned} H'(x) &= -\,f\!\bigl(g(x)\bigr)\cdot g'(x) \\[4pt] &= -\cos(x^2)\cdot 2x \\[4pt] &= -\,2x\cos(x^2) \end{aligned}$$核对一下。负号是因为交换上下限而出现的——不是被积函数带来的。如果跳过第一步，直接把链式公式套到下限变量的积分上，符号一定会算错。
Interpreting the Behavior of Accumulation Functions
解读累积函数的行为
If $g(x) = \int_a^x f(t)\,dt$, then the graph of $f$ tells you everything about the behavior of $g$. Since $g'(x) = f(x)$ by FTC Part 1, the graph of $f$ is literally the derivative graph of $g$ — so all the analysis techniques from Unit 5 apply directly.
若 $g(x) = \int_a^x f(t)\,dt$，则 $f$ 的图像就能告诉你 $g$ 的全部行为。因为由 FTC 第一部分 $g'(x) = f(x)$，$f$ 的图像其实就是 $g$ 的导数图——所以单元 5 的所有分析技巧都可以直接套用。
| If $f(t)$ is… | Then $g(x)$ is… | Because… |
|---|---|---|
| Positive | Increasing | $g' = f > 0$ → accumulation is growing |
| Negative | Decreasing | $g' = f < 0$ → accumulation is shrinking |
| Zero (sign change) | Relative extremum | $g' = 0$ and changes sign → first derivative test |
| Increasing | Concave up | $g'' = f' > 0$ |
| Decreasing | Concave down | $g'' = f' < 0$ |
| 若 $f(t)$ 为… | 则 $g(x)$ 为… | 原因… |
|---|---|---|
| 正 | 递增 | $g' = f > 0$ → 累积在增长 |
| 负 | 递减 | $g' = f < 0$ → 累积在减少 |
| 零（变号） | 相对极值 | $g' = 0$ 且变号 → 一阶导数判别法 |
| 递增 | 上凹 | $g'' = f' > 0$ |
| 递减 | 下凹 | $g'' = f' < 0$ |
accumulation function） $g$ 就是 $f$ 下方有符号面积的“累计总和”。$f$ 在 $x$ 轴上方时，$g$ 递增；$f$ 落到 $x$ 轴下方时，$g$ 递减。$f$ 的零点（变号点）对应 $g$ 的一个极值点。
Applying Properties of Definite Integrals
运用定积分的性质
| Property | Formula |
|---|---|
| Constant Multiple | $\displaystyle\int_a^b k\cdot f(x)\,dx = k\int_a^b f(x)\,dx$ |
| Sum / Difference | $\displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$ |
| Reversal of Limits | $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ |
| Adjacent Intervals | $\displaystyle\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$ |
| Same Limits | $\displaystyle\int_a^a f(x)\,dx = 0$ |
| 性质 | 公式 |
|---|---|
| 常数倍 | $\displaystyle\int_a^b k\cdot f(x)\,dx = k\int_a^b f(x)\,dx$ |
| 和/差 | $\displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$ |
| 上下限交换 | $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ |
| 相邻区间拼接 | $\displaystyle\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$ |
| 上下限相同 | $\displaystyle\int_a^a f(x)\,dx = 0$ |
Worked Example — Using Properties例题——运用积分性质
Given: $\displaystyle\int_1^4 f(x)\,dx = 7$ and $\displaystyle\int_1^4 g(x)\,dx = 3$.
Find $\displaystyle\int_1^4 [2f(x) - g(x)]\,dx$:
$$= 2\int_1^4 f(x)\,dx - \int_1^4 g(x)\,dx = 2(7) - 3 = 11$$Given additionally $\displaystyle\int_1^6 f(x)\,dx = 10$, find $\displaystyle\int_4^6 f(x)\,dx$:
$$\begin{aligned} \int_1^6 f(x)\,dx &= \int_1^4 f(x)\,dx + \int_4^6 f(x)\,dx \\[4pt] 10 &= 7 + \int_4^6 f(x)\,dx \\[4pt] \int_4^6 f(x)\,dx &= 3 \end{aligned}$$已知 $\displaystyle\int_1^4 f(x)\,dx = 7$，$\displaystyle\int_1^4 g(x)\,dx = 3$。
求 $\displaystyle\int_1^4 [2f(x) - g(x)]\,dx$：
$$= 2\int_1^4 f(x)\,dx - \int_1^4 g(x)\,dx = 2(7) - 3 = 11$$另已知 $\displaystyle\int_1^6 f(x)\,dx = 10$，求 $\displaystyle\int_4^6 f(x)\,dx$：
$$\begin{aligned} \int_1^6 f(x)\,dx &= \int_1^4 f(x)\,dx + \int_4^6 f(x)\,dx \\[4pt] 10 &= 7 + \int_4^6 f(x)\,dx \\[4pt] \int_4^6 f(x)\,dx &= 3 \end{aligned}$$The Fundamental Theorem of Calculus & Definite Integrals
微积分基本定理与定积分
where $F$ is any antiderivative of $f$ (i.e., $F'(x) = f(x)$)
其中 $F$ 是 $f$ 的任一原函数（antiderivative），即满足 $F'(x) = f(x)$
Worked Example — Evaluating a Definite Integral例题——求定积分
Evaluate $\displaystyle\int_1^3 (3x^2 + 2x)\,dx$.
Antiderivative: $F(x) = x^3 + x^2$.
$$\begin{aligned} F(3) - F(1) &= (27 + 9) - (1 + 1) \\[4pt] &= 36 - 2 = 34 \end{aligned}$$求 $\displaystyle\int_1^3 (3x^2 + 2x)\,dx$。
原函数：$F(x) = x^3 + x^2$。
$$\begin{aligned} F(3) - F(1) &= (27 + 9) - (1 + 1) \\[4pt] &= 36 - 2 = 34 \end{aligned}$$Finding Antiderivatives & Indefinite Integrals: Basic Rules
求原函数与不定积分：基本公式
indefinite integral）
$\displaystyle\int f(x)\,dx = F(x) + C$，其中 $F'(x) = f(x)$，$C$ 为任意常数。
不定积分必须写 $+C$。
| Function $f(x)$ | Antiderivative $F(x) + C$ |
|---|---|
| $x^n$ ($n \neq -1$) | $\dfrac{x^{n+1}}{n+1} + C$ |
| $\dfrac{1}{x}$ | $\ln|x| + C$ |
| $e^x$ | $e^x + C$ |
| $a^x$ | $\dfrac{a^x}{\ln a} + C$ |
| $\cos x$ | $\sin x + C$ |
| $\sin x$ | $-\cos x + C$ |
| $\sec^2 x$ | $\tan x + C$ |
| $\csc^2 x$ | $-\cot x + C$ |
| $\sec x \tan x$ | $\sec x + C$ |
| $\csc x \cot x$ | $-\csc x + C$ |
| $\dfrac{1}{\sqrt{1-x^2}}$ | $\arcsin x + C$ |
| $\dfrac{1}{1+x^2}$ | $\arctan x + C$ |
| 函数 $f(x)$ | 原函数 $F(x) + C$ |
|---|---|
| $x^n$ （$n \neq -1$） | $\dfrac{x^{n+1}}{n+1} + C$ |
| $\dfrac{1}{x}$ | $\ln|x| + C$ |
| $e^x$ | $e^x + C$ |
| $a^x$ | $\dfrac{a^x}{\ln a} + C$ |
| $\cos x$ | $\sin x + C$ |
| $\sin x$ | $-\cos x + C$ |
| $\sec^2 x$ | $\tan x + C$ |
| $\csc^2 x$ | $-\cot x + C$ |
| $\sec x \tan x$ | $\sec x + C$ |
| $\csc x \cot x$ | $-\csc x + C$ |
| $\dfrac{1}{\sqrt{1-x^2}}$ | $\arcsin x + C$ |
| $\dfrac{1}{1+x^2}$ | $\arctan x + C$ |
power rule）对 $\int \frac{1}{x}\,dx$ 不适用——答案是 $\ln|x| + C$，不是 $\frac{x^0}{0}$。
Integrating Using Substitution
用换元积分法
integrand）中同时出现某个函数和它的导数时，就用换元积分法（integration by substitution / u-substitution）——这是链式法则的逆运算。
Step-by-Step Process
分步流程
Step 1: Choose $u = g(x)$ (usually the "inside" function).
Step 2: Compute $du = g'(x)\,dx$.
Step 3: Rewrite the entire integral in terms of $u$ and $du$.
Step 4: Integrate with respect to $u$.
Step 5: Substitute back (replace $u$ with $g(x)$) — or, for definite integrals, change the limits to $u$-values.
第一步：选 $u = g(x)$（通常是“里层”的函数）。
第二步：算出 $du = g'(x)\,dx$。
第三步：把整个积分都改写成 $u$ 与 $du$。
第四步：对 $u$ 积分。
第五步：把 $u$ 换回 $g(x)$——或者，对定积分，直接把积分上下限换成 $u$ 值。
Worked Example — Indefinite Integral例题——不定积分
Evaluate $\displaystyle\int 2x\cos(x^2)\,dx$.
Let $u = x^2 \;\Longrightarrow\; du = 2x\,dx$ (already present).
$$\begin{aligned} \int 2x\cos(x^2)\,dx &= \int \cos(u)\,du \\[4pt] &= \sin(u) + C \\[4pt] &= \sin(x^2) + C \end{aligned}$$求 $\displaystyle\int 2x\cos(x^2)\,dx$。
令 $u = x^2 \;\Longrightarrow\; du = 2x\,dx$（被积函数里已经有了）。
$$\begin{aligned} \int 2x\cos(x^2)\,dx &= \int \cos(u)\,du \\[4pt] &= \sin(u) + C \\[4pt] &= \sin(x^2) + C \end{aligned}$$Worked Example — Definite Integral (Change Limits)例题——定积分（换上下限）
Evaluate $\displaystyle\int_0^2 x\,e^{x^2}\,dx$.
Let $u = x^2 \;\Longrightarrow\; du = 2x\,dx \;\Longrightarrow\; x\,dx = \frac{du}{2}$. Change limits: $x=0 \to u=0$; $x=2 \to u=4$.
$$\begin{aligned} \int_0^2 x\,e^{x^2}\,dx &= \frac{1}{2}\int_0^4 e^u\,du \\[4pt] &= \frac{1}{2}\bigl[e^u\bigr]_0^4 \\[4pt] &= \frac{1}{2}(e^4 - 1) \end{aligned}$$求 $\displaystyle\int_0^2 x\,e^{x^2}\,dx$。
令 $u = x^2 \;\Longrightarrow\; du = 2x\,dx \;\Longrightarrow\; x\,dx = \frac{du}{2}$。换上下限：$x=0 \to u=0$；$x=2 \to u=4$。
$$\begin{aligned} \int_0^2 x\,e^{x^2}\,dx &= \frac{1}{2}\int_0^4 e^u\,du \\[4pt] &= \frac{1}{2}\bigl[e^u\bigr]_0^4 \\[4pt] &= \frac{1}{2}(e^4 - 1) \end{aligned}$$Integrating Using Long Division & Completing the Square
用多项式长除法与配方法积分
Some rational integrands need algebraic manipulation before you can integrate.
有些有理函数被积式需要先做代数变形才能积分。
Long Division
多项式长除法
long division）。先做除法，再对得到的多项式部分 + 余式分式分别积分。
Worked Example — Long Division例题——多项式长除法
Evaluate $\displaystyle\int \frac{x^2 + 3}{x - 1}\,dx$.
Long division: $x^2 + 3 = (x-1)(x+1) + 4$, so:
$$\frac{x^2+3}{x-1} = x + 1 + \frac{4}{x-1}$$ $$\begin{aligned} \int \frac{x^2+3}{x-1}\,dx &= \int\!\left(x + 1 + \frac{4}{x-1}\right)dx \\[4pt] &= \frac{x^2}{2} + x + 4\ln|x-1| + C \end{aligned}$$求 $\displaystyle\int \frac{x^2 + 3}{x - 1}\,dx$。
长除法：$x^2 + 3 = (x-1)(x+1) + 4$，所以：
$$\frac{x^2+3}{x-1} = x + 1 + \frac{4}{x-1}$$ $$\begin{aligned} \int \frac{x^2+3}{x-1}\,dx &= \int\!\left(x + 1 + \frac{4}{x-1}\right)dx \\[4pt] &= \frac{x^2}{2} + x + 4\ln|x-1| + C \end{aligned}$$Completing the Square
配方法
Worked Example — Completing the Square例题——配方法
Evaluate $\displaystyle\int \frac{1}{x^2 + 4x + 13}\,dx$.
Complete the square: $x^2 + 4x + 13 = (x+2)^2 + 9$.
This matches $\displaystyle\int \frac{du}{u^2+a^2} = \frac{1}{a}\arctan\!\left(\frac{u}{a}\right)+C$ with $u = x+2$, $a = 3$:
$$= \frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$$求 $\displaystyle\int \frac{1}{x^2 + 4x + 13}\,dx$。
配方：$x^2 + 4x + 13 = (x+2)^2 + 9$。
对应公式 $\displaystyle\int \frac{du}{u^2+a^2} = \frac{1}{a}\arctan\!\left(\frac{u}{a}\right)+C$，取 $u = x+2$，$a = 3$：
$$= \frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$$Integrating Using Integration by Parts
用分部积分法
integration by parts）Worked Example — Integration by Parts例题——分部积分法
Evaluate $\displaystyle\int x\,e^x\,dx$.
Choose by LIATE: $u = x,\; dv = e^x\,dx \;\Longrightarrow\; du = dx,\; v = e^x$.
$$\begin{aligned} \int x\,e^x\,dx &= x\,e^x - \int e^x\,dx \\[4pt] &= x\,e^x - e^x + C \\[4pt] &= e^x(x - 1) + C \end{aligned}$$求 $\displaystyle\int x\,e^x\,dx$。
按 LIATE 选：$u = x,\; dv = e^x\,dx \;\Longrightarrow\; du = dx,\; v = e^x$。
$$\begin{aligned} \int x\,e^x\,dx &= x\,e^x - \int e^x\,dx \\[4pt] &= x\,e^x - e^x + C \\[4pt] &= e^x(x - 1) + C \end{aligned}$$Integrating Using Linear Partial Fractions
用一次因式的部分分式积分
partial fractions）。
Worked Example — Long Division First, Then Decompose例题——先做长除法，再分解
Evaluate $\displaystyle\int \frac{x^2 + 1}{x - 2}\,dx$. Note $\deg(\text{num}) = 2 \ge 1 = \deg(\text{denom})$ — improper. Long-divide first.
Step 1 — Polynomial long division of $x^2 + 1$ by $x - 2$.
- $x^2 \div x = x$. Multiply: $x(x-2) = x^2 - 2x$. Subtract: $(x^2 + 1) - (x^2 - 2x) = 2x + 1$.
- $2x \div x = 2$. Multiply: $2(x-2) = 2x - 4$. Subtract: $(2x + 1) - (2x - 4) = 5$.
So
$$\frac{x^2 + 1}{x - 2} \;=\; x + 2 + \frac{5}{x - 2}.$$The remainder $\dfrac{5}{x - 2}$ is now proper (numerator degree $0 < 1$).
Step 2 — Integrate term by term. The remainder is already a single linear factor — no further partial-fraction decomposition needed:
$$\begin{aligned} \int \frac{x^2 + 1}{x - 2}\,dx &= \int \!\bigl(x + 2 + \tfrac{5}{x - 2}\bigr)\,dx \\[4pt] &= \frac{x^2}{2} + 2x + 5\ln|x - 2| + C. \end{aligned}$$求 $\displaystyle\int \frac{x^2 + 1}{x - 2}\,dx$。注意分子次数为 $2$，不小于分母次数 $1$——假分式。先做长除法。
第一步——用 $x^2 + 1$ 除以 $x - 2$ 做多项式长除法。
- $x^2 \div x = x$。 乘回：$x(x-2) = x^2 - 2x$。 相减：$(x^2 + 1) - (x^2 - 2x) = 2x + 1$。
- $2x \div x = 2$。 乘回：$2(x-2) = 2x - 4$。 相减：$(2x + 1) - (2x - 4) = 5$。
因此
$$\frac{x^2 + 1}{x - 2} \;=\; x + 2 + \frac{5}{x - 2}.$$余项 $\dfrac{5}{x - 2}$ 现在是真分式（分子次数 $0 < 1$）。
第二步——逐项积分。余项已经是单个一次因式，不再需要部分分式分解：
$$\begin{aligned} \int \frac{x^2 + 1}{x - 2}\,dx &= \int \!\bigl(x + 2 + \tfrac{5}{x - 2}\bigr)\,dx \\[4pt] &= \frac{x^2}{2} + 2x + 5\ln|x - 2| + C. \end{aligned}$$Partial-fraction decomposition assumes the rational expression behaves like a sum of $\dfrac{A}{x-a}$ pieces (each of which has lower-degree numerator). If you skip the long division, you're trying to decompose a function that has a polynomial growth term — and the system $A = \cdots$ won't have a consistent solution.
部分分式分解的前提是这个有理式可以写成若干 $\dfrac{A}{x-a}$ 之和（每一项的分子次数都更低）。若跳过长除法，你实际上是在试图分解一个还带有多项式增长项的函数——求 $A = \cdots$ 的方程组就不会有相容解。
Worked Example — Partial Fractions例题——部分分式
Evaluate $\displaystyle\int \frac{5}{x^2 - x - 6}\,dx$.
Factor: $x^2 - x - 6 = (x-3)(x+2)$. Set up partial fractions:
$$\frac{5}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$$Multiply through: $5 = A(x+2) + B(x-3)$.
$$x=3:\; 5 = 5A \;\Longrightarrow\; A = 1 \qquad x=-2:\; 5 = -5B \;\Longrightarrow\; B = -1$$ $$\begin{aligned} \int \frac{5}{x^2-x-6}\,dx &= \int\!\left(\frac{1}{x-3} - \frac{1}{x+2}\right)dx \\[4pt] &= \ln|x-3| - \ln|x+2| + C \end{aligned}$$求 $\displaystyle\int \frac{5}{x^2 - x - 6}\,dx$。
分解：$x^2 - x - 6 = (x-3)(x+2)$。设部分分式：
$$\frac{5}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$$两边乘上分母：$5 = A(x+2) + B(x-3)$。
$$x=3:\; 5 = 5A \;\Longrightarrow\; A = 1 \qquad x=-2:\; 5 = -5B \;\Longrightarrow\; B = -1$$ $$\begin{aligned} \int \frac{5}{x^2-x-6}\,dx &= \int\!\left(\frac{1}{x-3} - \frac{1}{x+2}\right)dx \\[4pt] &= \ln|x-3| - \ln|x+2| + C \end{aligned}$$Evaluating Improper Integrals
计算反常积分
Case 1 — Infinite Limits of Integration
情形 1 —— 积分上下限为无穷
improper integral）—— 无穷上限If the limit exists and is finite, the integral converges. If the limit is infinite or does not exist, the integral diverges.
如果该极限存在且有限，则积分收敛（converges）。如果该极限为无穷或不存在，则积分发散（diverges）。
Worked Example — Convergent Improper Integral例题——收敛的反常积分
Evaluate $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$.
$$\begin{aligned} &= \lim_{b \to \infty} \int_1^b x^{-2}\,dx \\[4pt] &= \lim_{b \to \infty} \bigl[-x^{-1}\bigr]_1^b \\[4pt] &= \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) \\[4pt] &= 0 + 1 = 1 \end{aligned}$$The integral converges to $1$.
求 $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$。
$$\begin{aligned} &= \lim_{b \to \infty} \int_1^b x^{-2}\,dx \\[4pt] &= \lim_{b \to \infty} \bigl[-x^{-1}\bigr]_1^b \\[4pt] &= \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) \\[4pt] &= 0 + 1 = 1 \end{aligned}$$该积分收敛于 $1$。
Worked Example — Divergent Improper Integral例题——发散的反常积分
Evaluate $\displaystyle\int_1^{\infty} \frac{1}{x}\,dx$.
$$\begin{aligned} &= \lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx \\[4pt] &= \lim_{b \to \infty} \bigl[\ln|x|\bigr]_1^b \\[4pt] &= \lim_{b \to \infty} (\ln b - \ln 1) \\[4pt] &= \lim_{b \to \infty} \ln b = \infty \end{aligned}$$The integral diverges.
求 $\displaystyle\int_1^{\infty} \frac{1}{x}\,dx$。
$$\begin{aligned} &= \lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx \\[4pt] &= \lim_{b \to \infty} \bigl[\ln|x|\bigr]_1^b \\[4pt] &= \lim_{b \to \infty} (\ln b - \ln 1) \\[4pt] &= \lim_{b \to \infty} \ln b = \infty \end{aligned}$$该积分发散。
Case 2 — Discontinuity Inside the Interval
情形 2 —— 区间内部存在间断点
If the integrand has a vertical asymptote at an interior point $c$ with $a \lt c \lt b$, the integral is improper at $c$ and you must split it at the discontinuity before evaluating. Each piece becomes a one-sided limit.
若被积函数在内部某点 $c$（满足 $a \lt c \lt b$）有垂直渐近线，则该积分在 $c$ 处是反常的，计算前必须在该间断点处拆开。每一段都写成一个单侧极限。
The full integral converges only if both one-sided pieces converge. If either piece diverges, the whole integral diverges.
只有两段单侧极限都收敛，整体积分才收敛。任一段发散，整体就发散。
Worked Example — Splitting at an Interior Discontinuity ($\displaystyle\int_0^{2}\frac{1}{(x-1)^{1/3}}\,dx$)例题——在内部间断点处拆分（$\displaystyle\int_0^{2}\frac{1}{(x-1)^{1/3}}\,dx$）
Step 1 — Locate the discontinuity. The integrand $(x-1)^{-1/3}$ blows up as $x \to 1$. Since $0 \lt 1 \lt 2$, this is an interior singularity — split at $x=1$:
$$ \int_0^{2}(x-1)^{-1/3}\,dx \;=\; \underbrace{\lim_{t\to 1^{-}}\int_0^{t}(x-1)^{-1/3}\,dx}_{\text{Piece A}} \;+\; \underbrace{\lim_{s\to 1^{+}}\int_{s}^{2}(x-1)^{-1/3}\,dx}_{\text{Piece B}} $$Step 2 — Antiderivative. Using $u = x-1$, $du = dx$:
$$ \int (x-1)^{-1/3}\,dx \;=\; \tfrac{3}{2}\,(x-1)^{2/3} + C $$Step 3 — Evaluate Piece A.
$$\begin{aligned} \lim_{t\to 1^{-}} \tfrac{3}{2}(x-1)^{2/3}\Big|_0^{t} &= \lim_{t\to 1^{-}}\!\left[\tfrac{3}{2}(t-1)^{2/3} - \tfrac{3}{2}(-1)^{2/3}\right] \\[4pt] &= 0 - \tfrac{3}{2}(1) \;=\; -\tfrac{3}{2} \end{aligned}$$(Note: $(-1)^{2/3} = \bigl((-1)^2\bigr)^{1/3} = 1$, and $(t-1)^{2/3} \to 0$ since $2/3 \gt 0$.)
Step 4 — Evaluate Piece B.
$$\begin{aligned} \lim_{s\to 1^{+}} \tfrac{3}{2}(x-1)^{2/3}\Big|_s^{2} &= \lim_{s\to 1^{+}}\!\left[\tfrac{3}{2}(1)^{2/3} - \tfrac{3}{2}(s-1)^{2/3}\right] \\[4pt] &= \tfrac{3}{2} - 0 \;=\; \tfrac{3}{2} \end{aligned}$$Step 5 — Add the pieces. Both are finite, so the original integral converges:
$$ \int_0^{2}(x-1)^{-1/3}\,dx \;=\; -\tfrac{3}{2} + \tfrac{3}{2} \;=\; 0 $$Sanity check. The integrand is an odd function about $x=1$, so by symmetry the signed area on $[0,1]$ exactly cancels the signed area on $[1,2]$. The answer of $0$ matches that intuition.
第一步——找到间断点。被积函数 $(x-1)^{-1/3}$ 在 $x \to 1$ 时趋向无穷。由于 $0 \lt 1 \lt 2$，这是内部奇点——在 $x=1$ 处拆开：
$$ \int_0^{2}(x-1)^{-1/3}\,dx \;=\; \underbrace{\lim_{t\to 1^{-}}\int_0^{t}(x-1)^{-1/3}\,dx}_{\text{A}} \;+\; \underbrace{\lim_{s\to 1^{+}}\int_{s}^{2}(x-1)^{-1/3}\,dx}_{\text{B}} $$第二步——求原函数。令 $u = x-1$，$du = dx$：
$$ \int (x-1)^{-1/3}\,dx \;=\; \tfrac{3}{2}\,(x-1)^{2/3} + C $$第三步——算 A 段。
$$\begin{aligned} \lim_{t\to 1^{-}} \tfrac{3}{2}(x-1)^{2/3}\Big|_0^{t} &= \lim_{t\to 1^{-}}\!\left[\tfrac{3}{2}(t-1)^{2/3} - \tfrac{3}{2}(-1)^{2/3}\right] \\[4pt] &= 0 - \tfrac{3}{2}(1) \;=\; -\tfrac{3}{2} \end{aligned}$$（注意：$(-1)^{2/3} = \bigl((-1)^2\bigr)^{1/3} = 1$；因 $2/3 \gt 0$，$(t-1)^{2/3} \to 0$。）
第四步——算 B 段。
$$\begin{aligned} \lim_{s\to 1^{+}} \tfrac{3}{2}(x-1)^{2/3}\Big|_s^{2} &= \lim_{s\to 1^{+}}\!\left[\tfrac{3}{2}(1)^{2/3} - \tfrac{3}{2}(s-1)^{2/3}\right] \\[4pt] &= \tfrac{3}{2} - 0 \;=\; \tfrac{3}{2} \end{aligned}$$第五步——把两段加起来。两段都是有限值，故原积分收敛：
$$ \int_0^{2}(x-1)^{-1/3}\,dx \;=\; -\tfrac{3}{2} + \tfrac{3}{2} \;=\; 0 $$核对一下。被积函数关于 $x=1$ 是奇函数，由对称性，$[0,1]$ 上的有符号面积正好抵消 $[1,2]$ 上的有符号面积。答案 $0$ 与直觉吻合。
Selecting Techniques for Antidifferentiation
选择求原函数的方法
On the AP exam, you must quickly identify which technique to use. Here's a decision flowchart:
AP 考试中，你必须迅速判断该用哪种方法。下面是决策流程：
1. Basic rule? Check if the integrand matches a known antiderivative directly.
2. Algebraic simplification? Can you expand, factor, simplify, or split a fraction first?
3. U-substitution? Do you see a composite function with its derivative present (or nearly present)?
4. Long division? Is the numerator's degree ≥ the denominator's degree?
5. Completing the square? Irreducible quadratic in the denominator → $\arctan$ or $\arcsin$?
6. Integration by parts? BC Product of two different "types" (e.g., polynomial × exponential)?
7. Partial fractions? BC Proper fraction with factorable denominator?
1. 基本公式？看看被积函数是不是直接匹配一个已知原函数。
2. 代数化简？能不能先展开、因式分解、化简或拆分分式？
3. u 代换？是否看到一个复合函数，并且它的导数也（差不多）出现在被积式里？
4. 多项式长除法？分子次数是否 ≥ 分母次数？
5. 配方法？分母是不可分解的二次式 → $\arctan$ 或 $\arcsin$ 形式？
6. 分部积分？ BC 两种不同“类型”的乘积（例如多项式 × 指数）？
7. 部分分式？ BC 分母可因式分解的真分式？
How Unit 6 Appears on the AP Exam
单元 6 在 AP 考试中的考法
Evaluate a definite or indefinite integral using FTC, u-sub, or basic rules.
Find $F'(x)$ where $F(x)$ is defined by an integral with variable limits (FTC Part 1 + chain rule).
Approximate a definite integral using a Riemann sum or trapezoidal sum from a table.
Interpret accumulated change in a real-world context (rate × time = total).
Translate between a limit of a Riemann sum and a definite integral.
计算一个定积分或不定积分，用 FTC、u 代换或基本公式。
求 $F'(x)$，其中 $F(x)$ 是带变上下限的积分定义（FTC 第一部分 + 链式法则）。
近似计算定积分：由给定的数据表用黎曼和或梯形和近似。
解释真实情境中的累积变化（变化率 × 时间 = 总量）。
转换：在黎曼和的极限与定积分之间互相改写。
Accumulation in context: Given a rate function, interpret or compute total change over an interval.
Table problems: Approximate an integral using a Riemann or trapezoidal sum from given data, then determine if it's an over- or underestimate.
Graph of $f$: Given a graph of $f$, analyze the accumulation function $g(x) = \int_a^x f(t)\,dt$ — find where $g$ has extrema, inflection points, is increasing/decreasing.
情境中的累积：给出一个变化率函数，要求解释或计算某区间上的总变化（net change）。
表格题：用给定数据通过黎曼和或梯形和近似积分，然后判断它是高估还是低估。
$f$ 的图像题：给出 $f$ 的图像，分析累积函数 $g(x) = \int_a^x f(t)\,dt$——找 $g$ 的极值点、拐点（inflection point）、递增/递减区间。
Flashcards — Click to Flip
闪卡——点击翻转
$\frac{d}{dx}\int_a^x f(t)\,dt = \;?$FTC 第一部分：
$\frac{d}{dx}\int_a^x f(t)\,dt = \;?$
$$ f(x) $$
$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = \;?$FTC 第一部分 + 链式法则：
$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = \;?$
$$ f(g(x)) \cdot g'(x) $$
$\int_a^b f(x)\,dx = \;?$FTC 第二部分：
$\int_a^b f(x)\,dx = \;?$
$$ F(b) - F(a) $$
($n \neq -1$)
$$ \frac{x^{n+1}}{n+1} + C $$
$$ \ln|x| + C $$
$$f'' > 0 \;\Rightarrow\; T_n > \int_a^b f\,dx$$
BC Only分部积分公式？
仅 BC
$$ \int u\,dv = uv - \int v\,du $$
BC Only$\int_1^{\infty} \frac{1}{x^p}\,dx$ 何时收敛？
仅 BC
$$\text{converges} \iff p > 1$$
Unit 6 — Practice Quiz
单元 6——练习小测
Readiness Checklist
备考清单
Click each item you've mastered. Aim for 100% before exam day.
已掌握的逐条点勾。考前目标 100%。
- Interpret a definite integral as an accumulation of change把定积分解读为变化的累积
- Compute left, right, midpoint, and trapezoid Riemann sums计算左、右、中点、梯形黎曼和
- Express a Riemann sum in sigma notation and as an integral用求和记号和积分形式表示黎曼和
- State and apply the Fundamental Theorem of Calculus (Part 1)陈述并运用微积分基本定理（第一部分）
- Evaluate definite integrals using the FTC (Part 2)用 FTC（第二部分）求定积分
- Differentiate accumulation functions $F(x) = \int_a^{g(x)} f(t)\,dt$对累积函数 $F(x) = \int_a^{g(x)} f(t)\,dt$ 求导
- Apply linearity and splitting properties of definite integrals运用定积分的线性与区间拼接性质
- Compute antiderivatives of basic power, exponential, and trig functions求基本幂、指数、三角函数的原函数
- Apply u-substitution for definite and indefinite integrals对定积分与不定积分使用 u 代换
- Integrate rational functions using long division or completing the square用长除法或配方法对有理函数积分
- Apply integration by parts (BC)使用分部积分法（BC）
- Decompose into partial fractions and integrate (BC)做部分分式分解并积分（BC）
- Evaluate improper integrals with a limit definition (BC)用极限定义计算反常积分（BC）
- Determine convergence vs. divergence of an improper integral判断反常积分收敛或发散
- Select the appropriate technique for a given integral为给定积分选用合适的方法
- Use proper integral notation including limits of integration and $dx$使用规范的积分记号，含上下限和 $dx$
AP-Style Practice Problems
AP 风格练习题
Exam-level practice for this unit — multiple-choice plus extended-response items modeled on the AP rubric. Built for top-score prep; go here after you've worked through the notes and the in-page quizzes above.
本单元的考试级练习——按 AP 评分标准设计的选择题 + 解答题。冲刺顶分使用；先完成笔记和上方小测后再来。
AP Calculus AB & BC — Unit 6 Notes
AP 微积分 AB & BC——单元 6 笔记