High School Math

Introduction to Limits and Calculus极限与微积分入门

Calculus is the mathematics of "what happens as we zoom in." A limit describes the value a function approaches as the input slides toward a target; a derivative is the limit of a slope; an integral is a limit of a sum that totals up signed area. This unit is the AP/IB-feeder introduction at MCV4U level: it builds intuition for $\lim_{x \to c} f(x)$ from graphs and tables, walks the four standard algebraic techniques (direct substitution, factor-and-cancel, conjugate, rationalising), introduces one-sided limits and continuity, defines the derivative as $f'(x) = \lim_{h \to 0} (f(x+h)-f(x))/h$, drills the power rule, and closes with antiderivatives and the definite integral as signed area. The full $\epsilon$-$\delta$ definition, IVT/EVT/MVT, Riemann sums, and the formal Fundamental Theorem all live one step downstream in AP Calculus AB and IB Math HL E-cluster , the feeder links at the end of the guide point straight there.微积分(calculus)研究"无限放大时发生什么"。极限(limit)刻画自变量逼近某目标时函数的趋近值;导数(derivative)是斜率的极限;积分(integral)是把带符号面积累加起来的求和极限。本单元是 MCV4U 级别的 AP / IB 衔接入门:先用图像与数表建立 $\lim_{x \to c} f(x)$ 的直观,再走过四种标准代数技巧(直接代入、因式约分、共轭、有理化),引入单侧极限与连续性,定义导数 $f'(x) = \lim_{h \to 0} (f(x+h)-f(x))/h$,反复练习幂法则,最后以反导数与作为带符号面积的定积分收束。完整的 $\epsilon$-$\delta$ 定义、IVT / EVT / MVT、黎曼和、严格的微积分基本定理都属于下游内容,归 AP Calculus AB 与 IB Math HL E 簇 , 指南末尾的衔接链接直接指过去。

7 sections7 节内容 ON MCV4U · BC Calc 12 · AB Math 31 coreON MCV4U · BC Calc 12 · AB Math 31 核心 Honors / AP-feeder for US studentsUS 学生:荣誉 / AP 衔接

How to use this guide如何使用本指南

Calculus content lives in different places in each of the four curricula we map to. The US Common Core does not contain it at all , CCSSM stops at pre-calculus and AP Calculus AB / BC is where American students meet limits and derivatives. Ontario MCV4U (Grade 12, University) puts it at the centre of the course: Strand A is Rate of Change, Strand B is Derivatives. BC and Alberta both offer dedicated optional Grade 12 calculus courses (Calculus 12 and Math 31) that cover this material. The table below tells you which sections are on your syllabus right now and where to keep going. Treat this unit as the AP/IB feeder , the deeper material with $\epsilon$-$\delta$ rigor, IVT/EVT/MVT, full differentiation rules, Riemann sums, and the Fundamental Theorem lives in the AP Calculus and IB Math HL units cross-linked at the end.微积分内容在我们对照的四套大纲中位置各异。美国共同核心完全不含 , CCSSM 止于 pre-calculus,AP Calculus AB / BC 是美国学生接触极限与导数的地方。安大略 MCV4U(12 年级大学预备)把这部分放在课程核心:单元 A 为变化率,单元 B 为导数。BC 与阿尔伯塔都开设 12 年级专属选修微积分课程(Calculus 12、Math 31),覆盖同样内容。下表说明你的大纲下应学哪几节、以及深入路径。把本单元视为 AP / IB 衔接 , 严格的 $\epsilon$-$\delta$ 定义、IVT / EVT / MVT、完整的微分法则、黎曼和与微积分基本定理见末尾的 AP Calculus 与 IB Math HL 链接。

If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇺🇸 US Grade 11/12 , CCSSM Pre-Calc only美国 11/12 年级 , 仅 CCSSM Pre-Calc All seven sections as Honors / AP-feeder enrichment. Limits and derivatives are not on the SAT, but they are on the AP Calculus AB exam , this unit is your runway.全部 7 节皆为 Honors / AP 衔接拓展。极限与导数不在 SAT,但在 AP Calculus AB 考试 , 本单元为助跑道。 Nothing on the unit itself , but if you have not yet seen factoring (HS Math Unit 3) or rational expressions (Unit 4), do those first because §2 lives on them.单元本身无可跳过 , 但若尚未学过因式分解(HS Math Unit 3)或有理表达式(Unit 4),请先补完,因为 §2 依赖它们。 ccssm_hs_math.pdf , CCSSM HS standards contain no limits / derivatives / integrals. Closest standard: HSF-IF.B.6 (average rate of change). The full pathway is AP Calculus AB CED Units 1, 2, 6., CCSSM 高中标准不含极限 / 导数 / 积分。最接近的标准为 HSF-IF.B.6(平均变化率)。完整路径见 AP Calculus AB CED 的 Unit 1、2、6。
🇨🇦 ON Grade 12 , MCV4U安大略 12 年级 , MCV4U §1 through §5 are core. Strand A maps to §1, §2, §3; Strand B maps to §4 and §5. Treat §6 and §7 as enrichment that smooths the transition to first-year university calculus.§1 至 §5 为核心。单元 A 对应 §1、§2、§3;单元 B 对应 §4、§5。§6、§7 为拓展内容,帮助衔接大学一年级微积分。 §6 and §7 (antiderivatives and definite integrals are not in MCV4U , they live in first-year university calculus or AP Calc AB / BC).§6、§7(反导数与定积分不在 MCV4U , 属大学一年级微积分或 AP Calc AB / BC)。 math_grades_11-12_extract.md , MCV4U Strand A Rate of Change (limits, average vs instantaneous rate of change) and Strand B Derivatives (power rule, sum and constant multiple rules, derivative graphs). No integral strand., MCV4U 单元 A 变化率(极限、平均变化率与瞬时变化率)与单元 B 导数(幂法则、求和与常数倍法则、导数图像)。无积分单元。
🇨🇦 BC Grade 12 , Calculus 12 (optional)BC 12 年级 , Calculus 12(选修) All seven sections. Calculus 12 is BC's pre-university calculus course and covers limits, derivatives, and integrals at the depth of this unit and slightly beyond.全部 7 节。Calculus 12 是 BC 的大学前微积分课程,覆盖极限、导数与积分,深度与本单元一致并稍超出。 Nothing , treat as the recommended bridge before AP Calc AB or first-year university.无 , 视为 AP Calc AB 或大学一年级前的推荐桥梁。 calc12.pdf , BC Ministry, Calculus 12. Per-section citations pending until calc12_extract.md is produced., BC 教育厅 Calculus 12。逐节细化引用在 calc12_extract.md 产出前待补。
🇨🇦 BC Grade 11/12 , PC 11 / PC 12 only (no Calc 12)BC 11/12 年级 , 仅 PC 11 / PC 12 All seven sections as Honors / AP-feeder enrichment. Pre-Calc 12 contains no calculus content, so this unit is your only pre-AP exposure.全部 7 节作 Honors / AP 衔接拓展。Pre-Calc 12 不含微积分,本单元即为 AP 前唯一接触。 Nothing , but be honest about timing: if you have not yet finished PC 12 algebra / function chapters, do those first.无 , 但要量力而行:若尚未学完 PC 12 的代数 / 函数章节,请先完成。 pc12_elab_extract.md , PC 12 content elaborations contain no limits / derivatives / integrals., PC 12 内容细化不含极限 / 导数 / 积分。
🇨🇦 AB Grade 12 , Math 31 (optional)阿尔伯塔 12 年级 , Math 31(选修) All seven sections. Math 31 is Alberta's introductory calculus course , same scope as BC Calc 12.全部 7 节。Math 31 是阿尔伯塔的入门微积分课程 , 范围与 BC Calc 12 一致。 Nothing , use as a clean second pass before AP Calc AB or first-year university calculus.无 , AP Calc AB 或大学一年级微积分之前作为干净复习。 math31.pdf , Alberta Education, Math 31 Program of Studies. Per-section citations pending until math31_extract.md is produced., 阿尔伯塔教育厅 Math 31 课程大纲。逐节细化引用在 math31_extract.md 产出前待补。
🇨🇦 AB Grade 11/12 , Math 30-1 only (no Math 31)阿尔伯塔 11/12 年级 , 仅 Math 30-1 All seven sections as Honors / AP-feeder enrichment. Math 30-1 is pre-calculus , calculus is Math 31 only.全部 7 节作 Honors / AP 衔接拓展。Math 30-1 是 pre-calculus , 微积分仅限 Math 31。 Nothing on the unit. If you intend to sit AP Calc, study the AP feeders linked at the end too.单元本身无可跳过。若打算考 AP Calc,末尾的 AP 衔接也要看。 math30-1_standards_exemplars.pdf , Math 30-1 contains no limits / derivatives / integrals., Math 30-1 不含极限 / 导数 / 积分。
🇺🇸 US AP-feeder (Pre-Calc / Honors)美国 AP 衔接(Pre-Calc / 荣誉) All seven sections plus the feeder links. AP Calculus AB Unit 1 picks up exactly at §1-§3 with $\epsilon$-$\delta$ added; Unit 2 picks up at §4-§5; Unit 6 picks up at §6-§7 with Riemann sums and the Fundamental Theorem added.全部 7 节加末尾衔接链接。AP Calculus AB Unit 1 自 §1-§3 起接续并加入 $\epsilon$-$\delta$;Unit 2 自 §4-§5 起接续;Unit 6 自 §6-§7 起接续并加入黎曼和与微积分基本定理。 Nothing , this unit is exactly the right pre-read for AP Calc.无 , 本单元正是 AP Calc 的合适前置阅读。 AP Calculus AB CED (College Board), Units 1 (Limits and Continuity), 2 (Differentiation: Definition and Fundamental Properties), 6 (Integration and Accumulation of Change).AP Calculus AB CED(College Board)的 Unit 1(极限与连续)、Unit 2(微分:定义与基本性质)、Unit 6(积分与变化量的累积)。

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后,用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise five things: the meaning of $\lim_{x \to c} f(x) = L$ ("$f(x)$ gets arbitrarily close to $L$ as $x$ gets close to $c$"); the factor-and-cancel move for $0/0$ forms; the derivative definition $f'(x) = \lim_{h \to 0} (f(x+h)-f(x))/h$; the power rule $(x^n)' = n x^{n-1}$ and its reverse $\int x^n \, dx = x^{n+1}/(n+1) + C$; the picture of the definite integral as signed area between the graph and the $x$-axis. Read every cram-cheat box. Skip the derivations.背熟五件事:$\lim_{x \to c} f(x) = L$ 的含义("当 $x$ 趋于 $c$ 时 $f(x)$ 任意接近 $L$");$0/0$ 型用因式约分;导数定义 $f'(x) = \lim_{h \to 0} (f(x+h)-f(x))/h$;幂法则 $(x^n)' = n x^{n-1}$ 与其逆 $\int x^n \, dx = x^{n+1}/(n+1) + C$;定积分作为图像与 $x$ 轴间带符号面积的几何图像。读每个速记框,跳过推导。

*
If you are going for the top mark如果你目标顶分

Always justify a limit. "$\lim_{x \to 2} (x^2 - 4)/(x - 2) = 4$ because after factoring and cancelling we have $\lim_{x \to 2}(x + 2) = 4$" earns the method mark; the bare $4$ does not. Always write the derivative definition before invoking the power rule on a "from first principles" question (MCV4U Strand A expects exactly this). Memorise the antiderivative form with $+ C$ , missing the constant is the single most common AB / BC exam slip. Read the derivation linking the derivative to a secant slope and the derivation that turns the area-under-a-graph picture into $\int_a^b f(x) \, dx$.每个极限都要给出理由。"$\lim_{x \to 2} (x^2 - 4)/(x - 2) = 4$,因为因式约分后 $\lim_{x \to 2}(x + 2) = 4$"能拿到方法分;只写 $4$ 不能。"从定义出发"求导时务必先写导数定义(MCV4U 单元 A 正是此要求)。反导数务必带 $+ C$ , 漏掉常数是 AB / BC 考试最常见的单点失分。两条推导都要会:把割线斜率推为导数;把图像下方面积推为 $\int_a^b f(x) \, dx$。

Honors flag.荣誉级标记。 The entire unit carries the Honors chip for US students because limits, derivatives, and integrals are not in CCSSM , they belong to AP Calculus AB / BC. For students on a BC PC-12-only or AB Math 30-1-only track, the same chip applies: calculus is the optional course (Calculus 12 / Math 31), not the mainstream pre-calculus pathway. The unit is core, not honors, for Ontario MCV4U (§1-§5 are core; §6 and §7 are enrichment because integrals are not in MCV4U), and core for BC Calculus 12 and AB Math 31 throughout (§1-§7 all core).本单元在美国学生中整体标 Honors,因为极限、导数、积分不在 CCSSM , 它们属于 AP Calculus AB / BC。BC 仅修 PC 12 或阿尔伯塔仅修 Math 30-1 的学生同样适用:微积分是选修课程(Calculus 12 / Math 31),不在主流 pre-calculus 序列。在安大略 MCV4U 中本单元为核心而非荣誉(§1-§5 核心;§6、§7 为拓展,因积分不在 MCV4U);在 BC Calculus 12 与 AB Math 31 中 §1-§7 皆为核心。

Intuitive Limits: Graphs and Tables直观的极限:图像与数表

What $\lim_{x \to c} f(x) = L$ means in plain English.$\lim_{x \to c} f(x) = L$ 的白话含义。

As $x$ gets arbitrarily close to $c$ from either side (but never equal to $c$), the value of $f(x)$ gets arbitrarily close to $L$. The limit does not care what $f(c)$ is , only what $f$ does on a small neighbourhood around $c$.当 $x$ 从两侧任意接近 $c$(但不等于 $c$)时,$f(x)$ 任意接近 $L$。极限不在乎 $f(c)$ 本身的值 , 它只关心 $f$ 在 $c$ 附近邻域上的行为。

  • Two ways to read a limit.读极限的两种方式。 Graphical: zoom in on the graph near $x = c$; the $y$-value the curve approaches is the limit. Tabular: tabulate $f(x)$ at $x$-values that creep up on $c$ from both sides; the shared trend value is the limit.图像法:在 $x = c$ 附近放大图像,曲线趋近的 $y$ 值即极限。数表法:列出从两侧逼近 $c$ 的 $x$ 处的 $f(x)$,两侧共同趋近的值即极限。
  • "Hole" example."洞"的例子。 For $f(x) = (x^2 - 1)/(x - 1)$, the value $f(1)$ is undefined ($0/0$), but the limit $\lim_{x \to 1} f(x) = 2$ exists because $f(x) = x + 1$ for $x \ne 1$.对 $f(x) = (x^2 - 1)/(x - 1)$,$f(1)$ 未定义($0/0$),但极限 $\lim_{x \to 1} f(x) = 2$ 存在,因为 $x \ne 1$ 时 $f(x) = x + 1$。
  • Why this matters.为何重要。 The derivative ($\S 4$) and the definite integral ($\S 7$) are both limits. Without limits, no calculus. MCV4U Strand A spells this out: "make connections between average rate of change over an interval and instantaneous rate of change at a point, using the slopes of secants and tangents and the concept of the limit."导数(§4)与定积分(§7)都是极限。没有极限就没有微积分。MCV4U 单元 A 明确:"通过把区间上的平均变化率与某点的瞬时变化率联系起来,并使用割线、切线斜率与极限概念。"
Worked Example 1 · Estimate a limit from a table例题 1 · 从数表估计极限

Let $f(x) = \dfrac{\sin x}{x}$ with $x$ in radians. Estimate $\lim_{x \to 0} f(x)$ by tabulating $f$ at $x = \pm 0.1, \pm 0.01, \pm 0.001$.设 $f(x) = \dfrac{\sin x}{x}$,$x$ 为弧度。在 $x = \pm 0.1, \pm 0.01, \pm 0.001$ 处列表估计 $\lim_{x \to 0} f(x)$。

Why we cannot just plug in.为何不能直接代入。 $f(0) = 0/0$, which is indeterminate , not a value. So we tabulate on both sides instead.$f(0) = 0/0$ 是未定式 , 没有数值。所以从两侧列表逼近。

$x$$-0.1$$-0.01$$-0.001$$+0.001$$+0.01$$+0.1$
$f(x)$$0.99833$$0.99998$$1.00000$$1.00000$$0.99998$$0.99833$

Both sides agree.两侧吻合。 As $x \to 0$ from either side, $f(x)$ approaches $1$. So:$x$ 从两侧趋于 $0$ 时,$f(x)$ 趋近 $1$。故:

$$ \lim_{x \to 0} \frac{\sin x}{x} \;=\; 1. $$

Caveat.注意。 A table is a strong piece of evidence, not a proof. AP Calc and IB Math HL both have algebraic / squeeze-theorem proofs of this exact limit , see the AP Calc Unit 1 feeder.数表是有力证据,但不是证明。AP Calc 与 IB Math HL 都用代数 / 夹挤定理给出此极限的严格证明 , 见 AP Calc Unit 1 衔接。

The graph of $f$ has a hole at $x = 3$. From the graph, $f(x)$ approaches $5$ from both sides of $3$. What is $\lim_{x \to 3} f(x)$?$f$ 的图像在 $x = 3$ 处有一个洞。由图像可见 $f(x)$ 从两侧都趋近 $5$。$\lim_{x \to 3} f(x)$ 为何?
§1 · Q1
Does not exist (the hole)不存在(因为有洞)
$3$
$5$
Cannot tell without $f(3)$没有 $f(3)$ 无法判断
A limit ignores what happens at the point. Both one-sided approaches give $5$, so $\lim_{x \to 3} f(x) = 5$ regardless of whether $f(3)$ is defined or equal to a different value.极限不在乎该点处的值。两侧逼近都得 $5$,故 $\lim_{x \to 3} f(x) = 5$,无论 $f(3)$ 是否有定义或取其他值。
A hole at $x = 3$ does not prevent the limit from existing. The limit is the shared approach value, not $f(3)$.$x = 3$ 处有洞并不妨碍极限存在。极限是两侧共同趋近值,不是 $f(3)$。
Tabulating $g(x) = (1 + x)^{1/x}$ at $x = 0.1, 0.01, 0.001$ gives $\approx 2.5937, 2.7048, 2.7169$. What does $\lim_{x \to 0^+} g(x)$ appear to be?列表得 $g(x) = (1 + x)^{1/x}$ 在 $x = 0.1, 0.01, 0.001$ 处约为 $2.5937, 2.7048, 2.7169$。$\lim_{x \to 0^+} g(x)$ 看上去是?
§1 · Q2
$1$
$e \approx 2.71828$
$3$
$\infty$$\infty$
The values $2.5937 \to 2.7048 \to 2.7169$ creep up on $2.71828\ldots = e$, the natural-log base. This limit is one definition of $e$. (You'll see the algebraic / analytic proof in AP Calc Unit 1 or IB Math HL E1.)数值 $2.5937 \to 2.7048 \to 2.7169$ 逐步逼近 $2.71828\ldots = e$,即自然对数的底。此极限是 $e$ 的定义之一。(严格证明见 AP Calc Unit 1 或 IB Math HL E1。)
The numbers cluster around $2.718$, which is Euler's $e$ , one of its defining limits is $\lim_{x \to 0}(1 + x)^{1/x} = e$.数值聚集在 $2.718$ 附近 , 即欧拉数 $e$。$e$ 的定义之一就是 $\lim_{x \to 0}(1 + x)^{1/x} = e$。

Computing Limits Algebraically用代数方法计算极限

Four standard moves.四种标准操作。
  1. Direct substitution.直接代入。 If $f$ is "nice" (polynomial, or rational with non-zero denominator, etc.), then $\lim_{x \to c} f(x) = f(c)$. Always try this first.若 $f$ "好"(多项式,或分母不为零的有理式等),则 $\lim_{x \to c} f(x) = f(c)$。先试此法。
  2. Factor and cancel因式分解并约分 when substitution gives the indeterminate form $0/0$. Common when the numerator and denominator share a factor $(x - c)$.当代入得未定式 $0/0$ 时。常见于分子分母含公因式 $(x - c)$ 的情形。
  3. Conjugate technique共轭法 when the expression contains a $\sqrt{\,}$ that creates a $0/0$ form. Multiply numerator and denominator by the conjugate to rationalise.当表达式含 $\sqrt{\,}$ 且产生 $0/0$ 时。分子分母同乘共轭以有理化。
  4. Rationalise / common denominator通分 / 化为同分母 for expressions like $\lim_{x \to 0} (1/x)(1/(x+1) - 1)$ , combine the fractions, simplify, then substitute.对于 $\lim_{x \to 0} (1/x)(1/(x+1) - 1)$ 等表达式 , 先通分化简,再代入。
Indeterminate vs undefined. $0/0$ is indeterminate , the limit might exist, you have to do more work. $5/0$ with the numerator approaching a non-zero value is a vertical asymptote and the limit is typically $\pm \infty$, see §3.未定式与无定义的区别。$0/0$ 是未定式 , 极限可能存在,需进一步处理。$5/0$(分子趋于非零)则是竖直渐近线,极限通常为 $\pm \infty$,见 §3。
Worked Example 2a · Factor and cancel例题 2a · 因式约分

Evaluate $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.求 $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$。

Try direct substitution.先试直接代入。 $f(2) = (4 - 4)/(2 - 2) = 0/0$ , indeterminate. Cannot finish here.$f(2) = (4 - 4)/(2 - 2) = 0/0$ , 未定式,不能就此结束。

Factor the numerator.对分子因式分解。

$$ \frac{x^2 - 4}{x - 2} \;=\; \frac{(x - 2)(x + 2)}{x - 2} \;=\; x + 2 \quad \text{for } x \ne 2. $$

Now substitute into the simplified form.在化简形式下代入。

$$ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \;=\; \lim_{x \to 2} (x + 2) \;=\; 4. $$

Why cancelling is legal.约分为何合法。 The limit only looks at $x \ne c$, and on that set the original expression equals $x + 2$ identically. The function has a "hole" at $x = 2$ but the limit is $4$.极限只关注 $x \ne c$,在该集合上原表达式与 $x + 2$ 完全相同。函数在 $x = 2$ 处有"洞",但极限为 $4$。

Worked Example 2b · Conjugate technique例题 2b · 共轭法

Evaluate $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$.求 $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$。

Direct substitution.直接代入。 $(\sqrt{4} - 2)/0 = 0/0$ , indeterminate.$(\sqrt{4} - 2)/0 = 0/0$ , 未定式。

Multiply by the conjugate.同乘共轭式。

$$ \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} \;=\; \frac{(x + 4) - 4}{x (\sqrt{x + 4} + 2)} \;=\; \frac{x}{x (\sqrt{x + 4} + 2)}. $$

Cancel and substitute.约分并代入。

$$ \;=\; \frac{1}{\sqrt{x + 4} + 2} \;\xrightarrow{x \to 0}\; \frac{1}{\sqrt{4} + 2} \;=\; \frac{1}{4}. $$

Sanity-check.合理性核验。 A table at $x = 0.01$ gives $(\sqrt{4.01} - 2)/0.01 \approx 0.2497$, near $0.25 = 1/4$. ✓在 $x = 0.01$ 处数表:$(\sqrt{4.01} - 2)/0.01 \approx 0.2497$,接近 $0.25 = 1/4$ ✓。

Evaluate $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.求 $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$。
§2 · Q1
$0$
$3$
$6$
Does not exist不存在
Substitution gives $0/0$. Factor: $(x^2 - 9)/(x - 3) = (x - 3)(x + 3)/(x - 3) = x + 3$ for $x \ne 3$. Then $\lim_{x \to 3}(x + 3) = 6$.代入得 $0/0$。因式分解:$x \ne 3$ 时 $(x^2 - 9)/(x - 3) = (x - 3)(x + 3)/(x - 3) = x + 3$,故 $\lim_{x \to 3}(x + 3) = 6$。
$0/0$ is indeterminate, not "does not exist." Factor and cancel: $(x - 3)$ kills the trouble and leaves $x + 3 \to 6$.$0/0$ 是未定式,而非"不存在"。因式分解约分后得 $x + 3 \to 6$。
Evaluate $\lim_{x \to 0} \dfrac{\sqrt{x + 9} - 3}{x}$.求 $\lim_{x \to 0} \dfrac{\sqrt{x + 9} - 3}{x}$。
§2 · Q2
$1/6$
$1/3$
$0$
Does not exist不存在
Conjugate: multiply by $(\sqrt{x + 9} + 3)/(\sqrt{x + 9} + 3)$. Numerator becomes $(x + 9) - 9 = x$, cancelling with the denominator $x$. Remaining $1/(\sqrt{x + 9} + 3) \to 1/(3 + 3) = 1/6$.共轭法:同乘 $(\sqrt{x + 9} + 3)/(\sqrt{x + 9} + 3)$。分子变为 $(x + 9) - 9 = x$,与分母 $x$ 约消。剩余 $1/(\sqrt{x + 9} + 3) \to 1/(3 + 3) = 1/6$。
When $\sqrt{\,}$ appears in a $0/0$ form, multiply by the conjugate of the radical expression. Rationalising clears the surd and the indeterminacy.当 $0/0$ 中含 $\sqrt{\,}$ 时,对该根式同乘共轭。有理化后既去根号又去未定。

One-Sided Limits, Continuity, and Limits at Infinity单侧极限、连续性与无穷处的极限

Three definitions packed into one section.一节内装三个定义。
  • One-sided limits.单侧极限。 $\lim_{x \to c^-} f(x)$ means $x$ approaches $c$ from the left (smaller values); $\lim_{x \to c^+} f(x)$ from the right. The two-sided limit $\lim_{x \to c} f(x)$ exists if and only if both one-sided limits exist and are equal.$\lim_{x \to c^-} f(x)$ 表示 $x$ 从左侧(更小值)逼近 $c$;$\lim_{x \to c^+} f(x)$ 从右侧。双侧极限 $\lim_{x \to c} f(x)$ 存在当且仅当两个单侧极限都存在且相等。
  • Continuity at a point.在某点的连续性。 $f$ is continuous at $x = c$ when all three conditions hold: (1) $f(c)$ is defined; (2) $\lim_{x \to c} f(x)$ exists; (3) $\lim_{x \to c} f(x) = f(c)$.$f$ 在 $x = c$ 处连续,须三件事同时成立:(1) $f(c)$ 有定义;(2) $\lim_{x \to c} f(x)$ 存在;(3) $\lim_{x \to c} f(x) = f(c)$。
  • Limits at infinity无穷处的极限 describe end behaviour: $\lim_{x \to \infty} f(x) = L$ means $f(x)$ approaches $L$ for arbitrarily large $x$. Geometrically, $y = L$ is a horizontal asymptote.描述尾部行为:$\lim_{x \to \infty} f(x) = L$ 表示 $x$ 任意大时 $f(x)$ 趋近 $L$。几何上 $y = L$ 是水平渐近线。
  • Rational-function trick.有理函数技巧。 For $f(x) = (a_n x^n + \ldots)/(b_m x^m + \ldots)$ as $x \to \infty$: if $n < m$ the limit is $0$; if $n = m$ it is $a_n / b_m$; if $n > m$ it is $\pm \infty$.$x \to \infty$ 时对有理函数 $f(x) = (a_n x^n + \ldots)/(b_m x^m + \ldots)$:若 $n < m$ 极限为 $0$;若 $n = m$ 为 $a_n / b_m$;若 $n > m$ 为 $\pm \infty$。
Worked Example 3a · A jump discontinuity例题 3a · 跳跃间断

Let $f(x) = \begin{cases} x + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}$. Determine $\lim_{x \to 2^-} f(x)$, $\lim_{x \to 2^+} f(x)$, $\lim_{x \to 2} f(x)$, and decide whether $f$ is continuous at $x = 2$.设 $f(x) = \begin{cases} x + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}$。求 $\lim_{x \to 2^-} f(x)$、$\lim_{x \to 2^+} f(x)$、$\lim_{x \to 2} f(x)$,并判断 $f$ 在 $x = 2$ 处是否连续。

Left-side limit.左侧极限。 $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 3$.$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 3$。

Right-side limit.右侧极限。 $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x^2 = 4$.$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x^2 = 4$。

Two-sided limit.双侧极限。 Because $3 \ne 4$, $\lim_{x \to 2} f(x)$ does not exist. The graph has a jump of size $1$ at $x = 2$.因为 $3 \ne 4$,所以 $\lim_{x \to 2} f(x)$ 不存在。图像在 $x = 2$ 处有大小 $1$ 的跳跃。

Continuity.连续性。 Condition (2) of the three-part test fails, so $f$ is not continuous at $x = 2$. (It is continuous at every other real number.)三条件测试的第 (2) 条不成立,故 $f$ 在 $x = 2$ 不连续。(在其他实数处皆连续。)

Worked Example 3b · Horizontal asymptote例题 3b · 水平渐近线

Find $\lim_{x \to \infty} \dfrac{3x^2 + x - 5}{2x^2 + 7}$.求 $\lim_{x \to \infty} \dfrac{3x^2 + x - 5}{2x^2 + 7}$。

Divide numerator and denominator by the highest power of $x$.分子分母同除以 $x$ 的最高次幂。 Here that is $x^2$.此处即 $x^2$。

$$ \frac{3x^2 + x - 5}{2x^2 + 7} \;=\; \frac{3 + 1/x - 5/x^2}{2 + 7/x^2}. $$

Take the limit term by term.逐项求极限。 As $x \to \infty$, $1/x \to 0$ and $1/x^2 \to 0$.$x \to \infty$ 时 $1/x \to 0$,$1/x^2 \to 0$。

$$ \;\to\; \frac{3 + 0 - 0}{2 + 0} \;=\; \frac{3}{2}. $$

Geometric reading.几何解读。 The horizontal line $y = 3/2$ is a horizontal asymptote of the curve. Matches the cram-cheat shortcut: same degree, ratio of leading coefficients $= 3/2$.水平线 $y = 3/2$ 是曲线的水平渐近线。与速记一致:同次时取首项系数之比 $= 3/2$。

For $f(x) = |x|/x$ on $x \ne 0$, find $\lim_{x \to 0^-} f(x)$ and $\lim_{x \to 0^+} f(x)$.对 $x \ne 0$ 上的 $f(x) = |x|/x$,求 $\lim_{x \to 0^-} f(x)$ 与 $\lim_{x \to 0^+} f(x)$。
§3 · Q1
Both equal $0$, so $\lim_{x \to 0} f(x) = 0$两者皆为 $0$,故 $\lim_{x \to 0} f(x) = 0$
Both equal $1$, so $\lim_{x \to 0} f(x) = 1$两者皆为 $1$,故 $\lim_{x \to 0} f(x) = 1$
Both equal $-1$, so $\lim_{x \to 0} f(x) = -1$两者皆为 $-1$,故 $\lim_{x \to 0} f(x) = -1$
Left $= -1$, right $= 1$; two-sided limit does not exist左 $= -1$、右 $= 1$;双侧极限不存在
For $x > 0$, $|x| = x$ so $f(x) = 1$; for $x < 0$, $|x| = -x$ so $f(x) = -1$. The two one-sided limits at $0$ disagree, so the two-sided limit does not exist , jump discontinuity of size $2$.$x > 0$ 时 $|x| = x$,故 $f(x) = 1$;$x < 0$ 时 $|x| = -x$,故 $f(x) = -1$。两单侧极限不一致,双侧极限不存在 , 大小为 $2$ 的跳跃间断。
Split by sign of $x$. $|x|/x$ is $+1$ on the right and $-1$ on the left.按 $x$ 的正负分情形。$|x|/x$ 右侧为 $+1$、左侧为 $-1$。
Evaluate $\lim_{x \to \infty} \dfrac{5x + 1}{x^2 - 4}$.求 $\lim_{x \to \infty} \dfrac{5x + 1}{x^2 - 4}$。
§3 · Q2
$5$
$0$
$\infty$
$-1/4$
Degree of numerator ($1$) is less than degree of denominator ($2$), so the limit at $\infty$ is $0$. Algebraic check: divide by $x^2$ to get $(5/x + 1/x^2)/(1 - 4/x^2) \to 0/1 = 0$.分子次数($1$)小于分母次数($2$),故 $\infty$ 处极限为 $0$。代数验证:同除 $x^2$ 得 $(5/x + 1/x^2)/(1 - 4/x^2) \to 0/1 = 0$。
When the denominator grows faster than the numerator, the ratio shrinks to zero. Apply the degree-comparison rule from the cram cheat.分母增长快于分子时,比值趋于零。套用速记中的"次数比较"规则。

The Derivative as a Limit of a Slope作为斜率极限的导数

From secants to tangents.从割线到切线。

Pick a point $(x, f(x))$ and a nearby point $(x + h, f(x + h))$ on the graph. The line through them is the secant; its slope is the average rate of change:在图像上取一点 $(x, f(x))$ 与邻近点 $(x + h, f(x + h))$。两点连线为割线,其斜率即平均变化率:

$$ \frac{f(x + h) - f(x)}{h}. $$

As $h \to 0$, the secant rotates and limits onto the tangent line at $x$. The slope of that tangent is the derivative:当 $h \to 0$ 时,割线旋转并以 $x$ 处的切线为极限。该切线斜率即导数

$$ f'(x) \;=\; \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}. $$
  • Notation.记号。 $f'(x)$, $\dfrac{df}{dx}$, $\dfrac{dy}{dx}$, $Df(x)$ , all mean the same thing.$f'(x)$、$\dfrac{df}{dx}$、$\dfrac{dy}{dx}$、$Df(x)$ , 含义相同。
  • Geometric meaning.几何意义。 $f'(c)$ is the slope of the tangent line to $y = f(x)$ at $x = c$. Positive: curve rising. Negative: falling. Zero: horizontal tangent (often a max / min).$f'(c)$ 是 $y = f(x)$ 在 $x = c$ 处切线的斜率。正:曲线上升;负:下降;零:水平切线(常对应极值)。
  • "From first principles""从定义出发" means using this limit directly. MCV4U Strand A and AP Calc Unit 2 both require students to compute at least one derivative this way before reaching for shortcut rules.指直接用此极限求导。MCV4U 单元 A 与 AP Calc Unit 2 都要求至少有一次"从定义出发"求导,才能套用法则。
Worked Example 4 · Derivative from first principles例题 4 · 从定义出发求导

Let $f(x) = x^2$. Use the limit definition to find $f'(x)$.设 $f(x) = x^2$。用极限定义求 $f'(x)$。

Write the difference quotient.写出差商。

$$ \frac{f(x + h) - f(x)}{h} \;=\; \frac{(x + h)^2 - x^2}{h}. $$

Expand and simplify.展开并化简。

$$ \;=\; \frac{x^2 + 2 x h + h^2 - x^2}{h} \;=\; \frac{2 x h + h^2}{h} \;=\; 2 x + h \quad \text{for } h \ne 0. $$

Take the limit.取极限。

$$ f'(x) \;=\; \lim_{h \to 0} (2 x + h) \;=\; 2 x. $$

Geometric interpretation.几何解读。 At $x = 1$, $f'(1) = 2$ , the parabola $y = x^2$ has tangent slope $2$ at $(1, 1)$. At $x = 0$, $f'(0) = 0$ , horizontal tangent at the vertex. Matches the picture.$x = 1$ 时 $f'(1) = 2$ , 抛物线 $y = x^2$ 在 $(1, 1)$ 处切线斜率为 $2$。$x = 0$ 时 $f'(0) = 0$ , 顶点处水平切线,与图一致。

Using the limit definition, compute $f'(x)$ for $f(x) = 3x + 4$.用极限定义求 $f(x) = 3x + 4$ 的 $f'(x)$。
§4 · Q1
$0$
$3 + 4$
$3$
$3x$
$(f(x + h) - f(x))/h = ((3(x + h) + 4) - (3x + 4))/h = 3h/h = 3$. The limit as $h \to 0$ is $3$. (Geometric check: a line has constant slope.)$(f(x + h) - f(x))/h = ((3(x + h) + 4) - (3x + 4))/h = 3h/h = 3$。$h \to 0$ 时极限为 $3$。(几何验证:直线斜率恒定。)
For a linear function $mx + b$, the derivative is the slope $m$. Plug into the definition; the $h$'s cancel before the limit.线性函数 $mx + b$ 的导数是斜率 $m$。代入定义后 $h$ 在取极限前已约掉。
Using the limit definition, compute $f'(x)$ for $f(x) = x^3$.用极限定义求 $f(x) = x^3$ 的 $f'(x)$。
§4 · Q2
$3x$
$3x^2$
$x^2$
$3x^3$
$(x + h)^3 = x^3 + 3 x^2 h + 3 x h^2 + h^3$. So $((x + h)^3 - x^3)/h = 3 x^2 + 3 x h + h^2$. As $h \to 0$, this limits to $3 x^2$. (Power rule preview: $(x^3)' = 3 x^2$ matches.)$(x + h)^3 = x^3 + 3 x^2 h + 3 x h^2 + h^3$。故 $((x + h)^3 - x^3)/h = 3 x^2 + 3 x h + h^2$。$h \to 0$ 时极限为 $3 x^2$。(幂法则预演:$(x^3)' = 3 x^2$,吻合。)
Expand $(x + h)^3$ with the binomial theorem, cancel $x^3$, divide by $h$, then let $h \to 0$. The leading surviving term is $3 x^2$.用二项式定理展开 $(x + h)^3$,约掉 $x^3$,除以 $h$,再令 $h \to 0$。剩下的首项为 $3 x^2$。

The Power Rule and Linearity幂法则与线性性

Three rules that do 90% of HS differentiation.三条法则覆盖 90% 的高中求导。 $$ \frac{d}{dx}[x^n] \;=\; n\, x^{n - 1}, \qquad \frac{d}{dx}[c \cdot f(x)] \;=\; c \cdot f'(x), \qquad \frac{d}{dx}[f(x) \pm g(x)] \;=\; f'(x) \pm g'(x). $$
  • Power rule.幂法则。 Bring the exponent down as a coefficient, subtract one from the exponent. Works for every real $n$ (MCV4U verifies it for natural $n$; the AP Calc / IB feeders prove the general case).把指数变为系数,指数减一。对任意实数 $n$ 都成立(MCV4U 仅对自然数 $n$ 验证;AP Calc / IB 衔接证明一般情形)。
  • Constant multiple.常数倍法则。 Multiplying $f$ by a constant multiplies its derivative by the same constant. Constants pass through differentiation untouched.$f$ 乘常数后,导数也乘以同一常数。常数对求导是"透明"的。
  • Sum / difference rule.求和 / 差法则。 Differentiate term by term. Powerful in combination with the power rule for any polynomial.逐项求导。与幂法则结合可对任意多项式求导。
  • Derivative of a constant is zero.常数的导数为零。 $\dfrac{d}{dx}[k] = 0$. Geometrically, a horizontal line has slope zero everywhere.$\dfrac{d}{dx}[k] = 0$。几何上水平直线处处斜率为零。
Out of scope here. Product, quotient, and chain rules; derivatives of $\sin x$, $\cos x$, $e^x$, $\ln x$ , all live in AP Calc Unit 2 and IB Math HL E2.本节不覆盖。积法则、商法则、链式法则;$\sin x$、$\cos x$、$e^x$、$\ln x$ 的导数 , 均见 AP Calc Unit 2 与 IB Math HL E2。
Worked Example 5 · Polynomial derivative例题 5 · 多项式求导

Differentiate $f(x) = 5 x^4 - 3 x^2 + 7 x - 8$.对 $f(x) = 5 x^4 - 3 x^2 + 7 x - 8$ 求导。

Differentiate each term.逐项求导。

$$ \frac{d}{dx}[5 x^4] = 5 \cdot 4 x^3 = 20 x^3. $$ $$ \frac{d}{dx}[-3 x^2] = -3 \cdot 2 x = -6 x. $$ $$ \frac{d}{dx}[7 x] = 7. $$ $$ \frac{d}{dx}[-8] = 0. $$

Combine.合并。

$$ f'(x) \;=\; 20 x^3 - 6 x + 7. $$

Evaluate at a point.在某点取值。 For example $f'(1) = 20 - 6 + 7 = 21$ , the tangent to $y = f(x)$ at $x = 1$ has slope $21$.如 $f'(1) = 20 - 6 + 7 = 21$ , $y = f(x)$ 在 $x = 1$ 处切线斜率为 $21$。

Worked Example 5b · Negative and fractional exponents例题 5b · 负指数与分数指数

Differentiate $g(x) = \dfrac{1}{x^2} + \sqrt{x}$ by rewriting with exponents.将 $g(x) = \dfrac{1}{x^2} + \sqrt{x}$ 改写为指数形式后求导。

Rewrite.改写。

$$ g(x) \;=\; x^{-2} + x^{1/2}. $$

Apply the power rule term by term.逐项套用幂法则。

$$ g'(x) \;=\; -2 x^{-3} + \tfrac{1}{2} x^{-1/2}. $$

Re-express in clean form.化为整洁形式。

$$ g'(x) \;=\; -\frac{2}{x^3} + \frac{1}{2 \sqrt{x}}. $$

Sanity-check.合理性核验。 $1/x^2$ decreases as $x$ grows, so its derivative should be negative for $x > 0$ , matches the $-2/x^3$ term.$x > 0$ 时 $1/x^2$ 随 $x$ 增大而减小,其导数应为负 , 与 $-2/x^3$ 一致。

Differentiate $f(x) = x^7 - 2 x^3 + 9$.对 $f(x) = x^7 - 2 x^3 + 9$ 求导。
§5 · Q1
$7 x^6 - 2 x^2 + 9$
$7 x^6 - 6 x^2$
$x^6 - 6 x^2$
$7 x^6 - 6 x^2 + 9$
Power rule on each: $(x^7)' = 7 x^6$; $(-2 x^3)' = -6 x^2$; $(9)' = 0$. Sum: $7 x^6 - 6 x^2$. (The constant disappears , this is the most common single error on Q1-style problems.)逐项用幂法则:$(x^7)' = 7 x^6$;$(-2 x^3)' = -6 x^2$;$(9)' = 0$。求和:$7 x^6 - 6 x^2$。(常数消失 , 此类题最常见的单点错误就是漏掉这一点。)
Constants have derivative zero. Apply the power rule to each variable term and add.常数的导数为零。对每个变量项用幂法则后相加。
Differentiate $h(x) = 4\sqrt{x} - \dfrac{3}{x}$.对 $h(x) = 4\sqrt{x} - \dfrac{3}{x}$ 求导。
§5 · Q2
$\dfrac{4}{\sqrt{x}} + \dfrac{3}{x^2}$
$\dfrac{2}{\sqrt{x}} - \dfrac{3}{x^2}$
$\dfrac{2}{\sqrt{x}} + \dfrac{3}{x^2}$
$2 \sqrt{x} + \dfrac{3}{x^2}$
Rewrite: $h(x) = 4 x^{1/2} - 3 x^{-1}$. Power rule: $h'(x) = 4 \cdot \tfrac{1}{2} x^{-1/2} - 3 \cdot (-1) x^{-2} = 2 x^{-1/2} + 3 x^{-2} = 2/\sqrt{x} + 3/x^2$.改写:$h(x) = 4 x^{1/2} - 3 x^{-1}$。幂法则:$h'(x) = 4 \cdot \tfrac{1}{2} x^{-1/2} - 3 \cdot (-1) x^{-2} = 2 x^{-1/2} + 3 x^{-2} = 2/\sqrt{x} + 3/x^2$。
Watch the signs: $(-3 x^{-1})' = +3 x^{-2}$ because the exponent $-1$ becomes the coefficient, and $-(-1) = +1$.注意符号:$(-3 x^{-1})' = +3 x^{-2}$,因为指数 $-1$ 成为系数,且 $-(-1) = +1$。

Antiderivatives and the Indefinite Integral Honors — US (AP Calc feeder)反导数与不定积分 荣誉 — US(AP Calc 衔接)

Curriculum note.课纲提示。 Integrals are core content in BC Calculus 12 and AB Math 31, and in AP Calculus AB / BC. They are not in Ontario MCV4U (Ontario reserves them for first-year university). They are not in CCSSM at all.积分在 BC Calculus 12、AB Math 31 与 AP Calculus AB / BC 中为核心内容。在安大略 MCV4U 中不在大纲(安大略将其留给大学一年级)。CCSSM 中完全不含
Antidifferentiation is differentiation in reverse.反导数是求导的逆过程。

An antiderivative of $f(x)$ is any function $F(x)$ with $F'(x) = f(x)$. Antiderivatives are unique only up to an additive constant: if $F'(x) = f(x)$, then $(F + C)'(x) = f(x)$ too. The notation $\int f(x) \, dx$ stands for the entire family.$f(x)$ 的反导数是任何满足 $F'(x) = f(x)$ 的函数 $F(x)$。反导数仅在差一常数的意义下唯一:若 $F'(x) = f(x)$,则 $(F + C)'(x) = f(x)$ 也成立。记号 $\int f(x) \, dx$ 表示整族。

$$ \int x^n \, dx \;=\; \frac{x^{n + 1}}{n + 1} + C \quad \text{for } n \ne -1. $$
  • Reverse the power rule.幂法则倒过来。 Add one to the exponent, divide by the new exponent. Then add $+ C$.指数加一,除以新指数,再加 $+ C$。
  • Why $n \ne -1$.为何 $n \ne -1$。 If $n = -1$, the formula gives $x^0 / 0$ , undefined. The antiderivative of $1/x$ is $\ln |x| + C$, an exception that AP Calc Unit 6 and IB Math HL E3 treat carefully.若 $n = -1$,公式给出 $x^0 / 0$ , 未定义。$1/x$ 的反导数是 $\ln |x| + C$,AP Calc Unit 6 与 IB Math HL E3 专门处理此例外。
  • Linearity carries over.线性性同样保留。 $\int (c \cdot f) \, dx = c \int f \, dx$ and $\int (f \pm g) \, dx = \int f \, dx \pm \int g \, dx$.$\int (c \cdot f) \, dx = c \int f \, dx$;$\int (f \pm g) \, dx = \int f \, dx \pm \int g \, dx$。
  • Never forget $+ C$.永远别漏 $+ C$。 An indefinite integral without the constant is wrong by a whole infinite family of functions. AP Calc / IB markers deduct for this.不带常数的不定积分错了整整一族函数。AP Calc / IB 评卷会扣分。
Worked Example 6 · Polynomial antiderivative例题 6 · 多项式反导数

Find $\int (6 x^2 - 4 x + 5) \, dx$.求 $\int (6 x^2 - 4 x + 5) \, dx$。

Integrate term by term using the power rule in reverse.逐项用幂法则的逆运算积分。

$$ \int 6 x^2 \, dx = 6 \cdot \frac{x^3}{3} = 2 x^3. $$ $$ \int -4 x \, dx = -4 \cdot \frac{x^2}{2} = -2 x^2. $$ $$ \int 5 \, dx = 5 x. $$

Combine and add the constant.合并并加上常数。

$$ \int (6 x^2 - 4 x + 5) \, dx \;=\; 2 x^3 - 2 x^2 + 5 x + C. $$

Check by differentiating.求导验证。 $\dfrac{d}{dx}[2 x^3 - 2 x^2 + 5 x + C] = 6 x^2 - 4 x + 5$ ✓.$\dfrac{d}{dx}[2 x^3 - 2 x^2 + 5 x + C] = 6 x^2 - 4 x + 5$ ✓。

Find $\int (3 x^2 + 2) \, dx$.求 $\int (3 x^2 + 2) \, dx$。
§6 · Q1
$6x + C$
$x^3 + 2x + C$
$x^3 + 2x$
$3 x^3 + 2 x + C$
Reverse power rule: $\int 3 x^2 \, dx = 3 \cdot x^3/3 = x^3$; $\int 2 \, dx = 2 x$. Together with the constant: $x^3 + 2 x + C$. Check: $(x^3 + 2 x + C)' = 3 x^2 + 2$ ✓.幂法则倒过来:$\int 3 x^2 \, dx = 3 \cdot x^3/3 = x^3$;$\int 2 \, dx = 2 x$。加上常数:$x^3 + 2 x + C$。验证:$(x^3 + 2 x + C)' = 3 x^2 + 2$ ✓。
Add one to each exponent and divide by the new exponent , then add $+ C$. Choice (c) is missing the constant; choice (d) forgot to divide.指数加一并除以新指数 , 再加 $+ C$。选项 (c) 漏掉常数;选项 (d) 没除以新指数。
Find $\int \sqrt{x} \, dx$.求 $\int \sqrt{x} \, dx$。
§6 · Q2
$\dfrac{2}{3} x^{3/2} + C$
$\dfrac{1}{2} x^{1/2} + C$
$\dfrac{3}{2} x^{3/2} + C$
$\sqrt{x} \cdot x + C$
Rewrite $\sqrt{x} = x^{1/2}$. Reverse power: $\int x^{1/2} \, dx = x^{3/2}/(3/2) + C = \tfrac{2}{3} x^{3/2} + C$. Check: $(\tfrac{2}{3} x^{3/2})' = \tfrac{2}{3} \cdot \tfrac{3}{2} x^{1/2} = x^{1/2} = \sqrt{x}$ ✓.改写 $\sqrt{x} = x^{1/2}$。幂法则倒过来:$\int x^{1/2} \, dx = x^{3/2}/(3/2) + C = \tfrac{2}{3} x^{3/2} + C$。验证:$(\tfrac{2}{3} x^{3/2})' = \tfrac{2}{3} \cdot \tfrac{3}{2} x^{1/2} = x^{1/2} = \sqrt{x}$ ✓。
Treat $\sqrt{x}$ as $x^{1/2}$, then add one ($\to 3/2$) and divide by $3/2$ (i.e. multiply by $2/3$).把 $\sqrt{x}$ 看作 $x^{1/2}$,指数加一得 $3/2$,再除以 $3/2$(即乘 $2/3$)。

The Definite Integral as Signed Area Honors — US (AP Calc feeder)作为带符号面积的定积分 荣誉 — US(AP Calc 衔接)

Curriculum note.课纲提示。 Same as §6: core content in BC Calc 12, AB Math 31, AP Calc AB / BC. Not in MCV4U. Not in CCSSM. This section is the intuitive picture; the full Riemann-sum construction and the Fundamental Theorem of Calculus live in AP Calc Unit 6 and IB Math HL E3 / E4.同 §6:BC Calc 12、AB Math 31、AP Calc AB / BC 核心内容。不在 MCV4U。不在 CCSSM。本节给出直观图像;完整的黎曼和构造与微积分基本定理见 AP Calc Unit 6 与 IB Math HL E3 / E4。
$\int_a^b f(x) \, dx$ = signed area between the graph and the $x$-axis from $a$ to $b$.$\int_a^b f(x) \, dx$ = 图像与 $x$ 轴间从 $a$ 到 $b$ 的带符号面积。

"Signed" means areas above the $x$-axis count positive; areas below count negative. The notation comes from "summing infinitesimal slices": $\int$ is a stretched-out $S$ for "sum," and $dx$ is the width of each slice."带符号"指 $x$ 轴上方的面积为正,下方为负。记号源自"无穷小切片求和":$\int$ 是拉长的 $S$("sum"),$dx$ 是每片的宽度。

  • Fundamental Theorem of Calculus (intuitive form).微积分基本定理(直观形式)。 If $F$ is any antiderivative of $f$, then若 $F$ 是 $f$ 的任一反导数,则 $$\int_a^b f(x) \, dx \;=\; F(b) - F(a).$$ In words: definite integration and antidifferentiation are inverse operations. This is the theorem of all of calculus.即:定积分与反导数互为逆运算。这是整个微积分的核心定理。
  • No constant on the definite integral.定积分不带常数。 The $+ C$ cancels because $F(b) + C - (F(a) + C) = F(b) - F(a)$. Indefinite integrals carry $+ C$; definite integrals do not.$+ C$ 在差中消去:$F(b) + C - (F(a) + C) = F(b) - F(a)$。不定积分带 $+ C$,定积分不带。
  • Reversal sign-flip.交换上下限变号。 $\int_a^b f \, dx = -\int_b^a f \, dx$. Swapping the bounds flips the sign.$\int_a^b f \, dx = -\int_b^a f \, dx$。交换上下限改变符号。
Worked Example 7 · Area under $y = x^2$ from $0$ to $3$例题 7 · 从 $0$ 到 $3$ 在 $y = x^2$ 下的面积

Compute $\int_0^3 x^2 \, dx$ and interpret the result as area.计算 $\int_0^3 x^2 \, dx$ 并将结果解读为面积。

Find an antiderivative.求一个反导数。 By the reverse power rule, $F(x) = x^3 / 3$.由幂法则倒过来,$F(x) = x^3 / 3$。

Apply the Fundamental Theorem.套用微积分基本定理。

$$ \int_0^3 x^2 \, dx \;=\; F(3) - F(0) \;=\; \frac{27}{3} - \frac{0}{3} \;=\; 9. $$

Geometric reading.几何解读。 The region bounded by $y = x^2$, the $x$-axis, and the vertical lines $x = 0$ and $x = 3$ has area $9$ square units. Sanity-check: it sits inside a $3 \times 9$ rectangle (area $27$), and $9 = 27/3$ , a third of the rectangle, matching what the picture looks like.由 $y = x^2$、$x$ 轴及竖直线 $x = 0$、$x = 3$ 围成的区域面积为 $9$ 平方单位。验证:它在 $3 \times 9$ 矩形(面积 $27$)内,且 $9 = 27/3$ , 恰为矩形的三分之一,与图一致。

Compute $\int_1^4 (2 x + 1) \, dx$.计算 $\int_1^4 (2 x + 1) \, dx$。
§7 · Q1
$15$
$12$
$18$
$21$
Antiderivative $F(x) = x^2 + x$. Then $F(4) - F(1) = (16 + 4) - (1 + 1) = 20 - 2 = 18$. Geometric check: the region under $y = 2 x + 1$ from $1$ to $4$ is a trapezoid with parallel sides $f(1) = 3$ and $f(4) = 9$ and width $3$; area $= \tfrac{1}{2}(3 + 9)(3) = 18$ ✓.反导数 $F(x) = x^2 + x$。则 $F(4) - F(1) = (16 + 4) - (1 + 1) = 20 - 2 = 18$。几何验证:从 $1$ 到 $4$ 在 $y = 2 x + 1$ 下方的区域是梯形,平行边为 $f(1) = 3$、$f(4) = 9$,宽为 $3$;面积 $= \tfrac{1}{2}(3 + 9)(3) = 18$ ✓。
Find an antiderivative, then evaluate at the upper bound minus at the lower bound. No $+ C$ needed for a definite integral.先求反导数,再用上限值减下限值。定积分不带 $+ C$。
Compute $\int_{-1}^{1} x^3 \, dx$ and interpret the answer.计算 $\int_{-1}^{1} x^3 \, dx$ 并解释。
§7 · Q2
$0$ , the negative-area region below the $x$-axis on $[-1, 0]$ cancels the positive-area region above on $[0, 1]$$0$ , $[-1, 0]$ 上 $x$ 轴下方的负面积与 $[0, 1]$ 上方的正面积相消
$1/2$
$1$
$2$
Antiderivative $F(x) = x^4 / 4$. $F(1) - F(-1) = 1/4 - 1/4 = 0$. Geometric reading: $x^3$ is an odd function, so the signed areas on the two sides of $0$ have equal magnitude and opposite sign , they cancel.反导数 $F(x) = x^4 / 4$。$F(1) - F(-1) = 1/4 - 1/4 = 0$。几何解读:$x^3$ 是奇函数,故 $0$ 两侧的带符号面积绝对值相等、符号相反 , 互相抵消。
Use $F(x) = x^4 / 4$, evaluate at $1$ and $-1$, subtract. The odd-function symmetry forces the answer to zero.取 $F(x) = x^4 / 4$,在 $1$ 与 $-1$ 处取值后相减。奇函数对称性迫使结果为零。

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Limit hygiene求极限规范
  • Always try substitution first.先试直接代入。 It works for almost every limit at a point of continuity. Only if you get $0/0$ should you reach for factoring or the conjugate.在连续点处几乎总成立。只有得到 $0/0$ 时才用因式分解或共轭。
  • $0/0$ does not mean "does not exist."$0/0$ 等于"不存在"。 It means "do more work." Almost every §2 problem has a clean answer once you factor.它的意思是"还要继续做"。§2 几乎每道题在因式分解后都有干净答案。
  • Show the limit symbol through the cancellation.约分过程中保留极限符号。 Write $\lim_{x \to c}$ on every line until the moment you substitute. AP / MCV4U markers want the method, not just the number.在最终代入前,每一行都写 $\lim_{x \to c}$。AP / MCV4U 评卷重视方法过程,不只看数值。
Derivatives (§4-§5)导数(§4-§5)
  • "From first principles" means write the limit."从定义出发"意味着写出极限。 If the question says "from the definition" or "using the limit," start with $\lim_{h \to 0} (f(x + h) - f(x))/h$. Using the power rule directly will lose method marks.题目若要求"由定义"或"用极限",须从 $\lim_{h \to 0} (f(x + h) - f(x))/h$ 起步。直接套幂法则会丢方法分。
  • Constants disappear, exponents drop.常数消失,指数下降。 $(c)' = 0$ and $(x^n)' = n x^{n - 1}$. The most common slip on Q1-style problems is leaving a constant term untouched.$(c)' = 0$、$(x^n)' = n x^{n - 1}$。Q1 型题最常见错误是没动常数项。
  • Rewrite radicals and reciprocals as powers先把根式与倒数化为指数 before differentiating. $\sqrt{x} = x^{1/2}$, $1/x = x^{-1}$, $1/x^3 = x^{-3}$. The power rule then handles them mechanically.再求导。$\sqrt{x} = x^{1/2}$,$1/x = x^{-1}$,$1/x^3 = x^{-3}$。幂法则即可机械处理。
Integrals (§6-§7) , beyond MCV4U积分(§6-§7), MCV4U 之后
  • Never forget $+ C$ on indefinite integrals.不定积分务必带 $+ C$。 AP Calc explicitly deducts for missing constants. Indefinite has it; definite does not.AP Calc 对漏常数会扣分。不定积分带,定积分不带。
  • Reverse the power rule, then differentiate to check.幂法则倒过来后求导回检。 If $F'(x) = f(x)$, you're correct. This catches both arithmetic slips and the missing-$C$ trap (the $+ C$ check forces you to write it).若 $F'(x) = f(x)$ 即正确。此法同时抓住算术错误与漏常数(写下验证必带 $+ C$)。
  • Definite integrals can be negative.定积分可为负。 Area below the $x$-axis counts negative. Read the bounds and the sign of $f$ before you panic about a negative number.$x$ 轴下方面积为负。在为负数惊慌前,先看上下限和 $f$ 的符号。
Look ahead to AP / IB面向 AP / IB
  • $\epsilon$-$\delta$, IVT, EVT, MVT$\epsilon$-$\delta$、IVT、EVT、MVT are AP Calc Unit 1 content. Don't try them here , this unit is the geometric / tabular intuition that makes the formal version click.属 AP Calc Unit 1 内容。本单元不展开 , 这里是几何 / 数表直观,为后续严格化打底。
  • Product, quotient, chain rules; $\sin x$, $\cos x$, $e^x$, $\ln x$ derivatives积 / 商 / 链式法则;$\sin x$、$\cos x$、$e^x$、$\ln x$ 的导数 are AP Calc Unit 2 and IB Math HL E2. Bookmark the feeder links once you've nailed the power rule here.属 AP Calc Unit 2 与 IB Math HL E2。掌握幂法则后收藏末尾衔接链接。
  • Riemann sums and the formal Fundamental Theorem黎曼和与严格的微积分基本定理 are AP Calc Unit 6 and IB Math HL E3. This unit's signed-area picture is exactly the bridge between the intuitive "limit of a sum" idea and the formal Riemann construction.属 AP Calc Unit 6 与 IB Math HL E3。本单元的带符号面积图正是从"求和的极限"直观到严格黎曼构造的桥梁。

Flashcards闪卡

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Meaning of $\lim_{x \to c} f(x) = L$?$\lim_{x \to c} f(x) = L$ 的含义?
$f(x)$ approaches $L$ as $x$ gets arbitrarily close to $c$ from either side. Does not depend on $f(c)$.$x$ 从两侧任意接近 $c$ 时 $f(x)$ 趋近 $L$。与 $f(c)$ 无关。
Two-sided limit existence?双侧极限存在条件?
$$\lim_{x \to c} f(x) \;\text{exists} \iff \lim_{x \to c^-} f = \lim_{x \to c^+} f$$
Continuity at $x = c$ (three conditions)?$x = c$ 处连续(三条件)?
(1) $f(c)$ defined; (2) $\lim_{x \to c} f$ exists; (3) limit equals $f(c)$.(1) $f(c)$ 有定义;(2) $\lim_{x \to c} f$ 存在;(3) 极限 $= f(c)$。
Factor-and-cancel for $\lim_{x \to 2}(x^2 - 4)/(x - 2)$?用因式约分求 $\lim_{x \to 2}(x^2 - 4)/(x - 2)$?
$$\lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2}(x + 2) = 4$$
Limit at infinity for $(3x^2 + x)/(2x^2 - 5)$?$\infty$ 处 $(3x^2 + x)/(2x^2 - 5)$ 的极限?
$$\lim_{x \to \infty} = \frac{3}{2}$$ (same degree, ratio of leading coefficients)(同次时取首项系数之比)
Definition of the derivative?导数定义?
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
Geometric meaning of $f'(c)$?$f'(c)$ 的几何意义?
Slope of the tangent line to $y = f(x)$ at $x = c$.$y = f(x)$ 在 $x = c$ 处切线的斜率。
Power rule?幂法则?
$$\frac{d}{dx}[x^n] = n x^{n - 1}$$
Constant multiple and sum rules?常数倍法则与求和法则?
$$(c f)' = c f', \quad (f \pm g)' = f' \pm g'$$
Derivative of a constant?常数的导数?
$$\frac{d}{dx}[k] = 0$$
Reverse power rule (antiderivative)?幂法则的逆(反导数)?
$$\int x^n \, dx = \frac{x^{n + 1}}{n + 1} + C \;\; (n \ne -1)$$
Why $n \ne -1$ in the reverse power rule?为何 $n \ne -1$?
Formula gives $x^0/0$, undefined. Antiderivative of $1/x$ is $\ln |x| + C$.公式给出 $x^0/0$,未定义。$1/x$ 的反导数为 $\ln |x| + C$。
Definite integral as area?定积分即面积?
$$\int_a^b f(x) \, dx = \text{signed area between } y = f(x) \text{ and the } x\text{-axis}$$
Fundamental Theorem (intuitive)?微积分基本定理(直观形式)?
$$\int_a^b f(x) \, dx = F(b) - F(a) \text{ when } F' = f$$

Practice Quiz综合测验

Evaluate $\lim_{x \to 5} (3 x^2 - 7)$.求 $\lim_{x \to 5} (3 x^2 - 7)$。
Q1
$0$
$25$
$68$
Does not exist不存在
Polynomial $\Rightarrow$ continuous everywhere $\Rightarrow$ direct substitution: $3 \cdot 25 - 7 = 68$.多项式处处连续,直接代入:$3 \cdot 25 - 7 = 68$。
Polynomials are continuous, so the limit equals the value. Plug $x = 5$ in.多项式连续,极限即取值。代入 $x = 5$。
Evaluate $\lim_{x \to 4} \dfrac{x^2 - 16}{x - 4}$.求 $\lim_{x \to 4} \dfrac{x^2 - 16}{x - 4}$。
Q2
$0$
$8$
$4$
Does not exist不存在
$0/0$ ⇒ factor. $(x^2 - 16)/(x - 4) = (x - 4)(x + 4)/(x - 4) = x + 4 \to 8$ as $x \to 4$.$0/0$,因式分解。$(x^2 - 16)/(x - 4) = (x - 4)(x + 4)/(x - 4) = x + 4 \to 8$,当 $x \to 4$。
Factor the numerator as a difference of squares and cancel $x - 4$.分子用平方差分解,约掉 $x - 4$。
$f(x) = \begin{cases} 3 - x & x \le 1 \\ 2 x & x > 1 \end{cases}$. Is $f$ continuous at $x = 1$?$f(x) = \begin{cases} 3 - x & x \le 1 \\ 2 x & x > 1 \end{cases}$。$f$ 在 $x = 1$ 处连续吗?
Q3
Yes, both sides equal $2$.连续,两侧皆为 $2$。
Yes, both sides equal $3$.连续,两侧皆为 $3$。
No, $f(1)$ is undefined.不连续,$f(1)$ 无定义。
Yes , left $= 3 - 1 = 2$ and right $= 2 \cdot 1 = 2$, and $f(1) = 2$.连续 , 左 $= 3 - 1 = 2$、右 $= 2 \cdot 1 = 2$,且 $f(1) = 2$。
Left limit: $3 - 1 = 2$. Right limit: $2 \cdot 1 = 2$. Both agree, the two-sided limit is $2$, and $f(1) = 3 - 1 = 2$. All three continuity conditions hold ✓.左极限:$3 - 1 = 2$。右极限:$2 \cdot 1 = 2$。两侧一致,双侧极限为 $2$,且 $f(1) = 3 - 1 = 2$。三连续条件全满足 ✓。
Check both one-sided limits and the value $f(1)$. Continuity needs all three to match.检验两单侧极限与 $f(1)$。连续性要求三者一致。
Find $\lim_{x \to \infty} \dfrac{4 x^3 - x}{2 x^3 + 5}$.求 $\lim_{x \to \infty} \dfrac{4 x^3 - x}{2 x^3 + 5}$。
Q4
$2$
$1/2$
$0$
$\infty$
Same degree (3 / 3) $\Rightarrow$ limit equals ratio of leading coefficients $= 4 / 2 = 2$.同次(3 / 3)$\Rightarrow$ 极限为首项系数之比 $= 4 / 2 = 2$。
When numerator and denominator have the same degree, divide leading coefficients.分子分母同次时取首项系数之比。
Use the definition of the derivative to find $f'(x)$ for $f(x) = 1/x$. 🇨🇦 MCV4U Strand A "from first principles"用导数定义求 $f(x) = 1/x$ 的 $f'(x)$。🇨🇦 MCV4U 单元 A "从定义出发"
Q5
$1/x^2$
$-1/x$
$-1/x^2$
$0$
$(f(x + h) - f(x))/h = (1/(x + h) - 1/x)/h = ((x - (x + h))/(x(x + h)))/h = (-h)/(h \cdot x (x + h)) = -1/(x(x + h))$. As $h \to 0$, $\to -1/x^2$. (Matches the power rule: $(x^{-1})' = -x^{-2}$.)$(f(x + h) - f(x))/h = (1/(x + h) - 1/x)/h = ((x - (x + h))/(x(x + h)))/h = (-h)/(h \cdot x (x + h)) = -1/(x(x + h))$。$h \to 0$ 时 $\to -1/x^2$。(与幂法则一致:$(x^{-1})' = -x^{-2}$。)
Common-denominator the difference, simplify, cancel $h$, then take the limit.差通分、化简、约掉 $h$,再取极限。
Differentiate $f(x) = 2 x^5 + 3 \sqrt{x} - 1$.对 $f(x) = 2 x^5 + 3 \sqrt{x} - 1$ 求导。
Q6
$10 x^4 + \dfrac{3}{2\sqrt{x}} - 1$
$10 x^4 + \dfrac{3}{2\sqrt{x}}$
$10 x^5 + \dfrac{3}{2}\sqrt{x}$
$2 x^4 + 3\sqrt{x}$
$(2 x^5)' = 10 x^4$. $(3 x^{1/2})' = 3 \cdot \tfrac{1}{2} x^{-1/2} = 3/(2\sqrt{x})$. $(-1)' = 0$. Sum: $10 x^4 + 3/(2\sqrt{x})$.$(2 x^5)' = 10 x^4$。$(3 x^{1/2})' = 3 \cdot \tfrac{1}{2} x^{-1/2} = 3/(2\sqrt{x})$。$(-1)' = 0$。求和:$10 x^4 + 3/(2\sqrt{x})$。
Constants disappear. Rewrite $\sqrt{x} = x^{1/2}$ and apply the power rule.常数消失。把 $\sqrt{x}$ 改写为 $x^{1/2}$ 再用幂法则。
The curve $y = x^3 - 6 x^2 + 5$ has a horizontal tangent at which $x$-values?曲线 $y = x^3 - 6 x^2 + 5$ 在哪些 $x$ 值处切线水平?
Q7
$x = 0$ only
$x = 4$ only
$x = 0$ and $x = 4$
None
Horizontal tangent $\Leftrightarrow$ $f'(x) = 0$. Differentiate: $f'(x) = 3 x^2 - 12 x = 3 x (x - 4)$. Zeros: $x = 0$ and $x = 4$.水平切线 $\Leftrightarrow$ $f'(x) = 0$。求导:$f'(x) = 3 x^2 - 12 x = 3 x (x - 4)$。零点:$x = 0$、$x = 4$。
Differentiate the cubic, set the derivative to zero, factor the quadratic, read off both roots.对三次式求导,令导数为零,因式分解二次式,读出两根。
Find $\int (4 x^3 - 2 x + 1) \, dx$. Beyond MCV4U · BC Calc 12 / Math 31求 $\int (4 x^3 - 2 x + 1) \, dx$。MCV4U 之后 · BC Calc 12 / Math 31
Q8
$x^4 - x^2 + x + C$
$x^4 - x^2 + x$
$12 x^2 - 2 + C$
$4 x^4 - 2 x^2 + x + C$
$\int 4 x^3 \, dx = x^4$; $\int -2 x \, dx = -x^2$; $\int 1 \, dx = x$. Together: $x^4 - x^2 + x + C$. Check: $(x^4 - x^2 + x + C)' = 4 x^3 - 2 x + 1$ ✓.$\int 4 x^3 \, dx = x^4$;$\int -2 x \, dx = -x^2$;$\int 1 \, dx = x$。合并:$x^4 - x^2 + x + C$。验证:$(x^4 - x^2 + x + C)' = 4 x^3 - 2 x + 1$ ✓。
Reverse power rule term by term. Never forget $+ C$ on an indefinite integral.逐项用幂法则的逆。不定积分务必带 $+ C$。
Compute $\int_0^2 (3 x^2 + 1) \, dx$. Beyond MCV4U · BC Calc 12 / Math 31计算 $\int_0^2 (3 x^2 + 1) \, dx$。MCV4U 之后 · BC Calc 12 / Math 31
Q9
$8$
$9$
$10$
$12$
Antiderivative $F(x) = x^3 + x$. $F(2) - F(0) = (8 + 2) - 0 = 10$.反导数 $F(x) = x^3 + x$。$F(2) - F(0) = (8 + 2) - 0 = 10$。
Find $F$ with $F' = f$, then $F(\text{upper}) - F(\text{lower})$. No $+ C$ needed on the definite integral.先求 $F$ 使 $F' = f$,再用 $F(上限) - F(下限)$。定积分不带 $+ C$。
A particle moves with position $s(t) = t^3 - 9 t^2 + 24 t$ (in metres, $t$ in seconds). Find its velocity at $t = 2$ s.某质点位移 $s(t) = t^3 - 9 t^2 + 24 t$(米,$t$ 秒)。求 $t = 2$ s 时的速度。
Q10
$24$ m/s
$0$ m/s
$12$ m/s
$-12$ m/s
Velocity $v(t) = s'(t) = 3 t^2 - 18 t + 24$. At $t = 2$: $v(2) = 12 - 36 + 24 = 0$ m/s. The particle is momentarily at rest at $t = 2$ , a tell-tale of a local max / min of position. (This kinematic interpretation of the derivative is exactly the MCV4U Strand A motivating example.)速度 $v(t) = s'(t) = 3 t^2 - 18 t + 24$。$t = 2$ 时 $v(2) = 12 - 36 + 24 = 0$ m/s。$t = 2$ 时质点瞬时静止 , 位移的极值标志。(导数的运动学解读正是 MCV4U 单元 A 的引入例。)
Velocity = derivative of position. Differentiate, then plug in $t = 2$.速度 = 位移的导数。先求导,再代入 $t = 2$。

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成,再勾选每一项。

0 / 12 mastered已掌握 0 / 12

What This Feeds Into本单元的去向

This is the feeder unit. Everything you met here is treated formally and at greater depth in the AP Calculus and IB Math HL E-cluster units already shipped in this repo. AP Calculus Unit 1 picks up at §1-§3 with the $\epsilon$-$\delta$ definition, IVT/EVT/MVT, and continuity on closed intervals. AP Calculus Unit 2 picks up at §4-§5 with product, quotient, and chain rules, plus derivatives of $\sin x$, $\cos x$, $e^x$, $\ln x$. AP Calculus Unit 6 picks up at §6-§7 with Riemann sums, the formal Fundamental Theorem, and accumulation functions. The IB Math HL E-cluster covers the same span (E1 principles, E2 techniques, E3 integral techniques, E4 problem-solving, E5 ODEs, E6 Maclaurin) at HL depth.本单元是衔接单元。这里遇到的一切,在仓库已有的 AP Calculus 与 IB Math HL E 簇单元中都会得到严格、更深的处理。AP Calculus Unit 1 自 §1-§3 起接续,加入 $\epsilon$-$\delta$ 定义、IVT / EVT / MVT 与闭区间上的连续性。AP Calculus Unit 2 自 §4-§5 起接续,加入积 / 商 / 链式法则及 $\sin x$、$\cos x$、$e^x$、$\ln x$ 的导数。AP Calculus Unit 6 自 §6-§7 起接续,加入黎曼和、严格的微积分基本定理与累积函数。IB Math HL E 簇覆盖同等内容(E1 原理、E2 技巧、E3 积分技巧、E4 问题求解、E5 常微、E6 麦克劳林),但深度为 HL 级。

Within High School Math.在 HS Math 内部。

This unit is the last HS Math title in the sprint , it sits after Units 1-14 (linear, quadratic, polynomial, rational, exponential / log, sequences, right-triangle trig, unit-circle trig, trig identities, function transformations, combinatorics, probability, statistics, vectors). It draws on Unit 3 (polynomial factoring) and Unit 4 (rational and radical expressions) for the algebraic limit techniques in §2, and on Unit 9's identities for the limit $\lim_{x \to 0} \sin x / x$ touched on in §1.本单元是本冲刺的最后一个 HS Math 标题 , 排在 Unit 1-14 之后(线性、二次、多项式、有理与根式、指数 / 对数、数列、直角三角形三角学、单位圆三角学、三角恒等式、函数变换、排列组合、概率、统计、向量)。§2 的代数限技巧用到 Unit 3(多项式因式分解)与 Unit 4(有理与根式表达式);§1 中触及的 $\lim_{x \to 0} \sin x / x$ 用到 Unit 9 的恒等式。

Across the AP and IB feeders in this repo.本仓库中的 AP 与 IB 衔接单元。

AP Calculus Unit 1 · Limits and Continuity ($\epsilon$-$\delta$ definition, IVT / EVT / MVT, formal continuity)AP Calculus Unit 1 · 极限与连续($\epsilon$-$\delta$ 定义、IVT / EVT / MVT、严格连续性) AP Calculus Unit 2 · Differentiation (product, quotient, chain rules; trig / exp / log derivatives)AP Calculus Unit 2 · 微分(积 / 商 / 链式法则;三角 / 指数 / 对数函数导数) AP Calculus Unit 6 · Integration and Accumulation (Riemann sums, FTC, accumulation functions)AP Calculus Unit 6 · 积分与变化量的累积(黎曼和、微积分基本定理、累积函数) IB Math HL E1 · Principles of Differential Calculus (limit definition, first principles at HL depth)IB Math HL E1 · 微分学原理(极限定义,HL 级"从定义出发") IB Math HL E2 · Techniques of Differential Calculus (full rule kit at HL depth)IB Math HL E2 · 微分学技巧(HL 级完整法则) IB Math HL E3 · Techniques of Integral Calculus (substitution, parts, partial fractions)IB Math HL E3 · 积分学技巧(换元、分部、部分分式)

If you are aiming for AP Calculus AB: this unit covers the spine of Units 1, 2, and 6 of the AP CED at intuitive depth , treat the AP feeders as the rigorous second pass before exam season. If you are aiming for IB Math HL: this unit is the prerequisite to the E-cluster (E1-E6); E1 begins exactly where §4 leaves off. If you are aiming for first-year university calculus: this unit plus Calculus 12 / Math 31 (BC / AB) or any AP feeder is the standard runway.备考 AP Calculus AB:本单元在直观深度上覆盖 AP CED 的 Unit 1、Unit 2、Unit 6 主干 , 考试季前用 AP 衔接作严格的第二遍学习。备考 IB Math HL:本单元是 E 簇(E1-E6)的前置;E1 正自 §4 处接续。备考大学一年级微积分:本单元加上 Calculus 12 / Math 31(BC / AB)或任一 AP 衔接,即为标准助跑道。