Unit E3: Techniques
of Integral Calculus单元 E3：积分技巧
The companion to E2. Where E2 supplies four rules for differentiation, E3 inverts them: integration as anti-differentiation, the Fundamental Theorem of Calculus that ties integration to area, and the two main HL methods (substitution and parts) for integrals that do not yield to direct table lookup. Integration is partial inversion; many functions have no closed-form antiderivative, so technique selection matters far more than in differentiation.本单元是 E2 的对应。E2 给出四条求导规则，E3 把它们反过来：积分（integration）即反求导（anti-differentiation），微积分基本定理（Fundamental Theorem of Calculus）把积分与面积联系起来，HL 的两大方法（换元、分部）专攻无法直接查表的积分。积分是部分反演；许多函数没有封闭形式的原函数，因此选对技巧比求导时关键得多。
How to use this guide本指南使用说明
E3 has two layers. The SL layer (E3.1 through E3.4) is computational: learn the antiderivative table, apply the Fundamental Theorem, recognise reverse-chain patterns. The HL layer (E3.5, E3.6) is strategic: pick the right substitution or the right parts pairing.E3 分两层。SL 层（E3.1 至 E3.4）偏计算：记原函数表，套用基本定理，识别反向链式型。HL 层（E3.5、E3.6）偏策略：选对换元或选对分部配对。
Memorise the antiderivative table (E3.2): power, $\sin$, $\cos$, $e^{x}$, $1/x$. Practise four definite integrals using the Fundamental Theorem (E3.3). Skip substitution and parts if you are SL only.
背熟原函数表（E3.2）：幂、$\sin$、$\cos$、$e^{x}$、$1/x$。用基本定理（E3.3）做四道定积分。SL 学生可跳过换元与分部。
For substitution, build the reflex of "spot the inner function whose derivative also appears." For parts, memorise the LIATE priority order for choosing $u$. Practise problems that need both (parts followed by substitution, or substitution then parts).
换元法：训练"识别内函数及其导数同时出现"的条件反射。分部法：背熟 LIATE 选 $u$ 的优先序。多练需要"先分部再换元"或"先换元再分部"的复合题。
Anti-differentiation (Indefinite Integrals)反求导（不定积分） SL 5.5
Power rule for integration (the inverse of the power rule for differentiation): $$ \int x^{n} \, dx \;=\; \frac{x^{n+1}}{n + 1} + C, \qquad n \ne -1. $$ Linearity (sum and constant multiple) carries over from differentiation: $$ \int [a f(x) + b g(x)] \, dx \;=\; a \int f(x) \, dx + b \int g(x) \, dx. $$
constant of integration）。
积分的幂法则（求导幂法则的逆）： $$ \int x^{n} \, dx \;=\; \frac{x^{n+1}}{n + 1} + C, \qquad n \ne -1. $$ 线性性（求和与常数倍）由求导沿用： $$ \int [a f(x) + b g(x)] \, dx \;=\; a \int f(x) \, dx + b \int g(x) \, dx. $$
Find $\displaystyle\int \!\left( 4 x^{3} - 6 x^{2} + 2x - 5 \right) dx$.求 $\displaystyle\int \!\left( 4 x^{3} - 6 x^{2} + 2x - 5 \right) dx$。
Integrate term by term using the power rule.
逐项使用幂法则。
$$ \int 4 x^{3} \, dx \;=\; 4 \cdot \frac{x^{4}}{4} \;=\; x^{4}, \qquad \int 6 x^{2} \, dx \;=\; 2 x^{3}, $$ $$ \int 2x \, dx \;=\; x^{2}, \qquad \int 5 \, dx \;=\; 5x. $$Combine.
合并。
$$ \int \!\left( 4 x^{3} - 6 x^{2} + 2x - 5 \right) dx \;=\; x^{4} - 2 x^{3} + x^{2} - 5x + C. $$Verification. Differentiate the answer: $4 x^{3} - 6 x^{2} + 2x - 5$, recovering the integrand. Every indefinite integral can be checked this way.
验证。对答案求导：$4 x^{3} - 6 x^{2} + 2x - 5$，恢复了被积函数。每个不定积分都可这样验证。
The Standard Integral Table标准积分表 SL 5.5
| Integrand | Antiderivative (+ $C$) |
|---|---|
| $x^{n}\;\;(n \ne -1)$ | $\dfrac{x^{n+1}}{n+1}$ |
| $\dfrac{1}{x}$ | $\ln |x|$ |
| $e^{x}$ | $e^{x}$ |
| $a^{x}$ | $\dfrac{a^{x}}{\ln a}$ |
| $\sin x$ | $-\cos x$ |
| $\cos x$ | $\sin x$ |
| $\sec^{2} x$ | $\tan x$ |
The minus sign on $\int \sin x \, dx = -\cos x + C$ is the single most-forgotten entry. Cross-check by differentiating $-\cos x$: $\tfrac{d}{dx}(-\cos x) = -(-\sin x) = \sin x$.
| 被积函数 | 原函数（外加 $C$） |
|---|---|
| $x^{n}\;\;(n \ne -1)$ | $\dfrac{x^{n+1}}{n+1}$ |
| $\dfrac{1}{x}$ | $\ln |x|$ |
| $e^{x}$ | $e^{x}$ |
| $a^{x}$ | $\dfrac{a^{x}}{\ln a}$ |
| $\sin x$ | $-\cos x$ |
| $\cos x$ | $\sin x$ |
| $\sec^{2} x$ | $\tan x$ |
$\int \sin x \, dx = -\cos x + C$ 中的负号是最易漏的一项。验证：对 $-\cos x$ 求导得 $-(-\sin x) = \sin x$，正确。
Find $\displaystyle\int \!\bigl( 3 \cos x + 2 e^{x} - \tfrac{5}{x} \bigr) dx$.求 $\displaystyle\int \!\bigl( 3 \cos x + 2 e^{x} - \tfrac{5}{x} \bigr) dx$。
Term by term from the table.
逐项查表。
$$ \int 3 \cos x \, dx = 3 \sin x, \quad \int 2 e^{x} \, dx = 2 e^{x}, \quad \int \tfrac{5}{x} \, dx = 5 \ln |x|. $$Combine with the constant of integration.
合并并加积分常数。
$$ \int \!\bigl( 3 \cos x + 2 e^{x} - \tfrac{5}{x} \bigr) dx \;=\; 3 \sin x + 2 e^{x} - 5 \ln |x| + C. $$▸ Going deeper: why $\int \tfrac{1}{x} dx$ needs absolute value bars▸ 深入：$\int \tfrac{1}{x} dx$ 为何要加绝对值
The function $\ln x$ is only defined for $x > 0$, but $1/x$ is defined for all $x \ne 0$ (with $1/x$ negative when $x < 0$). For $x < 0$, the correct antiderivative is $\ln(-x)$, because $\tfrac{d}{dx} \ln(-x) = \tfrac{1}{-x} \cdot (-1) = \tfrac{1}{x}$. Combining the $x > 0$ and $x < 0$ branches gives $\ln |x|$, which is differentiable on $\mathbb{R} \setminus \{0\}$ with derivative $1/x$. Hence the absolute value bars are necessary, not stylistic.
$\ln x$ 仅在 $x > 0$ 处定义，但 $1/x$ 在所有 $x \ne 0$ 处定义（$x < 0$ 时 $1/x$ 为负）。当 $x < 0$ 时，正确的原函数是 $\ln(-x)$，因为 $\tfrac{d}{dx} \ln(-x) = \tfrac{1}{-x} \cdot (-1) = \tfrac{1}{x}$。将 $x > 0$ 与 $x < 0$ 两支合并即得 $\ln |x|$，它在 $\mathbb{R} \setminus \{0\}$ 上可导，导数为 $1/x$。因此绝对值是必须的，不是装饰。
Definite Integrals and the Fundamental Theorem of Calculus定积分与微积分基本定理 SL 5.6
Geometric reading. If $f(x) \ge 0$ on $[a, b]$, the definite integral equals the area under the curve between $x = a$ and $x = b$. If $f(x) < 0$ on part of the interval, the integral gives signed area: area above the $x$-axis counts positive, area below counts negative.
Three properties to use.
- $\displaystyle\int_{a}^{a} f \, dx = 0$.
- $\displaystyle\int_{a}^{b} f \, dx = -\int_{b}^{a} f \, dx$.
- $\displaystyle\int_{a}^{c} f \, dx = \int_{a}^{b} f \, dx + \int_{b}^{c} f \, dx$ (splitting at $b$, where $a < b < c$).
FTC）。若 $f$ 在 $[a, b]$ 上连续、$F$ 是 $f$ 的任一原函数（$F' = f$），则
$$ \int_{a}^{b} f(x) \, dx \;=\; F(b) - F(a). $$
记号 $\bigl[ F(x) \bigr]_{a}^{b}$ 即 $F(b) - F(a)$。
几何意义。若 $f(x) \ge 0$ 在 $[a, b]$ 上成立，则定积分等于曲线下从 $x = a$ 到 $x = b$ 的面积。若 $f(x)$ 在部分区间上为负，积分给出有符号面积：$x$ 轴上方记正、下方记负。
三条常用性质。
- $\displaystyle\int_{a}^{a} f \, dx = 0$。
- $\displaystyle\int_{a}^{b} f \, dx = -\int_{b}^{a} f \, dx$。
- $\displaystyle\int_{a}^{c} f \, dx = \int_{a}^{b} f \, dx + \int_{b}^{c} f \, dx$（在 $b$ 处分段，$a < b < c$）。
Evaluate $\displaystyle\int_{1}^{3} \!\bigl( x^{2} + 2 \bigr) dx$.求 $\displaystyle\int_{1}^{3} \!\bigl( x^{2} + 2 \bigr) dx$。
Find an antiderivative. $F(x) = \tfrac{x^{3}}{3} + 2x$.
求一个原函数。$F(x) = \tfrac{x^{3}}{3} + 2x$。
Apply the FTC.
套用 FTC。
$$ \int_{1}^{3} \!\bigl( x^{2} + 2 \bigr) dx \;=\; \Bigl[ \tfrac{x^{3}}{3} + 2x \Bigr]_{1}^{3} \;=\; \Bigl( \tfrac{27}{3} + 6 \Bigr) - \Bigl( \tfrac{1}{3} + 2 \Bigr) \;=\; 15 - \tfrac{7}{3} \;=\; \tfrac{38}{3}. $$Evaluate $\displaystyle\int_{0}^{\pi} \sin x \, dx$ and $\displaystyle\int_{0}^{2\pi} \sin x \, dx$. Comment on the difference.求 $\displaystyle\int_{0}^{\pi} \sin x \, dx$ 与 $\displaystyle\int_{0}^{2\pi} \sin x \, dx$，并说明区别。
First integral.
第一个积分。
$$ \int_{0}^{\pi} \sin x \, dx \;=\; \bigl[ -\cos x \bigr]_{0}^{\pi} \;=\; (-\cos \pi) - (-\cos 0) \;=\; 1 - (-1) \;=\; 2. $$Second integral.
第二个积分。
$$ \int_{0}^{2\pi} \sin x \, dx \;=\; \bigl[ -\cos x \bigr]_{0}^{2\pi} \;=\; -\cos(2\pi) + \cos 0 \;=\; -1 + 1 \;=\; 0. $$Remark. On $[0, \pi]$, $\sin x \ge 0$, so the integral measures area (which is $2$). On $[\pi, 2\pi]$, $\sin x \le 0$, contributing $-2$. The two cancel over $[0, 2\pi]$, giving zero. The integral is signed area, not geometric area. To get geometric area, integrate $|\sin x|$ instead, which gives $4$.
注。$[0, \pi]$ 上 $\sin x \ge 0$，积分等于面积（值为 $2$）。$[\pi, 2\pi]$ 上 $\sin x \le 0$，贡献 $-2$。两段在 $[0, 2\pi]$ 上抵消为 $0$。定积分给的是有符号面积，不是几何面积。若需几何面积，应积 $|\sin x|$，结果为 $4$。
▸ Going deeper: why the FTC holds▸ 深入：FTC 为何成立
Define the "area-so-far" function $A(x) = \int_{a}^{x} f(t) \, dt$. Then a small increase from $x$ to $x + h$ adds a thin strip of area approximately $f(x) \cdot h$, so $A(x + h) - A(x) \approx f(x) \cdot h$. Dividing by $h$ and taking $h \to 0$ gives $A'(x) = f(x)$. So $A$ is itself an antiderivative of $f$. Any other antiderivative $F$ differs from $A$ by a constant, and the FTC follows:
定义"截至 $x$ 的面积函数"$A(x) = \int_{a}^{x} f(t) \, dt$。从 $x$ 到 $x + h$ 的小增量大致是 $f(x) \cdot h$（一条窄条），故 $A(x + h) - A(x) \approx f(x) \cdot h$。两边除以 $h$ 并令 $h \to 0$ 得 $A'(x) = f(x)$。故 $A$ 本身是 $f$ 的原函数。任一原函数 $F$ 与 $A$ 相差常数，FTC 由此而来：
$$ \int_{a}^{b} f(x) \, dx \;=\; A(b) - A(a) \;=\; F(b) - F(a). $$This is the deepest theorem in single-variable calculus: it shows that differentiation and integration are inverse operations.
这是单变量微积分中最深刻的定理：它表明求导与积分互为逆运算。
Integration by Reverse Chain反向链式积分 SL 5.10
Three canonical templates (all worth memorising):
- $\displaystyle\int g'(x) \cdot \bigl[g(x)\bigr]^{n} \, dx = \dfrac{[g(x)]^{n+1}}{n + 1} + C, \quad n \ne -1$.
- $\displaystyle\int \dfrac{g'(x)}{g(x)} \, dx = \ln |g(x)| + C$.
- $\displaystyle\int g'(x) \cdot e^{g(x)} \, dx = e^{g(x)} + C$.
三条经典模板（都应背熟）：
- $\displaystyle\int g'(x) \cdot \bigl[g(x)\bigr]^{n} \, dx = \dfrac{[g(x)]^{n+1}}{n + 1} + C, \quad n \ne -1$。
- $\displaystyle\int \dfrac{g'(x)}{g(x)} \, dx = \ln |g(x)| + C$。
- $\displaystyle\int g'(x) \cdot e^{g(x)} \, dx = e^{g(x)} + C$。
Evaluate $\displaystyle\int 2x \,(x^{2} + 1)^{4} \, dx$.求 $\displaystyle\int 2x \,(x^{2} + 1)^{4} \, dx$。
Inner function. $g(x) = x^{2} + 1$ and $g'(x) = 2x$ appears exactly as a factor. Template 1 with $n = 4$:
内函数。$g(x) = x^{2} + 1$，$g'(x) = 2x$ 恰好作为因子出现。用模板 1，$n = 4$：
$$ \int 2x \,(x^{2} + 1)^{4} \, dx \;=\; \frac{(x^{2} + 1)^{5}}{5} + C. $$Evaluate $\displaystyle\int \dfrac{4 x^{3}}{x^{4} + 7} \, dx$.求 $\displaystyle\int \dfrac{4 x^{3}}{x^{4} + 7} \, dx$。
Inner function. $g(x) = x^{4} + 7$, so $g'(x) = 4 x^{3}$. The numerator is exactly $g'(x)$. Template 2:
内函数。$g(x) = x^{4} + 7$，$g'(x) = 4 x^{3}$。分子恰为 $g'(x)$。用模板 2：
$$ \int \frac{4 x^{3}}{x^{4} + 7} \, dx \;=\; \ln \bigl| x^{4} + 7 \bigr| + C \;=\; \ln(x^{4} + 7) + C. $$The absolute value bars can be dropped because $x^{4} + 7 > 0$ for all real $x$.
因 $x^{4} + 7 > 0$ 对所有实数成立，可去掉绝对值。
Integration by Substitution换元积分法 HL AHL 5.15
For definite integrals, change the limits as well: if the original is $\displaystyle\int_{a}^{b} f(g(x)) g'(x) \, dx$, the substituted form is $\displaystyle\int_{g(a)}^{g(b)} f(u) \, du$. The new limits are the values of $u$ at the old endpoints, not the old endpoints themselves.
How to choose $u$.
- Look for an inner function whose derivative also appears as a factor.
- If the integrand contains $\sqrt{\text{stuff}}$, try $u = \text{stuff}$.
- If the integrand contains $\ln(\text{stuff})$, try $u = \ln(\text{stuff})$ or $u = \text{stuff}$.
- If the integrand contains $\sin(\text{stuff})$ or $e^{\text{stuff}}$, try $u = \text{stuff}$.
对定积分，积分限也要换：原式 $\displaystyle\int_{a}^{b} f(g(x)) g'(x) \, dx$ 换元后为 $\displaystyle\int_{g(a)}^{g(b)} f(u) \, du$。新限是原端点处 $u$ 的值，不是原端点本身。
如何选 $u$。
- 找一个内函数，其导数也以因子形式出现。
- 被积式含 $\sqrt{\text{某式}}$ 时，试 $u = \text{某式}$。
- 被积式含 $\ln(\text{某式})$ 时，试 $u = \ln(\text{某式})$ 或 $u = \text{某式}$。
- 被积式含 $\sin(\text{某式})$ 或 $e^{\text{某式}}$ 时，试 $u = \text{某式}$。
Evaluate $\displaystyle\int x \sqrt{x^{2} + 4} \, dx$.求 $\displaystyle\int x \sqrt{x^{2} + 4} \, dx$。
Choose $u$. The inner $x^{2} + 4$ has derivative $2x$. Let $u = x^{2} + 4$, so $du = 2x \, dx$, hence $x \, dx = \tfrac{1}{2} du$.
选 $u$。内函数 $x^{2} + 4$ 的导数为 $2x$。令 $u = x^{2} + 4$，则 $du = 2x \, dx$，故 $x \, dx = \tfrac{1}{2} du$。
Substitute.
换元。
$$ \int x \sqrt{x^{2} + 4} \, dx \;=\; \int \sqrt{u} \cdot \tfrac{1}{2} \, du \;=\; \tfrac{1}{2} \int u^{1/2} \, du \;=\; \tfrac{1}{2} \cdot \tfrac{u^{3/2}}{3/2} \;=\; \tfrac{1}{3} u^{3/2}. $$Substitute back.
换回 $x$。
$$ \int x \sqrt{x^{2} + 4} \, dx \;=\; \tfrac{1}{3} (x^{2} + 4)^{3/2} + C. $$Evaluate $\displaystyle\int_{0}^{\pi/2} \sin^{3}(x) \cos x \, dx$.求 $\displaystyle\int_{0}^{\pi/2} \sin^{3}(x) \cos x \, dx$。
Choose $u$. Let $u = \sin x$, so $du = \cos x \, dx$.
选 $u$。令 $u = \sin x$，则 $du = \cos x \, dx$。
Change the limits. At $x = 0$: $u = \sin 0 = 0$. At $x = \pi/2$: $u = \sin(\pi/2) = 1$.
换积分限。$x = 0$：$u = \sin 0 = 0$。$x = \pi/2$：$u = \sin(\pi/2) = 1$。
Substitute and evaluate.
换元后求值。
$$ \int_{0}^{\pi/2} \sin^{3}(x) \cos x \, dx \;=\; \int_{0}^{1} u^{3} \, du \;=\; \Bigl[ \tfrac{u^{4}}{4} \Bigr]_{0}^{1} \;=\; \tfrac{1}{4}. $$Remark. Once the limits are changed, do not substitute back to $x$. Evaluate directly in $u$. This is the main advantage of changing limits versus leaving them in $x$.
注。积分限换好后，无需再换回 $x$；直接在 $u$ 下求值。这是"换限"相对于"保留 $x$"的主要优势。
▸ Going deeper: substitution as the chain rule in reverse▸ 深入：换元法即链式法则的逆
If $F$ is an antiderivative of $f$, the chain rule gives $\tfrac{d}{dx} F(g(x)) = f(g(x)) \cdot g'(x)$. Integrating both sides:
若 $F$ 是 $f$ 的原函数，链式法则给出 $\tfrac{d}{dx} F(g(x)) = f(g(x)) \cdot g'(x)$。两边积分：
$$ \int f(g(x)) \cdot g'(x) \, dx \;=\; F(g(x)) + C. $$The substitution $u = g(x)$ just relabels the antiderivative: the right side equals $F(u) + C = \int f(u) \, du$. So substitution is not a new technique; it is the chain rule applied in reverse, with $u$ as the bookkeeping device. The reason it feels different from "reverse chain" in E3.4 is that you write the $u$ explicitly and convert $dx$ to $du$, which makes the constant-multiple adjustments automatic rather than improvised.
换元 $u = g(x)$ 只是给原函数换个写法：右侧等于 $F(u) + C = \int f(u) \, du$。所以换元并非新方法，而是链式法则反过来用，$u$ 是记号。它与 E3.4 的"反向链式"感觉不同，是因为换元法显式写出 $u$ 并把 $dx$ 换成 $du$，常数倍调整自动而非临场。
Integration by Parts分部积分法 HL AHL 5.15
LIATE priority for choosing $u$. Pick $u$ in this order of preference, with the type higher on the list taking priority:
- Logarithmic ($\ln x$, $\log_{a} x$).
- Inverse trig ($\arcsin$, $\arctan$).
- Algebraic ($x^{n}$, polynomials).
- Trigonometric ($\sin x$, $\cos x$).
- Exponential ($e^{x}$).
LIATE 选 $u$ 优先序。按以下次序选 $u$，靠前者优先：
- L 对数（
Logarithmic）：$\ln x$、$\log_{a} x$。 - I 反三角（
Inverse trig）：$\arcsin$、$\arctan$。 - A 代数（
Algebraic）：$x^{n}$、多项式。 - T 三角（
Trig）：$\sin x$、$\cos x$。 - E 指数（
Exponential）：$e^{x}$。
Evaluate $\displaystyle\int x \, e^{x} \, dx$.求 $\displaystyle\int x \, e^{x} \, dx$。
Apply LIATE. Between $x$ (Algebraic) and $e^{x}$ (Exponential), Algebraic comes first, so $u = x$ and $dv = e^{x} \, dx$.
套 LIATE。$x$（A）与 $e^{x}$（E）中，A 先，故 $u = x$、$dv = e^{x} \, dx$。
Compute $du$ and $v$. $du = dx$ and $v = e^{x}$.
算 $du$ 与 $v$。$du = dx$，$v = e^{x}$。
Apply the parts formula.
代入分部公式。
$$ \int x \, e^{x} \, dx \;=\; x \, e^{x} \;-\; \int e^{x} \, dx \;=\; x e^{x} - e^{x} + C \;=\; (x - 1) e^{x} + C. $$Evaluate $\displaystyle\int \ln x \, dx$.求 $\displaystyle\int \ln x \, dx$。
This looks like a single function, but LIATE still applies. Treat the integrand as $\ln x \cdot 1$. Then $u = \ln x$ (L) and $dv = 1 \, dx$.
看似只有一个函数，但 LIATE 仍适用。把被积式视为 $\ln x \cdot 1$。$u = \ln x$（L），$dv = 1 \, dx$。
Compute $du$ and $v$. $du = \tfrac{1}{x} \, dx$ and $v = x$.
算 $du$ 与 $v$。$du = \tfrac{1}{x} \, dx$，$v = x$。
Apply parts.
套用分部。
$$ \int \ln x \, dx \;=\; x \ln x \;-\; \int x \cdot \tfrac{1}{x} \, dx \;=\; x \ln x - \int 1 \, dx \;=\; x \ln x - x + C. $$Remark. This is the standard derivation of $\int \ln x \, dx$, and the answer $x \ln x - x + C$ is worth memorising. It appears in almost every Paper 3 question that uses parts.
注。这是 $\int \ln x \, dx$ 的标准推导，答案 $x \ln x - x + C$ 值得背下。Paper 3 中几乎所有用分部的题目都会出现它。
▸ Going deeper: parts as the product rule in reverse▸ 深入：分部积分即乘积法则的逆
The product rule says $(uv)' = u' v + u v'$. Rearranging: $u v' = (uv)' - u' v$. Integrating both sides:
乘积法则：$(uv)' = u' v + u v'$。整理得 $u v' = (uv)' - u' v$。两边积分：
$$ \int u v' \, dx \;=\; uv \;-\; \int u' v \, dx. $$In differential notation, this is $\int u \, dv = uv - \int v \, du$. So parts is the integral version of the product rule, exactly as substitution is the integral version of the chain rule. The two HL integration techniques are the inverses of the two SL differentiation rules. Any time a derivative rule exists, an integration rule shadows it.
用微分记号即 $\int u \, dv = uv - \int v \, du$。所以分部积分是乘积法则的积分版本，正如换元是链式法则的积分版本。HL 的两条积分技巧分别对应 SL 的两条求导规则。求导有什么规则，积分就有什么对偶。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Always write $+ C$. Markschemes deduct A1 for missing constants. No exceptions.
- 必写 $+ C$。评分对漏写积分常数扣 A1，没有例外。
- Verify by differentiating. If you have time after evaluating an indefinite integral, differentiate the answer. If you do not recover the integrand exactly, find the error.
- 用求导验证。不定积分算完后有时间，对答案求导。若不能恢复被积函数，查错。
- Rewrite radicals and reciprocals as powers before integrating. $\sqrt{x} = x^{1/2}$, $\tfrac{1}{x^{n}} = x^{-n}$.
- 积分前先把根式与倒数化为幂。$\sqrt{x} = x^{1/2}$、$\tfrac{1}{x^{n}} = x^{-n}$。
- The "evaluate at upper minus lower" subtraction is two A1 marks. Write out $F(b) - F(a)$ in full before simplifying. Many students collapse it mentally and slip on a sign.
- "上限减下限"是两个 A1。简化前先完整写出 $F(b) - F(a)$。许多学生心算合并时弄错符号。
- Signed area is not the same as geometric area. If the question asks for "the area" (geometric) of a region where $f$ crosses zero, split the integral at the zero and add absolute values.
- 有符号面积 $\ne$ 几何面积。若题目问"区域的面积"且 $f$ 在区域内变号，应在零点处分段积分并取绝对值。
- No $+ C$ on definite integrals. The constants cancel in $F(b) - F(a)$. Including $+ C$ on a definite integral is at best wasted ink and at worst a markdown.
- 定积分不加 $+ C$。积分常数在 $F(b) - F(a)$ 中抵消。写了反倒可能扣分。
- Declare the substitution explicitly. Write "Let $u = \ldots$ Then $du = \ldots \, dx$." This is the M1 mark; skipping it forfeits the method credit.
- 显式声明换元。写出"令 $u = \ldots$，故 $du = \ldots \, dx$"。这是 M1，省略丢方法分。
- For definite substitutions, change the limits. Do not change limits and also substitute back to $x$. Pick one path.
- 定积分换元要换限。不要既换限又换回 $x$。二选一即可。
- For parts, write the LIATE choice and the resulting $du$ and $v$ before applying the formula. This is one M1 and one A1.
- 分部前先写 LIATE 选择与所得 $du$、$v$，再套公式。这相当于一个 M1 加一个 A1。
- If parts produces a new integral that needs parts again, do it. Some Paper 3 problems require two or three rounds. The "tabular" shortcut is acceptable if presented clearly.
- 若分部后的新积分仍需分部，继续做。有些 Paper 3 题要做两到三轮。"表格法"（
tabular method）若写得清晰也可接受。
Flashcards闪卡
Unit E3 Practice Quiz单元 E3 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, without the formula box, on your first attempt.
每一条都要"裸做"做对（不看笔记、不看公式框、一次过）才打勾。
- Apply the power rule for integration including negative and fractional exponents, with $+ C$使用积分幂法则（含负、分数指数），写 $+ C$
- Reproduce the standard integrals of $1/x$, $e^{x}$, $\sin x$, $\cos x$, $\sec^{2} x$复现 $1/x$、$e^{x}$、$\sin x$、$\cos x$、$\sec^{2} x$ 的标准积分
- Remember that $\int \sin x \, dx = -\cos x + C$ (with the minus sign)记住 $\int \sin x \, dx = -\cos x + C$（含负号）
- State the Fundamental Theorem of Calculus and apply it to a polynomial integrand陈述 FTC，并对多项式被积函数应用
- Distinguish signed area from geometric area; split at zeros when needed区分有符号面积与几何面积；必要时在零点处分段
- Recognise the three reverse-chain templates: $g'[g]^{n}$, $g'/g$, $g' e^{g}$识别三条反向链式模板：$g'[g]^{n}$、$g'/g$、$g' e^{g}$
- Adjust by a constant multiple when $g'$ is off by a factor$g'$ 差常数倍时用倒数调整
- HL Choose a substitution $u = g(x)$, compute $du$, and rewrite the integral in $u$选换元 $u = g(x)$、算 $du$、把积分改写为 $u$ 的形式
- HL Change limits when substituting in a definite integral定积分换元时同步换限
- HL Apply the LIATE rule to pick $u$ in integration by parts用 LIATE 规则在分部积分中选 $u$
- HL Compute $\int \ln x \, dx = x \ln x - x + C$ from parts用分部算 $\int \ln x \, dx = x \ln x - x + C$
- HL Apply parts twice when the result of one application still needs integration by parts当一轮分部后仍需分部时，连用两轮
- Verify any antiderivative by differentiation用求导验证任一原函数
IB Paper-Style PracticeIB 试卷风格练习
E3 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_E3_*.html with the bilingual built-in pattern. Meanwhile, the A2 and A5 Practice sets demonstrate the format.
E3 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_E3_*.html，采用双语内嵌格式。可参照 A2、A5 练习集了解模板。