📚 Review Set 24C: Mastering Integration Techniques | 复习集24C：掌握积分技巧
This article is designed as a comprehensive walkthrough of Review Set 24C, a typical end-of-topic exercise collection found in IB Mathematics Analysis & Approaches or Applications & Interpretation courses. The focus is on integration — both indefinite and definite — and its applications. Whether you are preparing for a unit test or consolidating your understanding before the final examination, working through these concepts methodically will strengthen your ability to handle a wide range of functions, from simple polynomials to rational, trigonometric, and exponential forms. Each section below mirrors the kind of problem you would encounter in a review set, breaking down techniques step by step and offering paired English–Chinese explanations to deepen conceptual clarity.
本文旨在对复习集24C进行详尽解析，该复习集常见于IB数学分析与方法或应用与解释课程中单元末的练习。核心内容围绕积分——包括不定积分和定积分——及其应用展开。无论你是在准备单元测验，还是在大考前巩固理解，有条理地梳理这些概念都能提升你处理各类函数的能力，从简单的多项式到有理函数、三角函数和指数函数。下面每一节都对应复习集中常见的问题类型，逐步拆解技巧，并以中英对照的形式加深概念理解。
A solid foundation in basic integration rules is essential. The power rule, the integral of 1/x, and the standard results for exponential and trigonometric functions must be memorised and applied fluently.
扎实掌握基本积分规则至关重要。幂函数积分法则、1/x的积分，以及指数函数和三角函数的标准化结果都必须熟记并能熟练应用。
Then ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C. This is the reverse of the power rule for differentiation.
设 f(x) = xⁿ, n ≠ -1，则 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C。这是微分幂函数法则的逆运算。
For the natural logarithm, ∫ (1/x) dx = ln|x| in is respected. This is crucial when dealing with definite integrals crossing the y-axis.
对于自然对数，∫ (1/x) dx = ln|x| + C，绝对值保证了定义域的合理性，这在处理穿越y轴的定积分时尤为重要。
Exponential and trigonometric integrals include: ∫ eˣ dx = eˣ + C, ∫ aˣ dx = aˣ , a≠1), ∫ sin x dx = -cos x + C, ∫ cos x dx = sin x + C. These must be recalled instantly.
指数与三角函数积分包括：∫ eˣ dx = eˣ + C，∫ aˣ dx = aˣ/ln a + C (a>0, a≠1)，∫ sin x dx = -cos x + C，∫ cos x dx = sin x + C。这些必须能够瞬间回忆。
2. Linearity and Sum Rules | 线性性质与和的积分法则
Integration is a linear operator. This means we can integrate term by term and factor out constants. In a review set, many mistakes arise from mishandling coefficients.
积分是一个线性算子，意味着我们可以逐项积分并将常数提取出来。在复习集中，许多错误都源于对系数的错误处理。
a constant k. This allows us to break down compound expressions such as 3x² - 2cos x + 5eˣ into simpler parts.
对于 ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx，以及 ∫ k·f(x) dx = k ∫ f(x) dx（k为常数）。这使我们能将诸如 3x² - 2cos x + 5eˣ 的复合表达式拆分成更简单的部分。
ntegrating 4x³ - 6/x + 7sin x. The solution is 4∫x³ dx - 6∫1/x dx + 7∫sin x dx = x⁴ - 6 ln|x| - 7 cos x + C.
复习集24C中的典型练习可能要求对 4x³ - 6/x + 7sin x 进行积分。解为 4∫x³ dx - 6∫1/x dx + 7∫sin x dx = x⁴ - 6 ln|x| - 7 cos x + C。
Always check by differentiation. Differentiating x⁴ - 6 ln|x| - 7 cos x yields 4x³ - 6/x + 7 sin x, confirming the result.
务必通过求导来检验。对 x⁴ - 6 ln|x| - 7 cos x 求导得到 4x³ - 6/x + 7 sin x，验证了结果的正确性。
3. Integration by Substitution (Reverse Chain Rule) | 换元积分法（逆链式法则）
stant factor), we can let u = inner function and transform the integral into a basic form.
换元法是IB数学中最常考的高级技巧。当被积函数包含一个复合函数乘以内层函数的导数（允许相差一个常数因子）时，我们可以令 u = 内层函数，将积分转化为基本形式。
For ∫ f(g(x))·g'(x) dx, set u = g(x), then du = g' ubstitute back to x for indefinite integrals, or change the limits for definite integrals.
对于 ∫ f(g(x))·g'(x) dx，设 u = g(x)，则 du = g'(x) dx，积分变为 ∫ f(u) du。在不定积分中，对 u 积分后再将 u 换回 x；在定积分中，则更换积分上下限。
Example from a typical 24C set: ∫ 2x·ex² dx. Let u = x², du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = ex² + C. Practice identifying the derivative factor: here 2x is the derivative of x².
来自典型24C复习集的例题：∫ 2x·ex² dx。令 u = x²，du = 2x dx，积分变为 ∫ eᵘ du = eᵘ + C = ex² + C。练习识别导数因子：这里的 2x 正是 x² 的导数。
For definite integrals, say ∫₀¹ 2x·ex² =0; when x=1, u=1. Then ∫₀¹ eᵘ du = e¹ - e⁰ = e - 1. This avoids back-substitution errors.
对于定积分，例如 ∫₀¹ 2x·ex² dx，变换上下限：x=0 时 u=0；x=1 时 u=1。则 ∫₀¹ eᵘ du = e¹ - e⁰ = e - 1。这避免了回代时出错。
4. Integration by Parts | 分部积分法
exponentials, or trigonometric functions. The formula is derived from d(uv) = u dv + v du.
分部积分法是乘积求导法则的逆运算，对于多项式与对数、指数或三角函数的乘积尤为重要。其公式可由 d(uv) = u dv + v du 推导而来。
∫ u dv = uv - ∫ v du
key. A mnemonic is LIATE (Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential) — differentiated. In Review Set 24C problems, you may encounter ∫ x·ln x dx or ∫ x·sin x dx.
巧妙选定 u 和 dv 是关键。助记符 LIATE（对数、反三角、代数、三角、指数）——通常 u 应选在求导后简化的函数。在复习集24C的问题中，你可能遇到 ∫ x·ln x dx 或 ∫ x·sin x dx。
Worked example: ∫ x·eˣ dx. Let u = x (algebraic), dv = eˣ dx. Then du = dx, v = eˣ. Applying the formula: ∫ x·eˣ dx = x·eˣ - ∫ eˣ dx = x·eˣ - eˣ + C = eˣ (x - 1) + C.
实际演算：∫ x·eˣ dx。令 u = x（代数函数），dv = eˣ dx，则 du = dx，v = eˣ。代入公式：∫ x·eˣ dx = x·eˣ - ∫ eˣ dx = x·eˣ - eˣ + C = eˣ (x - 1) + C。
in x dx. Each application reduces the power of x by one until the algebraic part vanishes.
有时需要反复应用分部积分，例如 ∫ x²·sin x dx。每次应用使 x 的幂降低一次，直到代数部分消失。
5. Definite Integrals and the Fundamental Theorem | 定积分与微积分基本定理
F(b) - F(a). This forms the backbone of area calculations and accumulated change problems.
微积分基本定理将微分与定积分联系起来。若 F 是 f 在 [a, b] 上的一个原函数，则 ∫ₐᵇ f(x) dx = F(b) - F(a)。这构成了面积计算和累积变化问题的核心。
ntiderivative is x² + 3x. Compute (4² + 3·4) - (1² + 3·1) = (16+12) - (1+3) = 28 - 4 = 24.
在复习集24C中，一个典型问题是：计算 ∫₁⁴ (2x + 3) dx。原函数为 x² + 3x。计算 (4² + 3·4) - (1² + 3·1) = (16+12) - (1+3) = 28 - 4 = 24。
f(x) dx = -∫ᵦₐ f(x) dx. Also, if evaluating area, you must consider where f(x) is negative — area = ∫ |f(x)| dx.
注意下限大于上限时的符号：∫ₐᵇ f(x) dx = -∫ᵦₐ f(x) dx。另外，在计算面积时必须考虑 f(x) 取负值的区间——面积 = ∫ |f(x)| dx。
Properties of definite integrals such as additivity over subintervals and the fact that integrals of odd functions over symmetric intervals evaluate to zero are often tested alongside the fundamental theorem.
定积分的性质，如对子区间的可加性，以及奇函数在对称区间上的积分为零，与基本定理一起常常被考查。
6. Area Between Curves | 曲线之间的面积
o correctly identify the upper and lower functions and find intersection points as limits.
计算两条曲线所围区域的面积是定积分的直接应用。对于 y = f(x) 和 y = g(x) 且在 [a, b] 上 f(x) ≥ g(x)，面积 A = ∫ₐᵇ [f(x) - g(x)] dx。你需要正确识别上、下函数，并求出交点作为积分限。
₀⁴ [(6x - x²) - (x² - 2x)] dx = ∫₀⁴ (8x - 2x²) dx = [4x² - (2/3)x³]₀⁴ = 64 - 128/3 = 64/3.
复习集24C可能会出这样的问题：给出 f(x) = 6x - x² 和 g(x) = x² - 2x。首先求交点：6x - x² = x² - 2x → 2x² - 8x = 0 → 2x(x - 4) = 0，得 x = 0, 4。在 (0,4) 上，f(x) ≥ g(x)。面积 ∫₀⁴ [(6x - x²) - (x² - 2x)] dx = ∫₀⁴ (8x - 2x²) dx = [4x² - (2/3)x³]₀⁴ = 64 - 128/3 = 64/3。
When curves intersect more than twice, you must split the integral into several regions and sum the absolute areas. Always sketch the region to avoid subtracting in the wrong order.
当曲线相交超过两次时，必须将积分分成多个区间，并将各面积绝对值相加。务必先画草图，避免上下函数顺序弄反。
7. Volumes of Revolution | 旋转体体积
r y-axis. The disk method formula for revolution about the x-axis is V = π ∫ₐᵇ [f(x)]² dx.
IB数学中另一个经典应用是计算区域绕 x 轴或 y 轴旋转生成的体积。绕 x 轴旋转的圆盘法公式为 V = π ∫ₐᵇ [f(x)]² dx。
Example from a review set: The region bounded by y = √x, y = 0, and x = 4 is revolved around the x-axis. Volume = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = π(16/2) = 8π.
复习集中的示例：由 y = √x, y = 0 和 x = 4 围成的区域绕 x 轴旋转。体积 = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = π(16/2) = 8π。
For revolution about the y-axis, rearrange to x = g(y) and use V = π ∫y=cy=d [g(y)]² dy. Alternatively, the shell method may be introduced, but the disk/washer method is the standard in IB.
对于绕 y 轴旋转的情形，将函数改写为 x = g(y)，并利用 V = π ∫y=cy=d [g(y)]² dy。另一种方法是柱壳法，但IB通常只要求圆盘/垫圈法。
d: V = π ∫ₐᵇ ([f(x)]² - [g(x)]²) dx, where f(x) ≥ g(x) ≥ 0 ensures the radii are positive.
当区域位于两曲线之间时，使用垫圈法：V = π ∫ₐᵇ ([f(x)]² - [g(x)]²) dx，其中 f(x) ≥ g(x) ≥ 0 以保证半径为正值。
8. Integration of Rational Functions by Partial Fractions | 用部分分式积分有理函数
When the integrand is a rational function where the degree of the numerator is less than the denominator, partial fraction decomposition allows us to break it into simpler fractions that can be integrated using logs and arctangents. This is a skill tested in higher-level IB sets.
当被积函数为有理函数且分子次数低于分母时，部分分式分解能将其拆分为更简单的分式，进而利用对数和反正切函数进行积分。这是IB高水平复习集中考查的一项技能。
= 1 → B = -1/3. Thus the integral becomes (1/3)∫ 1/(x-1) dx - (1/3)∫ 1/(x+2) dx = (1/3) ln|x-1| - (1/3) ln|x+2| + C.
对于相异线性因子，如 1/[(x-1)(x+2)]，设 A/(x-1) + B/(x+2)。解出 A 和 B：A(x+2) + B(x-1) = 1。令 x=1 得 3A = 1 → A = 1/3；x=-2 得 -3B = 1 → B = -1/3。因此积分成为 (1/3)∫ 1/(x-1) dx - (1/3)∫ 1/(x+2) dx = (1/3) ln|x-1| - (1/3) ln|x+2| + C。
If an irreducible quadratic factor like (x²+1) appears in the denominator, the corresponding term will be of the form (Bx+C)/(x²+1), leading to ln and arctan integrals.
如果分母中出现如 (x²+1) 的不可约二次因子，对应的分式为 (Bx+C)/(x²+1)，积分将涉及对数和反正切函数。
9. Trigonometric Integrals and Identities | 三角积分与恒等式
on identities: sin²x = (1 - cos 2x)/2, cos²x = (1 + cos 2x)/2, and sin 2x = 2 sin x cos x.
积分正弦和余弦的幂次常常需要利用三角恒等式降次。常用恒等式：sin²x = (1 - cos 2x)/2，cos²x = (1 + cos 2x)/2，以及 sin 2x = 2 sin x cos x。
tity: ∫ (1 - cos 2x)/2 dx = (1/2)x - (1/4) sin 2x + C. This is a staple in Review Set 24C.
计算 ∫ sin²x dx。利用恒等式：∫ (1 - cos 2x)/2 dx = (1/2)x - (1/4) sin 2x + C。这是复习集24C中的基础内容。
Integrals of tan x and sec²x appear: ∫ tan x dx = ln|sec x| + C, and ∫ sec²x dx = tan x + C. For ∫ sec x dx, the result is ln|sec x + tan x| + C, but this is less frequent at standard level.
正切和正割平方的积分也出现：∫ tan x dx = ln|sec x| + C，∫ sec²x dx = tan x + C。对于 ∫ sec x dx，结果为 ln|sec x + tan x| + C，但在标准水平中较少见。
10. Solving Simple Differential Equations | 求解简单微分方程
1/g(y)) dy = ∫ f(x) dx. The constant of integration is determined by an initial condition.
变量分离法是积分的直接延伸。形如 dy/dx = f(x)g(y) 的一阶微分方程可重新排列为 ∫ (1/g(y)) dy = ∫ f(x) dx。积分常数由初始条件确定。
Solve dy/dx = 2xy, given y(0) = 3. Separate: (1/y) dy = 2x dx → ∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C → y = ± ex²+C = Aex². Using y(0)=3 gives A=3, so particular solution y = 3ex².
例题：求解 dy/dx = 2xy，已知 y(0) = 3。分离变量：(1/y) dy = 2x dx → ∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C → y = ± ex²+C = Aex²。代入 y(0)=3 得 A=3，故特解为 y = 3ex²。
Review Set 24C often includes a differential equation modelling a real-world context, such as population growth or Newton's law of cooling, requiring interpretation of the solution.
复习集24C常常包含模拟真实世界场景的微分方程，如种群增长或牛顿冷却定律，要求解释解的实际意义。
11. Improper Integrals and Infinite Limits | 反常积分与无穷限
h integrals are called improper and are evaluated as limits. For ∫ₐ^∞ f(x) dx, compute limb→∞ f(x) dx. If the limit exists and is finite, the integral converges; otherwise it diverges.
有时积分区间为无穷，或被积函数在区间内有垂直渐近线。这类积分称为反常积分，需借助极限进行求值。对于 ∫ₐ^∞ f(x) dx，计算 limb→∞ ∫ₐᵇ f(x) dx。如果极限存在且有限，则积分收敛；否则发散。
A typical 24C question: Evaluate ∫₁∞ 1/x² dx. The antiderivative is -1/x, so limit as b→∞ of [-1/x]₁ᵇ = limb→∞ [-1/b + 1] = 1. Converges to 1. Contrast with ∫₁∞ 1/x dx which diverges because ln b → ∞.
典型的24C问题：计算 ∫₁∞ 1/x² dx。原函数为 -1/x，因此 b→∞ 时 [-1/x]₁ᵇ = limb→∞ [-1/b + 1] = 1。收敛于1。对比 ∫₁∞ 1/x dx，由于 ln b → ∞ 而发散。
For integrands with a discontinuity at c within [a,b], split into ∫ₐᶜ⁻ ε f(x) dx + ∫c+ε ε → 0⁺. Understanding the behaviour near asymptotes prevents mistaken convergence claims.
对于在 [a,b] 内 c 点有间断的被积函数，要拆分为 ∫ₐᶜ⁻ ε f(x) dx + ∫c+εᵇ f(x) dx 并在 ε → 0⁺ 时求极限。理解渐近线附近的行为，可避免误判收敛性。
12. Tips, Common Errors and Conclusion | 提示、常见错误与总结
As you complete Review Set 24C, keep a log of errors. The most common include forgetting the constant +C, misapplying the absolute value in log integrals, substituting back incorrectly, and confusing area with signed integral value. When computing areas, always check where the function crosses the x-axis and split accordingly.
在完成复习集24C时，请记录错误。最常见的包括忘记常数 +C、在対数积分中误用绝对值、回代错误，以及混淆面积与带符号积分值。在计算面积时，务必检查函数穿越 x 轴的位置并分段处理。
F(b) - F(a). A single misplaced sign can cost marks. Practice mental differentiation to quickly check indefinite integrals. If time permits, verify your definite integral with technology (GDC) but show full analytical working.
计算定积分时仔细使用方括号：[F(x)]ₐᵇ = F(b) - F(a)。一个符号错误就可能丢分。练习心算求导以快速检验不定积分。若时间允许，可用图形计算器验证定积分，但必须展示完整解析过程。
The integration techniques covered here—substitution, parts, partial fractions, and trigonometric identities—form the toolkit needed to tackle any IB Mathematics integration problem. Mastery of these methods, combined with conceptual understanding of area, volume, and differential equations, will prepare you thoroughly for the exam. Keep practicing the mixed exercises in Review Set 24C until the patterns become second nature.
本文涵盖的积分技巧——换元法、分部积分、部分分式和三角恒等式——构成了应对任何IB数学积分问题所需的工具箱。掌握这些方法，并结合对面积、体积和微分方程的概念理解，将为你充分备考。坚持练习复习集24C中的混合习题，直到各种模式成为你的第二天性。
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