📚 D – Integration | 积分
Integration is the second major pillar of calculus alongside differentiation. In IB Mathematics, particularly in the Analysis and Approaches course at both Standard and Higher Level, integration enables you to find areas, volumes, and solutions to differential equations, linking rates of change back to accumulated quantities. Mastering integration – from basic antiderivatives to sophisticated techniques – is essential for success in the calculus components of Paper 1 and Paper 2, as well as for internal assessments that involve analytical modeling. This article systematically covers the core integration topics required for IB, from indefinite integrals and the Fundamental Theorem of Calculus to volumes of revolution and kinematics, with clear bilingual explanations and exam-focused examples.
积分是与微分并列的微积分第二大支柱。在IB数学课程中，尤其是在分析与方法和应用与解释课程的标阶与高阶级别，积分可以帮助你计算面积、体积以及求解微分方程，将变化率与累积量联系起来。掌握积分——从基本的原函数到复杂的计算技巧——对于在Paper 1和Paper 2的微积分部分取得好成绩，以及需要分析建模的内部评估都至关重要。本文系统地覆盖了IB所需的积分核心专题，从不定积分和微积分基本定理到旋转体体积和运动学，提供清晰的中英双语解释和紧扣考试的示例。
1. What is Integration? | 什么是积分？
Integration, often referred to as antidifferentiation, is the process of recovering a function from its derivative. If differentiating a function F(x) yields f(x), then integrating f(x) gives back F(x) plus an arbitrary constant C. This is why the indefinite integral is expressed as ∫ f(x) dx = F(x) + C. The symbol ∫, an elongated ‘S’, was introduced by Leibniz and signifies the summation of infinitesimally small quantities. In IB terms, integration allows you to reverse the process of differentiation – to find a function when its gradient function is known.
积分，常被称为反微分，是从导数恢复原函数的过程。如果对函数 F(x) 求导得到 f(x)，那么对 f(x) 积分就会得到 F(x) 加上一个任意常数 C。这就是为什么不定积分表示为 ∫ f(x) dx = F(x) + C。符号 ∫ 是一个拉长的“S”，由莱布尼茨引入，代表无穷小量的累加。在IB术语中，积分让你能够逆转微分过程——在已知梯度函数的情况下找到原函数。
Integration also has a profound geometric meaning as the net area between a curve and the x-axis over an interval. A definite integral ∫ₐᵇ f(x) dx calculates the signed area from x = a to x = b. Areas above the x-axis contribute positively, while areas below the axis contribute negatively. This dual interpretation – as an antiderivative and as an accumulator of area – makes integration a versatile tool in both pure and applied mathematics.
积分还具有深刻的几何意义，即曲线与x轴之间在区间上的净面积。定积分 ∫ₐᵇ f(x) dx 计算从 x = a 到 x = b 的带符号面积。x轴上方的面积贡献正值，而轴下方的面积贡献负值。这种双重解释——作为原函数和作为面积累加器——使积分在纯数学和应用数学中都成为一个多功能工具。
2. Indefinite Integrals and Antiderivatives | 不定积分与原函数
An indefinite integral of a function f(x) is any function F(x) such that F'(x) = f(x). There are infinitely many antiderivatives differing by a constant, so we always add the constant of integration, usually denoted by C or k. For example, since the derivative of x³ is 3x², we write ∫ 3x² dx = x³ + C. The concept of the family of antiderivatives is fundamental: when solving an initial value problem, the given condition pins down the particular member of the family.
函数 f(x) 的不定积分是满足 F'(x) = f(x) 的任何函数 F(x)。存在无穷多个相差一个常数的原函数，因此我们总是加上积分常数，通常记为 C 或 k。例如，因为 x³ 的导数是 3x²，我们写成 ∫ 3x² dx = x³ + C。原函数族的这个概念是基础：在求解初值问题时，给定的条件会确定族中的特定成员。
IB candidates must be fluent in recognizing the antiderivatives of common functions: power functions, exponentials, trigonometric functions, and reciprocals. Memorising the forms ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ -1), ∫ eˣ dx = eˣ + C, ∫ 1/x dx = ln|x| + C, and the integrals of sin x and cos x is a prerequisite. The constant of integration must never be omitted in an indefinite integral answer on IB exams.
IB考生必须熟练识别常见函数的原函数：幂函数、指数函数、三角函数和倒数函数。记住公式 ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C（n ≠ -1），∫ eˣ dx = eˣ + C，∫ 1/x dx = ln|x| + C，以及 sin x 和 cos x 的积分是先决条件。在IB考试中，不定积分的答案绝不可省略积分常数。
3. Basic Integration Rules | 积分基本法则
Integration is linear, meaning that the integral of a sum is the sum of the integrals, and a constant factor can be pulled out: ∫ [a·f(x) + b·g(x)] dx = a·∫ f(x) dx + b·∫ g(x) dx. This linearity mirrors the sum rule and constant multiple rule of differentiation. For example, ∫ (4x² – 2eˣ) dx = 4·x³/3 – 2·eˣ + C. IB problems often combine polynomial, trigonometric, and exponential terms, requiring systematic term-by-term integration.
积分是线性的，意味着和的积分等于积分的和，常数因子可以提出去：∫ [a·f(x) + b·g(x)] dx = a·∫ f(x) dx + b·∫ g(x) dx。这种线性性质与微分的和法则及常数倍法则相对应。例如，∫ (4x² – 2eˣ) dx = 4·x³/3 – 2·eˣ + C。IB题目经常组合多项式、三角函数和指数项，需要系统地进行逐项积分。
Special attention must be paid to integrands that require algebraic simplification before integrating. Expanding brackets, simplifying fractions, or using trigonometric identities (like sin²x = (1 – cos 2x)/2) can transform an apparently complex expression into basic forms. The IB syllabus explicitly tests the skill of rewriting integrands. For instance, ∫ (x² + 1)/x dx becomes ∫ (x + 1/x) dx = x²/2 + ln|x| + C after simplifying.
需要特别注意那些在积分前需要代数化简的被积函数。展开括号、化简分式或使用三角恒等式（如 sin²x = (1 – cos 2x)/2）可以将看似复杂的表达式转化为基本形式。IB教学大纲明确考察改写被积函数的技能。例如，∫ (x² + 1)/x dx 在化简后成为 ∫ (x + 1/x) dx = x²/2 + ln|x| + C。
4. Integration by Substitution | 换元积分法
Integration by substitution is the reverse of the chain rule. When an integrand contains a composite function, we let u = inner function, find du/dx, and replace dx with du. The method transforms the integral into a simpler one in terms of u. For example, to integrate ∫ 2x·cos(x²) dx, set u = x² so du = 2x dx, giving ∫ cos u du = sin u + C = sin(x²) + C. IB examiners expect a clear demonstration of the substitution steps, especially at HL.
换元积分法是链式法则的逆运算。当被积函数包含复合函数时，我们令 u = 内部函数，求出 du/dx，并将 dx 替换为 du。该方法将积分转化为关于 u 的更简单积分。例如，要积分 ∫ 2x·cos(x²) dx，设 u = x²，则 du = 2x dx，得到 ∫ cos u du = sin u + C = sin(x²) + C。IB阅卷老师期望清楚地展示换元步骤，特别是在HL级别。
With definite integrals, substitution requires changing the limits of integration from x-values to u-values. When x = a, u = g(a); when x = b, u = g(b). This avoids the need to convert back to x after integrating. Carelessness with limits is a frequent source of errors. In IB HL, substitutions may involve trigonometric or logarithmic functions, and sometimes more than one substitution is needed if the derivative of u is not fully present; careful algebraic manipulation can compensate for missing constant factors.
对于定积分，换元法需要将积分上下限从 x 值变为 u 值。当 x = a 时，u = g(a)；当 x = b 时，u = g(b)。这样就避免了积分后再换回 x 的麻烦。上下限处理粗心是常见的错误来源。在IB HL中，换元可能涉及三角函数或对数函数，有时如果 u 的导数不完全存在，则需要多次换元；通过仔细的代数处理可以弥补缺失的常数因子。
5. Integration by Parts | 分部积分法
Integration by parts is derived from the product rule for differentiation and is a key HL technique. The formula is ∫ u dv = uv – ∫ v du. The art lies in choosing u (the part to differentiate) and dv (the part to integrate) wisely. A common mnemonic is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which suggests priority for u. For example, to integrate ∫ x·eˣ dx, let u = x (algebraic) and dv = eˣ dx, so du = dx, v = eˣ, giving x·eˣ – ∫ eˣ dx = x·eˣ – eˣ + C.
分部积分法源自微分的乘积法则，是一项关键的HL技巧。公式为 ∫ u dv = uv – ∫ v du。其艺术在于明智地选择 u（要微分的部分）和 dv（要积分的部分）。一个常用的记忆口诀是LIATE（对数、反三角、代数、三角、指数），它给出了 u 的优先顺序。例如，要积分 ∫ x·eˣ dx，令 u = x（代数），dv = eˣ dx，则 du = dx，v = eˣ，得到 x·eˣ – ∫ eˣ dx = x·eˣ – eˣ + C。
There are cases where integration by parts must be applied twice, or where the original integral reappears on the right-hand side, leading to an equation that can be solved for the integral. This is common with integrals of the form ∫ eˣ sin x dx. In IB HL exams, candidates may need to handle definite integration by parts, carefully evaluating the uv term at the limits: [uv]ₐᵇ – ∫ₐᵇ v du. Setting up a clear table for u, dv, du, v is highly recommended to avoid mistakes.
有些情况下，分部积分法需要应用两次，或者原积分在等式右边重新出现，从而可以通过解方程求出积分。这常见于 ∫ eˣ sin x dx 这类积分。在IB HL考试中，考生可能需要处理分部的定积分，仔细计算 uv 项在上下限的值：[uv]ₐᵇ – ∫ₐᵇ v du。强烈建议列出 u、dv、du、v 的清晰表格以避免错误。
6. Definite Integrals and the Fundamental Theorem of Calculus | 定积分与微积分基本定理
The Fundamental Theorem of Calculus (FTC) bridges differentiation and integration. Part 1 states that if F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). Part 2 provides the evaluation formula: ∫ₐᵇ f(x) dx = F(b) – F(a), where F is any antiderivative of f. IB questions often ask you to evaluate a definite integral using this result. Remember that the definite integral yields a number (no +C), and the answer represents a net signed area.
微积分基本定理（FTC）在微分和积分之间架起了桥梁。第一部分指出，如果 F(x) = ∫ₐˣ f(t) dt，那么 F'(x) = f(x)。第二部分给出了求值公式：∫ₐᵇ f(x) dx = F(b) – F(a)，其中 F 是 f 的任意一个原函数。IB题目经常要求你使用这一结果计算定积分。请记住，定积分得到一个数值（无 +C），并且答案代表净带符号面积。
The FTC also enables the differentiation of integral functions. For example, if you are asked to find the derivative of ∫ₓˣ² sin(t) dt, you would combine the FTC part 1 with the chain rule. This higher-order thinking is examined in IB HL papers. When evaluating definite integrals by calculator, IB candidates should still know how to set up the antiderivative manually, as technology-free Paper 1 demands symbolic manipulation.
FTC还能够对积分函数进行微分。例如，如果要求 ∫ₓˣ² sin(t) dt 的导数，你需要将FTC第一部分与链式法则结合。这种高阶思维在IB HL试卷中会考察。在使用计算器计算定积分时，IB考生仍应知道如何手动构造原函数，因为不允许使用技术的Paper 1需要符号操作能力。
7. Area Under a Curve | 曲线下方面积
The most direct application of definite integrals is finding the area between a curve y = f(x) and the x-axis from x = a to x = b, given by A = ∫ₐᵇ |f(x)| dx if we want total area without regard to sign. However, if the curve crosses the x-axis, the integral ∫ₐᵇ f(x) dx will subtract areas below the axis. To find total geometric area, you must split the interval at the roots and integrate the absolute value, or integrate each segment separately and add their absolute values.
定积分最直接的应用是求曲线 y = f(x) 与 x 轴之间在 x = a 到 x = b 所围的面积，公式为 A = ∫ₐᵇ |f(x)| dx（如果想得到不考虑符号的总面积）。但是，如果曲线穿过 x 轴，积分 ∫ₐᵇ f(x) dx 会减去轴下方的面积。要计算总的几何面积，你必须在根处分割区间并积分绝对值，或分别积分每段再将其绝对值相加。
IB exam questions frequently provide a graph and ask for the total area of shaded regions. Here, symmetry can sometimes reduce the work: if f(x) is even, the area from -a to a is twice the area from 0 to a. Always identify where f(x) = 0 within the interval, and compute ∫ |f(x)| dx by breaking the integral at those points. A common pitfall is to blindly use ∫ₐᵇ f(x) dx without checking for sign changes, resulting in an area that is too small or even zero.
IB考题经常给出一个图形，并要求计算阴影区域的总面积。此时，对称性有时可减少工作量：如果 f(x) 是偶函数，从 -a 到 a 的面积是从 0 到 a 面积的两倍。务必识别区间内 f(x) = 0 的位置，并在这些点处分解积分来计算 ∫ |f(x)| dx。一个常见的陷阱是没有检查符号变化就盲目使用 ∫ₐᵇ f(x) dx，导致算出的面积过小甚至为零。
8. Area Between Curves | 曲线间的面积
When a region is bounded by two curves y = f(x) (top) and y = g(x) (bottom) between x = a and x = b, the area is A = ∫ₐᵇ [f(x) – g(x)] dx. This formula holds provided f(x) ≥ g(x) on [a, b]. If the curves intersect, the interval must be divided where the upper and lower functions swap. Visualising the region and identifying the correct ‘top minus bottom’ or ‘right minus left’ (when integrating with respect to y) is crucial for IB success.
当区域由两条曲线 y = f(x)（上）和 y = g(x)（下）在 x = a 和 x = b 之间围成时，面积公式为 A = ∫ₐᵇ [f(x) – g(x)] dx。该公式成立的前提是在 [a, b] 上 f(x) ≥ g(x)。如果曲线相交，则必须在上下函数交换的位置拆分区间。在IB考试中，可视化区域并确定正确的“上减下”或“右减左”（当对 y 积分时）至关重要。
Some IB problems express boundaries as functions of y, e.g., x = h(y) and x = k(y). Then the area between them from y = c to y = d is A = ∫cᵈ [h(y) – k(y)] dy, assuming h(y) ≥ k(y). Choosing the orientation (dx or dy) that leads to simpler integrals is a strategic skill. A typical Higher Level question might involve a region bounded by a curve and a line, requiring you to find intersection points, sketch the region, and set up an integral for the area, sometimes evaluating it without a calculator.
有些IB题目将边界表示为 y 的函数，比如 x = h(y) 和 x = k(y)。那么从 y = c 到 y = d 之间的面积就是 A = ∫cᵈ [h(y) – k(y)] dy（假设 h(y) ≥ k(y)）。选择能导致更简单积分的取向（dx 或 dy）是一项策略性技能。一个典型的HL问题是，给定由一条曲线和一条直线围成的区域，要求你找到交点，勾勒出区域，并为面积建立积分，有时需要在不使用计算器的情况下求出值。
9. Volume of Revolution: Disk Method | 旋转体体积：圆盘法
When a region under a curve y = f(x) from x = a to x = b is rotated entirely about the x-axis, the solid generated has a volume given by V = π ∫ₐᵇ [f(x)]² dx. This is the disk method, where each cross-sectional disk has radius f(x) and thickness dx, resulting in volume element dV = π [f(x)]² dx. The formula assumes the region is bounded by the curve, the x-axis, and the vertical lines x = a, x = b. For rotation about the y-axis of a curve x = g(y) between y = c and y = d, V = π ∫cᵈ [g(y)]² dy.
当曲线 y = f(x) 下方从 x = a 到 x = b 的区域绕 x 轴旋转一周时，生成的立体体积公式为 V = π ∫ₐᵇ [f(x)]² dx。这就是圆盘法，每个横截面圆盘的半径为 f(x)，厚度为 dx，体积元素为 dV = π [f(x)]² dx。该公式假定区域由曲线、x 轴以及直线 x = a、x = b 所围成。对于绕 y 轴旋转，曲线 x = g(y) 在 y = c 到 y = d 之间的区域，体积为 V = π ∫cᵈ [g(y)]² dy。
IB exam candidates must be comfortable applying the disk method both symbolically and numerically. It is critical to correctly square the function before integrating. A common mistake is to integrate then square, which yields an entirely different and incorrect volume. For curves given parametrically or defined in pieces, the same principle applies; you may need to split the interval. Visualising a representative disk helps confirm the setup, especially when boundaries include the x-axis or y-axis.
IB考生必须能够熟练地以符号和数值方式应用圆盘法。关键是在积分之前正确地平方函数。一个常见的错误是先积分再平方，这会得到完全不同的错误体积。对于以参数形式给出或分段定义的曲线，原理相同；你可能需要分割区间。想象一个代表性的圆盘有助于确认设置，特别是在边界包含 x 轴或 y 轴时。
10. Volume of Revolution: Washer Method | 旋转体体积：垫圈法
When the region being rotated is bounded between two curves y = f(x) (outer) and y = g(x) (inner) from x = a to x = b, the rotation about the x-axis creates a solid with a hole. The volume is then V = π ∫ₐᵇ ( [f(x)]² – [g(x)]² ) dx. This is the washer method, where the cross-section is an annulus. The corresponding formula for rotation about the y-axis with functions of y is V = π ∫cᵈ ( [h(y)]² – [k(y)]² ) dy, where h(y) is the outer radius and k(y) the inner radius.
当旋转区域由两条曲线 y = f(x)（外）和 y = g(x)（内）在 x = a 到 x = b 间围成时，绕 x 轴旋转会生成一个带孔的立体。此时体积为 V = π ∫ₐᵇ ( [f(x)]² – [g(x)]² ) dx。这就是垫圈法，横截面是一个圆环。对于绕 y 轴旋转且以 y 的函数表示的情况，相应公式为 V = π ∫cᵈ ( [h(y)]² – [k(y)]² ) dy，其中 h(y) 为外半径，k(y) 为内半径。
In IB HL problems, regions are often bounded by a curve and a line, and identifying the outer and inner functions is a key step. A sketch is indispensable: the outer function is the one farther from the axis of revolution. For rotations about lines other than the axes, such as x = p or y = q, the radii become |f(x) – p| or |g(y) – q|, adjusting the washer formula accordingly. Careful geometry ensures the correct squared radii are subtracted.
在IB HL问题中，区域通常由一条曲线和一条直线围成，确定外函数和内函数是关键步骤。示意图必不可少：外函数是离旋转轴较远的那个。对于绕非坐标轴（如 x = p 或 y = q）的旋转，半径变为 |f(x) – p| 或 |g(y) – q|，并据此调整垫圈公式。仔细的几何分析可确保正确减去半径的平方。
11. Kinematic Applications of Integration | 运动学中的应用
Integration is a powerful tool in kinematics, the study of motion. Given an acceleration function a(t), the velocity v(t) is found by integrating: v(t) = ∫ a(t) dt + v₀, where v₀ is the initial velocity. Similarly, position s(t) = ∫ v(t) dt + s₀, with s₀ being the initial position. Definite integrals can directly compute displacement (change in position) over a time interval: displacement = ∫ₜ₁ᵗ² v(t) dt. Total distance travelled, however, requires integrating the speed |v(t)|, analogous to total area.
积分在运动学（对运动的研究）中是一个强大的工具。给定加速度函数 a(t)，速度 v(t) 通过积分求得：v(t) = ∫ a(t) dt + v₀，其中 v₀ 是初速度。类似地，位置 s(t) = ∫ v(t) dt + s₀，s₀ 为初始位置。定积分可以直接计算一段时间间隔内的位移（位置变化）：位移 = ∫ₜ₁ᵗ² v(t) dt。但是，总路程需要对速率 |v(t)| 积分，类似于总面积。
IB AA HL papers often present motion along a straight line with variable acceleration. A typical problem gives a(t) and initial conditions, asking for velocity, position, and the analysis of direction changes. Finding when the particle is at rest involves solving v(t) = 0; the total distance then integrates |v(t)| between turning points. Being able to switch seamlessly from a, v, s relationships using derivatives and integrals is a core competency.
IB AA HL试卷经常给出变加速度下的直线运动问题。一个典型题目会给出 a(t) 以及初始条件，要求求速度、位置，并分析方向变化。求粒子静止的时刻需要解 v(t) = 0；然后通过积分转折点之间的 |v(t)| 来求总路程。能够使用导数和积分在 a、v、s 的关系之间无缝切换是一项核心能力。
12. Tips and Common Mistakes | 技巧与常见错误
Always include the constant of integration for indefinite integrals; omitting it will lose a mark. When evaluating definite integrals, do not forget to substitute the limits, and use brackets to handle negative values correctly. In substitution, either change the limits or convert back to the original variable before applying original limits – never mix both approaches. For area and volume problems, a clear sketch with labelled boundaries and intersection points dramatically reduces setup errors.
对于不定积分，始终要加上积分常数；漏掉会失分。计算定积分时，不要忘记代入上下限，并使用括号正确处理负值。在换元法中，要么改变上下限，要么在应用原上下限之前换回原变量——切勿将两种方法混用。对于面积和体积问题，一张标注了边界和交点的清晰示意图会大大减少建立积分式的错误。
Be meticulous with signs: the integral of sin x is -cos x + C, and the derivative of cos x is -sin x. In volume problems, make sure the expression being squared is indeed the radius function – a shift of axis changes the radius significantly. Practice recognising integrals that match standard forms, but also build the flexibility to adapt through algebraic manipulation or simple substitution. Finally, in the technology-active Paper 2, verify your manual integrals with
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