Edexcel数学 积分Integration 定积分 换元法
积分是 Edexcel A-Level 数学 Pure Mathematics 的核心章节，贯穿 Paper 1 与 Paper 2 的必考内容。它和微分互为逆运算，同时又是计算曲线下面积、求解变化率问题的基本工具。本文面向初次系统学习积分的同学，用中英对照的方式讲清不定积分、定积分、换元法与分部积分的核心逻辑，并给出典型例题与常见失分点。
Integration is the core chapter of Edexcel A-Level Mathematics Pure Mathematics, appearing in both Paper 1 and Paper 2. It is the reverse process of differentiation and the fundamental tool for finding areas under curves and solving rate-of-change problems. This article is written for students meeting integration for the first time, explaining indefinite integrals, definite integrals, substitution and integration by parts with worked examples and common pitfalls, in both Chinese and English.
一、微分与积分的关系
微分的任务是求变化率：已知位移函数 s(t)，求速度 v(t)=s'(t)。积分则反过来：已知速度，如何还原位移？更一般地说，如果 F'(x)=f(x)，那么 F 就是 f 的一个原函数。例如 (x³)'=3x²，所以 3x² 的一个原函数是 x³。由于常数的导数为零，x³+1、x³−5 的导数同样是 3x²，因此原函数必须写作 x³+C，其中 C 为任意常数。
Differentiation finds rates of change: given a displacement function s(t), velocity is v(t)=s'(t). Integration works the other way round. Given velocity, how do we recover displacement? In general, if F'(x)=f(x), then F is called an antiderivative of f. For instance, the derivative of x³ is 3x², so one antiderivative of 3x² is x³. Because the derivative of any constant is zero, the antiderivatives x³+1 and x³−5 also differentiate to 3x². So the general antiderivative must be written as x³+C, where C is an arbitrary constant.
二、不定积分与基本公式
不定积分记作 ∫f(x)dx=F(x)+C，结果是一族函数而不是一个数。Edexcel 考纲要求掌握的核心公式如下：幂法则 ∫xⁿdx=xⁿ⁺¹/(n+1)+C（n≠−1）；∫1/x dx=ln|x|+C；∫eˣdx=eˣ+C；以及 ∫sinx dx=−cosx+C、∫cosx dx=sinx+C。这里的绝对值符号 |x| 极易被忽略：ln x 只对正数有定义，而 ln|x| 在 x<0 时同样可导，答案必须写 |x| 才能拿全分。
An indefinite integral is written ∫f(x)dx=F(x)+C; it describes a family of functions rather than a single number. The core rules required by the Edexcel specification are: the power rule ∫xⁿdx=xⁿ⁺¹/(n+1)+C for n≠−1; ∫1/x dx=ln|x|+C; ∫eˣdx=eˣ+C; together with ∫sinx dx=−cosx+C and ∫cosx dx=sinx+C. The modulus sign in ln|x| is easy to forget: ln x is only defined for positive x, whereas ln|x| can also be differentiated when x is negative. Writing |x| is essential for full marks.
例题：求 ∫(3x²−2x+5)dx。逐项积分得 3·x³/3 −2·x²/2 +5x+C，化简为 x³−x²+5x+C。注意每一项都要检查：x 的幂指数加一再除以新指数，常数项 5 直接写成 5x。最后务必补上积分常数 C，这是 A-Level 阅卷中最常见的一分失分点。
Worked example: find ∫(3x²−2x+5)dx. Integrating term by term gives 3·x³/3 −2·x²/2 +5x+C, which simplifies to x³−x²+5x+C. Check every term: raise the power by one and divide by the new index, while the constant 5 simply becomes 5x. Always add the constant C at the end, as missing C is one of the most common single-mark losses in A-Level marking.
三、定积分与曲线下面积
定积分 ∫ₐᵇf(x)dx 给出一个数值，它等于曲线 y=f(x) 在区间 [a,b] 上与 x 轴围成的有向面积。计算方法是微积分基本定理：先求原函数 F(x)，再代入上下限 F(b)−F(a)，书写格式为 [F(x)]ₐᵇ。若曲线在 x 轴下方，积分值为负，此时求实际面积需要取绝对值，或者先求零点分段再分别积分。
A definite integral ∫ₐᵇf(x)dx produces a single number: the signed area enclosed between the curve y=f(x) and the x-axis over the interval [a,b]. It is evaluated with the Fundamental Theorem of Calculus: find an antiderivative F(x), then substitute the limits to get F(b)−F(a), written as [F(x)]ₐᵇ. When the curve lies below the x-axis the integral is negative, so for a real area you must take the absolute value, or split the interval at the roots and integrate each part separately.
例题：求曲线 y=4x−x² 与 x 轴围成的面积。先求零点 4x−x²=x(4−x)=0，得 x=0 与 x=4，故面积为 ∫₀⁴(4x−x²)dx=[2x²−x³/3]₀⁴=32−64/3=32/3≈10.67。若题目要求两曲线之间的面积，则用上曲线减下曲线：∫(上−下)dx，例如 y=4x−x² 与 y=x 所围区域面积为 ∫₀³(3x−x²)dx=27/2−9=9/2。
Worked example: find the area enclosed by the curve y=4x−x² and the x-axis. Setting 4x−x²=x(4−x)=0 gives roots x=0 and x=4, so the area is ∫₀⁴(4x−x²)dx=[2x²−x³/3]₀⁴=32−64/3=32/3≈10.67. For the area between two curves, integrate top minus bottom: ∫(upper−lower)dx. For example the region bounded by y=4x−x² and y=x has area ∫₀³(3x−x²)dx=27/2−9=9/2.
除解析求积外，Edexcel 还要求掌握梯形法则这一数值方法。当被积函数难以求出原函数时，把区间 [a,b] 等分成 n 份，每份宽 h=(b−a)/n，则 ∫ₐᵇf(x)dx 近似为 (h/2)[f(x₀)+f(xₙ)+2(f(x₁)+...+f(xₙ₋₁))]。分割份数越多，近似越精确。考试常以表格形式给出各分点的 y 值，代入公式即可，注意首尾端点只算一次、中间点都要乘以 2。
Besides exact integration, Edexcel also requires the trapezium rule, a numerical method. When an antiderivative is hard to find, split [a,b] into n strips of equal width h=(b−a)/n, and approximate the integral ∫ₐᵇf(x)dx by (h/2)[f(x₀)+f(xₙ)+2(f(x₁)+...+f(xₙ₋₁))]. More strips give a more accurate estimate. In exams the y values are usually supplied in a table, so you only substitute them into the formula, remembering that the two endpoints are counted once while every interior value is doubled.
四、换元法
换元法对应微分的链式法则，专门处理复合函数。核心模式是形如 ∫f(g(x))·g'(x)dx 的积分：令 u=g(x)，则 du=g'(x)dx，原积分化为 ∫f(u)du。例如 ∫2x·e^(x²)dx：令 u=x²，du=2xdx，得到 ∫eᵘdu=eᵘ+C=e^(x²)+C。对定积分换元时，上下限也要同步换成 u 的值，即 x=a 时 u=g(a)，省去换回 x 的步骤。
Substitution is the integral version of the chain rule and handles composite functions. The key pattern is an integrand of the form ∫f(g(x))·g'(x)dx: set u=g(x), then du=g'(x)dx and the integral becomes ∫f(u)du. For example ∫2x·e^(x²)dx: with u=x² and du=2xdx we obtain ∫eᵘdu=eᵘ+C=e^(x²)+C. When substituting in a definite integral, convert the limits at the same time: when x=a, u=g(a). This avoids the need to substitute back in terms of x afterwards.
判断是否可用换元法的信号：被积函数中出现一个函数与它的导数相乘。比如 ∫x(x²+1)⁵dx 中，x²+1 的导数是 2x，只差常数因子 2，于是令 u=x²+1，du=2xdx，积分化为 (1/2)∫u⁵du=u⁶/12+C，最后代回得 (x²+1)⁶/12+C。注意检查常数因子是否配平，这是换元失分的高发点。
How do you recognise substitution? Look for a function multiplied by its own derivative inside the integrand. In ∫x(x²+1)⁵dx the derivative of x²+1 is 2x, so we are only missing a factor of 2. Set u=x²+1, du=2xdx, and the integral becomes (1/2)∫u⁵du=u⁶/12+C, then substitute back to get (x²+1)⁶/12+C. Balancing the constant factor is where many students lose marks.
五、分部积分
分部积分处理两个不同类函数相乘的积分，如 x·eˣ、x·sinx、x·lnx。公式为 ∫u·(dv/dx)dx=uv−∫v·(du/dx)dx。使用口诀：把容易求导的选作 u，容易积分的选作 dv/dx。例如 ∫x·eˣdx：取 u=x（求导后变简单），dv/dx=eˣ，则 v=eˣ，代入得 x·eˣ−∫eˣdx=x·eˣ−eˣ+C。若选反，∫eˣ 的积分会越算越复杂。
Integration by parts handles products of two different types of function, such as x·eˣ, x·sinx and x·lnx. The formula is ∫u·(dv/dx)dx=uv−∫v·(du/dx)dx. The rule of thumb: choose u as the factor that becomes simpler when differentiated, and dv/dx as the factor that is easy to integrate. For ∫x·eˣdx choose u=x, since differentiation simplifies it, and dv/dx=eˣ, giving v=eˣ. Substituting yields x·eˣ−∫eˣdx=x·eˣ−eˣ+C. Choosing the other way round leaves a product that never gets simpler.
分部积分可能要用两次以上，称为循环分部积分，典型如 ∫eˣsinx dx：连续分部两次后原积分会再次出现，可把等式看作关于 I 的方程解出。对定积分使用分部积分时，记得 uv 项要代入上下限。Edexcel Paper 2 常在最后一道大题中以分部积分配合换元法综合考查，务必写出 u、dv/dx、v、du/dx 的完整过程，步骤分与答案分兼得。
Sometimes integration by parts must be applied twice or more, which leads to cyclic by-parts. A classic example is ∫eˣsinx dx: after two applications the original integral reappears, and the equation can be solved for I as an unknown. When using by-parts on a definite integral, remember to substitute the limits into the uv term. Edexcel Paper 2 often combines substitution with by-parts in the final question, so always write out u, dv/dx, v and du/dx clearly to secure both method marks and the final answer.
六、积分的应用与微分方程入门
积分在力学中的应用：已知加速度 a=dv/dt，积分一次得速度 v，再积分得位移 s，积分常数由初始条件 v(0)、s(0) 确定。这种由导数还原函数的过程，本质上就是解一阶微分方程。Edexcel 要求掌握可分离变量型：dy/dx=f(x)g(y) 时，写成 (1/g(y))dy=f(x)dx，两边分别积分即可。例如 dy/dx=2xy：分离得 (1/y)dy=2xdx，积分得 ln|y|=x²+C，故 y=Ae^(x²)。
Integration in mechanics: given acceleration a=dv/dt, integrate once to obtain velocity v and again to obtain displacement s, with the constants fixed by the initial conditions v(0) and s(0). Recovering a function from its derivative in this way is exactly what it means to solve a first-order differential equation. Edexcel requires the separable-variable type: if dy/dx=f(x)g(y), rearrange to (1/g(y))dy=f(x)dx and integrate both sides. For example dy/dx=2xy: separating gives (1/y)dy=2xdx, integrating yields ln|y|=x²+C, hence y=Ae^(x²).
运动学中的完整例子最能说明积分常数的意义：设物体加速度 a=6t，t=0 时初速度 v=2、初始位移 s=0。对加速度积分得 v=∫6t dt=3t²+C，代入 v(0)=2 得 C=2，所以 v=3t²+2；再积分得 s=∫(3t²+2)dt=t³+2t+D，代入 s(0)=0 得 D=0，故 s=t³+2t。每一步的常数都由初始条件唯一确定，缺一步就无法得到确定的运动方程。
A complete kinematics example best shows the role of the constant: suppose acceleration is a=6t, with initial velocity v=2 and initial displacement s=0 at t=0. Integrating the acceleration gives v=∫6t dt=3t²+C, and v(0)=2 forces C=2, so v=3t²+2. Integrating again gives s=∫(3t²+2)dt=t³+2t+D, and s(0)=0 forces D=0, hence s=t³+2t. Each constant is fixed by an initial condition, and skipping any step leaves the equation of motion undetermined.
经济与生物中的应用同样高频：边际成本函数积分还原总成本，人口增长率积分预测种群数量。判断积分常数时，题目给出的初始条件 x=0 时 y 的值是唯一的线索，代入求解 C 后再写最终表达式。若题目要求求曲线方程本身，记得最后保留显式形式 y=f(x)。
Applications in economics and biology are equally common: integrating a marginal cost function recovers total cost, and integrating a population growth rate predicts the population size. The initial condition given in the question, such as the value of y when x=0, is the only way to determine C, so substitute it and solve before writing the final expression. When a question asks for the equation of a curve, finish with y written explicitly as f(x).
七、考点分布与备考建议
按近年 Edexcel 真题统计，积分相关分值约占 Pure 部分的 15% 到 20%。Paper 1 侧重基本公式与定积分计算，Paper 2 侧重换元、分部与微分方程综合题。计算器政策下，积分结果必须写成精确形式，例如 32/3 而非 10.6667，除非题目明确要求近似值。订正错题时建议把错误按类型归类：漏 C、忘 |x|、上下限代错、换元未配平常数，这四类占了积分失分的大头。
According to recent Edexcel papers, integration accounts for roughly 15% to 20% of the Pure Mathematics marks. Paper 1 focuses on basic rules and definite integrals, while Paper 2 features substitution, by-parts and differential equation problems. Under the calculator policy, results must be left in exact form such as 32/3 rather than 10.6667 unless the question asks for an approximation. When reviewing mistakes, classify them by type: missing C, forgetting |x|, substituting limits incorrectly, and failing to balance the constant in substitution account for the majority of marks lost on integration.
复习顺序建议：先默写全部基本积分公式并限时完成 20 道基础题，再集中练习曲线下面积与两曲线间面积，随后攻克换元法与分部积分，最后用历年真题 Paper 1、Paper 2 的积分大题做限时模拟。把每道错题的完整解答重写一遍，比盲目刷新题更能稳定提分。
A sensible revision order: memorise every basic integral formula and complete 20 basic questions under time pressure, then practise areas under and between curves, then tackle substitution and by-parts, and finally simulate with integration questions from real Paper 1 and Paper 2 papers. Rewriting the full solution to every wrong question builds far more reliable marks than rushing through new questions blindly.
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