📚 Integration Techniques for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学积分技巧
Integration is one of the largest and most skill-based topics in Edexcel A-Level Pure Mathematics. It appears not only as a standalone topic but also inside differential equations, area problems, volumes of revolution, and modelling questions. This article brings together the standard integrals, substitution, integration by parts, partial fractions, numerical applications, and exam tips in one revision guide.
积分是 Edexcel A-Level 纯数学中篇幅最大、最依赖计算技巧的主题之一。它不仅单独出现，还贯穿于微分方程、面积问题、旋转体体积和建模题中。本文将标准积分、代换法、分部积分、部分分式、数值应用和考试提示整合成一份完整复习指南。
1. Standard Integrals and Reversing Differentiation | 标准积分与微分的逆运算
At A-Level, integration is first introduced as the reverse process of differentiation. You are expected to recall a set of standard results instantly. The general power rule is ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1, and the exceptional case is ∫ 1/x dx = ln|x| + C.
在 A-Level 阶段，积分首先作为微分的逆运算引入。你需要熟练掌握一组标准结果。一般幂函数法则为 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，其中 n ≠ −1；特殊情况是 ∫ 1/x dx = ln|x| + C。
Other compulsory standard integrals include ∫ eˣ dx = eˣ + C, ∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C, ∫ sec²x dx = tan x + C, ∫ cosec x cot x dx = −cosec x + C, and ∫ sec x tan x dx = sec x + C. These should be memorised because many harder questions rely on recognising them inside a larger expression.
其他必记标准积分包括 ∫ eˣ dx = eˣ + C、∫ sin x dx = −cos x + C、∫ cos x dx = sin x + C、∫ sec²x dx = tan x + C、∫ cosec x cot x dx = −cosec x + C 以及 ∫ sec x tan x dx = sec x + C。这些结果必须熟记，因为许多较难的题目都依赖在复杂表达式中识别这些形式。
- ∫ x⁴ dx = x⁵/5 + C
- ∫ (3x² + 2x − 5) dx = x³ + x² − 5x + C
- ∫ 2 sin x dx = −2 cos x + C
2. Integration of Trigonometric Functions | 三角函数的积分
Edexcel papers frequently test integrals involving powers and products of sine and cosine. To integrate sin²x or cos²x, use the double-angle identities: sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. These convert the square into a simple cosine term that can be integrated directly.
Edexcel 试卷经常考查正弦和余弦的幂及乘积的积分。要积分 sin²x 或 cos²x，应使用二倍角恒等式：sin²x = (1 − cos 2x)/2，cos²x = (1 + cos 2x)/2。这些恒等式将平方项转化为简单的余弦项，从而可以直接积分。
For products such as sin mx cos nx, the product-to-sum formulae are useful: 2 sin A cos B = sin(A+B) + sin(A−B). This changes the product into a sum of sine terms, which is easier to handle. For odd powers such as sin³x, write sin³x = sin x(1 − cos²x) and use substitution or recognition.
对于 sin mx cos nx 这类乘积，积化和差公式非常有用：2 sin A cos B = sin(A+B) + sin(A−B)。这样可将乘积转化为正弦之和，更易于处理。对于 sin³x 这类奇次幂，可写成 sin³x = sin x(1 − cos²x)，然后使用代换或直接识别。
∫ sin²x dx = x/2 − sin 2x/4 + C
∫ cos²x dx = x/2 + sin 2x/4 + C
3. Integration by Substitution (Change of Variable) | 代换积分法（变量替换）
Substitution is the first major integration technique beyond standard results. The idea is to replace a complicated inner function with a single variable u, then integrate with respect to u. If u = g(x), then du/dx = g′(x), so du = g′(x) dx. The integral is transformed into ∫ f(u) du.
代换法是超越标准积分的第一种主要技巧。其思想是将复杂的内层函数替换为单个变量 u，然后对 u 积分。如果 u = g(x)，则 du/dx = g′(x)，因此 du = g′(x) dx。积分被转换为 ∫ f(u) du。
For a definite integral, you must also change the limits. If x goes from a to b, then u goes from g(a) to g(b). After integrating with respect to u, substitute the new limits directly; do not convert back to x. This saves time and reduces sign errors.
对于定积分，还必须更换积分限。若 x 从 a 变化到 b，则 u 从 g(a) 变化到 g(b)。对 u 积分后直接代入新限即可，不必换回 x。这样做可节省时间并减少符号错误。
Common substitutions include u = ax + b, u = x² + 1, u = ln x, u = sin x, and trigonometric substitutions such as x = a sin θ for expressions involving √(a² − x²).
常见代换包括 u = ax + b、u = x² + 1、u = ln x、u = sin x，以及涉及 √(a² − x²) 的三角代换，例如 x = a sin θ。
∫ x(x² + 1)⁵ dx = (x² + 1)⁶/12 + C
4. Integration by Parts | 分部积分法
Integration by parts is the reverse of the product rule. The formula is ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. The key skill is choosing u and dv/dx correctly. A useful order of priority for u is logarithmic, inverse trigonometric, algebraic, trigonometric, exponential — often remembered as LIATE.
分部积分法是乘法法则的逆运算。公式为 ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。关键技巧在于正确选择 u 和 dv/dx。选择 u 的一个有效优先级是对数函数、反三角函数、代数函数、三角函数、指数函数，通常记为 LIATE。
Typical exam questions include ∫ x eˣ dx, ∫ x ln x dx, ∫ x sin x dx, and ∫ ln x dx. For ∫ ln x dx, write ln x as u and dx as dv, giving u = ln x, v = x, so the result is x ln x − x + C. Sometimes integration by parts must be applied twice, especially with eˣ sin x or eˣ cos x.
典型考题包括 ∫ x eˣ dx、∫ x ln x dx、∫ x sin x dx 和 ∫ ln x dx。对于 ∫ ln x dx，可令 u = ln x，dv = dx，则 v = x，结果为 x ln x − x + C。有时需要两次分部积分，尤其是 eˣ sin x 或 eˣ cos x 的积分。
∫ x eˣ dx = x eˣ − eˣ + C
5. Integration Using Partial Fractions | 部分分式积分法
Rational functions with denominators that factorise can often be integrated by splitting them into partial fractions. For example, an expression of the form (2x + 1)/((x − 1)(x + 2)) can be written as A/(x − 1) + B/(x + 2), where A and B are constants found by equating coefficients or substituting convenient values.
分母可因式分解的有理函数通常可以通过拆分为部分分式来积分。例如，形如 (2x + 1)/((x − 1)(x + 2)) 的表达式可写成 A/(x − 1) + B/(x + 2)，其中 A 和 B 是通过比较系数或代入适当值求出的常数。
Once the partial fractions are found, each term integrates to a natural logarithm: ∫ A/(x − 1) dx = A ln|x − 1| + C. If the denominator contains a repeated linear factor, you must include both A/(x + a) and B/(x + a)². For improper fractions, divide first so the degree of the numerator is less than the denominator.
得到部分分式后，每一项都可以积分为自然对数：∫ A/(x − 1) dx = A ln|x − 1| + C。如果分母含有重复线性因子，必须同时包含 A/(x + a) 和 B/(x + a)²。对于假分式，应先做长除法，使分子次数低于分母次数。
∫ 2/(x² − 1) dx = ln|x − 1| − ln|x + 1| + C
6. Definite Integrals and Area Under a Curve | 定积分与曲线下面积
A definite integral ∫ₐᵇ f(x) dx gives the signed area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b. If the curve lies above the x-axis, the integral is positive; if below, it is negative. To find the total area, split the interval at the points where the curve crosses the x-axis.
定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x)、x 轴以及直线 x = a、x = b 之间所夹的有向面积。若曲线位于 x 轴上方，则积分为正；若位于下方，则为负。求总面积时，应在曲线与 x 轴交点处拆分区间。
When calculating areas, always sketch the curve and identify the roots of f(x) within the interval. For example, if f(x) = x² − 4 on [0,3], the curve crosses the x-axis at x = 2. The total area is |∫₀² (x² − 4) dx| + ∫₂³ (x² − 4) dx because the first segment lies below the axis.
计算面积时，应始终先画出曲线草图，并确定区间内 f(x) 的零点。例如，若 f(x) = x² − 4 在 [0,3] 上，曲线在 x = 2 处穿过 x 轴。总面积为 |∫₀² (x² − 4) dx| + ∫₂³ (x² − 4) dx，因为第一段位于 x 轴下方。
Area = ∫ₐᵇ |f(x)| dx
7. Area Between Two Curves | 两曲线之间的面积
If two curves y = f(x) and y = g(x) intersect at x = a and x = b, the area enclosed between them is ∫ₐᵇ [f(x) − g(x)] dx, assuming f(x) ≥ g(x) throughout the interval. Before integrating, find the intersection points by solving f(x) = g(x).
若两条曲线 y = f(x) 与 y = g(x) 在 x = a 和 x = b 处相交，则它们之间所围成的面积为 ∫ₐᵇ [f(x) − g(x)] dx，前提是在该区间上 f(x) ≥ g(x)。积分前应通过解方程 f(x) = g(x) 求出交点。
If the curves cross within the interval, the upper and lower functions change. In that case, split the integral at each crossing point and take the absolute value of each section. A common exam fail is integrating f(x) − g(x) over the whole interval without checking which function is on top.
如果曲线在区间内相交，上下函数会发生变化。此时应在每个交点处拆分积分，并对每一段取绝对值。常见考试失分点是未检查哪条曲线在上方，就对整个区间直接积分 f(x) − g(x)。
Enclosed area = ∫ₐᵇ [upper curve − lower curve] dx
8. Volumes of Revolution Around the x-axis | 绕 x 轴旋转体的体积
When a curve y = f(x) between x = a and x = b is rotated through 360° about the x-axis, it forms a solid of revolution. The volume is given by V = π ∫ₐᵇ [f(x)]² dx. This formula comes from summing thin disks of radius y and thickness dx.
当曲线 y = f(x) 在 x = a 到 x = b 之间绕 x 轴旋转 360° 时，会形成一个旋转体。其体积公式为 V = π ∫ₐᵇ [f(x)]² dx。该公式来源于对半径为 y、厚度为 dx 的薄圆盘进行求和。
Before applying the formula, always expand [f(x)]². For instance, if y = x² + 1, then y² = (x² + 1)² = x⁴ + 2x² + 1. After integrating term by term, multiply the whole result by π. Leaving the square outside the integral is a very common mistake.
应用公式前，务必先展开 [f(x)]²。例如，若 y = x² + 1，则 y² = (x² + 1)² = x⁴ + 2x² + 1。逐项积分后，再将整个结果乘以 π。把平方留在积分号外是非常常见的错误。
V = π ∫ₐᵇ y² dx
9. Volumes of Revolution Around the y-axis | 绕 y 轴旋转体的体积
For rotation around the y-axis, the curve must be arranged as x = g(y). The volume is V = π ∫₋ₐᵇ [g(y)]² dy, where c and d are the y-limits. If the original equation is y = f(x), rearrange it to make x the subject before squaring and integrating.
绕 y 轴旋转时，曲线必须整理成 x = g(y) 的形式。体积公式为 V = π ∫₋ₐᵇ [g(y)]² dy，其中 c 和 d 是 y 的上下限。如果原方程是 y = f(x)，应先将其整理为 x 关于 y 的表达式，再平方并积分。
Limits must also be in terms of y, not x. Find them from the given intersection points or by substituting the x-limits into the rearranged equation. A parameterised curve may require converting both x and y into the parameter and using dx/dt or dy/dt accordingly.
积分限也必须用 y 表示，而不是 x。可通过已知交点求出，或将 x 限制代入整理后的方程。参数方程曲线可能需要将 x、y 都转化为参数形式，并相应使用 dx/dt 或 dy/dt。
V = π ∫₋ₐᵇ x² dy
10. Solving Differential Equations by Separation of Variables | 分离变量法解微分方程
Many Edexcel A-Level exam questions end with a first-order differential equation of the form dy/dx = f(x)g(y). To solve it, separate the variables so that all y terms are on one side and all x terms on the other: ∫ 1/g(y) dy = ∫ f(x) dx. Then integrate both sides and add one constant of integration.
许多 Edexcel A-Level 考题以一阶微分方程 dy/dx = f(x)g(y) 结尾。解这类方程需分离变量，使所有含 y 的项在一侧，所有含 x 的项在另一侧：∫ 1/g(y) dy = ∫ f(x) dx。然后对两边积分，并添加一个积分常数。
If an initial condition is given, substitute it after the general solution is obtained to find the particular value of C. When the general solution is implicit, you may need to rearrange it into the form y = … for the final answer. Always check whether the question asks for an exact form or a specified number of decimal places.
若给出初始条件，应在得到通解后代入该条件，以求出 C 的具体值。当通解为隐式时，可能需要将其整理成 y = … 的形式作为最终答案。务必检查题目是否要求精确形式或指定小数位数。
dy/dx = 3x²y ⇒ ∫ 1/y dy = ∫ 3x² dx ⇒ ln|y| = x³ + C
11. Common Exam Mistakes and Tips | 常见考试错误与提示
The most frequent mistake in integration is forgetting the constant of integration or dropping it after a substitution. Always write + C for indefinite integrals. For definite integrals, the constant is not needed, but you must change the limits when using substitution or integration by parts.
积分中最常见的错误是忘记积分常数或代换后丢失常数。不定积分一定要写 + C。对于定积分，不需要常数，但在使用代换法或分部积分法时必须更换积分限。
Another common error is sign confusion with trigonometric integrals and areas below the x-axis. Always sketch the graph before any area or volume question. Also, when using partial fractions, do not forget to divide first if the fraction is improper. Finally, show sufficient working because method marks are heavily weighted in Edexcel maths exams.
另一个常见错误是三角积分和 x 轴下方面积的符号混淆。做任何面积或体积题之前，都应先画出图像。此外，使用部分分式时，若为假分式，不要忘记先做除法。最后，务必展示足够步骤，因为 Edexcel 数学考试中方法分占比很大。
| Common mistake | How to avoid it |
| Forgetting + C | Write + C immediately after integrating every indefinite integral |
| Not changing limits in substitution | Convert x-limits to u-limits before substituting |
| Using the lower curve as upper | Sketch or test one point to see which function is larger |
| Square outside the π formula | Expand y² completely before integrating |
12. Worked Example Summary | 例题总结
Consider the integral ∫ x cos x dx. Using integration by parts with u = x and dv/dx = cos x gives du/dx = 1 and v = sin x. The formula gives x sin x − ∫ sin x dx = x sin x + cos x + C. This example shows how choosing u as the algebraic factor simplifies the remaining integral.
考虑积分 ∫ x cos x dx。使用分部积分法，令 u = x，dv/dx = cos x，则 du/dx = 1，v = sin x。根据公式可得 x sin x − ∫ sin x dx = x sin x + cos x + C。这个例子说明将代数因子选作 u 可以简化余下积分。
For a volume question, suppose y = √x from x = 0 to x = 4 is rotated about the x-axis. Then V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π. Notice that squaring √x first removes the square root, making the integration straightforward.
对于体积题，假设 y = √x 在 x = 0 到 x = 4 之间绕 x 轴旋转。则 V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π。注意先对 √x 平方可以消去根号，使积分变得直接。
Finally, for area between curves, find the intersection points first. If y = x² and y = x + 2 intersect at x = −1 and x = 2, the area is ∫₋₁² [(x + 2) − x²] dx. Evaluate this as [x²/2 + 2x − x³/3]₋₁² = 9/2 square units. Always keep the upper function minus the lower function inside the integral.
最后，对于两曲线之间的面积，应首先求出交点。若 y = x² 与 y = x + 2 在 x = −1 和 x = 2 处相交，则面积为 ∫₋₁² [(x + 2) − x²] dx。计算得 [x²/2 + 2x − x³/3]₋₁² = 9/2 平方单位。始终将上方函数减下方函数放在积分内。
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