📚 Mastering Integration: IGCSE WJEC Mathematics Revision | IGCSE WJEC 数学：积分 考点精讲
Integration is a cornerstone of the IGCSE WJEC Mathematics syllabus, often introduced as the reverse of differentiation. It enables us to recover original functions from their derivatives, compute areas bounded by curves, and solve practical problems in kinematics and other real-world contexts. This revision guide walks you through every key topic — from indefinite integrals and the constant of integration to definite integrals, area calculations, and exam-style pitfalls. By mastering these concepts, you will confidently tackle WJEC integration questions and earn top marks.
积分是 IGCSE WJEC 数学大纲的核心内容之一，通常被介绍为微分的逆运算。它使我们能够从导函数还原出原函数、计算曲线围成的面积，并解决运动学及其他实际情境中的问题。本复习指南将带你逐一掌握所有关键考点 —— 从不定积分和积分常数，到定积分、面积计算以及考试中常见的失分陷阱。熟练掌握这些概念后，你将能够自信地应对 WJEC 的积分题目并取得高分。
1. Understanding Integration | 理解积分
Integration is the reverse process of differentiation. If differentiating a function f(x) yields its derivative f'(x), then integrating f'(x) recovers the original function f(x), plus an arbitrary constant. This constant appears because differentiation loses constant terms. In notation, the indefinite integral of f(x) with respect to x is written as ∫ f(x) dx. For instance, since d/dx (x²) = 2x, we have ∫ 2x dx = x² + C, where C is the constant of integration.
积分是微分的逆过程。如果对函数 f(x) 求导得到导数 f'(x)，那么对 f'(x) 积分将恢复原函数 f(x)，并且会多出一个任意常数。这个常数之所以出现，是因为求导过程中常数项会丢失。用符号表示，f(x) 关于 x 的不定积分写作 ∫ f(x) dx。例如，由于 d/dx (x²) = 2x，我们有 ∫ 2x dx = x² + C，其中 C 是积分常数。
2. Indefinite Integration: The Power Rule | 不定积分：幂法则
The most fundamental rule for integration is the power rule, which reverses the differentiation power rule. For any real number n except –1, the integral of x to the power n is x raised to n+1 divided by n+1, all plus C. This rule also applies to constant multiples. The two key forms are:
积分最基本的法则是幂法则，它逆转了微分的幂法则。对于任意实数 n（n ≠ –1），x 的 n 次方的积分等于 x 的 n+1 次方除以 n+1，再加上 C。该法则同样适用于常数倍。两种主要形式如下：
∫ xn dx = (xn+1)/(n+1) + C (n ≠ –1)
∫ k dx = kx + C (where k is a constant)
Always check that the exponent is not –1. For n = –1, the integral would involve the natural logarithm, which is beyond the scope of IGCSE WJEC. A simple example: ∫ x³ dx = x&sup4;/4 + C. Notice how the power increases by one and we divide by the new power.
务必检查指数是否不为 –1。当 n = –1 时，积分将涉及自然对数，这超出了 IGCSE WJEC 的范围。一个简单例子：∫ x³ dx = x&sup4;/4 + C。注意指数加 1，然后除以这个新的指数。
3. Integrating Polynomials and Linear Functions | 多项式与线性函数的积分
To integrate a polynomial, apply the power rule term by term. The integral of a sum is the sum of the integrals. For example, ∫ (3x² + 2x + 1) dx = x³ + x² + x + C. For linear expressions of the form (ax + b)n, a slightly extended rule is used that accounts for the coefficient of x inside the bracket. The formula is:
积分多项式时，逐项运用幂法则即可。和的积分等于各部分积分之和。例如，∫ (3x² + 2x + 1) dx = x³ + x² + x + C。对于形如 (ax + b)n 的线性表达式，则需要使用一个略加扩展的法则，以处理括号内 x 的系数。公式如下：
∫ (ax + b)n dx = (ax + b)n+1 / [a(n+1)] + C (n ≠ –1)
This is derived by reversing the chain rule. As an example, ∫ (2x + 1)³ dx = (2x + 1)&sup4; / (2×4) + C = (2x + 1)&sup4;/8 + C. Many WJEC questions combine polynomial terms with such linear brackets, so being able to integrate both types efficiently is essential.
该公式由链式法则的逆推得来。举例来说，∫ (2x + 1)³ dx = (2x + 1)&sup4; / (2×4) + C = (2x + 1)&sup4;/8 + C。许多 WJEC 题目会将多项式项与这类线性括号结合在一起考查，因此能够高效地对这两类函数积分至关重要。
4. The Constant of Integration | 积分常数
The constant C, known as the constant of integration, arises because differentiation removes constant terms. Every indefinite integral must include ‘+ C’, unless additional information (an initial condition) is provided to determine its specific value. Forgetting to write ‘+ C’ is a common reason for losing marks. If a question states that the curve y = F(x) passes through a point, say (1, 5), substitute x = 1 and F(1) = 5 into your integrated expression to find C.
常数 C，即积分常数，产生于微分会消除常数项这一事实。任何一个不定积分都必须包含 “+ C”，除非题目提供了额外信息（初始条件）以确定其具体数值。忘记写 “+ C” 是丢分的常见原因。如果题目指出曲线 y = F(x) 经过某点，例如 (1, 5)，则将 x=1 和 F(1)=5 代回积分所得表达式，即可求出 C。
Example: If f'(x) = 4x + 1 and f(0) = 3, then f(x) = ∫ (4x + 1) dx = 2x² + x + C. Substituting x = 0 gives C = 3, hence f(x) = 2x² + x + 3. Always show this working clearly in the exam.
例子：若 f'(x) = 4x + 1 且 f(0) = 3，则 f(x) = ∫ (4x+1) dx = 2x² + x + C。代入 x=0 得 C=3，因此 f(x) = 2x² + x + 3。考试中必须清楚写出这一求解过程。
5. Definite Integration | 定积分
A definite integral calculates the net accumulation of a function between two limits a and b. It is written as ∫ab f(x) dx, and its value is found by evaluating an antiderivative F(x) at the upper and lower limits, then subtracting: F(b) – F(a). There is no ‘+ C’ in the final numerical answer because the constants cancel out. The fundamental theorem of calculus underpins this process.
定积分计算函数在两个界限 a 和 b 之间的净累积量。它写作 ∫ab f(x) dx，其值通过找出一个原函数 F(x) 并代入上限和下限相减求得：F(b) – F(a)。最终的数值结果中不需要 “+ C”，因为常数会相互抵消。微积分基本定理正是这一过程的数学基础。
∫ab f(x) dx = F(b) – F(a)
For example, ∫12 2x dx = [x²]12 = 4 – 1 = 3. When working with definite integrals, be meticulous with signs when substituting, especially if the antiderivative contains negative terms. Use brackets to avoid arithmetic errors.
例如，∫12 2x dx = [x²]12 = 4 – 1 = 3。处理定积分时，代值时要特别注意符号，尤其当原函数包含负项时。使用括号可以避免算术错误。
6. Area Under a Curve | 曲线下的面积
The definite integral ∫ab f(x) dx gives the signed area between the curve y = f(x) and the x-axis from x = a to x = b. Regions above the x-axis yield positive contributions; regions below the axis give negative contributions. If a question asks for the total geometric area (not the signed area), you must split the integral at the points where the curve crosses the x-axis and take the absolute value of each portion.
定积分 ∫ab f(x) dx 给出的是曲线 y = f(x) 与 x 轴之间从 x=a 到 x=b 的代数和面积。x 轴上方的区域贡献正值，下方的区域贡献负值。如果题目要求计算总的几何面积（而非代数和面积），则必须在曲线与 x 轴的交点处将积分分段，并对每一部分取绝对值。
For instance, consider y = x – 2 between x = 0 and x = 4. The curve crosses the x-axis at x = 2. The area above the axis is ∫24 (x – 2) dx = 2, and the area below is |∫02 (x – 2) dx| = 2. The total geometric area is 4 square units. Always sketch the graph to visualise which parts lie above or below the axis.
例如，考虑在 x=0 至 x=4 之间的曲线 y = x – 2。该曲线在 x=2 处穿过 x 轴。x 轴上方面积为 ∫24 (x – 2) dx = 2，下方面积取绝对值 |∫02 (x – 2) dx| = 2。几何总面积是 4 平方单位。始终画草图，以便直观地看出哪些部分位于 x 轴上方或下方。
7. Area Between Two Curves | 两曲线之间的面积
When you need to find the area enclosed between two curves y = f(x) and y = g(x) from x = a to x = b, and f(x) lies above g(x) on that interval, the area is given by ∫ab [f(x) – g(x)] dx. If the curves cross, you must split the interval so that the ‘top minus bottom’ rule applies separately. The formula can also be expressed using absolute value: Area = ∫ab |f(x) – g(x)| dx, but in practice it is easier to determine the top curve and subtract without absolute value signs.
当需要计算两条曲线 y = f(x) 与 y = g(x) 在 x=a 至 x=b 之间所围成的面积，且在该区间内 f(x) 位于 g(x) 之上时，面积由 ∫ab [f(x) – g(x)] dx 给出。若两条曲线相交，则必须将区间分段，使每一段都满足“上方减下方”的规则。该公式也可写成绝对值形式：面积 = ∫ab |f(x) – g(x)| dx，但在实际计算中，判断哪条曲线在上方然后直接相减更容易操作。
Area = ∫ab (top – bottom) dx
A classic WJEC example: find the area between y = x and y = x² from x = 0 to x = 1. On this interval, x ≥ x², so the area is ∫01 (x – x²) dx = [x²/2 – x³/3]01 = 1/2 – 1/3 = 1/6. Always double-check which function is larger before setting up the integral.
一个 WJEC 经典例子：求曲线 y = x 与 y = x² 在 x=0 到 x=1 之间围成的面积。在该区间内，x ≥ x²，因此面积为 ∫01 (x – x²) dx = [x²/2 – x³/3]01 = 1/2 – 1/3 = 1/6。列积分式前务必先确认哪个函数值更大。
8. Applications to Kinematics | 运动学应用
Integration appears frequently in WJEC kinematics problems. Given the velocity v(t) as the derivative of displacement s(t), the displacement over a time interval [t&sub1;, t&sub2;] is s(t&sub2;) – s(t&sub1;) = ∫t&sub1;t&sub2; v(t) dt. Similarly, if acceleration a(t) is known, velocity is v(t) = ∫ a(t) dt + C, where the constant can be found using initial velocity. When the question asks for total distance travelled rather than displacement, you must integrate the absolute value of velocity, i.e. split the interval wherever velocity changes sign.
积分经常出现在 WJEC 运动学问题中。若速度 v(t) 是位移 s(t) 的导数，则在时间区间 [t&sub1;, t&sub2;] 内的位移为 s(t&sub2;) – s(t&sub1;) = ∫t&sub1;t&sub2; v(t) dt。类似地，若已知加速度 a(t)，则速度 v(t) = ∫ a(t) dt + C，其中的常数可通过初始速度求出。当题目要求计算总路程而非位移时，必须对速度的
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