📚 GCSE CIE Mathematics: Integration Key Points | GCSE CIE 数学：积分 考点精讲
Integration is a fundamental concept in calculus that reverses differentiation. In the CIE GCSE Mathematics syllabus, you are expected to find antiderivatives, evaluate definite integrals, and compute areas under curves. This comprehensive guide walks you through every key point, from indefinite integrals to kinematics applications, ensuring you are fully prepared for your exam.
积分是微积分中的一个基本概念，它是微分的逆运算。在 CIE GCSE 数学考纲中，你需要会求反导数、计算定积分，并能求出曲线下方的面积。本文全面地梳理了不定积分、定积分以及运动学应用中的每一个核心考点，帮助你充分备考。
1. What is Integration? The Reverse of Differentiation | 什么是积分？微分的逆运算
Integration is often called anti-differentiation. If differentiating a function F(x) gives f(x), then integrating f(x) retrieves F(x), up to an unknown constant.
积分常被称为反微分。如果对函数 F(x) 求导得到 f(x)，那么对 f(x) 积分就能回到 F(x)，只是会多出一个未知常数。
If d/dx [F(x)] = f(x), then ∫ f(x) dx = F(x) + c
如果 d/dx [F(x)] = f(x)，那么 ∫ f(x) dx = F(x) + c
For example, the derivative of x3 is 3x2, therefore the integral of 3x2 with respect to x is x3 + c.
例如，x3 的导数是 3x2，所以 3x2 关于 x 的积分就是 x3 + c。
In the GCSE exam, you will see both the ‘integral of’ notation and the instruction to ‘find an expression for y in terms of x’ given dy/dx.
在 GCSE 考试中，你会看到“求……的积分”的记号，也会遇到已知 dy/dx 求 y 关于 x 的表达式这类问题。
2. The Indefinite Integral and the Constant of Integration | 不定积分与积分常数
When we perform indefinite integration, we write ∫ f(x) dx = F(x) + c. The term ‘c’ is called the constant of integration. It is essential because differentiating a constant gives zero, so many different functions can have the same derivative.
进行不定积分时，我们写成 ∫ f(x) dx = F(x) + c。其中的“c”称为积分常数。这一项必不可少，因为常数的导数为零，所以许多不同的函数可以有相同的导数。
For instance, all of the functions y = x2 + 3, y = x2 – 5, and y = x2 + ½ share the same derivative dy/dx = 2x. Hence, ∫ 2x dx = x2 + c captures the whole family of solutions.
例如，函数 y = x2 + 3、y = x2 – 5 以及 y = x2 + ½ 都有相同的导数 dy/dx = 2x。因此，∫ 2x dx = x2 + c 就抓住了整个函数族。
You must always add ‘+ c’ unless the question provides an additional condition (such as a point on the curve) that allows you to determine its value.
除非题目给出了额外的条件（比如曲线经过某一点），否则你必须加上“+ c”。
3. Basic Rule: Integrating Powers of x | 基本法则：x 的幂次积分
The most important rule for GCSE integration is the power rule. For any real number n except -1, we have:
GCSE 积分中最重要的法则是幂函数法则。对于任何不等于 -1 的实数 n，有：
∫ xn dx = xn+1/(n+1) + c, n ≠ -1
This rule is obtained by reversing the differentiation rule for powers: add one to the exponent and divide by the new exponent.
这条法则来自对幂函数求导法则的逆推：指数加 1，再除以新的指数。
For example, ∫ x3 dx = x4/4 + c. ∫ x-2 dx = x-1/(-1) + c = -1/x + c, since x-1 = 1/x.
例如，∫ x3 dx = x4/4 + c。 ∫ x-2 dx = x-1/(-1) + c = -1/x + c，因为 x-1 = 1/x。
This rule also works for fractional and negative powers, such as x½, which integrates to (2/3)x3/2 + c. Simply rewrite square roots and cube roots using indices before integrating.
这条法则同样适用于分数指数和负指数，比如 x½，积分为 (2/3)x3/2 + c。遇到平方根、立方根等，只需先把它们写成指数形式再积分。
4. Integrating Polynomials and Linear Combinations | 多项式与线性组合的积分
Integration is a linear operation, meaning we can integrate term by term and factor out constant multipliers.
积分是线性运算，意味着我们可以逐项积分，并把常数因子提到积分号外面。
For any functions f(x) and g(x) and constants a and b:
对于任意函数 f(x) 和 g(x) 以及常数 a 和 b：
∫ [a f(x) + b g(x)] dx = a ∫ f(x) dx + b ∫ g(x) dx
To integrate a polynomial, apply the power rule to each term separately. For instance:
对多项式积分时，对每一项分别使用幂法则。例如：
Example: ∫ (4x3 – 6x2 + 2x – 5) dx
示例： ∫ (4x3 – 6x2 + 2x – 5) dx
= 4·x4/4 – 6·x3/3 + 2·x2/2 – 5x + c = x4 – 2x3 + x2 – 5x + c.
Always check your answer by differentiating to see if you get back the original expression.
务必通过求导来检验答案，看看能否回到原函数。
5. Finding the Equation of a Curve from its Derivative | 通过导数求曲线方程
A common exam question gives you dy/dx and a point (x₁, y₁) on the curve, asking you to find the equation y = f(x). First integrate dy/dx to obtain an expression containing + c, then substitute the coordinates to solve for c.
一种常见考题会给出 dy/dx 以及曲线上某点 (x₁, y₁)，要求你求出曲线方程 y = f(x)。先对 dy/dx 积分，得到带有 + c 的表达式，然后代入坐标求出 c。
Example: A curve passes through (2, 7) and has gradient function dy/dx = 3x2 – 2. Find the equation of the curve.
示例： 一条曲线经过点 (2, 7)，且其斜率函数为 dy/dx = 3x2 – 2。求该曲线的方程。
Integrate: y = ∫ (3x2 – 2) dx = x3 – 2x + c.
积分得：y = ∫ (3x2 – 2) dx = x3 – 2x + c。
Substitute x = 2, y = 7: 7 = 23 – 2·2 + c → 7 = 8 – 4 + c → c = 3.
代入 x = 2, y = 7：7 = 8 – 4 + c → c = 3。
Thus the equation is y = x3 – 2x + 3.
因此曲线方程为 y = x3 – 2x + 3。
This skill links differentiation and integration seamlessly, so practice both directions.
这一技巧将微分与积分无缝衔接，建议双向练习。
6. Definite Integration: The Fundamental Theorem | 定积分：微积分基本定理
A definite integral has limits of integration and produces a number rather than a function. It is defined as the difference of the antiderivative evaluated at the upper and lower limits.
定积分带有积分上下限，其结果是一个数而非函数。它被定义为反导数在上限和下限处的差值。
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
Here F(x) is any antiderivative of f(x). The constant of integration cancels out, so you do not need to add + c for definite integrals.
其中 F(x) 是 f(x) 的任意一个反导数。积分常数会互相抵消，因此计算定积分时无需添加 + c。
Example: Evaluate ∫13 (2x + 1) dx.
示例： 计算 ∫13 (2x + 1) dx。
Antiderivative: F(x) = x2 + x. Then F(3) = 9 + 3 = 12, F(1) = 1 + 1 = 2, so the value is 12 − 2 = 10.
反导数：F(x) = x2 + x。于是 F(3) = 9 + 3 = 12，F(1) = 1 + 1 = 2，结果为 12 − 2 = 10。
Definite integrals form the basis for area calculations and motion problems.
定积分是计算面积和运动问题的基础。
7. Calculating the Area Under a Curve | 计算曲线下的面积
When the graph of y = f(x) lies above the x-axis between x = a and x = b, the area enclosed by the curve, the x-axis, and the vertical lines x = a and x = b is given by:
当 y = f(x) 的图像在 x = a 与 x = b 之间位于 x 轴上方时，由曲线、x 轴以及直线 x = a、x = b 围成的面积为：
Area = ∫ab f(x) dx
Example: Find the area under the curve y = x2 + 1 from x = 0 to x = 2.
示例： 求曲线 y = x2 + 1 在 x = 0 到 x = 2 之间的面积。
Integrate: ∫02 (x2 + 1) dx = [x3/3 + x]02 = (8/3 + 2) − (0) = 14/3 square units.
积分：∫02 (x2 + 1) dx = [x3/3 + x]02 = (8/3 + 2) − (0) = 14/3 平方单位。
Always sketch the region or check that f(x) ≥ 0 on the interval, because a negative integral corresponds to an area below the axis.
务必先画出示意图或确认在该区间内 f(x) ≥ 0，因为负的积分值对应的是 x 轴下方的面积。
8. Areas Above and Below the x-axis: Handling Negative Parts | x 轴上方和下方的面积：处理负值部分
If the curve dips below the x-axis, the definite integral ∫ab f(x) dx will be negative for that portion. However, area is always a positive quantity, so you must split the interval at the roots and take the absolute value of each negative part.
如果曲线有一部分落到 x 轴下方，那么该部分的定积分 ∫ab f(x) dx 会得到负值。但面积永远为正，因此必须在曲线与 x 轴的交点处拆分区间，并对负值部分取绝对值。
For example, suppose f(x) crosses the x-axis at x = k between a and b, with f(x) ≥ 0 on [a, k] and f(x) ≤ 0 on [k, b]. Then total area = ∫ak f(x) dx + |∫kb f(x) dx|.
例如， 设 f(x) 在 x = k 处穿过 x 轴，且 a < k < b，f(x) 在 [a, k] 上非负，在 [k, b] 上非正。那么总面积 = ∫ak f(x) dx + |∫kb f(x) dx|。
In the exam, you will typically be asked to find the total shaded area when the region consists of several parts. Write down the separate definite integrals, evaluate them, and add their absolute values.
在考试中，当阴影区域由若干部分组成时，通常会让你求出总的阴影面积。应当分别列出各个定积分，求出它们的值，然后取绝对值相加。
Never just blindly integrate from a to b if the function changes sign, as the positive and negative parts can cancel each other out, giving an incorrect area.
如果函数正负号发生变化，千万不要简单地从 a 到 b 做一次积分，因为正负部分会相互抵消，导致面积计算错误。
9. Area Between a Curve and a Straight Line | 曲线与直线之间的面积
To find the area enclosed between a curve y = f(x) and a line y = g(x), first identify the x-coordinates where they intersect. Assuming f(x) is the upper curve and g(x) the lower curve between the intersection points x = a and x = b, the area is:
要求曲线 y = f(x) 与一条直线 y = g(x) 围成的面积，首先确定两条线的交点 x 坐标。假设在交点 x = a 与 x = b 之间，f(x) 在上方而 g(x) 在下方，则面积为：
Area = ∫ab [f(x) − g(x)] dx
You must ensure the correct order: top minus bottom. If the curves cross within the interval, split it at the crossing points and treat each segment separately, always subtracting the lower curve from the upper one.
必须保证顺序正确：上方曲线的表达式减去下方曲线的表达式。如果在区间内两线相交，则要在交点处拆分，每一段都确保用上方的减去下方的。
Example: Find the area bounded by the curve y = x2 and the line y = x + 2. First solve x2 = x + 2 → x2 − x − 2 = 0 → x = -1, x = 2. In [-1, 2], the line y = x + 2 is above y = x2. Thus area = ∫-12 [(x + 2) − x2] dx.
示例： 求曲线 y = x2 与直线 y = x + 2 围成的面积。先解 x2 = x + 2 → x = -1, x = 2。在 [-1, 2] 上，直线 y = x + 2 在 y = x2 的上方。因此面积 = ∫-12 [(x + 2) − x2] dx。
Evaluating: ∫ (x + 2 − x2) dx = x2/2 + 2x − x3/3. Plugging limits gives (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = (10/3) − (−7/6) = 9/2 square units.
计算：∫ (x + 2 − x2) dx = x2/2 + 2x − x3/3。代入上下限得到 9/2 平方单位。
This type of question appears frequently, so practice finding intersection points and accurately identifying which function is above the other.
这类问题出现频率很高，建议多练习求交点以及正确判断哪条曲线在上方。
10. Kinematics Applications:
Published by TutorHao | GCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导