📚 IGCSE OCR Maths: Basics of Calculus | IGCSE OCR 数学：微积分基础 考点精讲
Calculus is one of the most powerful branches of mathematics, allowing us to describe how quantities change and accumulate. In the IGCSE OCR syllabus, you are introduced to differentiation and integration – the two fundamental ideas of calculus – along with their geometric and real-world interpretations. This article covers all key points you need to master for the exam, from the gradient of a curve and the power rule, to stationary points and kinematic applications, and finally to basic integration and area under a curve.
微积分是数学最强大的分支之一，它让我们能够描述量是如何变化和累积的。在 IGCSE OCR 课程大纲中，你会学习到微分和积分这两个微积分的核心概念，以及它们在几何和现实世界中的解释。本文涵盖了考试中需要掌握的所有关键考点，从曲线的梯度和幂法则，到驻点和运动学应用，最后到基本积分和曲线下的面积。
1. What is Calculus? | 什么是微积分？
Calculus is essentially the study of change. It is split into two main areas: differentiation, which finds the rate at which something changes, and integration, which finds the total accumulation of a quantity. In OCR IGCSE, you focus mainly on polynomial functions and learn how to differentiate to find gradients of curves, velocities, and turning points, and integrate to recover original quantities or calculate areas.
微积分本质上是研究变化的学科。它主要分为两个领域：微分，用来寻找事物变化的速率；和积分，用来寻找某个量的累积总和。在 OCR IGCSE 课程中，你主要学习多项式函数，并掌握如何通过微分来求曲线的梯度、速度和转折点，以及通过积分来还原原函数或计算面积。
2. The Gradient of a Curve | 曲线的梯度
ient changes from point to point. The gradient at any specific point is given by the derivative, f' just touches the curve at that point. You can estimate the gradient by drawing a tangent and calculating rise over run, but differentiation gives an exact value.
对于直线，梯度是恒定的。对于曲线 y = f(x)，梯度会随着点的不同而变化。任一点处的梯度由导数 f'(x) 或 dy/dx 给出。从几何上讲，这就是在该点处刚好与曲线相切的切线的梯度。你可以通过画切线并计算纵差/横差来估算梯度，但微分可以给出精确值。
3. The Power Rule for Differentiation | 求导的幂法则
erm by term. Remember that a constant term differentiates to zero. For example, if y = 3x⁴ – 5x² + 2x – 7, then dy/dx = 12x³ – 10x + 2.
IGCSE 最重要的求导规则是幂法则：如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹。对多项式求导时，逐项应用该法则。记住常数项求导后为零。例如，若 y = 3x⁴ – 5x² + 2x – 7，则 dy/dx = 12x³ – 10x + 2。
The power rule also applies to fractional and negative powers, such as √x written as x¹/², giving derivative (1/2)x⁻¹/². Be ready to rewrite expressions like 1/x² as x⁻² before differentiating.
幂法则同样适用于分数和负指数，例如 √x 写作 x¹/²，其导数为 (1/2)x⁻¹/²。要准备好先将 1/x² 这类表达式改写为 x⁻² 再求导。
4. Finding the Equation of a Tangent and Normal | 求切线与法线方程
ven point. First, substitute the x-coordinate into dy/dx to get the gradient m. Then use y – y₁ = m(x – point (x₁, y₁). For a normal line, which is perpendicular to the tangent, its gradient is –1/m, provided m ≠ 0.
一旦知道了导数，你就可以求出曲线上给定点处的切线方程。首先，将 x 坐标代入 dy/dx 得到梯度 m。然后利用点斜式 y – y₁ = m(x – x₁)，代入点 (x₁, y₁)。对于与切线垂直的法线，其梯度为 –1/m（假设 m ≠ 0）。
= 2. We have dy/dx = 2x, so at x=2, gradient m=4. The point is (2, 5). Tangent equation: y – 5 = 4(x – 2) → y = 4x – 3. Normal gradient is –1/4, so normal equation: y – 5 = –1/4 (x – 2).
示例：求 y = x² + 1 在 x = 2 处的切线。dy/dx = 2x，因此在 x=2 处，梯度 m=4。该点为 (2, 5)。切线方程为：y – 5 = 4(x – 2) → y = 4x – 3。法线梯度为 –1/4，法线方程为：y – 5 = –1/4 (x – 2)。
5. Stationary Points and Their Nature | 驻点及其性质
tion. To find stationary points, set the derivative equal to zero and solve for x. Then substitute back into y = f(x) to find the y-coordinate.
驻点出现在曲线梯度为零的地方：dy/dx = 0。驻点有三种类型：局部极大值、局部极小值和拐点。要找到驻点，令导数等于零并解出 x。然后代回 y = f(x) 求出 y 坐标。
s a minimum; if d²y/dx² < 0, it is a maximum. If d²y/dx² = 0, the test is inconclusive and you may need to check the gradient sign on either side.
要判断驻点的性质，可以使用二阶导数检验法：对 dy/dx 再次求导得到 d²y/dx²。代入 x 值：若 d²y/dx² > 0，该点是极小值点；若 d²y/dx² < 0，该点是极大值点。若 d²y/dx² = 0，则该方法无法判断，你可能需要检查该点两侧导数的符号。
6. Sketching Graphs with Derivatives | 利用导数绘制函数草图
Derivatives help you sketch graphs accurately. Intercepts with axes give basic points. The sign of dy/dx tells you where the function is increasing (dy/dx > 0) or decreasing (dy/dx < 0). Stationary points mark peaks and troughs. The second derivative indicates concavity: if d²y/dx² > 0, the curve is concave up (∪ shaped); if d²y/dx² < 0, concave down (∩ shaped).
导数有助于你准确地绘制函数图像。与坐标轴的截距给出了基础点。dy/dx 的符号告诉你函数在何处递增（dy/dx > 0）或递减（dy/dx < 0）。驻点标记了波峰和波谷。二阶导数表示凹凸性：如果 d²y/dx² > 0，曲线是下凸的（∪ 形）；如果 d²y/dx² < 0，曲线是上凸的（∩ 形）。
Putting all this together, you can plot key points, determine rising/falling intervals, and draw a smooth curve without needing a table of many values. This is particularly useful for cubic functions where you can locate the two turning points quickly.
综合所有这些信息，你可以绘制关键点、确定递增/递减区间，并画出光滑曲线，而无需计算大量数值的表格。这对于三次函数尤其有用，你可以快速定位两个转折点。
7. Kinematics: Displacement, Velocity and Acceleration | 运动学：位移、速度与加速度
ds/dt (the first derivative) and acceleration a = dv/dt = d²s/dt² (the second derivative). You will often be given a displacement formula and asked to find the velocity or acceleration at a specific time, or to determine when the particle is at rest (v=0).
IGCSE 中微积分最实际的应用之一是运动学。如果 s(t) 表示位移随时间 t 变化的函数，那么速度 v = ds/dt（一阶导数），加速度 a = dv/dt = d²s/dt²（二阶导数）。你经常会遇到给定位移公式，要求找出某个时刻的速度或加速度，或者判断质点何时静止（v=0）的问题。
For example, if s = t³ – 6t² + 9t, then v = 3t² – 12t + 9, and a = 6t – ges. Understanding these relationships helps you describe complete motion without a graph.
例如，若 s = t³ – 6t² + 9t，则 v = 3t² – 12t + 9，a = 6t – 12。令 v = 0 得到静止时刻，这可能是运动方向改变的时刻。理解这些关系有助于你无需图像就能全面描述运动过程。
8. Introduction to Integration | 积分入门
Integration is the reverse process of differentiation. If you know the derivative f' ginal function f(x), called the antiderivative or indefinite integral. The notation is ∫ f' f(x) + c, where c is an arbitrary constant because differentiating a constant gives zero.
积分是微分的逆过程。如果已知导数 f'(x)，可以求出原函数 f(x)，它被称为原函数或不定积分。记作 ∫ f'(x) dx = f(x) + c，其中 c 是一个任意常数，因为对常数求导会得到零。
The power rule for integration reverses differentiation: for n ≠ – = x⁴ + c. OCR expects you to handle polynomials and simple fractional or negative indices.
积分的幂法则是求导的逆过程：对于 n ≠ –1，∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + c。对于不定积分，务必加上常数 c。例如，∫ 4x³ dx = x⁴ + c。OCR 考试希望你能够处理多项式以及简单的分数或负指数函数。
9. Finding the Constant of Integration | 求积分常数
If you have additional information, such as a point on the original curve, you can find the value of c. Substitute the coordinates into the integrated expression and solve for c. This yields the specific equation of the curve. This is a very common exam question bridging differentiation and integration.
如果你掌握额外的信息，比如曲线上的一个特定点，你就可以求出常数 c 的值。将点的坐标代入积分后的表达式，解出 c。这样就得到了具体的曲线方程。这是连接微分和积分的一种非常常见的考题。
Example: A curve has derivative dy/dx = 6x² – 2 and passes through (1, 4). Integrate to get y = 2x³ – 2x + c. Substitute x=1, y=4: 4 = 2(1) – 2(1) + c → 4 = 0 + c → c=4. So the curve is y = 2x³ – 2x + 4.
示例：某曲线的导数 dy/dx = 6x² – 2 且经过点 (1, 4)。积分得到 y = 2x³ – 2x + c。代入 x=1, y=4：4 = 2(1) – 2(1) + c → 4 = 0 + c → c=4。因此该曲线方程为 y = 2x³ – 2x + 4。
10. Definite Integration and Area Under a Curve | 定积分与曲线下的面积
is ∫ₐᵇ f(x) dx. To evaluate it, first find the indefinite integral F(x), then compute F(b) – nd the total physical area, you must split the interval at roots and take absolute values.
定积分计算的是曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的净面积。记作 ∫ₐᵇ f(x) dx。计算时，先求出不定积分 F(x)，然后计算 F(b) – F(a)。在此计算中，x 轴下方的面积会被计为负值。要计算总的实际面积，你必须在根处分割区间，并对各部分取绝对值。
y = x² + 1 from x=0 to x=2. The integral is ∫₀² (x²+1) dx = [ (1/3)x³ + x ]₀² = (8/3 + 2) – he constant for indefinite integrals only; omit c in definite integrals as it cancels out.
示例：求 y = x² + 1 下从 x=0 到 x=2 的面积。积分式为 ∫₀² (x²+1) dx = [ (1/3)x³ + x ]₀² = (8/3 + 2) – (0) = 14/3 平方单位。记住，只在不定期积分中加 c；在定积分中省略 c，因为它会在相减时消去。
When the curve crosses the x-axis, calculate the positive and negative areas separately using absolute values or splitting the integral at the intercepts. This avoids cancellation that would give an incorrect physical area.
当曲线穿越 x 轴时，需要使用绝对值或在截点处分割积分，分别计算正负部分的面积之后再相加。这样可以避免正负抵销，导致实际物理面积计算错误。
11. Summary of Key Concepts and Exam Tips | 核心概念总结与考试技巧
To succeed in IGCSE OCR calculus questions, you must be confident with: (1) differentiating polynomials using the power rule, (2) finding gradients of curves and equations of tangents/normals, (3) identifying stationary points and their nature using the second derivative, (4) solving kinematics problems linking s, v, a, (5) integrating polynomials and finding the constant c, and (6) calculating areas using definite integrals, with careful attention to sign.
要在 IGCSE OCR 微积分试题中取得成功，你必须熟练掌握以下内容：(1) 使用幂法则对多项式求导，(2) 求曲线的梯度以及切线/法线方程，(3) 识别驻点并利用二阶导数判断其性质，(4) 解决关联 s、v、a 的运动学问题，(5) 对多项式积分并求出常数 c，(6) 用定积分计算面积，并仔细处理符号。
Common pitfalls include forgetting the constant of integration, mishandling negative areas, and misapplying the power rule when rewriting radicals. Practice by doing past paper questions, and always check that differentiation and integration are inverse operations in your answers to catch algebraic mistakes.
常见陷阱包括忘记积分常数、错误处理负面积，以及在改写根式时误用幂法则。建议多做历年考题进行练习，并利用微分与积分互为逆运算的性质，时常检验你的答案是否有代数错误。
Published by TutorHao | Math Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图