📚 Mastering Calculus: Essential Worked Examples for IB & Edexcel Maths | IB与Edexcel数学微积分典型例题精讲
Whether you are preparing for the IB Mathematics: Analysis and Approaches (AA) course or sitting the Edexcel A Level Mathematics examinations, a strong command of calculus is non-negotiable. Differentiation and integration form the backbone of countless problems, from optimisation to kinematics. This tutorial walks you through carefully selected, fully worked examples that reflect the style and rigour of both syllabi, helping you bridge the gap between theory and exam success.
无论你是在备考 IB 数学：分析与方法 (AA) 课程，还是准备参加 Edexcel A Level 数学考试，扎实地掌握微积分都是必须的。微分与积分构成了无数习题的骨干，从最优化问题到运动学问题。本教程将带你逐一解析精心挑选的典型例题，这些题目反映了两种课程体系的风格与难度，帮助你在理论与考试高分之间架起桥梁。
1. Understanding Limits: The Foundation of Calculus | 理解极限：微积分的基石
A sound understanding of limits is crucial for tackling both basic differentiation from first principles and asymptotic behaviour in functions. The following example tests your ability to evaluate limits analytically, a skill often examined in IB and Edexcel exams.
对极限的透彻理解对于处理从第一原理进行基本微分以及函数的渐近行为至关重要。以下例题考查的是你分析求解极限的能力，这也是 IB 和 Edexcel 考试中常常涉及的技能。
Example 1: Evaluate limx→2 (x² − 4) / (x − 2).
例题 1： 计算 limx→2 (x² − 4) / (x − 2)。
Solution (step-by-step): Direct substitution gives 0/0, so we factor the numerator.
解答（逐步）： 直接代入得到 0/0，因此我们对分子进行因式分解。
(x² − 4) = (x − 2)(x + 2), so the expression becomes (x − 2)(x + 2) / (x − 2) = x + 2, for x ≠ 2. Taking the limit as x → 2 yields 2 + 2 = 4.
(x² − 4) = (x − 2)(x + 2)，因此表达式变为 (x − 2)(x + 2) / (x − 2) = x + 2，其中 x ≠ 2。当 x → 2 时取极限得到 2 + 2 = 4。
This technique—simplifying the expression before substituting—is the most common way to resolve indeterminate forms. Students often lose marks by forgetting to state that the simplification holds only when x ≠ 2, though the limit itself is unaffected.
这种在代入之前先化简表达式的技巧，是解决不定式最常用的方法。学生们常常会因为忘记说明该化简仅在 x ≠ 2 时成立而丢分，尽管极限本身并不受影响。
2. Basic Differentiation Rules: Power Rule & Beyond | 基本微分法则：幂法则及其它
Many differentiation problems begin with the power rule and extend to sums and multiples. A classic exam question asks for the derivative of a polynomial function and then evaluates it at a particular point.
许多微分问题都从幂法则开始，并扩展到和与常数倍。经典的考试题型会要求求一个多项式函数的导数，并在某一点处求值。
Example 2: If f(x) = 3x⁵ − 2x³ + 7x − 9, find f'(x) and the slope at x = 1.
例题 2： 若 f(x) = 3x⁵ − 2x³ + 7x − 9，求 f'(x) 以及在 x = 1 处的斜率。
Using the power rule d/dx (xⁿ) = n xⁿ⁻¹, we differentiate term by term.
使用幂法则 d/dx (xⁿ) = n xⁿ⁻¹，我们逐项求导。
f'(x) = 3·5x⁴ − 2·3x² + 7 + 0 = 15x⁴ − 6x² + 7.
f'(x) = 3·5x⁴ − 2·3x² + 7 + 0 = 15x⁴ − 6x² + 7。
At x = 1, f'(1) = 15(1)⁴ − 6(1)² + 7 = 15 − 6 + 7 = 16. The gradient of the curve at that point is 16.
在 x = 1 处，f'(1) = 15(1)⁴ − 6(1)² + 7 = 15 − 6 + 7 = 16。曲线在该点的斜率为 16。
Always double-check your arithmetic; a simple mistake in the exponent or coefficient can cost valuable marks in both IB and Edexcel papers.
务必反复检查算术；指数或系数上的一个小错误，在 IB 和 Edexcel 试卷中都会让你丢分不少。
3. Product, Quotient & Chain Rules | 积法则、商法则与链式法则
When functions are multiplied, divided, or composed, straightforward term‑by‑term differentiation no longer works. Mastering the product rule, quotient rule, and chain rule is essential for handling trigonometric, exponential, and logarithmic functions.
当函数相乘、相除或复合时，直接的逐项求导就不再适用了。掌握积法则、商法则和链式法则对于处理三角函数、指数函数和对数函数至关重要。
Example 3: Differentiate y = (3x² + 1)² · eˣ with respect to x.
例题 3： 对 y = (3x² + 1)² · eˣ 求关于 x 的导数。
This requires a combination of the chain rule and the product rule. Let u = (3x² + 1)², v = eˣ. Then y = u·v, so dy/dx = u’v + u v’.
这需要结合链式法则和积法则。令 u = (3x² + 1)²，v = eˣ。则 y = u·v，所以 dy/dx = u’v + u v’。
First, find u’ via the chain rule: u = (3x² + 1)² → u’ = 2(3x² + 1)·(6x) = 12x(3x² + 1). v’ = eˣ.
首先，通过链式法则求 u’：u = (3x² + 1)² → u’ = 2(3x² + 1)·(6x) = 12x(3x² + 1)。v’ = eˣ。
Now substitute: dy/dx = 12x(3x² + 1)·eˣ + (3x² + 1)²·eˣ = eˣ(3x² + 1)[12x + (3x² + 1)].
现在代入：dy/dx = 12x(3x² + 1)·eˣ + (3x² + 1)²·eˣ = eˣ(3x² + 1)[12x + (3x² + 1)]。
Factorising is good practice: it often reveals stationary points or simplifies further integration.
进行因式分解是一个很好的习惯：它常常能揭示驻点或简化后续的积分运算。
4. Implicit Differentiation | 隐函数求导
When an equation mixes x and y in a way that makes it difficult to solve for y explicitly, implicit differentiation is employed. This is a staple of IB HL and Edexcel A Level coverage on related rates and curve gradients.
当一个方程以难以显式解出 y 的方式混合了 x 和 y 时，就会用到隐函数求导。这是 IB 高水平 (HL) 和 Edexcel A Level 在相关变化率与曲线斜率部分的核心内容。
Example 4: Given x²y + y³ = 2x, find dy/dx in terms of x and y.
例题 4： 已知 x²y + y³ = 2x，求出用 x 和 y 表示的 dy/dx。
Differentiate both sides with respect to x, treating y as a function of x. For x²y, use the product rule: d/dx (x²y) = 2x·y + x²·(dy/dx).
对方程两边关于 x 求导，将 y 视为 x 的函数。对于 x²y，使用积法则：d/dx (x²y) = 2x·y + x²·(dy/dx)。
The derivative of y³ is 3y²(dy/dx) by the chain rule; the right side differentiates to 2.
y³ 的导数为 3y²(dy/dx)（依据链式法则）；右侧求导得到 2。
So: 2xy + x²(dy/dx) + 3y²(dy/dx) = 2. Collect dy/dx terms: (x² + 3y²)(dy/dx) = 2 − 2xy. Therefore dy/dx = (2 − 2xy) / (x² + 3y²).
因此：2xy + x²(dy/dx) + 3y²(dy/dx) = 2。整理 dy/dx 项：(x² + 3y²)(dy/dx) = 2 − 2xy。于是 dy/dx = (2 − 2xy) / (x² + 3y²)。
Always remember to multiply by dy/dx every time you differentiate a function of y.
务必记住，每次对 y 的函数求导时都要乘以 dy/dx。
5. Applications of Derivatives: Tangents & Normals | 导数的应用：切线与法线
One of the most straightforward yet frequently examined applications is finding equations of tangents and normals to a curve at a given point.
最直接但也最常考的应用之一，就是求曲线在某一点处的切线和法线方程。
Example 5: Find the equation of the tangent to the curve y = ln(2x) at the point where x = 1.
例题 5： 求曲线 y = ln(2x) 在 x = 1 处的切线方程。
First compute the y-coordinate: when x = 1, y = ln(2). Next, differentiate: dy/dx = 1/(2x) · 2 = 1/x. At x = 1, gradient m = 1.
首先计算 y 坐标：当 x = 1 时，y = ln(2)。接着求导：dy/dx = 1/(2x) · 2 = 1/x。在 x = 1 处，斜率 m = 1。
Using the point-slope form: y − ln(2) = 1·(x − 1), so y = x − 1 + ln(2).
使用点斜式：y − ln(2) = 1·(x − 1)，因此 y = x − 1 + ln(2)。
The normal line (perpendicular) would have gradient −1, leading to y = −x + 1 + ln(2). Both IB and Edexcel mark schemes expect exact forms with ln, not decimal approximations.
法线（垂直）的斜率将为 −1，得出 y = −x + 1 + ln(2)。IB 和 Edexcel 的阅卷评分标准都要求保留带 ln 的精确形式，而不是小数近似值。
6. Optimization Problems | 最优化问题
Optimisation requires you to use differentiation to find maximum or minimum values of a function, often modelled from a geometric or contextual scenario.
最优化问题要求你利用微分求函数的最大值或最小值，通常由一个几何图形或实际情境建模而来。
Example 6: An open box is made from a square sheet of side 12 cm, by cutting small squares of side x cm from each corner and folding up the sides. Show that the volume V = 4x(6 − x)², and find the maximum volume.
例题 6： 用一块边长为 12 cm 的正方形纸板制作一个无盖盒子，方法是从每个角剪去边长为 x cm 的小正方形，然后折起各边。证明体积 V = 4x(6 − x)²，并求最大体积。
After cutting, the base is (12 − 2x) by (12 − 2x). Volume V = x(12 − 2x)² = x·4(6 − x)² = 4x(6 − x)².
剪去后，底面为 (12 − 2x) 乘 (12 − 2x)。体积 V = x(12 − 2x)² = x·4(6 − x)² = 4x(6 − x)²。
To maximise, differentiate V with respect to x. V = 4x(36 − 12x + x²) = 144x − 48x² + 4x³. dV/dx = 144 − 96x + 12x² = 12(12 − 8x + x²) = 12(x² − 8x + 12) = 12(x − 2)(x − 6).
为求最大值，对 V 关于 x 求导。V = 4x(36 − 12x + x²) = 144x − 48x² + 4x³。dV/dx = 144 − 96x + 12x² = 12(12 − 8x + x²) = 12(x² − 8x + 12) = 12(x − 2)(x − 6)。
Set dV/dx = 0: x = 2 or x = 6 (reject x=6 as domain is 0 < x < 6). Second derivative test or sign test confirms x = 2 gives a maximum. V_max = 4·2·(6−2)² = 8·16 = 128 cm³.
令 dV/dx = 0：x = 2 或 x = 6（舍去 x=6，因为定义域为 0 < x < 6）。二阶导数检验或符号检验确认 x = 2 时取得最大值。V_max = 4·2·(6−2)² = 8·16 = 128 cm³。
This problem appears in both IB AI and Edexcel Applied Maths, testing modelling and calculus simultaneously.
这类问题同时出现在 IB AI 和 Edexcel 应用数学中，一道题同时考查了建模与微积分。
7. Introduction to Integration: Indefinite Integrals | 积分入门：不定积分
Integration is the reverse of differentiation. The indefinite integral includes the constant of integration ‘+ c’, which must never be omitted in a general antiderivative.
积分是微分的逆运算。不定积分包含积分常数“+ c”，在一般反导数中绝不能遗漏。
Example 7: Find ∫ (4x³ − 6x + 3/√x) dx, simplifying your answer.
例题 7： 计算 ∫ (4x³ − 6x + 3/√x) dx，并化简答案。
Rewrite 3/√x as 3x⁻¹⁄². Then integrate termwise: ∫ 4x³ dx = 4·(x⁴/4) = x⁴; ∫ −6x dx = −6·(x²/2) = −3x²; ∫ 3x⁻¹⁄² dx = 3·(x¹⁄² / (1/2)) = 6x¹⁄².
将 3/√x 改写为 3x⁻¹⁄²。然后逐项积分：∫ 4x³ dx = 4·(x⁴/4) = x⁴；∫ −6x dx = −6·(x²/2) = −3x²；∫ 3x⁻¹⁄² dx = 3·(x¹⁄² / (1/2)) = 6x¹⁄²。
Therefore, the integral is x⁴ − 3x² + 6√x + c.
因此，积分结果为 x⁴ − 3x² + 6√x + c。
Examiners consistently penalise missing +c, so make it a habit to add it immediately after every indefinite integration.
考官们总是会对漏写 +c 的行为扣分，所以每次做完不定积分后，要养成立刻加上 +c 的习惯。
8. Definite Integrals & Area Under a Curve | 定积分与曲线下方面积
Definite integrals compute the net area between a curve and the x-axis, using the Fundamental Theorem of Calculus. Special care is needed when the function crosses the axis.
定积分利用微积分基本定理，计算曲线与 x 轴之间的净面积。当函数穿越坐标轴时，需要特别注意。
Example 8: Evaluate ∫₀³ (3x² − 6x) dx and interpret the result geometrically.
例题 8： 计算 ∫₀³ (3x² − 6x) dx，并从几何角度解释结果。
First find antiderivative: F(x) = x³ − 3x². Then evaluate F(3) − F(0): (27 − 27) − (0 − 0) = 0. The net area is 0.
首先求反导数：F(x) = x³ − 3x²。然后计算 F(3) − F(0)：(27 − 27) − (0 − 0) = 0。净面积为 0。
However, f(x) = 3x(x − 2) is negative between 0 and 2 and positive between 2 and 3. The negative area cancels the positive area exactly, leading to zero net area. If asked for total enclosed area, you would split the integral at x = 2 and take absolute values.
然而，f(x) = 3x(x − 2) 在 0 到 2 之间为负，在 2 到 3 之间为正。负面积恰好抵消了正面积，导致净面积为 0。若要求的是总包围面积，你需要在 x = 2 处将积分拆开，并取绝对值。
This distinction between net area and total area is a classic IB/Edexcel trap.
这种净面积与总面积的区别，是 IB 和 Edexcel 考试中一个经典的陷阱。
9. Integration by Substitution | 换元积分法
Substitution is a powerful method for transforming complicated integrals into simpler ones. It is extensively tested in both exams, especially with trigonometric or exponential integrands.
换元法是一种将复杂积分转化为简单积分的强大方法。这在两个考试中都有大量考查，尤其是被积函数含有三角函数或指数函数时。
Example 9: Find ∫ x·cos(x² + 1) dx.
例题 9： 求 ∫ x·cos(x² + 1) dx。
Let u = x² + 1. Then du/dx = 2x, so dx = du/(2x). Substitute: ∫ x·cos(u) · (du/(2x)) = (1/2) ∫ cos(u) du = (1/2) sin(u) + c = (1/2) sin(x² + 1) + c.
令 u = x² + 1。则 du/dx = 2x，所以 dx = du/(2x)。代入：∫ x·cos(u) · (du/(2x)) = (1/2) ∫ cos(u) du = (1/2) sin(u) + c = (1/2) sin(x² + 1) + c。
The key is to recognise that the derivative of the inner function (x² + 1) appears as a factor. Always rewrite everything in terms of the new variable and remember to express your final answer in the original variable.
关键在于识别出内层函数 (x² + 1) 的导数正好作为一个因子出现。一定要将所有内容用新变量改写，并记住最终答案要重新用原变量表示。
10. Applications of Integration: Kinematics & Volumes | 积分的应用：运动学与体积
Integration is the tool of choice for recovering displacement from velocity or velocity from acceleration, as well as for calculating volumes of revolution—topics heavily favoured in both IB HL and Edexcel.
积分是从速度还原到位移或从加速度还原到速度，以及计算旋转体体积的首选工具——这些专题在 IB HL 和 Edexcel 中都很受青睐。
Example 10: A particle moves along a straight line with velocity v(t) = 3t² − 8t + 5 m/s, where t is in seconds. Given initial displacement s(0) = 2 m, find the displacement function s(t) and the acceleration a(t).
例题 10： 一质点沿直线运动，速度 v(t) = 3t² − 8t + 5 m/s，其中 t 以秒为单位。已知初始位移 s(0) = 2 m，求位移函数 s(t) 和加速度 a(t)。
Displacement is the integral of velocity: s(t) = ∫ v(t) dt = ∫ (3t² − 8t + 5) dt = t³ − 4t² + 5t + c. Plug in s(0) = 2 → c = 2, so s(t) = t³ − 4t² + 5t + 2.
位移是速度的积分：s(t) = ∫ v(t) dt = ∫ (3t² − 8t + 5) dt = t³ − 4t² + 5t + c。代入 s(0) = 2 → c = 2，因此 s(t) = t³ − 4t² + 5t + 2。
Acceleration is the derivative: a(t) = d/dt (3t² − 8t + 5) = 6t − 8 m/s².
加速度是导数：a(t) = d/dt (3t² − 8t + 5) = 6t − 8 m/s²。
This example ties together differentiation and integration, reflecting the unified approach required in both IB and Edexcel papers.
这道例题将微分和积分数理结合在一起，体现了 IB 和 Edexcel 考卷要求的那种统一性的解题思路。
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