📚 G Answers & Index 338-341: A-Level Maths Calculus Worked Examples | G答案与索引338-341：A-Level数学微积分例题解析
This revision guide covers the worked examples and key methods behind the answer set labelled G Answers & Index 338-341. The focus is on A-Level calculus: differentiation, integration, and the link between them. Each section pairs the core idea with a worked-style explanation so you can use it either before or after attempting the questions.
本复习指南涵盖了标为“G Answers & Index 338-341”的答案集背后的例题与核心方法。重点是 A-Level 微积分：微分、积分及其联系。每一节将核心思想与例题式讲解配对，方便你在做题前后使用。
1. Understanding the Index Range | 理解索引范围
In many A-Level textbooks, index numbers 338-341 usually point to a block of questions on differentiation and basic integration. Before checking the answers, make sure you can identify the rule required by the form of the function.
在许多 A-Level 教材中，索引 338-341 通常指向一组关于微分和基本积分的习题。在对照答案之前，请先确保你能根据函数形式判断所需的运算法则。
- Power rule, chain rule, product rule and quotient rule for differentiation / 微分的幂法则、链式法则、乘积法则和商法则
- Reverse power rule, substitution and parts for integration / 积分的逆幂法则、换元法和分部积分法
- Definite integrals used for area and net change / 用于面积和净变化的定积分
Treat each index range as a mini syllabus. If you can classify the question type quickly, the algebra becomes much easier.
把每个索引范围当作一个小型考试大纲。如果你能快速判断题型，代数运算就会容易得多。
2. Differentiation Basics | 微分基础
ation questions. It states that if y = xⁿ, then dy/dx = nxⁿ⁻¹, where n is any real number.
幂法则是大多数微分题的起点。它指出，如果 y = xⁿ，那么 dy/dx = nxⁿ⁻¹，其中 n 是任意实数。
d/dx (xⁿ) = nxⁿ⁻¹
nt terms differentiate to zero, and constant multiples are carried through the derivative.
例如，d/dx (x⁵) = 5x⁴。常数项微分为零，常数倍数在求导时保持不变。
Always rewrite roots and reciprocals as powers before differentiating. For instance, √x = x^½ and 1/x² = x⁻², so the same rule applies directly.
求导前一定要把根式和倒数改写为幂的形式。例如，√x = x^½，1/x² = x⁻²，这样就能直接应用同一条法则。
3. Chain Rule | 链式法则
and u = g(x), then dy/dx = dy/du × du/dx. This rule is essential for composite functions.
当一个函数嵌套在另一个函数中时，就需要使用链式法则。如果 y = f(u) 且 u = g(x)，那么 dy/dx = dy/du × du/dx。该法则对复合函数至关重要。
dy/dx = dy/du × du/dx
+ 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x. Multiplying gives dy/dx = 24x(3x² + 5)³.
例题：对 y = (3x² + 5)⁴ 求导。设 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x。相乘得到 dy/dx = 24x(3x² + 5)³。
In answers for index questions, a common error is to forget the inner derivative du/dx. Always check that the derivative of the bracket appears as a factor.
在索引题的答案中，一个常见错误是忘记内层导数 du/dx。务必检查括号的导数是否作为因子出现。
4. Product and Quotient Rules | 乘积法则与商法则
When two functions are multiplied, use the product rule: d/dx (uv) = u'v + uv', where u' and v' are the derivatives of u and v with respect to x.
当两个函数相乘时，使用乘积法则：d/dx (uv) = u'v + uv'，其中 u' 和 v' 分别是 u 和 v 关于 x 的导数。
d/dx (uv) = u'v + uv'
When one function is divided by another, use the quotient rule: d/dx (u/v) = (u'v − uv') / v². The numerator order matters: u'v comes first, then minus uv'.
当一个函数除以另一个函数时，使用商法则：d/dx (u/v) = (u'v − uv') / v²。分子的顺序很重要：先写 u'v，再减 uv'。
d/dx (u/v) = (u'v − uv') / v²
For example, if y = x²eˣ, let u = x² and v = eˣ. Then u' = 2x and v' = eˣ, so dy/dx = 2xeˣ + x²eˣ = xeˣ(2 + x).
例如，如果 y = x²eˣ，设 u = x² 且 v = eˣ。则 u' = 2x，v' = eˣ，所以 dy/dx = 2xeˣ + x²eˣ = xeˣ(2 + x)。
5. Integration as Reverse Differentiation | 积分作为微分的逆运算
1) + C, provided n ≠ -1. The constant C is added because derivative of a constant is zero.
不定积分是微分的逆运算。逆幂法则为 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，前提是 n ≠ -1。由于常数的导数为零，所以要加上常数 C。
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, n ≠ -1
For n = -1, the special result is ∫ 1/x dx = ln|x| + C. Use the absolute value to keep the logarithm defined for negative x.
当 n = -1 时，特殊结果是 ∫ 1/x dx = ln|x| + C。使用绝对值可以保证对数对负数 x 也有定义。
∫ 1/x dx = ln|x| + C
Basic integration questions in this index range often ask you to integrate polynomials such as 3x² − 4x + 7. Integrate term by term and then add a single constant C.
该索引范围内基础的积分题常要求你积分多项式，如 3x² − 4x + 7。要逐项积分，然后只加一个常数 C。
6. Integration by Substitution | 换元积分法
in rule. If an integral contains a function and its derivative, set u = g(x) and replace g'(x)dx by du. Then integrate with respect to u.
换元法是链式法则的逆运算。如果被积函数包含一个函数及其导数，就设 u = g(x)，并用 du 替换 g'(x)dx。然后对 u 积分。
∫ f(g(x)) g'(x) dx = ∫ f(u) du
Let u = x² + 1, so du = 2x dx. The integral becomes ∫ u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C.
例题：求 ∫ 2x(x² + 1)³ dx。设 u = x² + 1，则 du = 2x dx。积分变为 ∫ u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C。
For definite integrals, remember to change the limits when you substitute. If x goes from a to b, then u goes from g(a) to g(b).
对于定积分，换元时要记得改变积分限。如果 x 从 a 到 b，那么 u 从 g(a) 到 g(b)。
7. Integration by Parts | 分部积分法
Integration by parts is used for products of different types of functions, such as xeˣ or x ln x. The formula is ∫ u dv = uv − ∫ v du.
分部积分法用于不同类型函数的乘积，例如 xeˣ 或 x ln x。公式为 ∫ u dv = uv − ∫ v du。
∫ u dv = uv − ∫ v du
Choose u as the part that becomes simpler when differentiated, and dv as the part that can be integrated easily. A common order is logarithmic, inverse trig, algebraic, trigonometric, exponential.
选择 u 时，应选微分后变得更简单的部分；选择 dv 时，应选容易积分的部分。常见顺序是对数、反三角、代数、三角、指数。
For example, ∫ x eˣ dx: let u = x and dv = eˣ dx. Then du = dx and v = eˣ. The result is x eˣ − ∫ eˣ dx = x eˣ − eˣ + C.
例如，∫ x eˣ dx：设 u = x，dv = eˣ dx。则 du = dx，v = eˣ。结果是 x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。
8. Definite Integrals and Area | 定积分与面积
s a numerical value. If F(x) is an antiderivative of f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a).
定积分带有上下限，结果是一个数值。如果 F(x) 是 f(x) 的一个原函数，那么 ∫ₐᵇ f(x) dx = F(b) − F(a)。
∫ₐᵇ f(x) dx = F(b) − F(a)
When using a definite integral to find the area between a curve and the x-axis, remember that areas below the x-axis contribute negative values to the integral. Split the region and use absolute values where necessary.
用定积分求曲线与 x 轴之间的面积时，要记住 x 轴下方的区域对积分的贡献为负值。应拆分区域，并在必要时使用绝对值。
Worked example: find ∫₁³ 2x dx. Since F(x) = x², the value is F(3) − F(1) = 9 − 1 = 8.
例题：求 ∫₁³ 2x dx。因为 F(x) = x²，其值为 F(3) − F(1) = 9 − 1 = 8。
9. Common Mistakes in Index 338-341 | 索引338-341中的常见错误
Many marks are lost through small errors rather than a lack of method. One frequent mistake is forgetting the constant of integration in indefinite integrals.
许多失分来自小错误，而不是方法不懂。一个常见错误是在不定积分中忘记积分常数。
Another common error is losing the inner derivative when using the chain rule. If you differentiate (2x + 1)⁵, the factor 2 must appear in the answer.
另一个常见错误是在使用链式法则时丢失内层导数。如果对 (2x + 1)⁵
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