📚 Integration Techniques for Prelims Paper I-III: Chapter 9C Homework | 预科试卷 I-III：第9C章作业 – 积分技巧
Integration is a core topic in A-level Pure Mathematics, appearing across Papers I, II and III. In Chapter 9C homework, students often meet indefinite integrals, definite integrals, areas between curves, and the first formal methods such as substitution and integration by parts. This revision guide gives you the exact tools needed to complete the homework accurately and to prepare for Prelims-style questions.
积分是 A-level 纯数学的核心主题，出现在试卷 I、II 和 III 中。在第 9C 章作业中，学生通常会遇到不定积分、定积分、曲线之间的面积，以及换元积分法和分部积分法等首个正式方法。本复习指南为你提供完成作业并准备预科考试题型所需的精确工具。
1. Learning Objectives | 学习目标
The Chapter 9C homework on integration tests three main skills: reversing differentiation, evaluating definite integrals, and applying integration to find areas. It also introduces the first advanced techniques, substitution and integration by parts, which appear in Papers II and III.
第 9C 章积分作业测试三项主要技能：逆转微分、计算定积分，以及应用积分求面积。它还介绍了最初的进阶技巧——换元积分法和分部积分法，这些出现在试卷 II 和 III 中。
- Integrate powers, exponential and trigonometric functions. 对幂函数、指数函数和三角函数积分。
- Evaluate definite integrals using antiderivatives. 使用原函数计算定积分。
- Find areas under curves and between curves. 求曲线下方及曲线之间的面积。
- Apply substitution and integration by parts. 应用换元积分法和分部积分法。
By the end of the homework you should be able to recognise which method a question demands within a few seconds of reading it, and to show every intermediate step so that method marks are never lost.
完成作业后，你应当能在读完题目的几秒内判断出该用哪种方法，并完整写出每一步中间过程，从而不失方法分。
2. The Basic Integration Rules | 基本积分法则
The power rule is the foundation of most Chapter 9C questions. You must memorise the single exception n = −1, where the integral is the natural logarithm of the absolute value of x.
幂法则是大多数第 9C 章题目的基础。你必须记住唯一的例外情况 n = −1，此时积分为 x 的绝对值的自然对数。
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, n ≠ −1
∫ (1 / x) dx = ln|x| + C
The table below collects the standard results you should be able to quote instantly. Notice that every indefinite integral carries the constant C, and that the reciprocal rule uses the modulus sign because ln is only defined for positive arguments.
下表汇总了你应当能脱口而出的标准结果。注意每个不定积分都带有常数 C，而倒数法则使用绝对值符号，因为 ln 只对正数自变量有定义。
| Function | Integral | 中文说明 |
| xⁿ (n ≠ −1) | xⁿ⁺¹ / (n + 1) + C | 幂函数，指数加一再除以新指数 |
| 1 / x | ln|x| + C | 倒数函数，结果为自然对数 |
| eˣ | eˣ + C | 指数函数积分后不变 |
| eᵏˣ | (1 / k) eᵏˣ + C | 线性代换要除以系数 k |
| sin x | −cos x + C | 注意负号，角度必须用弧度 |
| cos x | sin x + C | 结果为正弦，无负号 |
| sec² x | tan x + C | 源自 tan 的导数 |
| (ax + b)ⁿ | (ax + b)ⁿ⁺¹ / (a(n + 1)) + C | 线性内函数，结果除以 a |
| 1 / (ax + b) | (1 / a) ln|ax + b| + C | 线性内函数，结果除以 a |
3. Exponential, Trigonometric and Reciprocal Functions | 指数、三角与倒数函数的积分
Exponential and trigonometric integrals are almost always combined with a linear inner function, so the chain-rule reversal factor 1 / a is the single most common source of lost marks. Write the factor down before you simplify anything else.
指数与三角函数的积分几乎总是与线性内函数组合出现，因此"链式法则反推"产生的 1 / a 系数是最常见的失分点。在化简任何其他内容之前，先把该系数写下来。
∫ sin(ax + b) dx = −(1 / a) cos(ax + b) + C
∫ cos(ax + b) dx = (1 / a) sin(ax + b) + C
∫ eᵃˣ⁺ᵇ dx = (1 / a) eᵃˣ⁺ᵇ + C
Trigonometric functions in calculus always work in radians. If your calculator is set to degrees, the numerical answers for definite integrals of trigonometric functions will be wrong, even though your method looks perfect.
微积分中的三角函数一律使用弧度。如果你的计算器设置为角度制，三角函数的定积分数值答案就会出错，即使方法看起来完全正确。
- Always check the inner function is linear before applying the 1 / a rule. 应用 1 / a 规则前，确认内函数是线性的。
- If the inner function is quadratic or trigonometric, use substitution instead. 若内函数是二次或三角函数形式，改用换元积分法。
- Reciprocal integrals of the form 1 / (ax + b) always produce a logarithm with a modulus. 形如 1 / (ax + b) 的倒数积分总是产生带绝对值的对数。
4. Indefinite Integrals and the Constant of Integration | 不定积分与积分常数
An indefinite integral represents a whole family of curves that differ only by a vertical translation. That is why the constant C must appear in every answer, and why an examiner can deduct a mark for omitting it even when the algebra is flawless.
不定积分代表的是一族彼此仅相差竖直平移的曲线。这就是为什么常数 C 必须出现在每一个答案中，也是为什么即使代数过程毫无瑕疵，遗漏 C 仍会被扣分。
The language of the homework is important: "find the general solution" means include C, while "find the particular solution" means use a given point to determine the numerical value of C.
作业中的措辞很重要："求通解"意味着必须写出 C；"求特解"意味着要用给定点来确定 C 的具体数值。
- General solution: F(x) + C for an unknown constant. 通解：F(x) + C，常数为未知量。
- Particular solution: substitute the given coordinates to solve for C. 特解：代入已知坐标求出 C 的值。
- Never absorb C into a logarithm without stating it, for example ln|x| + C is not the same as ln|kx| unless you explain the change. 不要把 C 悄悄并入对数，例如 ln|x| + C 与 ln|kx| 并不等价，除非你做出说明。
constants is itself just another constant, so write C₁ − C₂ = C at the end of the working.
当题目要求比较两个原函数时，请记住两个常数之差本身仍是常数，因此在解题末尾应写成 C₁ − C₂ = C。
5. Definite Integrals and the Fundamental Theorem | 定积分与微积分基本定理
A definite integral is evaluated by finding any antiderivative and subtracting its value at the lower limit from its value at the upper limit. The constant C cancels, which is why it is normally dropped in square-bracket notation.
定积分的计算方式是：先求任一原函数，再用上、下限处的取值相减。常数 C 会互相抵消，这就是使用方括号记号时通常省略 C 的原因。
∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)
Three properties of definite integrals appear repeatedly in Papers I to III. Reversing the limits changes the sign, a zero-width interval gives zero, and an integral can be split at an interior point.
定积分的三条性质在试卷 I 至 III 中反复出现：交换上下限会改变符号；上下限相同的区间积分为零；积分区间可以在内部分点处拆分。
- ∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx. 交换上下限，符号改变。
- ∫ₐᵃ f(x) dx = 0. 上下限相同，结果为零。
- ∫ₐᵇ + ∫ᵦᶜ = ∫ₐᶜ. 区间可以相加拆分。
Always write the square bracket with the limits before substituting. Students who substitute directly into the antiderivative without showing the bracket frequently lose the accuracy mark when a sign slips.
代入之前一定要先写出带上下限的方括号。那些不写出方括号、直接代入原函数的学生，一旦符号出错，往往就会丢掉精确分。
6. Areas Under Curves | 曲线下方的面积
The definite integral of a positive function gives the area between the curve and the x-axis on the given interval. Set up the integral with the correct limits before evaluating, and sketch the curve whenever the question provides enough information.
正函数的定积分给出曲线与 x 轴之间在给定区间上的面积。计算前要先用正确的上下限建立积分式，只要题目信息足够，就应画出草图。
Area = ∫ₐᵇ y dx, when y ≥ 0 on [a, b]
If part of the curve lies below the x-axis, that portion contributes a negative value. You must either split the integral at the root or take the absolute value of each piece; simply integrating straight through will underestimate the total area.
如果曲线的一部分位于 x 轴下方，那部分会贡献负值。你必须要么在零点处拆分积分，要么对每一段取绝对值；直接一次积完会低估总面积。
- Find the roots first and split the interval at each root. 先求零点，并在每个零点处拆分区间。
- Integrate each piece separately and add the absolute values. 分段积分后取绝对值再相加。
- State the area as a positive number with units squared if units are given. 面积应写成正数，若题目给出单位则用平方单位。
A common Prelims twist is to give the area and ask for an unknown limit. In that case, set up the definite integral with the unknown limit, evaluate, and solve the resulting equation for that limit.
预科考试中常见的变形是：给出面积，反求未知的积分限。此时应把未知限写成一般形式建立定积分，算出结果，再解方程求出该限。
7. Areas Between Curves | 曲线之间的面积
When two curves enclose a region, the area is found by integrating the upper curve minus the lower curve between the intersection points. Identify which curve is on top by substituting a convenient test value between the intersections.
当两条曲线围成一个区域时，面积等于在交点之间对"上曲线减下曲线"进行积分。可在两个交点之间取一个方便检验的 x 值代入，判断哪条曲线在上方。
Area = ∫ₐᵇ (ytop − ybottom) dx
The limits a and b are the x-coordinates of the intersection points, so finding the intersections is usually the first mark of the question. Set the two equations equal to each other and solve for x.
积分限 a 和 b 就是交点的横坐标，因此求交点通常是本题的第一个得分点。令两式相等并解出 x 即可。
- Set y₁ = y₂ and solve to find both limits. 令 y₁ = y₂ 解出两个积分限。
- Test a midpoint value to decide the order of subtraction. 用中点值检验，确定相减顺序。
- Integrate term by term, keeping the bracket and limits visible. 逐项积分，保留方括号与上下限。
If the curves intersect more than twice, the region must be split at every intersection, because the curve that is on top changes at each crossing. Drawing a quick sketch makes this obvious and saves a great deal of time.
若两曲线交点超过两个，必须在每个交点处分割区域，因为每经过一个交点，位于上方的曲线就发生改变。快速画一张草图能一眼看出这一点，并节省大量时间。
8. Integration by Substitution | 换元积分法
Substitution reverses the chain rule. Choose u to be the inner function whose derivative also appears in the integrand, then replace both the variable and the differential before integrating.
换元积分法是链式法则的逆过程。选择 u 为内函数，使其导数也出现在被积函数中，然后在积分前同时替换变量与微分。
∫ f(g(x)) g′(x) dx = ∫ f(u) du, with u = g(x)
For definite integrals you have two valid routes: change the limits to u-values, or integrate in u and convert back to x before substituting the original limits. Changing the limits is usually faster but must be done carefully.
对于定积分有两条可行路线：把上下限换成 u 的值，或者先以 u 积分再换回 x 后代入原上下限。换限通常更快，但必须仔细。
- Choose u, then compute du / dx and rearrange to get dx. 选定 u，求 du / dx，再整理得到 dx。
- Replace every x in the integrand, including the limits if it is a definite integral. 替换被积函数中所有的 x，若为定积分则连同上下限一起替换。
- Write the final answer back in terms of x unless the limits have been changed. 除非已经换了限，否则最终答案要换回 x。
x² + 1, so du = 2x dx, and the integral becomes ∫ u⁵ du = u⁶ / 6 + C = (x² + 1)⁶ / 6 + C.
一个经典例题是 ∫ 2x(x² + 1)⁵ dx。令 u = x² + 1，则 du = 2x dx，积分化为 ∫ u⁵ du = u⁶ / 6 + C = (x² + 1)⁶ / 6 + C。
9. Integration by Parts | 分部积分法
Integration by parts reverses the product rule and is used when the integrand is a product of two different types of function, such as x eˣ, x sin x or ln x. The choice of which factor to call u is made with the LIATE order.
分部积分法是乘积法则的逆过程，用于被积函数是两类不同函数之积的情形，例如 x eˣ、x sin x 或 ln x。选择哪个因子作为 u 要依据 LIATE 顺序。
∫ u (dv / dx) dx = uv − ∫ v (du / dx) dx
∫ₐᵇ u (dv / dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du / dx) dx
LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric and Exponential. Whichever type appears earliest in that list becomes u; the remaining factor becomes dv / dx.
LIATE 依次代表对数函数、反三角函数、代数函数、三角函数和指数函数。在该顺序中出现得最早的类型作为 u，剩下的因子作为 dv / dx。
| Integral | Choice of u | Result |
| ∫ x eˣ dx | u = x | x eˣ − eˣ + C |
| ∫ x sin x dx | u = x | −x cos x + sin x + C |
| ∫ ln x dx | u = ln x, dv = dx | x ln x − x + C |
| ∫ x² eˣ dx | u = x² | apply twice, then simplify |
Some integrals, such as ∫ eˣ is is a favourite Paper III question, so practise the full cycle until it becomes routine.
某些积分（例如 ∫ eˣ sin x dx）需要连续使用两次分部积分，再对原积分建立方程求解。这是试卷 III 的常考题型，请把整个循环练到熟练为止。
10. Common Errors and Marking Points | 常见错误与评分要点
Most lost marks in Chapter 9C homework come from a small set of repeated mistakes rather than from genuine misunderstanding. Reviewing these before you submit the homework is far more efficient than redoing whole questions.
第 9C 章作业中的失分大多来自一小批反复出现的错误，而不是真正的理解偏差。交作业前先复查这些要点，比重新做整道题高效得多。
- Omitting the constant C in an indefinite integral. 不定积分遗漏常数 C。
- Forgetting to divide by the inner coefficient a. 忘记除以内函数系数 a。
- Sign errors in ∫ sin x dx = −cos x + C. ∫ sin x dx = −cos x + C 的符号错误。
- Not changing the limits when using substitution on a definite integral. 对定积分换元时未更换上下限。
- Dropping the modulus sign in ln|x|. ln|x| 中遗漏绝对值符号。
- Leaving a calculator in degree mode. 计算器仍处于角度制模式。
- Reporting a negative area. 面积写成负数。
- Subtracting the curves in the wrong order. 曲线相减顺序颠倒。
Method marks are awarded for a correct substitution or a correct choice of u, even if the arithmetic later fails. Always write the substitution line explicitly: state u, state du, and show the transformed integral before you evaluate anything.
即使后续运算出错，正确的换元或正确的 u 选择仍可获得方法分。务必明确写出换元这一行：写出 u，写出 du，并在计算之前先展示变换后的积分式。
11. Worked Examples from the Homework | 作业例题精讲
Example 1 is a straightforward polynomial integral. Write each term separately and apply the power rule, remembering that the integral of the constant term is that constant multiplied by x.
例 1 是基础的多项式积分。逐项书写并应用幂法则，注意常数项的积分等于该常数乘以 x。
∫ (3x² + 2x − 5) dx = x³ + x² − 5x + C
Example 2 evaluates a definite integral of a linear function, showing the bracket notation that earns the accuracy mark.
例 2 计算线性函数的定积分，展示能拿到精确分的方括号写法。
∫₁³ (2x + 1) dx = [x² + x]₁³ = 12 − 2 = 10
= x + 2, that is x = −1 and x = 2, and the line lies above the parabola on that interval.
例 3 求 y = x² 与 y = x + 2 之间的面积。两曲线在 x² = x + 2 处相交，即 x = −1 与 x = 2，且在该区间上直线位于抛物线之上。
Area = ∫₋₁² (x + 2 − x²) dx = [x² / 2 + 2x − x³ / 3]₋₁² = 9 / 2
² the limits change from 0 and 1 to 0 and 1, and the factor ½ appears from dx = du / (2x).
例 4 对定积分使用换元法。令 u = x²，上下限由 0、1 变为 0、1，而由 dx = du / (2x) 产生系数 ½。
∫₀¹ x ex² dx = ½ ∫₀¹ eᵘ du = ½ (e − 1)
12. Summary Checklist and Formula Sheet | 总结清单与公式表
Use this checklist before you close the homework. If every box is ticked, your work is at the standard expected in Prelims Papers I to III.
合上作业本之前，请使用下面这份清单。如果每一项都已打勾，你的作业就达到了预科试卷 I 至 III 的预期水准。
- Every indefinite integral ends with + C
广告 / Sponsored预售 英国CGP教材 GCSE English KS4 中学中考英语写作技巧指导备考教辅2册 单词拼写语法 标点符号 …英语 备考教辅 · 正版现货 · 学习备考通用查看详情 / View on Taobao广告 / Sponsored英文原版 Cambridge IGCSE Co-ordinated Sciences Chemistry 柯林斯剑桥I…化学 备考教辅 · 正版现货 · 学习备考通用查看详情 / View on Taobao
更多咨询请联系16621398022（同微信）
备考资讯每日更新，关注不错过考纲变化 · 科目分析 · 学习方法
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图公众号：tutorhao