📚 Exercise 18A.2: Basic Integration Techniques | 练习18A.2：基本积分技巧
Exercise 18A.2 in the IB Mathematics course builds your foundational skills in indefinite integration. It focuses on reversing differentiation to find antiderivatives, applying the power rule, handling constants, and integrating sums of simple functions. Mastering these core techniques is essential before moving on to definite integrals, areas under curves, and more advanced integration methods such as substitution or integration by parts.
IB 数学课程中的练习 18A.2 旨在帮助你建立不定积分的基础技能。该练习着重于逆向运用微分来寻找原函数，应用幂法则，处理常数，以及对简单函数的和进行积分。在进入定积分、曲线下方面积以及代入法或分部积分法等更高级的积分方法之前，掌握这些核心技巧至关重要。
1. Antiderivatives: The Reverse of Differentiation | 原函数：微分的逆运算
An antiderivative of a function f(x) is another function F(x) such that F' iderivatives is called integration. Unlike differentiation, which yields a unique result, integration produces a family of functions that differ by a constant, because the derivative of any constant is zero. This is why we always add a constant of integration, traditionally denoted by C.
函数 f(x) 的原函数是另一个函数 F(x)，满足 F'(x) = f(x)。寻找原函数的过程称为积分。与微分产生唯一结果不同，积分会生成一族相差一个常数的函数，因为任意常数的导数均为零。这就是为什么我们总是要加上一个积分常数，通常记作 C。
(x) = 2x, then one antiderivative is F(x) = x² because d/dx (x²) = 2x. However, x² + 1, x² – Thus, the indefinite integral is written as ∫ 2x dx = x² + C, capturing the entire family.
例如，如果 f(x) = 2x，那么它的一个原函数是 F(x) = x²，因为 d/dx (x²) = 2x。然而，x² + 1、x² – 5 以及 x² + 100 也都是它的原函数。因此，不定积分记作 ∫ 2x dx = x² + C，以此来涵盖整个函数族。
2. The Power Rule for Integration | 积分的幂法则
rule is the most frequently used integration tool. It states that for any real number n ≠ – . This rule directly reverses the power rule in differentiation, where d/dx (xⁿ) = n xⁿ⁻¹.
幂法则是使用最频繁的积分工具。它指出，对于任何实数 n ≠ –1，xⁿ 关于 x 的积分等于 xⁿ⁺¹ 除以 (n + 1)，再加上积分常数。这一法则直接反转了微分中的幂法则，在微分中 d/dx (xⁿ) = n xⁿ⁻¹。
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ –1
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, 当 n ≠ –1 时
the xⁿ form before integrating. A classic case that does not follow the power rule is n = –1, which leads to the natural logarithm function.
在将这一法则应用于常数项时，请记住常数 k 可以写成 k x⁰，因此 ∫ k dx = k x + C。对于根式和倒数形式，应该先利用指数律将表达式写为 xⁿ 的形式，然后再积分。一个不遵循幂法则的经典情形是 n = –1，该情形的积分结果是自然对数函数。
3. Key Integration Formulas Table | 关键积分公式表
IB students often find it helpful to memorise a compact set of integrals. The following table summarises the fundamental indefinite integrals that form the backbone of Exercise 18A.2 and many subsequent topics.
IB 学生常常发现，熟记一组简洁的积分公式非常有帮助。下表总结了构成练习 18A.2 以及后续许多主题基础的基本不定积分公式。
| Function f(x) | Indefinite Integral ∫ f(x) dx |
|---|---|
| xⁿ, n ≠ –1 | xⁿ⁺¹/(n + 1) + C |
| 1/x (or x⁻¹) | ln |x| + C |
| eˣ | eˣ + C |
| aˣ (a > 0, a ≠ 1) | aˣ / ln a + C |
| sin x | –cos x + C |
| cos x | sin x + C |
| sec² x | tan x + C |
In Exercise 18A.2, the focus is typically on algebraic functions, so you will rely heavily on the first two rows. The trigonometric and exponential entries become more prominent in later mixed exercises, but knowing them early builds confidence.
练习 18A.2 通常聚焦于代数函数，因此你会大量依赖前两行公式。三角函数和指数函数的条目在后续的混合练习中会更突出，但尽早了解它们有助于建立信心。
4. Constant Multiple and Sum Rules | 常数乘法和加法法则
Integration is a linear operation, which means you can pull constant factors outside the integral sign and integrate sums term by term. These two properties allow you to break complicated expressions into smaller, manageable pieces that can be handled using the power rule and known formulas.
积分是一种线性运算，这意味着你可以将常数因子提到积分号外，并且可以逐项对和进行积分。这两个性质使你能够将复杂表达式拆分成更小、易处理的部分，再利用幂法则和已知公式逐一处理。
∫ k · f(x) dx = k ∫ f(x) dx, where k is a constant
∫ k · f(x) dx = k ∫ f(x) dx, 其中 k 是常数
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
When you encounter an expression like 5x³ – 2x + 7, apply these rules: integrate 5x³, then –2x, then 7 as separate tasks, and finally combine the results. Do not forget to add only one constant C at the end; there is no need to write a separate constant for each term, as they collectively represent an arbitrary constant.
当你遇到像 5x³ – 2x + 7 这样的表达式时，可以应用这些法则：分别对 5x³、–2x 和 7 进行积分，然后将结果合并。不要忘记只需在最后加上一个常数 C；无需为每一项单独写一个常数，因为它们共同表示一个任意常数。
5. Integrating Polynomial Functions | 多项式函数的积分
Polynomials are the simplest class of functions to integrate because they are finite sums of power functions. To integrate a polynomial, increase the exponent of each term by 1, divide by the new exponent, and keep the coefficients intact. This step-by-step approach appears throughout Exercise 18A.2 and is often the first skill tested in IB integration problems.
多项式是最简单的一类可以积分的函数，因为它们是幂函数的有限和。要对多项式积分，只需将每一项的指数增加 1，除以新的指数，并保持系数不变。这种逐步操作的方法在练习 18A.2 中随处可见，并且通常是 IB 积分题中考查的第一项技能。
For instance, to integrate 6x² – 4x + 3, apply the power rule to each term: ∫ 6x² dx = 6 · (x³/3) = 2x³; ∫ –4x dx = –4 · (x²/2) = –2x²; ∫ 3 dx = 3x. Then combine to get 2x³ – + 3x + C. Always check that differentiating your answer gives back the original integrand.
例如，要对 6x² – 4x + 3 积分，可以对每一项应用幂法则：∫ 6x² dx = 6 · (x³/3) = 2x³；∫ –4x dx = –4 · (x²/2) = –2x²；∫ 3 dx = 3x。然后合并得到 2x³ – 2x² + 3x + C。务必检查，当你对答案求导后，是否得到了原来的被积函数。
6. Dealing with Rational and Negative Exponents | 处理有理指数和负指数
Many functions in Exercise 18A.2 are not presented as clean polynomials. You will often see expressions with square roots, cube roots, or variables in the denominator. The key strategy is to rewrite every term in the form k xⁿ before using the power rule. This means converting √x into x¹⁄², 1/x² into x⁻², and 1/√x into x⁻¹⁄².
练习 18A.2 中的许多函数并不是以整齐的多项式形式出现的。你常常会看到带有平方根、立方根或分母中含有变量的表达式。关键策略是，在使用幂法则之前，将每一项都改写为 k xⁿ 的形式。这意味着将 √x 转化为 x¹⁄²，将 1/x² 转化为 x⁻²，以及将 1/√x 转化为 x⁻¹⁄²。
o the exponent and divide by the new exponent. For example, ∫ (1/x²) dx = ∫ x⁻² dx = x⁻¹/(–1) + C = –1/x + C. Similarly, ∫ √x = (2/3) x³⁄² + C. Neat algebraic manipulation avoids unnecessary mistakes with fractions.
改写后，就可以应用标准法则：指数加 1，再除以新的指数。例如，∫ (1/x²) dx = ∫ x⁻² dx = x⁻¹/(–1) + C = –1/x + C。类似地，∫ √x dx = ∫ x¹⁄² dx = x³⁄² / (3/2) + C = (2/3) x³⁄² + C。熟练的代数操作能够避免在处理分数时出现不必要的错误。
7. The Special Case of 1/x | 1/x 的特殊情况
The power rule explicitly excludes the exponent n = – ult is fundamental and is tested frequently, including in contexts that go beyond Exercise 18A.2, such as when integrating rational functions using partial fractions.
幂法则明确排除了指数 n = –1 的情形，因为它会导致除以零。作为替代，1/x 关于 x 的积分是 x 的绝对值的自然对数。这一结果是基础性的，并且考查频率很高，包括在超越练习 18A.2 的情境中，例如利用部分分式对有理函数进行积分时。
∫ 1/x dx = ln |x| + C, for x ≠ 0
∫ 1/x dx = ln |x| + C, 当 x ≠ 0
you encounter a constant multiple, such as 5/x, simply carry the constant: ∫ 5/x dx = 5 ln |x| ny IB problems, the context may allow you to drop the absolute value if x is known to be positive, but it is safer to include it in an indefinite integral.
当你遇到常数倍，例如 5/x 时，只需将常数提出来：∫ 5/x dx = 5 ln |x| + C。绝对值符号确保定义域包含了负的 x 值，因为自然对数仅对正数有定义。在许多 IB 问题中，如果已知 x 为正，上下文可能允许你省略绝对值，但在不定积分中保留它更为稳妥。
8. The Constant of Integration and Its Importance | 积分常数及其重要性
The '+ C' at the end of an indefinite integral is not optional; it is a critical part of the answer. Omitting it implies that you have found a specific antiderivative rather than the complete family. In IB exams, marks are routinely allocated for including the constant, and missing it can cost you a significant number of points over a paper.
不定积分末尾的 “+ C” 不是可选项；它是答案的关键组成部分。省略它就意味着你找到了一个特定的原函数，而不是完整的函数族。在 IB 考试中，包含常数通常会占分，遗漏它可能让你在一份试卷中丢失相当可观的分数。
e given a point on the curve of an antiderivative, you can solve for C. For instance, if f' 5, then f(x) = x³ + C, and substituting x = 1 gives 1 + C = 5, so C = 4 and f(x) = x³ + 4.
理解常数的一种有效方法是借助初始条件。如果给定了原函数曲线上的一个点，你就可以解出 C。例如，若 f'(x) = 3x² 且 f(1) = 5，那么 f(x) = x³ + C，代入 x = 1 得到 1 + C = 5，因此 C = 4，f(x) = x³ + 4。
Treat the constant of integration as a reminder that integration undoes differentiation only up to an additive constant. Once you begin applying integration to real-world problems, such as finding displacement from velocity, the constant gains physical meaning, representing initial conditions.
把积分常数看作一个提示：积分只能将微分还原到一个加法常数。一旦你开始将积分应用于实际问题，例如从速度求位移，这个常数就具有了物理意义，代表着初始条件。
9. Verifying Answers Through Differentiation | 通过微分验证答案
Because differentiation is straightforward and mechanical, you can always check your integration work by differentiating your result. If you get back the original integrand, your answer is correct. This verification habit is highly recommended in IB assessments; it catches sign errors, exponent mistakes, and forgotten constants.
由于微分操作直接且机械，你总可以通过对结果求导来检查你的积分是否正确。如果你得到了原来的被积函数，那么你的答案就是正确的。强烈建议在 IB 评估中养成这种验证习惯；它能帮你发现符号错误、指数错误以及被遗忘的常数。
Suppose your answer to ∫ (4x³ – 6x) dx is x⁴ – 3x² + C. Differentiating gives 4x³ – 6x, which matches the original integrand. If instead you had written x⁴ – 3x + C, differentiating would yield 4x³ – 3, immediately revealing the error in the middle term.
假设你对 ∫ (4x³ – 6x) dx 的答案是 x⁴ – 3x² + C。求导得到 4x³ – 6x，这与原被积函数相符。如果你错误地写成了 x⁴ – 3x + C，求导后就会得到 4x³ – 3，从而立刻揭示出中间项的差错。
This checking method also reinforces your understanding of the inverse relationship between differentiation and integration, which lies at the heart of the Fundamental Theorem of Calculus to be studied later in the course.
这种检验方法还能加深你对微分和积分之间互为逆运算关系的理解，而这一关系正是后续课程中将学习的微积分基本定理的核心。
10. Common Pitfalls and How to Avoid Them | 常见陷阱及其避免方法
ding or using an appropriate substitution. In the context of Exercise 18A.2, where expressions are usually presented in expanded form, the biggest dangers are algebraic slips with fractions and negative exponents.
即便是成绩优秀的学生在基础积分上也可能失手。典型错误包括：在除以新指数前忘记给指数加 1、错误处理负号、对 1/x 应用幂法则，以及写出像 ∫ (2x + 1)² dx = (2x + 1)³/3 + C 这样的式子，却没有先展开或使用合适的代换。在练习 18A.2 的语境中，表达式通常以展开的形式呈现，因此最大的危险在于处理分数和负指数时的代数失误。
To avoid errors, always rewrite the integrand in a simplified, exponent-friendly form before integrating. Keep a mental checklist: (1) Are all terms in the form k xⁿ? (2) Have I increased each exponent by 1? (3) Did I divide by the new exponent correctly? (4) Did I include '+ C'? And finally, differentiate mentally to confirm.
为避免错误，务必在积分前将被积函数改写为简化且便于使用指数的形式。心里要有一个检查清单：(1) 所有项是否都已整理为 k xⁿ 的形式？(2) 我是否已将每个指数都加了 1？(3) 我是否正确除以了新的指数？(4) 我是否包含了 “+ C”？最后，在心里求导确认一下。
11. Worked Examples Inspired by Exercise 18A.2 | 练习18A.2 典型例题解析
Let’s work through problems similar to those you will meet in Exercise 18A.2. These examples incorporate polynomial terms, fractional exponents, and negative powers, reflecting the typical mix found in IB practice sets.
让我们来解答一些与练习 18A.2 中相似的题目。这些例题涵盖了多项式项、分数指数和负指数，反映了 IB 习题集中常见的混合题型。
Example 1: Find ∫ (3x⁴ – 2x² + 5) dx.
例题 1：求 ∫ (3x⁴ – 2x² + 5) dx。
Solution: Integrate term by term. ∫ 3x⁴ dx = 3 · (x⁵/5) = (3/5)x⁵. ∫ –2x² dx = –2 · (x³/3) = –(2/3)x³. ∫ 5 dx = 5x. Combining, we obtain (3/5)x⁵ – (2/3)x³ + 5x + C. Always write the constant.
解答：逐项积分。∫ 3x⁴ dx = 3 · (x⁵/5) = (3/5)x⁵。∫ –2x² dx = –2 · (x³/3) = –(2/3)x³。∫ 5 dx = 5x。合并后得到 (3/5)x⁵ – (2/3)x³ + 5x + C。务必写上常数。
Example 2: Evaluate ∫ (4/√x – 3/x³) dx.
例题 2：计算 ∫ (4/√x – 3/x³) dx。
Solution: Rewrite as ∫ (4 x⁻¹⁄² – 3 x⁻³) dx. For the first term, exponent –1/2 + 1 = 1/2, so ∫ 4 x⁻¹⁄² dx = 4 · (x¹⁄² / (1/2)) = 8 √x. For the second term, exponent –3 + 1 = –2, so ∫ –3 x⁻³ dx = –3 · (x⁻²/(–2)) = (3/2) x⁻² = 3/(2x²). Answer: 8√x + 3/(2x²) + C.
解答：改写为 ∫ (4 x⁻¹⁄² – 3 x⁻³) dx。对于第一项，指数 –1/2 + 1 = 1/2，因此 ∫ 4 x⁻¹⁄² dx = 4 · (x¹⁄² / (1/2)) = 8 √x。对于第二项，指数 –3 + 1 = –2，因此 ∫ –3 x⁻³ dx = –3 · (x⁻²/(–2)) = (3/2) x⁻² = 3/(2x²)。答案：8√x + 3/(2x²) + C。
Example 3: Given that dy/dx = 6x – 2 and y = 5 when x = 1, find y in terms of x.
例题 3：已知 dy/dx = 6x – 2，并且当 x = 1 时 y = 5，求 y 关于 x 的表达式。
Solution: Integrate to get y = ∫ (6x – 2) dx = 3x² – 2x + C. Using the condition y(1) = 5: 3(1)² – 2(1) + C = 5 → 3 – 2 + C = 5 → C = 4. Hence, y = 3x² – + 4. This form of problem links the constant of integration to a specific solution curve.
解答：积分得到 y = ∫ (6x – 2) dx = 3x² – 2x + C。利用条件 y(1) = 5：3(1)² – 2(1) + C = 5 → 3 – 2 + C = 5 → C = 4。因此，y = 3x² – 2x + 4。这类问题将积分常数与特定的解曲线联系了起来。
12. Connecting to the Bigger Picture in IB Mathematics | 与 IB 数学更广阔图景的联系
Basic integration techniques are not an isolated topic. They lay the groundwork for definite integrals, which allow you to calculate exact areas between curves and the x-axis. They are also essential for solving differential equations, modelling population growth, and analysing motion in physics. Exercise 18A.2 may feel repetitive, but it cements the mechanical fluency you need when later chapters introduce u-substitution and integration by parts.
基本积分技巧并非孤立的主题。它们为定积分打下了基础，而定积分能让你计算曲线与 x 轴之间的精确面积。它们也是求解微分方程、模拟人口增长以及分析物理运动所不可或缺的。练习 18A.2 可能感觉有些重复，但它能巩固你在后续章节引入 u 代换和分部积分时所需的机械熟练度。
In the IB Analysis and Approaches course, integration is heavily assessed in both Paper 1 and Paper 2, often combined with functions, trigonometry, and exponentials. The confidence you build now by smoothly integrating polynomials and rational powers will pay dividends when tackling multi-step, context-rich problems in your final exams.
在 IB 分析与方法的课程中，积分在试卷一和试卷二中都有大量考查，常常与函数、三角学和指数函数相结合。你现在通过流畅地积分多项式和有理指数所建立起来的信心，将在你解决期末考试中那些多步骤、背景丰富的问题时带来丰厚的回报。
Remember that mathematics is cumulative. Every correctly integrated expression, every carefully added '+ C', and every verified step strengthens the neural pathways that make advanced calculus feel intuitive. Embrace the drill of Exercise 18A.2 as an investment in your long-term mathematical fluency.
请记住，数学是累积性的。每一个正确积分的表达式、每一个仔细加上的 “+ C”，以及每一个经过验证的步骤，都在强化那些让高等微积分变得直观的神经通路。请把练习 18A.2 的反复训练视为对你长期数学流畅度的一种投资。
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