Exercise 21I: Integration by Substitution | 练习21I：换元积分法

📚 Exercise 21I: Integration by Substitution | 练习21I：换元积分法

Exercise 21I in the IB Mathematics HL or SL syllabus is designed to build proficiency in one of the most powerful integration techniques: integration by substitution. This method, often introduced as the reverse of the chain rule for differentiation, allows students to transform complicated integrals into simpler forms that can be evaluated directly. Mastering substitution is essential not only for exam success but also for tackling advanced calculus topics encountered in further study. The following guide breaks down the fundamental concepts, walks through worked examples, and highlights common pitfalls — all tailored to help you complete Exercise 21I with confidence.

IB 数学 HL 或 SL 课程中的练习 21I 旨在帮助学生熟练掌握一种最强大的积分技巧：换元积分法。该方法通常被介绍为微分链式法则的逆运算，可以将复杂的积分转化为可以直接求解的简单形式。掌握换元法不仅是考试成功的关键，也是进一步学习高等微积分的基础。以下指南将分解基本概念、逐步讲解典型例题，并重点提示常见错误——所有内容都为了帮助你自信地完成练习 21I。

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1. What is Integration by Substitution? | 什么是换元积分法？

Integration by substitution is a technique used to evaluate integrals by changing the variable of integration. The idea is to replace a complicated expression with a new variable, typically u, so that the integrand becomes a standard form we know how to integrate. This method works because it reverses the chain rule of differentiation: if F'(x) = f(g(x))·g'(x), then integrating f(g(x))·g'(x) gives F(x). In substitution notation, we set u = g(x) so that du = g'(x) dx, transforming the integral into ∫ f(u) du.

换元积分法是通过改变积分变量来求解积分的一种方法。其核心思想是用一个新变量（通常是 u）替换掉复杂的表达式，从而将被积函数转化为我们知道如何积分的标准形式。这种方法之所以成立，是因为它逆向运用了微分的链式法则：如果 F'(x) = f(g(x))·g'(x)，那么对 f(g(x))·g'(x) 积分就会得到 F(x)。用换元符号表示就是，设 u = g(x)，则有 du = g'(x) dx，从而将积分变为 ∫ f(u) du。

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2. The Reverse Chain Rule | 逆向链式法则

To understand why substitution works, recall that the chain rule states that the derivative of a composite function y = F(u) with u = g(x) is dy/dx = F'(u)·g'(x). When we integrate, we are looking for an antiderivative. Therefore, if we see an integrand of the form k·f(g(x))·g'(x), we can “guess” that the antiderivative is F(g(x)), where F’ = f. Substitution formalises this intuition: by setting u = g(x), the integral becomes ∫ f(u) du, which is simply F(u) + C. This is why substitution is often called the “reverse chain rule.”

要理解换元法为什么有效，需要回忆复合函数 y = F(u) （其中 u = g(x)）的导数是 dy/dx = F'(u)·g'(x)。在进行积分时，我们寻找的就是原函数。因此，如果被积函数的形式为 k·f(g(x))·g'(x)，我们就可以“猜测”原函数是 F(g(x))，其中 F’ = f。换元法将这种直觉规范化：设 u = g(x)，积分就变成了 ∫ f(u) du，结果就是 F(u) + C。这就是为什么换元法常被称为“逆向链式法则”。

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3. Identifying the Inner Function | 识别内层函数

The first and most critical step in substitution is choosing the right expression for u. Typically, you look for an “inner function” — an expression inside brackets, under a root, or in the exponent of an exponential function. The derivative of this inner function should appear elsewhere in the integrand, up to a constant factor. For instance, in ∫ 2x·(x²+1)³ dx, the inner function is x²+1, and its derivative 2x is present. Picking u = x²+1 immediately simplifies the integral.

换元过程中第一个也是最关键的一步是选择合适的 u。通常，你可以寻找一个“内层函数”——即括号内的表达式、根号下或指数函数指数中的表达式。这个内层函数的导数应当以常数倍数的形式出现在被积函数的其他地方。例如，在 ∫ 2x·(x²+1)³ dx 中，内层函数是 x²+1，而它的导数 2x 正好出现。选择 u = x²+1 就能立刻简化积分。

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4. Step-by-Step Substitution Procedure | 逐步换元步骤

Following a structured approach reduces errors. Here are the recommended steps:

• Step 1: Identify the inner function u = g(x) and compute du = g'(x) dx.
• Step 2: Solve for dx if necessary: dx = du / g'(x).
• Step 3: Substitute all x expressions and dx in the integral to change it entirely to terms of u.
• Step 4: Simplify the new integrand and integrate with respect to u.
• Step 5: Substitute back u = g(x) to express the answer in terms of the original variable.
• Step 6: For definite integrals, change the limits of integration to u-values to avoid back-substitution.

This systematic method ensures that no algebraic manipulation is missed.

采用结构化的方法可以减少错误。推荐步骤如下：

• 第 1 步：识别内层函数 u = g(x) 并计算 du = g'(x) dx。
• 第 2 步：如有必要，解出 dx = du / g'(x)。
• 第 3 步：将积分中所有含 x 的表达式和 dx 全部替换为关于 u 的式子。
• 第 4 步：化简新的被积函数并对 u 积分。
• 第 5 步：将 u = g(x) 代回，用原变量表示答案。
• 第 6 步：对于定积分，将积分上下限转换为 u 的值，这样就可以避免代回。

这种系统化的方法可以确保不遗漏任何代数操作。

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5. Example 1: Basic Polynomial Integral | 例题 1：基本多项式积分

Evaluate ∫ 2x√(x²+1) dx.
Let u = x² + 1, then du = 2x dx. The integral contains exactly 2x dx, so we can substitute directly: ∫ √(u) du = ∫ u^{1/2} du. Integrating gives (2/3) u^{3/2} + C. Finally, replace u with x²+1 to obtain the answer: (2/3)(x²+1)^{3/2} + C. This example illustrates the simplest case where the derivative of the inner function is fully present.

计算 ∫ 2x√(x²+1) dx。
设 u = x² + 1，则 du = 2x dx。积分中恰好含有 2x dx，所以可直接代入：∫ √(u) du = ∫ u^{1/2} du。积分后得到 (2/3) u^{3/2} + C。最后将 u 替换为 x²+1 即得答案：(2/3)(x²+1)^{3/2} + C。这个例子展示了内层函数的导数完全匹配的最简情形。

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6. Example 2: Trigonometric Substitution | 例题 2：三角函数积分

Find ∫ sin³x cos x dx.
Let u = sin x, so du = cos x dx. The integral becomes ∫ u³ du = (1/4) u⁴ + C. Substituting back yields (1/4) sin⁴x + C. Notice that the factor cos x dx was perfectly available for du. When you encounter an odd power of sine or cosine multiplied by the derivative of the other, substitution is often the quickest route.

求 ∫ sin³x cos x dx。
设 u = sin x，于是 du = cos x dx。积分变为 ∫ u³ du = (1/4) u⁴ + C。代回后得 (1/4) sin⁴x + C。请注意，因子 cos x dx 恰好提供了 du。当遇到某个三角函数的奇次幂乘以其导数中出现的另一个三角函数时，换元法往往是最快捷的路径。

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7. Example 3: Exponential and Logarithmic Integrals | 例题 3：指数与对数积分

Compute ∫ e^(x²)·x dx.
Here the inner function is x², with derivative 2x. We have only x dx, so we set u = x², du = 2x dx, hence x dx = (1/2) du. The integral becomes ∫ e^u · (1/2) du = (1/2) e^u + C = (1/2) e^(x²) + C. For logarithmic examples, try ∫ (ln x)/x dx: let u = ln x, then du = (1/x) dx, giving ∫ u du = (1/2) u² + C = (1/2)(ln x)² + C.

计算 ∫ e^(x²)·x dx。
这里内层函数是 x²，导数为 2x。被积函数中只有 x dx，因此设 u = x²，du = 2x dx，从而 x dx = (1/2) du。积分变为 ∫ e^u · (1/2) du = (1/2) e^u + C = (1/2) e^(x²) + C。对于含对数的题目，试试 ∫ (ln x)/x dx：设 u = ln x，则 du = (1/x) dx，得到 ∫ u du = (1/2) u² + C = (1/2)(ln x)² + C。

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8. Handling Definite Integrals | 处理定积分

For a definite integral, you have two options: substitute back to x and then apply the original limits, or — preferably — change the limits to u-values right after substitution. For ∫₀¹ 2x e^(x²) dx, set u = x². When x=0, u=0; when x=1, u=1. With du = 2x dx, the integral becomes ∫₀¹ e^u du = [e^u]₀¹ = e – 1. Changing limits saves effort and avoids the mistake of forgetting to adjust them.

对于定积分，有两种处理方式：代回 x 后再代入原上下限，或者——更可取的做法——在换元后立即将积分上下限转换为 u 的值。例如，对于 ∫₀¹ 2x e^(x²) dx，设 u = x²。当 x=0 时，u=0；当 x=1 时，u=1。结合 du = 2x dx，积分变为 ∫₀¹ e^u du = [e^u]₀¹ = e – 1。改变积分限能节省工作量，并避免忘记调整上下限的常见错误。

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9. Substitution with Linear Functions | 线性函数的换元

A special case occurs when the inner function is linear, such as u = ax + b. Then du = a dx, so dx = du/a. For instance, ∫ cos(2x+3) dx becomes (1/2)∫ cos u du = (1/2) sin u + C = (1/2) sin(2x+3) + C. Even though the integrand looks simple, this substitution is invaluable in avoiding sign errors. It is also the basis for integration of reciprocal functions and other standard results.

当内层函数为线性函数时是一种特殊情况，例如 u = ax + b。此时 du = a dx，所以 dx = du/a。比如，∫ cos(2x+3) dx 可化为 (1/2)∫ cos u du = (1/2) sin u + C = (1/2) sin(2x+3) + C。尽管被积函数看起来简单，这种换元对于避免符号错误非常有价值。它也是积分倒数函数和其他标准结果的基础。

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10. Common Mistakes to Avoid | 常见错误要避免

Mistake 1: Forgetting to replace dx fully. When du = g'(x) dx, you must solve for dx if the factor is not exactly present. Mistake 2: Substituting a function whose derivative does not appear, leading to a dead end. Always check that du can be matched with part of the integrand. Mistake 3: In definite integrals, neglecting to change the limits or leaving the answer in terms of u. Mistake 4: Attempting to integrate ∫ f(u) du but forgetting the chain rule factor when differentiating to check the answer.

错误 1：忘记完全替换 dx。当 du = g'(x) dx 时，如果因子没有恰好出现，必须解出 dx。错误 2：代入了一个其导数没有出现的函数，导致走入死胡同。务必检查 du 能否与被积函数的一部分匹配。错误 3：在定积分中，忽略改变积分限，或者将答案留在 u 的式子中。错误 4：试图积分 ∫ f(u) du，但在通过求导检查答案时，忘记了链式法则中的因子。

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11. Tips for Mastering Exercise 21I | 掌握练习 21I 的技巧

To tackle Exercise 21I efficiently, practice recognising patterns that invite substitution. Create a checklist: Is there a composite function? Does the derivative of the inner function appear nearby? If the integrand is a fraction, could the numerator be the derivative of the denominator (leading to a log form)? Use substitution regularly even for simple integrals to build fluency, and always verify your answer by differentiation. Working through all problems in the exercise without skipping will solidify your intuition.

要高效地完成练习 21I，有意识地训练识别呼唤换元的模式。做一个检查清单：是否有复合函数？内层函数的导数是否在附近出现？如果被积函数是分式，分子是否为分母的导数（这将导致对数形式）？即使对简单的积分也经常使用换元法，以培养熟练度，并始终通过求导验证你的答案。不跳过练习中的任何一道题，从头到尾做完，将巩固你的直觉。

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12. Real-World Applications and IB Exam Focus | 实际应用与 IB 考试重点

Integration by substitution is not just an academic exercise; it is used in physics to calculate work done by a variable force, in probability to find cumulative distribution functions, and in economics to model growth. In the IB examination, substitution questions often appear in Paper 1 and Paper 2, sometimes embedded in kinematics or area/volume problems. Marks are awarded for correct choice of u, correct du, and accurate substitution. Show all steps clearly to gain method marks even if the final answer has a minor slip.

换元积分法不仅是学术练习；它在物理学中用于计算变力做的功，在概率论中求累积分布函数，在经济学中模拟增长。在 IB 考试中，换元题目经常出现在试卷一和试卷二中，有时嵌套在运动学或面积/体积问题中。评分会给在正确选择 u、正确写出 du 以及准确代入上。即使最终答案有小失误，清晰地展示所有步骤仍可获得步骤分。

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