📚 Exercise 11C.1: Integration by Substitution | 练习 11C.1：换元积分法
Integration by substitution is one of the core analytical skills tested in IB Mathematics: Analysis and Approaches, as well as Applications and Interpretation. Exercise 11C.1 typically introduces the method where an integral is simplified by replacing a function of x with a single variable u. This article explains the theory, walks through worked examples, highlights common mistakes, and provides targeted practice to help you master u-substitution and succeed in your IB exams.
换元积分法是 IB 数学：分析与方法以及应用与解释中考查的核心分析技能之一。练习 11C.1 通常引入的方法是通过将 x 的函数替换为单一变量 u 来简化积分。本文讲解理论，逐步演示例题，指出常见错误，并提供有针对性的练习，帮助你掌握 u-代换并在 IB 考试中取得成功。
1. Introduction to Integration by Substitution | 换元积分法简介
In IB Mathematics, you will frequently meet integrals that cannot be tackled using standard rules like the power rule alone. Integration by substitution, also called u-substitution, transforms a complicated integrand into a simpler one by introducing a new variable. It is the reverse process of the chain rule in differentiation, and it allows us to handle compositions of functions efficiently.
在 IB 数学中，你经常会遇到无法仅用幂法则等标准规则解决的积分。换元积分法，也称为 u-代换，通过引入一个新变量将复杂的被积函数转化为更简单的形式。它是微分链式法则的逆过程，使我们能够高效地处理函数的复合。
2. The Reverse Chain Rule | 反向链式法则
Recall the chain rule: if y = f(g(x)), then dy/dx = f'(g(x)) g'(x). Integrating both sides gives ∫ f'(g(x)) g'(x) dx = f(g(x)) + C. If we set u = g(x), the differential du = g'(x) dx replaces the g'(x) dx term, and the integral becomes ∫ f'(u) du = f(u) + C. This is the essence of substitution – it undoes the chain rule.
回顾链式法则：如果 y = f(g(x))，则 dy/dx = f'(g(x)) g'(x)。对两边积分得到 ∫ f'(g(x)) g'(x) dx = f(g(x)) + C。如果我们设 u = g(x)，微分 du = g'(x) dx 替换掉 g'(x) dx 部分，积分就变成 ∫ f'(u) du = f(u) + C。这就是代换的本质——它是链式法则的逆运算。
3. Choosing the Substitution u | 选择换元 u
The main difficulty in u-substitution is picking the right u. Look for an 'inner function' ssion inside a root, a denominator, or a quantity raised to a power. For example, in ∫ 3x² √x³+1 dx, we would set u = x³+1 because du/dx = 3x², which matches the 3x² factor exactly.
u-代换的主要困难在于选择合适的 u。寻找一个“内层函数”，其导数也在积分中出现（可能相差一个常数因子）。通常，u 被选为复合函数的自变量、根号内部、分母或是被升幂的表达式。例如，在 ∫ 3x² √x³+1 dx 中，我们设 u = x³+1，因为 du/dx = 3x² 恰好与 3x² 因子匹配。
4. Step-by-Step Substitution for Indefinite Integrals | 不定积分的换元步骤
Step 1: Choose u = g(x) – pick the inner function. Step 2: Differentiate to find du/dx and write du = g'(x) dx. Step 3: Rewrite the whole integral in terms of u and du; all x's must disappear. Step 4: Integrate with respect to u. Step 5: e final answer in terms of x. Always add the constant of integration +C for indefinite integrals.
步骤 1：选择 u = g(x) – 挑选内层函数。步骤 2：求导得到 du/dx，并写出 du = g'(x) dx。步骤 3：将整个积分用 u 和 du 重新表达；所有 x 必须消失。步骤 4：对 u 积分。步骤 5：将 u = g(x) 代回，用 x 写出最终答案。对于不定积分，始终加上积分常数 +C。
5. Worked Example 1: ∫ 2x cos(x²) dx | 例题1：∫ 2x cos(x²) dx
Evaluate the integral ∫ 2x cos(x²) dx.
u, which integrates to sin(u) + C. Replacing u with x² gives the final answer sin(x²) + C.
计算积分 ∫ 2x cos(x²) dx。
设 u = x²，则 du/dx = 2x，故 du = 2x dx。积分变为 ∫ cos(u) du，积分结果为 sin(u) + C。将 u 替换为 x²，得到最终答案 sin(x²) + C。
∫ 2x cos(x²) dx = sin(x²) + C
6. Worked Example 2: ∫ x√x+1 dx | 例题2：∫ x√x+1 dx
This integral is trickier because the derivative of x+1 is simply 1, and an extra x remains. We handle this by expressing x in terms of u.
Let u = x+1. Then x = u - 1 and du = dx. The integral becomes ∫ (u - 1) √u du = ∫ (u3/2 - u1/2) du.
Integrate term-by-term: (2/5) u5/2 - (2/3) u3/2 + C. Finally, substitute back: (2/5)(x+1)5/2 - (2/3)(x+1)3/2 + C.
这个积分稍微棘手，因为 x+1 的导数仅仅是 1，而多余的 x 仍然存在。我们通过用 u 表示 x 来处理。
设 u = x+1，则 x = u - 1，且 du = dx。积分变为 ∫ (u - 1) √u du = ∫ (u3/2 - u1/2) du。
逐项积分：(2/5) u5/2 - (2/3) u3/2 + C。最后代回：(2/5)(x+1)5/2 - (2/3)(x+1)3/2 + C。
∫ x√x+1 dx = (2/5)(x+1)5/2 - (2/3)(x+1)3/2 + C
7. Substitution in Definite Integrals | 定积分中的换元
When evaluating a definite integral using substitution, you have two options. Method A (change of limits): mpute the new limits u(a) and u(b). Then integrate with respect to u and evaluate directly – no need to return to x. Method B (keep original limits): find the antiderivative in u, substitute back to x, and then use the original x-limits. Method A is often cleaner and reduces algebraic errors.
当用代换法计算定积分时，有两种选择。方法 A（改变积分限）：选择 u = g(x) 后，计算新的下限 u(a) 和上限 u(b)。然后对 u 积分并直接求值——无需返回 x。方法 B（保留原积分限）：求出用 u 表示的原函数，代回 x，然后使用原来的 x 上下限。方法 A 通常更简洁，能减少代数错误。
8. Worked Example 3: Definite Integral | 定积分例题
Evaluate ∫01 2x (x²+1)³ dx using substitution.
u=2. Differentiating gives du = 2x dx, which matches exactly. The integral transforms to ∫12 u³ du.
Integrate: [¼ u⁴]12 = ¼(16 - 1) = 15/4. No back-substitution is needed.
使用代换法计算 ∫01 2x (x²+1)³ dx。
设 u = x²+1。当 x=0 时，u=1；当 x=1 时，u=2。求导得 du = 2x dx，恰好匹配。积分转化为 ∫12 u³ du。
积分：[¼ u⁴]12 = ¼(16 - 1) = 15/4。无需代回。
∫01 2x (x²+1)³ dx = 15/4
9. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
olutions for Exercise 11C.1. First, forgetting to change the limits in a definite integral – erms of u instead of substituting back to x. Third, misidentifying du: always write du = g' must end with du or dx. Finally, beware of sign errors when handling negative derivatives.
在 IB 练习 11C.1 的解答中，几个错误经常出现。第一，定积分中忘记改变积分限——如果你在 u-积分中使用 x 值，结果将是错误的。第二，不定积分最终保留为 u 的形式，而没有代回 x。第三，错误识别 du：始终写出 du = g'(x) dx，并检查是否需要调整常数因子。第四，在重写过程中丢失微分 dx；每个积分必须以 du 或 dx 结尾。最后，处理负导数时要小心符号错误。
10. Practice Exercise 11C.1 (Sample) | 练习11C.1 样例
is typical problem mirroring Exercise 11C.1: Use the substitution u = x³ + 2 to find ∫ 6x² √x³+2 dx.
³+2, then du/dx = 3x², so du = 3x² dx. Notice that 6x² dx = 2 du. The integral becomes ∫ 2 √u du = 2 ∫ u1/2 du = 2 × (2/3) u3/2 + C = (4/3) u3/2 + C. Substitute back: (4/3)(x³+2)3/2 + C.
试试这道反映练习 11C.1 的典型问题：利用换元 u = x³ + 2 求 ∫ 6x² √x³+2 dx。
设 u = x³+2，则 du/dx = 3x²，所以 du = 3x² dx。注意到 6x² dx = 2 du。积分变为 ∫ 2 √u du = 2 ∫ u1/2 du = 2 × (2/3) u3/2 + C = (4/3) u3/2 + C。代回：(4/3)(x³+2)3/2 + C。
∫ 6x² √x³+2 dx = (4/3)(x³+2)3/2 + C
11. Tips for IB Exam Success | IB考试成功技巧
In IB exams, always show your substitution clearly. Write 'Let u = ... then du/dx = ..., so du = ... dx' lifies easily, this working earns method marks. Practice recognising classic patterns: ∫ f'(x)/f(x) dx = ln|f(x)| ps. Time yourself on Exercise 11C.1 questions to build speed without sacrificing accuracy.
在 IB 考试中，始终清晰地展示你的代换过程。写出“设 u = ... 则 du/dx = ...，故 du = ... dx”。即使积分很容易简化，这样的步骤也能获得过程分。练习识别经典模式：∫ f'(x)/f(x) dx = ln|f(x)| + C 就是直接代换 (u=f(x))。此外，处理定积分时，使用清晰的积分限变换表格以避免算术失误。对练习 11C.1 的题目计时，在保证准确度的同时提高速度。
12. Summary | 总结
Integration by substitution is a cornerstone technique in IB Mathematics. By reversing the chain rule, it turns challenging integrals into manageable forms. The key steps – choose u, find du, rewrite, integrate, and (for indefinite integrals) back-substitute – must be practised until they become automatic. Use the guidance and worked examples in this article alongside your textbook's Exercise 11C.1 to build confidence and proficiency. With consistent practice, you will handle any substitution problem the IB exam presents.
换元积分法是 IB 数学的基石技巧。通过逆用链式法则，它将棘手的积分转化为易于处理的形式。关键步骤——选择 u、计算 du、重写、积分，以及（对于不定积分）代回——必须练习到成为下意识的反应。将本文的指导和例题与教材中的练习 11C.1 结合使用，以建立信心和熟练度。通过持续练习，你将能应对 IB 考试中出现的任何代换问题。
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