Exercise 12I | 练习12I

📚 Exercise 12I | 练习12I

Welcome to the focused revision guide for Exercise 12I from the IB Mathematics: Analysis and Approaches HL course. This exercise is designed to deepen your command of integration techniques, including substitution, integration by parts, partial fractions, and special trigonometric methods. Each problem in Exercise 12I tests your ability to select and apply the right strategy under time pressure—exactly what is required in Paper 1 and Paper 2. In this article, we will work through representative questions, highlight common pitfalls, and reinforce effective problem-solving habits. Let’s turn integration into one of your strongest topics.

欢迎来到IB数学：分析与方法HL课程中练习12I的专项复习指南。该练习旨在强化你对积分技巧的掌握，包括换元法、分部积分法、部分分式法和专门的三角方法。练习12I中的每一道题都考验你在时间压力下选择并运用正确策略的能力——这正是试卷一和试卷二所要求的。本文将通过典型题目进行讲解，指出常见错误，并强化有效的解题习惯。让积分成为你最有把握的主题之一。

---

1. Revision of Basic Integration Rules | 基本积分法则回顾

Before tackling the exercises, ensure you can instantly recall the fundamental antiderivatives. These are the building blocks for every advanced method.

在开始做练习之前，确保你能立刻回忆起基本的原函数。它们是所有高级方法的基础构件。

Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1.

幂函数法则：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，当 n ≠ −1。

Reciprocal: ∫ 1/x dx = ln|x| + C.

倒数积分：∫ 1/x dx = ln|x| + C。

Exponential: ∫ eˣ dx = eˣ + C; ∫ aˣ dx = aˣ/ln a + C.

指数函数：∫ eˣ dx = eˣ + C；∫ aˣ dx = aˣ/ln a + C。

Trigonometric: ∫ sin x dx = −cos x + C; ∫ cos x dx = sin x + C; ∫ sec² x dx = tan x + C.

三角函数：∫ sin x dx = −cos x + C；∫ cos x dx = sin x + C；∫ sec² x dx = tan x + C。

: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx and ∫ k·f(x) dx = k·∫ f(x) dx for constant k.

线性性质：∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx，且对于常数 k 有 ∫ k·f(x) dx = k·∫ f(x) dx。

---

2. Integration by Substitution – Simple Cases | 简单换元积分

Substitution is the chain rule in reverse. Look for a function and its derivative appearing together.

换元法就是链式法则的逆运算。留意同时出现的函数及其导数。

Example 1: Find ∫ 2x·cos(x²) dx.

例题1：计算 ∫ 2x·cos(x²) dx。

so du = 2x dx. The integral becomes ∫ cos u du = sin u + C. Substitute back: sin(x²) + C.

令 u = x²，则 du/dx = 2x，因此 du = 2x dx。原积分变为 ∫ cos u du = sin u + C。代回得 sin(x²) + C。

Example 2: Evaluate ∫ e³ˣ⁺¹ dx.

例题2：计算 ∫ e³ˣ⁺¹ dx。

Set u = 3x + 1, so du = 3 dx, giving dx = du/3. ∫ e³ˣ⁺¹ dx = (1/3)∫ eᵘ du = (1/3)eᵘ + C = (1/3)e³ˣ⁺¹ + C.

设 u = 3x + 1，则 du = 3 dx，得到 dx = du/3。∫ e³ˣ⁺¹ dx = (1/3)∫ eᵘ du = (1/3)eᵘ + C = (1/3)e³ˣ⁺¹ + C。

---

3. Integration by Substitution – Definite Integrals | 定积分的换元

For definite integrals, change the limits to avoid back-substitution.

对于定积分，改变上下限以避免代回原变量。

Example: Compute ∫₀^√π x·cos(x²) dx.

例题：计算 ∫₀^√π x·cos(x²) dx。

dx = du/2. The integral becomes (1/2) ∫₀^π cos u du = (1/2)[sin u]₀^π = (1/2)(0 − 0) = 0.

令 u = x²。当 x = 0 时，u = 0；当 x = √π 时，u = π。且 du = 2x dx，故 x dx = du/2。积分变为 (1/2) ∫₀^π cos u du = (1/2)[sin u]₀^π = (1/2)(0 − 0) = 0。

---

4. Integration by Parts – Basic | 分部积分法基础

he function that simplifies when differentiated, and dv to be something easily integrated.

分部积分法使用公式 ∫ u dv = uv − ∫ v du。选择微分后简化的函数作为 u，容易积分的部分作为 dv。

∫ u dv = uv − ∫ v du

∫ u dv = uv − ∫ v du

Example: Evaluate ∫ x·eˣ dx.

例题：计算 ∫ x·eˣ dx。

Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Applying the formula: ∫ x·eˣ dx = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C.

令 u = x，dv = eˣ dx。则 du = dx，v = eˣ。代入公式：∫ x·eˣ dx = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C。

---

5. Integration by Parts – Repeated Application | 重复应用分部积分

When a power of x multiplies an exponential or trigonometric function, you may need to apply integration by parts twice.

当 x 的幂次与指数函数或三角函数相乘时，可能需要使用两次分部积分。

Example: Find ∫ x²·eˣ dx.

例题：求 ∫ x²·eˣ dx。

First, let u = x², dv = eˣ dx ⇒ du = 2x dx, v = eˣ. ∫ x²·eˣ dx = x²·eˣ − ∫ 2x·eˣ dx. Now apply parts again to ∫ 2x·eˣ dx: let u = 2x, dv = eˣ dx, giving ∫ 2x·eˣ dx = 2x·eˣ − 2∫ eˣ dx = 2x·eˣ − 2eˣ. Combine: x²·eˣ − (2x·eˣ − 2eˣ) + C = eˣ(x² − 2x + 2) + C.

第一次，令 u = x²，dv = eˣ dx ⇒ du = 2x dx，v = eˣ。∫ x²·eˣ dx = x²·eˣ − ∫ 2x·eˣ dx。再对 ∫ 2x·eˣ dx 使用分部积分：令 u = 2x，dv = eˣ dx，得 ∫ 2x·eˣ dx = 2x·eˣ − 2∫ eˣ dx = 2x·eˣ − 2eˣ。合并：x²·eˣ − (2x·eˣ − 2eˣ) + C = eˣ(x² − 2x + 2) + C。

---

广告 / Sponsored

6. Integration Using Partial Fractions | 用部分分式积分

When integrating rational functions, decompose the fraction into simpler terms if the denominator factorises.

积分有理函数时，若分母可分解，将分式拆分成更简单的部分。

Example: Evaluate ∫ (2x+1)/(x²+x−2) dx.

例题：计算 ∫ (2x+1)/(x²+x−2) dx。

x=1: 3 = 3B → B=1. Set x=−2: −3 = −3A → A=1. Thus integral = ∫ [1/(x+2) + 1/(x−1)] dx = ln|x+2| + ln|x−1| + C = ln|(x+2)(x−1)| + C.

分母因式分解：x²+x−2 = (x+2)(x−1)。设 (2x+1)/[(x+2)(x−1)] = A/(x+2) + B/(x−1)。通分：2x+1 = A(x−1) + B(x+2)。令 x=1：3 = 3B → B=1；令 x=−2：−3 = −3A → A=1。因此积分 = ∫ [1/(x+2) + 1/(x−1)] dx = ln|x+2| + ln|x−1| + C = ln|(x+2)(x−1)| + C。

---

7. Trigonometric Integrals | 三角函数的积分

Use trigonometric identities to simplify integrands like sin²x, cos²x, or products of sines and cosines.

利用三角恒等式化简被积函数，如 sin²x、cos²x 或正弦与余弦的乘积。

Example 1: n ∫ sin²x dx = (1/2)∫ (1 − cos 2x) dx = (1/2)[x − (1/2)sin 2x] + C = x/2 − (sin 2x)/4 + C.

例题1：∫ sin²x dx。使用恒等式 sin²x = (1 − cos 2x)/2。则 ∫ sin²x dx = (1/2)∫ (1 − cos 2x) dx = (1/2)[x − (1/2)sin 2x] + C = x/2 − (sin 2x)/4 + C。

Example 2: , du = −sin x dx. Then integral = −∫ (1 − u²) du = −u + u³/3 + C = −cos x + (cos³x)/3 + C.

例题2：∫ sin³x dx。改写 sin³x = sin x·sin²x = sin x(1 − cos²x)。令 u = cos x，du = −sin x dx。则积分 = −∫ (1 − u²) du = −u + u³/3 + C = −cos x + (cos³x)/3 + C。

---

8. Integration by Trigonometric Substitution | 三角换元积分

For integrands containing √a² − x², √a² + x², or √x² − a², trigonometric substitution simplifies the radical.

对于含有 √a² − x²、√a² + x² 或 √x² − a² 的被积函数，三角换元可以消去根号。

Example: Evaluate ∫ dx / √9 − x². Let x = 3 sin θ, so dx = 3 cos θ dθ, and √9 − x² 3 cos θ) dθ = ∫ 1 dθ = θ + C. Since sin θ = x/3, θ = arcsin(x/3). Result: arcsin(x/3) + C.

例题：计算 ∫ dx / √9 − x²。设 x = 3 sin θ，则 dx = 3 cos θ dθ，且 √9 − x² = 3 cos θ。积分变为 ∫ (3 cos θ)/(3 cos θ) dθ = ∫ 1 dθ = θ + C。由 sin θ = x/3，得 θ = arcsin(x/3)。结果为 arcsin(x/3) + C。

---

9. Mixed Practice Problems | 混合练习

In Exercise 12I, questions often require you to choose the best technique from several options. Here are three quick mixed examples with hints.

在练习12I中，题目经常要求从多种技巧中选出最佳方法。下面三个快速混合例题并附提示。

Problem A: nt: use integration by parts with u = ln x, dv = x dx. You will get (x²/2)ln x − x²/4 + C.

题目A：∫ x·ln x dx。提示：用分部积分法，令 u = ln x，dv = x dx。得到 (x²/2)ln x − x²/4 + C。

Problem B: dx. Hint: note that denominator is (x−2)²; use substitution u = x−2, then split numerator.

题目B：∫ (3x−1)/(x²−4x+4) dx。提示：注意分母为 (x−2)²；用换元 u = x−2，然后拆分分子。

Problem C: t: use u = sin x, change limits from x=0→u=0, x=π/2→u=1; integral becomes ∫₀¹ u² du = 1/3.

题目C：∫₀^π/² cos x·sin²x dx。提示：设 u = sin x，变更上下限：x=0→u=0，x=π/2→u=1；积分变为 ∫₀¹ u² du = 1/3。

---

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Even strong calculus students lose marks on Exercise 12I by missing these details. Avoid them systematically.

即使微积分很强的学生也会因漏掉这些细节而在练习12I中失分。请系统地避免这些错误。

Forgetting the constant +C: indefinite integrals must include the constant of integration. In an exam, dropping +C costs a mark every time.

忘记常数 +C：不定积分必须包含积分常数。考试中，每次漏写 +C 都会被扣分。

Not adjusting limits in definite substitution: when you change the variable, you must change the limits accordingly; otherwise the numerical answer will be wrong.

定积分换元时不调整上下限：更换变量时，必须相应地改变积分限；否则数值答案将出错。

Incorrect choice of u in integration by parts: a poor choice of u can make the integral harder rather than simpler. Use the LIATE rule as a guide: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential – choose u higher in the list.

分部积分中 u 选择不当：选择不当的 u 会使积分更复杂，而非更简单。可使用 LIATE 法则：对数、反三角、代数、三角、指数——选择列表中较前的函数作为 u。

Misapplying trigonometric identities: dentities like sin²x = (1−cos 2x)/2; using the wrong sign or coefficient is a common slip.

错误使用三角恒等式：仔细检查如 sin²x = (1−cos 2x)/2 这类恒等式；符号或系数出错是常见失误。

Exam tip: When you finish a problem, differentiate your answer to check it matches the original integrand. This quick mental check can save you several marks.

考试技巧：做完一道题后，对你的答案求导，检查是否等于原被积函数。这个快速的心算检验能帮你避免丢分。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）