📚 Exercise 26A: Mastering Basic Integration | 练习26A：掌握基本积分
Integration is a cornerstone of calculus, and Exercise 26A provides a thorough practice of basic integration techniques essential for IB Mathematics. This article will guide you through each type of integral you might encounter, with step-by-step explanations and exam-style tips.
积分是微积分的基石，练习26A为IB数学提供了基本积分技巧的全面训练。本文将带你逐一攻克可能遇到的各种积分类型，附上逐步解析与贴近考试的技巧。
1. Integration as Anti-differentiation | 积分作为反求导
Integration reverses differentiation. If F' te integral of f(x) is the family of functions F(x) + C, where C is an arbitrary constant.
积分是微分的逆运算。如果 F'(x) = f(x)，则 f(x) 的不定积分就是函数族 F(x) + C，其中 C 为任意常数。
The notation ∫ f(x) dx reads 'the integral of f of x with respect to x'. The symbol ∫ is an elongated S, representing a sum of infinitesimal areas.
记号 ∫ f(x) dx 读作“f 关于 x 的积分”。符号 ∫ 是拉长的 S，代表无穷小面积的求和。
3x² dx = x³ + C. Every differentiation formula gives a corresponding integration formula.
例如，由于 d/dx (x³) = 3x²，我们写成 ∫ 3x² dx = x³ + C。每一条微分公式都对应一条积分公式。
2. The Power Rule for Integration | 积分的幂法则
For any real number n ≠ –1, the integral of xⁿ is xⁿ⁺¹/(n+1) + C. This is the reverse of the power rule for differentiation.
对于任意实数 n ≠ –1，xⁿ 的积分是 xⁿ⁺¹/(n+1) + C。这就是微分幂法则的逆运算。
xample 1: ∫ x⁴ dx = x⁵/5 + C. Simply add 1 to the exponent and divide by the new exponent.
例1：∫ x⁴ dx = x⁵/5 + C。只需将指数加1，再除以新指数。
Example 2: ∫ √x dx = ∫ x¹ᐟ² dx = x³ᐟ² / (3/2) + C = (2/3)x³ᐟ² + C.
例2：∫ √x dx = ∫ x¹ᐟ² dx = x³ᐟ² / (3/2) + C = (2/3)x³ᐟ² + C。
e is in the denominator, rewrite it with a negative exponent: ∫ 1/x³ dx = ∫ x⁻³ dx = x⁻²/(–2) + C = –1/(2x²) + C.
若变量在分母，先用负指数改写：∫ 1/x³ dx = ∫ x⁻³ dx = x⁻²/(–2) + C = –1/(2x²) + C。
The power rule fails when n = –, because the antiderivative of x⁻¹ is the natural logarithm, which we will discuss later.
当 n = –1 时幂法则失效，因为 x⁻¹ 的原函数是自然对数，稍后会讲解。
3. Integrating Exponential Functions | 指数函数的积分
The natural exponential function is its own antiderivative: ∫ eˣ dx = eˣ + C. For a general exponential function aˣ (a > 0, a ≠ 1), the rule is ∫ aˣ dx = aˣ / ln a + C.
自然指数函数就是自身的原函数：∫ eˣ dx = eˣ + C。对于一般的指数函数 aˣ (a > 0, a ≠ 1)，积分公式为 ∫ aˣ dx = aˣ / ln a + C。
Example: ∫ 5eˣ dx = 5eˣ + C. Constants factor out of the integral: ∫ k·f(x) dx = k ∫ f(x) dx.
例：∫ 5eˣ dx = 5eˣ + C。常数可以提到积分号外：∫ k·f(x) dx = k ∫ f(x) dx。
When the exponent is a linear function, use a simple adjustment: ∫ eᵏˣ dx = (1/k)eᵏˣ + C. For instance, ∫ e²ˣ dx = (1/2)e²ˣ + C.
当指数为线性函数时，作简单调整：∫ eᵏˣ dx = (1/k)eᵏˣ + C。例如，∫ e²ˣ dx = (1/2)e²ˣ + C。
Be careful: ∫ eˣ sed in terms of elementary functions; this is a common trick in multiple-choice questions.
注意：∫ eˣ² dx 无法用初等函数表示，这是选择题中常见的陷阱。
4. Trigonometric Integrals | 三角函数积分
The six basic trigonometric integrals that appear in IB Mathematics are summarised in the table below.
IB数学中出现的六个基本三角函数积分总结于下表。
| f(x) | ∫ f(x) dx |
|---|---|
| sin x | –cos x + C |
| cos x | sin x + C |
| sec² x | tan x + C |
| csc² x | –cot x + C |
| sec x tan x | sec x + C |
| csc x cot x | –csc x + C |
记住这些积分的关键，是要意识到它们正是相应三角函数导数的逆向结果。
当处理 sin(kx) 或 cos(kx) 时，除以 k：∫ sin(3x) dx = –(1/3)cos(3x) + C；∫ cos(2x) dx = (1/2)sin(2x) + C。
For squared sine or cosine, use double-angle identities: ∫ sin² x dx = ∫ (1 – cos 2x)/2 dx = x/2 – (1/4)sin 2x + C.
对于正弦或余弦的平方，使用倍角公式：∫ sin² x dx = ∫ (1 – cos 2x)/2 dx = x/2 – (1/4)sin 2x + C。
5. Integrals Leading to Logarithms | 导出对数函数的积分
The special case n = –1 from the power rule gives ∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C. The absolute value ensures the logarithm is defined for negative x as well.
幂法则中 n = –1 的特殊情况给出 ∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C。绝对值保证了当 x 为负时对数仍有定义。
A more general form is ∫ f'(x)/f(x) dx = ln|f(x)| + C. This is extremely useful for recognising logarithmic integrals.
更一般的形式是 ∫ f'(x)/f(x) dx = ln|f(x)| + C。这非常有利于识别对数型积分。
x)/(x² + 1) dx. Notice that the derivative of the denominator is 2x, so the integral is ln|x² + 1| + C = ln(x² + 1) + C (since x²+1 > 0).
例：∫ (2x)/(x² + 1) dx。注意到分母的导数是 2x，因此积分为 ln|x² + 1| + C = ln(x² + 1) + C（因为 x²+1 > 0）。
Another example: ∫ tan x dx = ∫ sin x / cos x dx = –ln|cos x| + C = ln|sec x| + C.
另一个例子：∫ tan x dx = ∫ sin x / cos x dx = –ln|cos x| + C = ln|sec x| + C。
6. Integration by Substitution | 换元积分法
bstitution) simplifies an integrand by introducing a new variable u = g(x), so that du = g'(x) dx.
换元积分法（常称为 u 替换）通过引入新变量 u = g(x)，使得 du = g'(x) dx，从而简化被积函数。
e whole integral in terms of u. (4) Integrate with respect to u. (5) Substitute back to x.
步骤：(1) 选取 u = 内层函数。(2) 计算 du/dx，写成 du = 导数 · dx。(3) 把整个积分用 u 表示。(4) 对 u 积分。(5) 回代成 x 的表达式。
x²) dx. Let u = x², du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x²) + C.
例：∫ 2x·cos(x²) dx。令 u = x², du = 2x dx。积分化为 ∫ cos u du = sin u + C = sin(x²) + C。
In definite integrals, remember to change the limits of integration when performing a substitution. If x goes from a to b, then u goes from g(a) to g(b).
对于定积分，换元时务必变更积分上下限。若 x 从 a 到 b，则 u 从 g(a) 到 g(b)。
7. Definite Integrals and the Fundamental Theorem | 定积分与微积分基本定理
ndamental Theorem of Calculus connects differentiation and integration: ∫ₐᵇ f(x) dx = F(b) – F(a), where F is any antiderivative of f.
微积分基本定理将微分与积分联系起来：∫ₐᵇ f(x) dx = F(b) – F(a)，其中 F 是 f 的任一原函数。
To evaluate ∫₀² 3x² dx, first find the antiderivative x³, then compute x³|₀² = 2³ – 0³ = 8.
计算 ∫₀² 3x² dx，先求原函数 x³，再计算 x³|₀² = 2³ – 0³ = 8。
Properties of definite integrals: reversing limits changes the sign ∫ₐᵇ f(x) dx = –∫ₐᵇ? corrected: ∫ₐᵇ f(x) dx = –∫ᵇₐ f(x) dx. Also, integrals over adjacent intervals add: ∫ₐᶜ f = ∫ₐᵇ f + ∫ᵇᶜ f.
定积分的性质：颠倒上下限会变号 ∫ₐᵇ f(x) dx = –∫ᵇₐ f(x) dx。相邻区间上的积分可加：∫ₐᶜ f = ∫ₐᵇ f + ∫ᵇᶜ f。
Be cautious when the integrand changes sign within the interval; the definite integral represents net area, not total area.
当被积函数在区间内变号时要小心；定积分表示净面积，而非总面积。
8. Area Under a Curve | 曲线下的面积
[a,b], the area enclosed by y = f(x), the x-axis, and x = a, x = b is exactly ∫ₐᵇ f(x) dx.
若在 [a,b] 上 f(x) ≥ 0，则由 y = f(x)、x 轴以及 x = a, x = b 所围的面积恰好是 ∫ₐᵇ f(x) dx。
area, split the interval where f(x) changes sign and use absolute values: Total area = ∫ₐᵇ |f(x)| dx.
当曲线落到 x 轴下方时，积分会给出负值。要求总面积，需按 f(x) 变号处分割区间并使用绝对值：总面积 = ∫ₐᵇ |f(x)| dx。
Example: Find the area bounded by y = x² – 4 and the x-axis between x = –2 and x = 2. The function is negative on (–2,2). Area = –∫₋₂² (x² – 4) dx = [–(x³/3) + 4x]₋₂² = ( –8/3 + 8 ) – ( 8/3 – 8 ) = 32/3.
例：求 y = x² – 4 与 x 轴在 x = –2 到 x = 2 之间围成的面积。该函数在 (–2,2) 内为负。面积 = –∫₋₂² (x² – 4) dx = [–(x³/3) + 4x]₋₂² = 32/3。
Area between two curves y = f(x) and y = g(x) from a to b, with f(x) ≥ g(x), is ∫ₐ
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