📚 Basic Integration Rules and Operations | 积分基本运算法则归纳
Integration is one of the two central operations in calculus, alongside differentiation. In the IB Mathematics curriculum—whether you are studying Analysis and Approaches (AA) or Applications and Interpretation (AI)—a solid command of integration techniques is essential for solving problems in kinematics, area, volume, and many other applied contexts. This article systematically summarises the fundamental rules and operations of integration that you must master for your IB exams.
积分是微积分学中与微分并列的两大核心运算之一。无论你学习的是IB数学中的分析与方法（AA）还是应用与解释（AI），熟练掌握积分技巧都是解决运动学、面积、体积以及众多应用问题的关键。本文系统归纳了IB考试中你必须掌握的积分基本运算法则与操作。
1. The Indefinite Integral as Antiderivative | 不定积分与原函数
x) is the family of all antiderivatives of f. It is written as ∫f(x)dx = F(x) + C, where F' cause differentiating any constant gives zero, so the derivative of F(x) + C is also f(x).
函数 f(x) 的不定积分是 f 的所有原函数的集合，记作 ∫f(x)dx = F(x) + C，其中 F'(x) = f(x)，C 为积分常数。常数 C 必不可少，因为任何常数的导数为零，所以 F(x) + C 的导数同样是 f(x)。
ocess of integration is therefore the reverse of differentiation, and verification of any integration result can be achieved by differentiating the answer.
例如，由于 x³ 的导数为 3x²，因此 ∫3x²dx = x³ + C。积分过程因此是微分的逆运算，任何积分结果都可以通过求导来验证。
2. Basic Power Rule | 基本幂法则
The power rule for integration is the inverse of the power rule for differentiation. For any real number n ≠ −1,
积分中的幂法则是微分幂法则的逆运算。对于任意实数 n ≠ −1：
∫xⁿ dx = xⁿ⁺¹ / (n + 1) + C
hen n = −1, the integrand is 1/x, and the integral is the natural logarithm: ∫(1/x)dx = ln|x| + C. The absolute value ensures the logarithm is defined for both positive and negative x.
该法则适用于除 n = −1 以外的所有实数指数。当 n = −1 时，被积函数为 1/x，其积分为自然对数：∫(1/x)dx = ln|x| + C。绝对值符号确保了在对数在 x 为正或负时均有定义。
Common examples include ∫x⁵dx = x⁶/6 + C, ∫√x dx = ∫x1/2dx = (2/3)x3/2 + C, and ∫(1/x²)dx = ∫x⁻²dx = −x⁻¹ + C = −1/x + C.
常见例子包括 ∫x⁵dx = x⁶/6 + C，∫√x dx = ∫x1/2dx = (2/3)x3/2 + C，以及 ∫(1/x²)dx = ∫x⁻²dx = −x⁻¹ + C = −1/x + C。
3. Linearity of Integration | 积分的线性运算
nstant. Second, the integral of a sum or difference: ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx.
积分是线性算子，满足两个关键性质。第一，常数倍法则：∫k·f(x)dx = k·∫f(x)dx，其中 k 为任意常数。第二，和差法则：∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx。
These properties allow us to break complicated integrals into simpler parts. For example:
这些性质允许我们将复杂积分拆分为更简单的部分。例如：
∫(4x³ − 2x + 7)dx = 4·(x⁴/4) − 2·(x²/2) + 7x + C = x⁴ − x² + 7x + C
Note that we only need to write one constant C at the end, not one per term, since the sum of several arbitrary constants is still an arbitrary constant.
注意：我们只需在最后写一个常数 C，而不必每项都写，因为多个任意常数的和仍然是任意常数。
4. Integrating Standard Functions | 基本初等函数的积分
Beyond powers, the IB syllabus requires fluency in integrating a set of standard elementary functions. These should be memorised:
除幂函数外，IB 教学大纲还要求熟练掌握一组基本初等函数的积分，这些公式需要牢记：
| Function | Integral | 函数 | 积分 |
| sin x | −cos x + C | sin x | −cos x + C |
| cos x | sin x + C | cos x | sin x + C |
| eˣ | eˣ + C | eˣ | eˣ + C |
| 1/x | ln|x| + C | 1/x | ln|x| + C |
| sec²x | tan x + C | sec²x | tan x + C |
For functions of the form sin(kx), cos(kx), or ekx, a chain-rule reversal applies:
对于形如 sin(kx)、cos(kx) 或 ekx 的函数，可以应用链式法则的逆运算：
∫sin(kx)dx = −(1/k)cos(kx) + C, ∫cos(kx)dx = (1/k)sin(kx) + C, ∫ekxdx = (1/k)ekx + C
5. The Substitution Method | 换元积分法
Substitution is the integral counterpart of the chain rule in differentiation. When the integrand contains a composite function multiplied by the derivative of its inner function, we use substitution to simplify.
换元积分法是微分中链式法则在积分中的对应。当被积函数包含复合函数乘以其内层函数的导数时，我们使用换元法进行化简。
The general procedure is: let u = g(x), so du = g'(x)dx. Then ∫f(g(x))·g' After integrating with respect to u, substitute back to express the answer in terms of x.
一般步骤如下：令 u = g(x)，则 du = g'(x)dx。于是 ∫f(g(x))·g'(x)dx = ∫f(u)du。对 u 积分后，将结果代回 x 的表达式。
x²) + C. The key is to identify the inner function and ensure its derivative appears in the integrand.
例如，计算 ∫2x·cos(x²)dx 时，令 u = x²，则 du = 2xdx。积分变为 ∫cos(u)du = sin(u) + C = sin(x²) + C。关键在于识别内层函数，并确认其导数出现在被积函数中。
6. Integration by Parts | 分部积分法
Integration by parts is derived from the product rule and is used to integrate products of functions. The formula in IB is:
分部积分法由乘积极导出，用于对函数乘积进行积分。IB 中的公式为：
∫u dv = uv − ∫v du
A helpful mnemonic for choosing u is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u as the function that appears earliest in this list, and dv as the remaining part.
选择 u 的常用口诀是 LIATE：对数的（Logarithmic）、反三角的（Inverse trigonometric）、代数的（Algebraic）、三角的（Trigonometric）、指数的（Exponential）。选择列表中靠前的函数作为 u，其余部分作为 dv。
For example, to compute ∫x·eˣdx, choose u = x (algebraic) and dv = eˣdx. Then du = dx and v = eˣ. Applying the formula: ∫x·eˣdx = x·eˣ − ∫eˣdx = x·eˣ − eˣ + C = eˣ(x − 1) + C.
例如，计算 ∫x·eˣdx 时，取 u = x（代数函数），dv = eˣdx。则 du = dx，v = eˣ。应用公式得：∫x·eˣdx = x·eˣ − ∫eˣdx = x·eˣ − eˣ + C = eˣ(x − 1) + C。
7. The Definite Integral and the Fundamental Theorem | 定积分与微积分基本定理
x = b. The Fundamental Theorem of Calculus links the definite integral to antiderivatives:
定积分 ∫ₐᵇf(x)dx 表示曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的有向面积。微积分基本定理将定积分与原函数联系起来：
∫ₐᵇf(x)dx = F(b) − F(a)
where F is any antiderivative of f. This means we evaluate the antiderivative at the upper limit, subtract its value at the lower limit, and there is no constant of integration in the final answer.
其中 F 是 f 的任意一个原函数。这意味着我们需要计算原函数在上限的值，减去其在下限的值，最终答案中不包含积分常数。
For instance, ∫₀¹x²dx = [x³/3]₀¹ = 1³/3 − 0³/3 = 1/3. The definite integral's value is a number, not a family of functions.
例如，∫₀¹x²dx = [x³/3]₀¹ = 1³/3 − 0³/3 = 1/3。定积分的值是一个确定的数，而非一族函数。
8. Properties of Definite Integrals | 定积分的运算法则
Several properties of definite integrals are frequently tested in IB papers. Mastering them simplifies computation significantly.
定积分有若干常用性质，在 IB 考试中经常出现。掌握这些性质可以显著简化计算。
ₐᵇf(x)dx = −∫ᵦₐf(x)dx. Reversing the order of the limits changes the sign of the integral.
交换上下限：∫ₐᵇf(x)dx = −∫ᵦₐf(x)dx。交换积分上下限会改变积分的符号。
Splitting the interval: ∫ₐᵇf(x)dx = ∫ₐᶜf(x)dx + ∫꜀ᵇf(x)dx, for any c in the interval [a, b].
区间可加性：∫ₐᵇf(x)dx = ∫ₐᶜf(x)dx + ∫꜀ᵇf(x)dx，其中 c 是区间 [a, b] 内任意一点。
ro-length interval: ∫ₐₐf(x)dx = 0. Integrating over an interval of zero length gives zero.
零长度区间：∫ₐₐf(x)dx = 0。在长度为零的区间上积分为零。
f(x), then ∫₋ₐᵃf(x)dx = 2∫₀ᵃf(x)dx. If f is odd, i.e., f(−x) = −f(x), then ∫₋ₐᵃf(x)dx = 0.
奇函数与偶函数：若 f 为偶函数，即 f(−x) = f(x)，则 ∫₋ₐᵃf(x)dx = 2∫₀ᵃf(x)dx。若 f 为奇函数，即 f(−x) = −f(x)，则 ∫₋ₐᵃf(x)dx = 0。
9. Area between Curves | 曲线之间的面积
between two curves. If f(x) ≥ g(x) on the interval [a, b], the area between the curves is:
定积分的一个重要应用是计算两条曲线之间围成的面积。如果在区间 [a, b] 上 f(x) ≥ g(x)，则两曲线之间的面积为：
A = ∫ₐᵇ [f(x) − g(x)] dx
This formula gives a positive result since the upper curve is subtracted by the lower curve. When curves intersect, the interval must be split at each intersection point, and the appropriate difference must be calculated for each sub-interval.
该公式始终给出正值，因为是用上方曲线减去下方曲线。当曲线相交时，需要在每个交点处分割区间，并在每个子区间中计算正确的被减表达式。
1, note that x ≥ x² on [0, 1]. Thus A = ∫₀¹(x − x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
例如，求 y = x² 与 y = x 在 x = 0 到 x = 1 之间围成的面积。在 [0, 1] 上 x ≥ x²，因此 A = ∫₀¹(x − x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6。
10. Common Pitfalls and Examination Tips | 常见误区与考试技巧
Students frequently make avoidable errors in integration. Being aware of these pitfalls will help you secure full marks on exam questions.
学生在积分中经常犯一些可以避免的错误。注意这些常见陷阱能帮助你在考试中拿到满分。
Forgetting the constant of integration C in indefinite integrals: Without C, the answer is incomplete and marks will be deducted.
忘记不定积分中的积分常数 C：缺少 C 的答案是不完整的，考试中会被扣分。
er rule to n = −1: The formula ∫xⁿdx = xⁿ⁺¹/(n+1) is invalid for n = −1. Use ∫(1/x)dx = ln|x| + C instead.
对 n = −1 使用幂法则：公式 ∫xⁿdx = xⁿ⁺¹/(n+1) 不适用于 n = −1，此时应使用 ∫(1/x)dx = ln|x| + C。
Forgetting to substitute the limits in definite integrals when using substitution: Convert the limits of integration when you change the variable, or substitute back to x before evaluating.
使用换元法时忘记转换定积分的上下限：修改变量时需要同时转换积分上下限，或者先代回 x 再求值。
the minus sign in ∫sin(x)dx: The integral of sine is negative cosine, not positive cosine.
忽略 ∫sin(x)dx 中的负号：正弦的积分是负余弦，而不是正余弦。
Confusing (ln x) with ln|x|: The required answer for 1/x is ln|x| + C, since the domain of 1/x excludes zero.
混淆 ln x 与 ln|x|：1/x 的积分应为 ln|x| + C，因为 1/x 的定义域不包括零。
Always verify your answer by differentiation. If the derivative of your result equals the original integrand, your integration is correct. This simple check takes only seconds and can prevent many errors.
始终通过求导来验证你的答案。如果结果的导数等于原被积函数，则你的积分是正确的。这个简单检查只需几秒钟，可以避免许多错误。
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