📚 pdfjoiner_(4)-010 | 数学复习：积分技巧
Welcome to this concise yet comprehensive revision guide on integration techniques for Edexcel A-Level Mathematics. Mastering integration is essential for tackling a wide range of problems, from finding areas under curves to solving differential equations. This article will walk you through substitution, integration by parts, partial fractions, and more, with clear examples and paired English-Chinese explanations.
欢迎阅读本简明全面的 Edexcel A-Level 数学积分技巧复习指南。掌握积分对于解决从曲线下面积到微分方程的各种问题至关重要。本文将通过代换法、分部积分法、部分分式法等技巧，配以清晰的例子和中英文对照解释，带您逐一掌握。
1. Basic Integration Rules and Standard Integrals | 基本积分规则与标准积分
el, the reverse of differentiation gives: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1), ∫ 1/x dx = ln|x| + C, ∫ eˣ dx = eˣ -cos x + C. Remember to always add the constant of integration C for indefinite integrals.
在深入学习高级技巧之前，请确保你熟悉基本的积分结果。对于 Edexcel A-Level，微分的逆运算给出：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C（n ≠ -1），∫ 1/x dx = ln|x| + C，∫ eˣ dx = eˣ + C，以及 ∫ sin x dx = -cos x + C。对于不定积分，请务必加上积分常数 C。
2. Integration by Substitution (u-substitution) | 代换积分法（u-代换）
function multiplied by the derivative of its inner function. Set u = g(x), then du/dx = g'(x), so du = g'(x) dx. The integral ∫ f(g(x)) g' sin u + C = sin(x² + 1) + C. Always replace u back with the original variable at the end.
当代积函数为复合函数乘以其内层函数的导数时，可使用代换法。设 u = g(x)，则 du/dx = g'(x)，从而 du = g'(x) dx。积分 ∫ f(g(x)) g'(x) dx 变为 ∫ f(u) du。例如，求 ∫ 2x cos(x² + 1) dx，令 u = x² + 1，则 du = 2x dx。积分转化为 ∫ cos u du = sin u + C = sin(x² + 1) + C。最后务必将 u 换回原变量。
3. Integrating Rational Functions using Partial Fractions | 使用部分分式积分有理函数
ample, (3x + 1)/((x - 2)(x + 1)) can be written as A/(x - 2) + B/(x + 1). Solve for A and B, then integrate each term separately: ∫ A/(x - 2) dx + ∫ B/(x + 1) dx = A ln|x - 2| + B ln|x + 1| + C. If the numerator' degree is equal to or higher than the denominator, perform polynomial long division first.
当积分形式为 P(x)/Q(x) 的有理函数且分子次数小于分母次数时，可将分式表达为若干简单分式之和。例如，(3x + 1)/((x - 2)(x + 1)) 可写为 A/(x - 2) + B/(x + 1)。解出 A 和 B，然后逐项积分：∫ A/(x - 2) dx + ∫ B/(x + 1) dx = A ln|x - 2| + B ln|x + 1| + C。如果分子次数等于或高于分母，需先进行多项式长除法。
4. Integration by Parts | 分部积分法
u dv = uv - ∫ v du, where u and dv are chosen appropriately. Typically, choose u as a function that simplifies on differentiation (like ln x or xⁿ) and dv as a function that is easy to integrate (like eˣ, sin x). For example, ∫ x eˣ dx: let u = x (so du = dx), dv = eˣ dx (so v = eˣ). Then ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C. The LIATE rule (Log, Inverse trig, Algebraic, Trig, Exponential) helps decide u.
分部积分法基于微分的乘法法则，适用于被积函数为两个函数乘积的情况。公式为 ∫ u dv = uv - ∫ v du，其中需合理选择 u 和 dv。通常，选择求导后简化的函数作为 u（如 ln x 或 xⁿ），选择易于积分的函数作为 dv（如 eˣ、sin x）。例如，∫ x eˣ dx：令 u = x（则 du = dx），dv = eˣ dx（则 v = eˣ）。那么 ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C。LIATE 法则（对数、反三角、代数、三角、指数）有助于确定 u 的选择。
5. Integration of Trigonometric Functions | 三角函数的积分
os 2x)/2 to rewrite the integrand: ∫ sin² x dx = ∫ (1/2) dx - ∫ (1/2) cos 2x dx = x/2 - (1/4) sin 2x + C. Similarly, cos² x = (1 + cos 2x)/2. For odd powers like ∫ sin³ x dx, factor out sin x and use the identity cos² x + sin² x = 1 to substitute.
除了基本的 sin 和 cos 积分，你还需要处理三角函数的幂次和乘积。对于 ∫ sin² x dx，使用恒等式 sin² x = (1 - cos 2x)/2 改写被积函数：∫ sin² x dx = ∫ (1/2) dx - ∫ (1/2) cos 2x dx = x/2 - (1/4) sin 2x + C。类似地，cos² x = (1 + cos 2x)/2。对于奇数次幂如 ∫ sin³ x dx，提取出一个 sin x 并利用恒等式 cos² x + sin² x = 1 进行代换。
6. Integration using Trigonometric Identities | 利用三角恒等式积分
m: ∫ (1/2) sin 5x dx = -(1/10) cos 5x, and ∫ (1/2) sin x dx = -(1/2) cos x. So the original integral becomes -(1/10) cos 5x - (1/2) cos x + C. Familiarise yourself with the identities: 2 sin A cos B = sin(A+B) + sin(A-B), 2 cos A cos B = cos(A-B) + cos(A+B), and 2 sin A sin B = cos(A-B) - cos(A+B).
像 sin mx cos nx 这样的乘积需要用到积化和差公式。例如，sin 3x cos 2x = (1/2)[sin(5x) + sin x]。然后逐项积分：∫ (1/2) sin 5x dx = -(1/10) cos 5x，∫ (1/2) sin x dx = -(1/2) cos x。因此原积分变为 -(1/10) cos 5x - (1/2) cos x + C。请熟悉以下恒等式：2 sin A cos B = sin(A+B) + sin(A-B)，2 cos A cos B = cos(A-B) + cos(A+B)，以及 2 sin A sin B = cos(A-B) - cos(A+B)。
7. Definite Integration and Area Problems | 定积分与面积问题
is within the interval; areas below the axis contribute negative values when integrating directly. To find the total area enclosed, split the integral at the roots and take absolute values, or integrate the absolute value of the function. Additionally, area between two curves y = f(x) and y = g(x) from a to b is ∫ₐᵇ (f(x) - g(x)) dx, assuming f(x) ≥ g(x) on [a, b].
计算定积分 ∫ₐᵇ f(x) dx 时，先找到原函数 F(x)，然后计算 F(b) - F(a)。务必检查函数在区间内是否跨越 x 轴；直接积分时，x 轴下方的区域会贡献负值。要计算所围总面积，应在根处拆分积分并取绝对值，或者直接对函数的绝对值进行积分。此外，两条曲线 y = f(x) 与 y = g(x) 在 a 到 b 之间的面积由 ∫ₐᵇ (f(x) - g(x)) dx 给出，前提是在 [a, b] 上 f(x) ≥ g(x)。
8. Solving Differential Equations by Integration | 通过积分求解微分方程
For example, dy/dx = xy can be rewritten as (1/y) dy = x dx, then integrate both sides: ln|y| = x²/2 + C, so y = ± eC ex²/2 = A ex²/2 to find the constant A. This technique links integration directly to modelling real-world situations.
A-Level 数学中的许多一阶微分方程都可通过分离变量法求解。例如，dy/dx = xy 可重写为 (1/y) dy = x dx，然后两边积分：ln|y| = x²/2 + C，于是 y = ± eC ex²/2 = A ex²/2。积分后需立即加上积分常数。对于边界值问题，利用给定条件（如 y(0) = 2）求出常数 A。这一技巧将积分直接与建模现实情境联系起来。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图