📚 Integration Formulas and Their Applications | 积分计算公式与应用
Integration is one of the two fundamental operations in calculus, alongside differentiation. This article consolidates the essential integration formulas, explains the key techniques, and demonstrates their practical applications in solving A-Level mathematics problems.
积分是微积分学中与微分并列的两大基本运算之一。本文系统梳理积分的基本公式，阐释核心计算方法，并通过典型例题展示其在A-Level数学解题中的实际应用。
1. Basic Integration Formulas | 基本积分公式
The most fundamental integration formulas form the backbone of every integral calculation. For a constant a and variable x, the power rule states: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ −1. When n = −1, the result is the natural logarithm: ∫x⁻¹ dx = ln|x| + C.
最基本的积分公式是所有积分运算的基石。对于常数 a 和变量 x，幂函数法则给出：∫xⁿ dx = xⁿ⁺¹/(n+1) + C，其中 n ≠ −1。当 n = −1 时，结果为自然对数：∫x⁻¹ dx = ln|x| + C。
- ∫a dx = ax + C
- ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
- ∫sin x dx = −cos x + C
- ∫cos x dx = sin x + C
- ∫sec² x dx = tan x + C
- ∫eˣ dx = eˣ + C
- ∫(1/x) dx = ln|x| + C
2. Linearity Rules of Integration | 积分的线性运算法则
Integration is a linear operation. This means that the integral of a sum equals the sum of the integrals, and a constant factor can be pulled out of the integral sign.
积分是线性运算。这意味着积分的和等于和的积分，常数因子可以从积分号中提取出来。
∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
∫k·f(x) dx = k·∫f(x) dx (k is a constant)
For example, to integrate 3x² + 5x − 2, we apply the rules term by term: ∫(3x² + 5x − 2) dx = 3·(x³/3) + 5·(x²/2) − 2x + C = x³ + (5/2)x² − 2x + C.
例如，对 3x² + 5x − 2 积分，逐项运用法则：∫(3x² + 5x − 2) dx = 3·(x³/3) + 5·(x²/2) − 2x + C = x³ + (5/2)x² − 2x + C。
3. Integration by Substitution | 换元积分法
Integration by substitution reverses the chain rule of differentiation. This technique is particularly useful when the integrand contains a composite function multiplied by the derivative of its inner function.
换元积分法是微分链式法则的逆运算。当被积函数包含复合函数乘以其内层函数导数时，这种方法尤为有效。
General procedure | 一般步骤：
- Identify a suitable substitution u = g(x) and compute du = g'(x) dx
- Rewrite the entire integral in terms of u and du
- Integrate with respect to u, then substitute back to x
Example: Evaluate ∫2x·e^(x²) dx. Let u = x², then du = 2x dx. The integral becomes ∫eᵘ du = eᵘ + C = e^(x²) + C.
例题：计算 ∫2x·e^(x²) dx。令 u = x²，则 du = 2x dx。原积分变为 ∫eᵘ du = eᵘ + C = e^(x²) + C。
4. Integration by Parts | 分部积分法
Integration by parts is derived from the product rule of differentiation. It is used when the integrand is a product of two functions that cannot be integrated directly.
分部积分法源于微分学中的乘积法则，适用于被积函数为两个无法直接积分的函数之积的情形。
∫u dv = uv − ∫v du
For definite integrals, the formula becomes:
∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du
The acronym LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) helps choose which function should be u — the earlier in the list, the better choice for u.
口诀 LIATE（对数函数、反三角函数、代数函数、三角函数、指数函数）帮助选取 u —— 列表越靠前的函数，越适合作为 u。
Example: Evaluate ∫x·cos x dx. Choose u = x and dv = cos x dx, then du = dx and v = sin x. Apply the formula: ∫x·cos x dx = x·sin x − ∫sin x dx = x·sin x + cos x + C.
例题：计算 ∫x·cos x dx。选取 u = x，dv = cos x dx，则 du = dx，v = sin x。代入公式：∫x·cos x dx = x·sin x − ∫sin x dx = x·sin x + cos x + C。
5. Definite Integrals and the Fundamental Theorem of Calculus | 定积分与微积分基本定理
Definite integrals provide the signed area between the curve and the x-axis over a specified interval. The Fundamental Theorem of Calculus links differentiation and integration: to evaluate a definite integral, find an antiderivative and evaluate it at the boundaries.
定积分表示曲线与 x 轴在指定区间上的有向面积。微积分基本定理将微分与积分联系起来：要计算定积分，先求原函数，再代入上下限求值。
∫ₐᵇ f(x) dx = F(b) − F(a), where F'(x) = f(x)
Key properties of definite integrals include:
定积分的关键性质包括：
- ∫ₐᵇ f(x) dx = −∫ᵦₐ f(x) dx
- ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx (where a < c < b)
- If f(x) ≥ g(x) on [a, b], then ∫ₐᵇ f(x) dx ≥ ∫ₐᵇ g(x) dx
6. Area Under a Curve | 曲线下的面积
One of the most common applications of definite integrals is computing the area under a curve. For a function f(x) ≥ 0 on the interval [a, b], the area bounded by the curve, the x-axis, and the lines x = a and x = b is given by:
定积分最普遍的应用之一是计算曲线下的面积。对于区间 [a, b] 上满足 f(x) ≥ 0 的函数 f(x)，由曲线、x 轴以及直线 x = a 和 x = b 所围成的面积为：
A = ∫ₐᵇ f(x) dx
When the function is negative over part of the interval, the definite integral yields a negative value. To find the total area, split the integral at the zeros of the function and take the absolute value of each portion.
当函数在区间的一部分上取负值时，定积分会给出负值。要计算总面积，需在函数的零点处分段积分，并对每一段取绝对值。
7. Area Between Two Curves | 两曲线之间的面积
When calculating the area between two curves y = f(x) and y = g(x), where f(x) ≥ g(x) on [a, b], the area is found by integrating the difference of the functions.
计算两条曲线 y = f(x) 与 y = g(x) 之间的面积时，若在 [a, b] 上有 f(x) ≥ g(x)，则面积为两函数之差的积分。
A = ∫ₐᵇ [f(x) − g(x)] dx
Example: Find the area enclosed by y = x² and y = x + 2. First solve for intersection points: x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0, giving x = −1 and x = 2. The area is ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2.
例题：求 y = x² 与 y = x + 2 所围图形的面积。先求交点：x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0，解得 x = −1 和 x = 2。面积为 ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2。
8. Volume of Revolution | 旋转体的体积
When a region under a curve y = f(x) between x = a and x = b is rotated about the x-axis, it generates a solid of revolution. The volume is calculated using the disc method:
当曲线 y = f(x) 在 x = a 与 x = b 之间的区域绕 x 轴旋转时，形成旋转体。使用圆盘法计算其体积：
V = π∫ₐᵇ [f(x)]² dx
If the region is instead rotated about the y-axis, express x as a function of y and integrate with respect to y:
V = π∫ₐᵇ [x(y)]² dy
Example: Find the volume generated by rotating the curve y = √x from x = 0 to x = 4 about the x-axis. V = π∫₀⁴ x dx = π·[x²/2]₀⁴ = π·(16/2) = 8π cubic units.
例题：求曲线 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转所得体积。V = π∫₀⁴ x dx = π·[x²/2]₀⁴ = π·(16/2) = 8π 立方单位。
9. Integration with Linear Substitution | 线性换元积分
For integrals of the form ∫(ax + b)ⁿ dx where n ≠ −1, we can use the linear substitution u = ax + b. Since du = a dx, the integral simplifies significantly.
对于形如 ∫(ax + b)ⁿ dx（n ≠ −1）的积分，可使用线性换元 u = ax + b。由于 du = a dx，积分可大幅简化。
∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/[a(n + 1)] + C
This pattern generalises to other functions: ∫sin(ax + b) dx = −(1/a)·cos(ax + b) + C, and ∫e^(ax + b) dx = (1/a)·e^(ax + b) + C.
此模式可推广到其他函数：∫sin(ax + b) dx = −(1/a)·cos(ax + b) + C，∫e^(ax + b) dx = (1/a)·e^(ax + b) + C。
10. Solving Differential Equations by Separation of Variables | 变量分离法求解微分方程
Differential equations of the form dy/dx = f(x)·g(y) can be solved by separating the variables. Rearrange to group all y-terms with dy and all x-terms with dx, then integrate both sides.
形如 dy/dx = f(x)·g(y) 的微分方程可通过分离变量求解。将含 y 的项与 dy 放在一侧，含 x 的项与 dx 放在另一侧，然后对两边同时积分。
Example: Solve dy/dx = x/y with y(0) = 2. Separate: y dy = x dx. Integrate both sides: ∫y dy = ∫x dx → y²/2 = x²/2 + C → y² = x² + 2C. Using y(0) = 2, we get 4 = 0 + 2C → C = 2, so y² = x² + 4, or y = √(x² + 4) (taking the positive root since y(0) > 0).
例题：解微分方程 dy/dx = x/y，初始条件 y(0) = 2。分离变量：y dy = x dx。两边积分：∫y dy = ∫x dx → y²/2 = x²/2 + C → y² = x² + 2C。代入 y(0) = 2，得 4 = 0 + 2C → C = 2。因此 y² = x² + 4，即 y = √(x² + 4)（因 y(0) > 0，取正根）。
11. Trapezium Rule for Numerical Integration | 梯形法则数值积分
When a function cannot be integrated analytically, numerical methods are required. The trapezium rule approximates the definite integral by dividing the interval [a, b] into n equal sub-intervals of width h = (b − a)/n, then approximating the area under the curve as a series of trapezia.
当函数无法解析积分时，需要借助数值方法。梯形法则将区间 [a, b] 等分为 n 个宽度 h = (b − a)/n 的子区间，用一系列梯形近似曲线下的面积。
∫ₐᵇ f(x) dx ≈ (h/2)·[y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]
where yᵢ = f(xᵢ) and xᵢ = a + ih. The error of the trapezium rule decreases as n increases, but more intervals mean more computation.
其中 yᵢ = f(xᵢ)，xᵢ = a + ih。梯形法则的误差随 n 增大而减小，但区间数越多，计算量越大。
Example: Approximate ∫₀¹ e^(x²) dx using the trapezium rule with n = 4. Here h = 0.25 and the y-values are computed at x = 0, 0.25, 0.5, 0.75, 1: y₀ = 1, y₁ = e^(0.0625) ≈ 1.0645, y₂ = e^(0.25) ≈ 1.2840, y₃ = e^(0.5625) ≈ 1.7551, y₄ = e ≈ 2.7183. Thus the integral is approximately (0.25/2)·[1 + 2.7183 + 2(1.0645 + 1.2840 + 1.7551)] = 0.125·(3.7183 + 8.2072) ≈ 1.4907.
例题：用梯形法则以 n = 4 近似计算 ∫₀¹ e^(x²) dx。此处 h = 0.25，在各点 x = 0, 0.25, 0.5, 0.75, 1 处计算 y 值：y₀ = 1，y₁ = e^(0.0625) ≈ 1.0645，y₂ = e^(0.25) ≈ 1.2840，y₃ = e^(0.5625) ≈ 1.7551，y₄ = e ≈ 2.7183。因此积分近似值为 (0.25/2)·[1 + 2.7183 + 2(1.0645 + 1.2840 + 1.7551)] = 0.125·(3.7183 + 8.2072) ≈ 1.4907。
12. Strategy Summary for Integration Problems | 积分问题解题策略总结
Selecting the correct technique is often the greatest challenge. Use this decision framework:
选择正确的积分技巧往往是最大的挑战。可参考以下决策框架：
| Type of Integrand | 被积函数类型 | Recommended Method | 推荐方法 |
| Polynomial | 多项式 | Power rule term by term | 逐项幂法则 |
| Composite function with inner derivative | 复合函数且含有内层导数 | Substitution | 换元法 |
| Product of unrelated functions | 不相关函数的乘积 | Integration by parts | 分部积分法 |
| Linear composite (ax + b)ⁿ | 线性复合函数 (ax + b)ⁿ | Linear substitution pattern | 线性换元公式 |
| Rational function | 有理函数 | Partial fractions then integrate | 部分分式后积分 |
| Non-integrable analytically | 无法解析积分 | Trapezium rule or numerical methods | 梯形法则或数值方法 |
Mastery of integration requires recognising patterns, selecting the right method, and practising carefully with attention to constants and limits. Always remember to include the constant of integration C for indefinite integrals, and to evaluate boundary values correctly for definite integrals.
掌握积分需要识别题型模式、选择合适方法，并通过精心练习关注常数与上下限。切记：不定积分需添加积分常数 C，定积分需正确代入上下限求值。
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