📚 IGCSE Mathematics: Integration Key Points | IGCSE 数学：积分 考点精讲
Integration is one of the two central operations in calculus, working as the reverse of differentiation. In IGCSE Mathematics, you are expected to master basic integration techniques, apply the power rule to polynomial functions, handle simple trigonometric and exponential integrals, and use definite integrals to find areas under curves and between curves. This revision guide breaks down every core concept into clear, paired explanations so you can approach integration with confidence.
积分是微积分中的两大核心运算之一，作为微分的逆运算。在IGCSE数学中，你需要掌握基本积分技巧，将幂函数法则应用于多项式，处理简单的三角函数与指数函数积分，并利用定积分求曲线下方及曲线之间的面积。本复习指南将每个核心概念分解为清晰的中英对照解释，让你自信应对积分题。
1. Understanding Indefinite Integration | 理解不定积分
Indefinite integration finds a family of functions whose derivative is the given function. If F'(x) = f(x), we write ∫ f(x) dx = F(x) + C, where C is an arbitrary constant called the constant of integration. The symbol ∫ is the integral sign, and dx indicates the variable of integration.
不定积分找出导数为给定函数的一族函数。若 F'(x) = f(x)，我们记作 ∫ f(x) dx = F(x) + C，其中 C 是任意常数，称为积分常数。符号 ∫ 是积分号，dx 表示积分变量。
Because the derivative of any constant is zero, adding C covers all possible antiderivatives. For example, ∫ 2x dx = x² + C, since d/dx (x² + C) = 2x.
由于任何常数的导数均为零，加上 C 涵盖了所有可能的原函数。例如，∫ 2x dx = x² + C，因为 d/dx (x² + C) = 2x。
2. The Power Rule for Integration | 幂函数积分法则
For any real number n ≠ -1, the integral of xⁿ with respect to x is given by ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C. Simply increase the exponent by 1 and divide by the new exponent. This rule is the reverse of the power rule for derivatives.
对于任何实数 n ≠ -1，xⁿ 的积分公式为 ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C。只需将指数加 1，再除以新指数。该法则正是幂函数求导法则的逆运算。
Examples: ∫ x³ dx = x⁴/4 + C; ∫ x⁻² dx = -x⁻¹ + C; ∫ √x dx = ∫ x½ dx = (x³/²)/(3/2) + C = (2/3)x³/² + C. Rewrite roots and negative exponents as powers before integrating.
示例：∫ x³ dx = x⁴/4 + C；∫ x⁻² dx = -x⁻¹ + C；∫ √x dx = ∫ x½ dx = (x³/²)/(3/2) + C = (2/3)x³/² + C。积分前先将根式和负指数改写为幂的形式。
3. Integrating sin x, cos x, and eˣ | 积分 sin x、cos x 和 eˣ
Basic trigonometric and exponential integrals come directly from differentiation formulas. Since d/dx (sin x) = cos x, we have ∫ cos x dx = sin x + C. Since d/dx (cos x) = -sin x, it follows that ∫ sin x dx = -cos x + C.
基本的三角函数和指数函数积分直接源自求导公式。因为 d/dx (sin x) = cos x，所以 ∫ cos x dx = sin x + C。因为 d/dx (cos x) = -sin x，可得 ∫ sin x dx = -cos x + C。
The exponential function eˣ is its own derivative, so ∫ eˣ dx = eˣ + C. These three integrals frequently appear in IGCSE problems alongside polynomial terms.
指数函数 eˣ 的导数等于它本身，因此 ∫ eˣ dx = eˣ + C。这三种积分经常与多项式项一同出现在IGCSE试题中。
4. Integrating Sums and Constant Multiples | 积分和差与常数倍法则
Integration is linear: the integral of a sum is the sum of the integrals, and constants factor out. ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx, and ∫ k·f(x) dx = k ∫ f(x) dx, where k is a constant.
积分具有线性性质：和差的积分等于积分的和差，常数因子可提出。∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx，且 ∫ k·f(x) dx = k ∫ f(x) dx，其中 k 为常数。
For instance, ∫ (4x³ – 2 cos x) dx = 4 ∫ x³ dx – 2 ∫ cos x dx = x⁴ – 2 sin x + C. Always add a single C at the end, not one for each term.
例如，∫ (4x³ – 2 cos x) dx = 4 ∫ x³ dx – 2 ∫ cos x dx = x⁴ – 2 sin x + C。注意最终只加一个常数 C，而不是每一项都加。
5. Finding the Equation of a Curve | 求曲线方程
Given the derivative f'(x) and a point on the original curve (x₁, y₁), you can recover f(x). First integrate f'(x) to get f(x) + C. Then substitute x₁ and y₁ into the equation and solve for C.
已知导数 f'(x) 及原曲线上一点 (x₁, y₁)，可以还原 f(x)。先对 f'(x) 积分得到 f(x) + C，再将 x₁ 和 y₁ 代入方程解出 C。
Example: f'(x) = 6x – 4, and the curve passes through (1, 5). Integrate: f(x) = 3x² – 4x + C. Substitute: 5 = 3(1)² – 4(1) + C → C = 6. Hence f(x) = 3x² – 4x + 6.
示例：f'(x) = 6x – 4，曲线经过 (1, 5)。积分得 f(x) = 3x² – 4x + C。代入得 5 = 3(1)² – 4(1) + C → C = 6，因此 f(x) = 3x² – 4x + 6。
6. Introduction to Definite Integrals | 定积分导论
A definite integral computes the net signed area under a curve between two limits a and b. It is written as ∫ₐᵇ f(x) dx and evaluated as F(b) – F(a), where F'(x) = f(x). Unlike indefinite integrals, the constant C cancels out and is not written.
定积分计算曲线在上下限 a 与 b 之间的净有向面积。写作 ∫ₐᵇ f(x) dx，求值为 F(b) – F(a)，其中 F'(x) = f(x)。与不定积分不同，常数 C 会抵消，因此不写出来。
Evaluation uses square bracket notation: [F(x)]ₐᵇ = F(b) – F(a). Example: ∫₁³ 2x dx = [x²]₁³ = 9 – 1 = 8. Always substitute the upper limit first, then subtract the lower limit result.
求值时使用方括号记法：[F(x)]ₐᵇ = F(b) – F(a)。示例：∫₁³ 2x dx = [x²]₁³ = 9 – 1 = 8。务必先代入上限，再减去下限的计算结果。
7. Area Between a Curve and the x-axis | 曲线与 x 轴之间的面积
When f(x) ≥ 0 for a ≤ x ≤ b, the area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is given exactly by the definite integral A = ∫ₐᵇ f(x) dx. No absolute value is needed because the function stays above the axis.
当在区间 a ≤ x ≤ b 内 f(x) ≥ 0 时，由曲线 y = f(x)、x 轴及直线 x = a、x = b 所围成的面积可以直接用定积分 A = ∫ₐᵇ f(x) dx 表示。因函数保持在轴上方，无需取绝对值。
Example: Find the area under y = x² + 1 from x = 0 to x = 2. Area = ∫₀² (x² + 1) dx = [x³/3 + x]₀² = (8/3 + 2) – 0 = 14/3 square units.
示例：求 y = x² + 1 从 x = 0 到 x = 2 下方的面积。面积 = ∫₀² (x² + 1) dx = [x³/3 + x]₀² = (8/3 + 2) – 0 = 14/3 平方单位。
8. Dealing with Areas Below the x-axis | 处理 x 轴下方的面积
If f(x) ≤ 0 on an interval, the definite integral yields a negative value. The physical area must be positive, so we take the absolute value: Area = -∫ₐᵇ f(x) dx or write |∫ₐᵇ f(x) dx|. When the curve crosses the axis, split the integral at the roots and add the absolute values of each part’s integral.
若在某一区间内 f(x) ≤ 0，定积分将得到负值。实际的面积必须为正，因此需要取绝对值：面积 = -∫ₐᵇ f(x) dx 或写作 |∫ₐᵇ f(x) dx|。当曲线穿越 x 轴时，应在交点处拆分积分，并将各段积分的绝对值相加。
For y = x² – 4 from x = 0 to x = 3, the graph is below the axis from 0 to 2. Calculate ∫₀² (x² – 4) dx = [x³/3 – 4x]₀² = (8/3 – 8) = -16/3. Area₁ = 16/3. From 2 to 3 it is above: ∫₂³ (x² – 4) dx = [x³/3 – 4x]₂³ = (9 – 12) – (8/3 – 8) = -3 – (-16/3) = 7/3. Total area = 16/3 + 7/3 = 23/3.
对于 y = x² – 4 从 x = 0 到 x = 3，图像在 0 到 2 部分位于轴下方。计算 ∫₀² (x² – 4) dx = [x³/3 – 4x]₀² = (8/3 – 8) = -16/3。区域面积为 16/3。从 2 到 3 位于轴上方：∫₂³ (x² – 4) dx = [x³/3 – 4x]₂³ = (9 – 12) – (8/3 – 8) = -3 – (-16/3) = 7/3。总面积 = 16/3 + 7/3 = 23/3。
9. Area Between Two Curves | 两曲线之间的面积
To find the area enclosed between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b, integrate the difference: A = ∫ₐᵇ [f(x) – g(x)] dx. Identify which function is on top by checking values or sketching, and find the intersection points for the limits if not given.
求两曲线 y = f(x)（上方）与 y = g(x)（下方）在 x = a 到 x = b 之间所围的面积时，对差值积分：A = ∫ₐᵇ [f(x) – g(x)] dx。需通过代入数值或画草图确定哪条曲线在上方，若未直接给出积分限，则先求交点作为上下限。
Example: Find the area bounded by y = x² and y = x + 2. Intersections: x² = x + 2 → x² – x – 2 = 0 → x = -1, 2. On [-1,2], line is above parabola. Area = ∫₋₁² [(x + 2) – x²] dx = [ -x³/3 + x²/2 + 2x ]₋₁² = (-8/3 + 2 + 4) – (1/3 + 1/2 – 2) = (10/3) – (-7/6) = 9/2.
示例：求 y = x² 和 y = x + 2 所围面积。求交点：x² = x + 2 → x² – x – 2 = 0 → x = -1, 2。在 [-1,2] 上，直线在抛物线之上。面积 = ∫₋₁² [(x + 2) – x²] dx = [ -x³/3 + x²/2 + 2x ]₋₁² = (-8/3 + 2 + 4) – (1/3 + 1/2 – 2) = (10/3) – (-7/6) = 9/2。
10. Integrating (ax + b)ⁿ | 积分 (ax + b)ⁿ
When the integrand is a linear function raised to a power, apply the reverse chain rule. For n ≠ -1, ∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n + 1)] + C. Dividing by the derivative of the inner function ‘a’ compensates for the chain rule that ‘a’ would bring when differentiating.
当被积函数为线性函数的幂次时，应用逆链式法则。对于 n ≠ -1，∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n + 1)] + C。除以内部函数导数 a，可以抵消求导时链式法则产生的系数 a。
Example: ∫ (3x – 1)⁴ dx = (3x – 1)⁵ / (3 × 5) + C = (1/15)(3x – 1)⁵ + C. Always check by differentiating: d/dx [(1/15)(3x – 1)⁵] = (1/15)·5(3x – 1)⁴·3 = (3x – 1)⁴.
示例：∫ (3x – 1)⁴ dx = (3x – 1)⁵ / (3 × 5) + C = (1/15)(3x – 1)⁵ + C。可用求导验证：d/dx [(1/15)(3x – 1)⁵] = (1/15)·5(3x – 1)⁴·3 = (3x – 1)⁴。
This method extends to definite integrals too. For ∫₀¹ (2x + 1)³ dx, find antiderivative (2x + 1)⁴ / 8, then substitute limits: [ (2×1+1)⁴/8 ] – [ (2×0+1)⁴/8 ] = (81/8) – (1/8) = 10.
此方法同样适用于定积分。如 ∫₀¹ (2x + 1)³ dx，先求原函数 (2x + 1)⁴ / 8，代入上下限：[(2×1+1)⁴/8] – [(2×0+1)⁴/8] = (81/8) – (1/8) = 10。
11. Common Mistakes to Avoid | 常见错误避免
One frequent error is forgetting the ‘+ C’ in indefinite integrals. Although the constant cancels in definite integrals, every antiderivative requires the constant for full marks.
常见错误之一是在不定积分中漏掉 ‘+ C’。尽管常数在定积分中会抵消，但每个原函数都必须写出常数才能获得满分。
Another pitfall is misapplying the power rule to x⁻¹. ∫ x⁻¹ dx is not x⁰/0; it is ln|x| + C, which is beyond IGCSE scope but highlights why n = -1 is excluded. Also, when computing areas below the axis, many students forget to take absolute values, yielding a negative or zero area.
另一个易错点是对 x⁻¹ 误用幂法则。∫ x⁻¹ dx 不是 x⁰/0，而是 ln|x| + C，此内容虽超出IGCSE范围，但说明了为何 n = -1 被排除。此外，计算轴下方区域时，许多学生忘记取绝对值，导致面积出现负数或零。
In definite integrals, always subtract in the order F(b) – F(a); reversing gives the wrong sign. For composite linear functions, students often omit dividing by the inner derivative ‘a’, so the result is off by a constant factor.
在定积分中，务必依 F(b) – F(a) 的顺序相减；颠倒顺序将导致符号错误。对于线性复合函数，学生常常忘记除以内部导数 a，导致结果相差一个常数因子。
Finally, when finding areas between curves, ensure you identify the correct upper function on the entire interval. If the curves cross, split the integral at the intersection and treat each part separately with the correct upper-lower order.
最后，在求曲线之间的面积时，确保在整个区间上确认正确的上方函数。若曲线相交，需在交点处拆分积分，对每一段分别按正确的上下顺序处理。
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