📚 Definite Integrals | 定积分
The definite integral lies at the heart of A-Level calculus, transforming the idea of accumulation into a precise numerical tool. In Edexcel Mathematics, mastering definite integrals means you can compute exact areas, solve kinematic problems, and build a bridge between algebraic expressions and geometric meaning. This article revisits every essential aspect of the topic, from fundamental definitions to common exam pitfalls.
定积分是A-Level微积分的核心，它把累积的思想转化为精确的数值工具。在爱德思数学中，掌握定积分意味着你能计算精确的面积、解决运动学问题，并在代数表达式与几何意义之间架起桥梁。本文重新梳理该主题的每一个关键层面，从基本定义到常见的考试陷阱。
1. Introduction to Definite Integrals | 定积分简介
f(x) dx, where a is the lower limit and b is the upper limit. It represents the net signed accumulation of the quantity f(x) over the interval [a, b]. While an indefinite integral returns a family of functions plus a constant, a definite integral evaluates to a single real number.
定积分写作 ∫ₐᵇ f(x) dx，其中 a 是下限，b 是上限。它表示函数 f(x) 在区间 [a, b] 上的净有号累积量。与不定积分得到一族函数加上积分常数不同，定积分计算出的是一个实数。
In the context of a graph, if f(x) stays above the x‑axis, the definite integral gives the area trapped between the curve, the x‑axis and the lines x = a and x = b. The notation deliberately resembles an elongated ‘ reminding us that integration is built from summing infinitely many infinitesimal slices.
在图像语境中，若 f(x) 保持在 x 轴上方，定积分给出的就是曲线、x 轴以及直线 x = a 和 x = b 所围成的面积。积分符号特意设计成拉长的“S”，提醒我们积分的本质是对无穷多个无穷小切片求和。
2. The Fundamental Theorem of Calculus | 微积分基本定理
The Fundamental Theorem of Calculus is the engine that makes evaluation practical. It states that if F is any antiderivative of f on [a, b], then
微积分基本定理是使计算成为可能的引擎。它指出，如果 F 是 f 在 [a, b] 上的任意一个反导数，那么
∫ₐᵇ f(x) dx = F(b) − F(a)
This elegant result frees us from needing to compute limits of Riemann sums. All we require is one antiderivative F, which we often find using standard integration techniques such as reverse power rule, trigonometric integrals, or exponential rules. The subtraction F(b) − F(a) neatly removes the constant of integration, which is why we never need to write +C in a definite integral.
这个优美的结果让我们不必再去计算黎曼和的极限。我们只需要一个反导数 F，通常利用反幂法则、三角积分或指数法则等标准积分技巧就能求出。减法 F(b) − F(a) 巧妙地消去了积分常数，因此我们在定积分中永远不需要写 +C。
For example, because the derivative of sin x is cos x, we immediately have ∫₀π/2 e theorem turns differentiation and integration into inverse processes, which is the conceptual backbone of calculus.
例如，因为 sin x 的导数是 cos x，我们马上得到 ∫₀π/2 cos x dx = sin(π/2) − sin(0) = 1 − 0 = 1。该定理将微分与积分变成互逆的过程，这正是整个微积分的概念支柱。
3. Evaluating a Definite Integral | 计算定积分
To evaluate a definite integral by hand, first find the antiderivative F(x) of the integrand. Then substitute the upper limit and lower limit, always writing the antiderivative inside square brackets with the limits to the right: [F(x)]ₐᵇ. Finally compute F(b) − F(a).
要手算一个定积分，首先求出被积函数的反导数 F(x)，然后把反导数写在方括号内并标注上下限：[F(x)]ₐᵇ，最后计算出 F(b) − F(a)。
) dx. The antiderivative of 3x² is x³, so we write [x³]₁² = 2³ − 1³ = 8 − 1 = 7. For ∫₀¹ eˣ dx, [eˣ nomials, always increase the power by one and divide by the new power before substituting.
考虑 ∫₁² (3x²) dx，3x² 的反导数是 x³，因此写作 [x³]₁² = 2³ − 1³ = 8 − 1 = 7。对于 ∫₀¹ eˣ dx，[eˣ]₀¹ = e¹ − e⁰ = e − 1。处理多项式时，务必先将指数加一，再除以新指数，然后代入上下限。
als with fractional or negative powers follow exactly the same rule; just be careful when substituting limits that yield zero or negative values.
当被积函数是函数的和时，逐项积分即可。例如 ∫₁⁴ (2x + 3) dx = [x² + 3x]₁⁴ = (16 + 12) − (1 + 3) = 28 − 4 = 24。含有分数或负指数幂的定积分遵循完全相同的方法；只是在代入得到零或负值时需要格外小心。
4. Properties of Definite Integrals | 定积分的性质
dx = −∫ᵇₐ f(x) dx. The integral over an interval of zero length is zero: ∫ₐᵃ f(x) dx = 0.
若干代数性质使定积分灵活且易于处理。颠倒上下限会改变符号：∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx。在零长度区间上的积分为零：∫ₐᵃ f(x) dx = 0。
Additivity over intervals means ∫ₐᶜ f(x) dx = ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx, which is particularly useful when a curve crosses the x‑ the sum or difference of the integrals: ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx.
区间的可加性意味着 ∫ₐᶜ f(x) dx = ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx，这在曲线穿过 x 轴时特别有用。常数可以提到积分号外：∫ₐᵇ k f(x) dx = k ∫ₐᵇ f(x) dx。和或差的积分等于积分的和或差：∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx。
you already know the individual integrals. This saves time and reduces the risk of algebraic errors.
这些性质常常组合使用。例如，要计算 ∫₁³ (4f(x) − g(x)) dx，你可以把它写成 4∫₁³ f(x) dx − ∫₁³ g(x) dx，前提是你已经知道各个单独的积分值。这既节省时间又降低代数出错的概率。
5. Area Under a Curve | 曲线下的面积
], the definite integral ∫ₐᵇ f(x) dx gives the true area between the curve y = f(x), the x‑ es x = a and x = b. No absolute values are needed because the integral itself is positive.
当 f(x) 在 [a, b] 上恒非负时，定积分 ∫ₐᵇ f(x) dx 直接给出曲线 y = f(x)、x 轴以及垂直线 x = a 和 x = b 之间的真实面积。因为积分值本身就为正，所以无需取绝对值。
epresents a physical quantity if the axes have appropriate scales, but in pure mathematics we simply quote the number of square units.
例如，y = x² 在 x = 0 到 x = 2 之间的下方面积为 ∫₀² x² dx = [⅓ x³]₀² = 8/3 − 0 = 8/3。在带单位的示意图中，如果坐标轴有适当的尺度，这样的面积也代表物理量，但在纯数学中我们只需给出平方单位的数目。
Always sketch the curve or at least check its sign over the interval. If the curve dips below the x‑axis even once, the integral alone will give a signed area, not the physical total area. This is the most common trap in exam questions on area.
永远要画出曲线草图，或至少检查它在给定区间上的正负。若曲线哪怕只有一次落到 x 轴下方，仅用积分得到的就是有号面积，而不是物理总面积。这是考试中面积计算题最常见的陷阱。
6. Area Between a Curve and the x‑axis (including negative parts) | 曲线与x轴之间的面积（含负部分）
and take the absolute value of each piece’s integral. Mathematically, the total area = ∫ₐᵇ |f(x)| dx.
如果 f(x) 在 [a, b] 上既有正值也有负值，我们必须在每一个 f(x)=0 的点处将积分拆开，并对每一段的积分取绝对值。数学上，总面积 = ∫ₐᵇ |f(x)| dx。
A classic exam example: find the area bounded by y = x² − 4, the x‑axis, x = 0 and x = 3. The curve meets the x‑ negative; from 2 to 3 it is positive. Therefore Area = −∫₀² (x² − 4) dx + ∫₂³ (x² − 4) dx.
一个经典考题：求 y = x² − 4、x 轴、x = 0 和 x = 3 所围成的面积。曲线与 x 轴相交于 x² − 4 = 0 → x = 2（x = −2 不在区间内）。在 0 到 2 上函数为负；在 2 到 3 上为正。因此面积 = −∫₀² (x² − 4) dx + ∫₂³ (x² − 4) dx。
− 12) − (8/3 − 8) = (−3) − (−16/3) = 7/3. The total area is 16/3 + 7/3 = 23/3. Never simply plug the limits into one integral — you must break the interval at the roots.
分别计算：∫₀² (x² − 4) dx = [⅓x³ − 4x]₀² = (8/3 − 8) − 0 = −16/3，因此第一部分的贡献为 +16/3。第二部分：∫₂³ (x² − 4) dx = [⅓x³ − 4x]₂³ = (9 − 12) − (8/3 − 8) = (−3) − (−16/3) = 7/3。总面积为 16/3 + 7/3 = 23/3。永远不要简单地把上下限代入一个积分——你必须在零点处拆分区间。
7. Area Between Two Curves | 两条曲线之间的面积
area enclosed between two curves y = f(x) and y = g(x) from x = a to x = b is given by ∫ₐᵇ |f(x) − g(x)| dx. In questions where the curves do not cross, we can simply integrate ‘top minus bottom’.
由两条曲线 y = f(x) 和 y = g(x) 在 x = a 到 x = b 之间所围成的面积为 ∫ₐᵇ |f(x) − g(x)| dx。在曲线不相交的问题中，只需对“上曲线减下曲线”进行积分即可。
², so the upper curve is y = x. Thus Area = ∫₀¹ (x − x²) dx = [½x² − ⅓x³]₀¹ = ½ − ⅓ = 1/6.
例如，求 y = x 与 y = x² 在 x = 0 到 x = 1 之间围成的面积。在该区间上 x ≥ x²，所以上曲线为 y = x。因此面积 = ∫₀¹ (x − x²) dx = [½x² − ⅓x³]₀¹ = ½ − ⅓ = 1/6。
the interval, you must find all intersection points, split the interval, and for each sub‑interval determine which curve is on top. The absolute value formula ∫ |f − g| dx automatically accounts for any crossing, but in practice a sketch is invaluable.
如果曲线在区间内相交，则必须先找出所有交点，拆分区间，并在每一段子区间上确定哪条曲线在上方。公式 ∫ |f − g| dx 自动处理了任何交叉的情形，但在实际运算中，画一张草图仍然非常有用。
Sometimes the bounds are given in y rather than x; the same principle applies by expressing x in terms of y and integrating with respect to y. The Edexcel specification expects you to handle both orientations.
有时边界是以 y 给出的而非 x；此时只需把 x 表示成 y 的函数并对 y 积分，同样的原理适用。爱德思考试大纲要求你能够处理两种方向的情形。
8. Numerical Integration: The Trapezium Rule | 数值积分：梯形法则
roximate value. Divide the interval [a, b] into n strips of equal width h = (b − a)/n. Then
当积分无法用标准的解析方法求出，或我们处理的是经验数据时，梯形法则能够提供一个近似值。将区间 [a, b] 划分为 n 个等宽条带，宽度为 h = (b − a)/n。那么
∫ₐᵇ f(x) dx ≈ ½h [y₀ + 2(y₁ + y₂ + ... + yₙ₋₁) + yₙ]
works by replacing the curve with a series of straight line segments, calculating the area of each trapezium, and summing them. The more strips you use, the smaller the error tends to be.
其中 yᵢ = f(xᵢ) 且 xᵢ = a + i h。该法则用一系列直线段替代曲线，计算每个梯形的面积并求和。使用的条带越多，误差通常越小。
), (0.5, 4.2), (1, 5.1), (1.5, 5.8), (2, 6.0). Here h = 0.5, n = 4. The formula gives 0.5/2 × [3 + 2(4.2 + 5.1 + 5.8) + 6.0] = 0.25 × (3 + 30.2 + 6) = 0.25 × 39.2 = 9.8. Always show the substitution clearly to gain method marks.
考试中经常给出一张数值表格，要求你应用梯形法则。例如，利用 (0, 3), (0.5, 4.2), (1, 5.1), (1.5, 5.8), (2, 6.0) 估计 ∫₀² f(x) dx。这里 h = 0.5，n = 4。公式给出 0.5/2 × [3 + 2(4.2 + 5.1 + 5.8) + 6.0] = 0.25 × (3 + 30.2 + 6) = 0.25 × 39.2 = 9.8。务必清晰写出代入过程以取得方法分。
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