Definite Integrals: Key Formulas and Applications | 定积分公式及其应用

📚 Definite Integrals: Key Formulas and Applications | 定积分公式及其应用

The definite integral is one of the most powerful tools in A-Level mathematics. It allows us to calculate exact areas under curves, volumes of revolution, and accumulated quantities across intervals. This article covers the essential formulas, standard techniques, and practical applications you need for exam success.

定积分是 A-Level 数学中最强大的工具之一。它使我们能够精确计算曲线下方的面积、旋转体的体积，以及区间内的累积量。本文涵盖考试必备的核心公式、标准技巧和实际应用。

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1. The Fundamental Theorem of Calculus | 微积分基本定理

If f(x) is continuous on [a, b] and F(x) is an antiderivative of f(x), then the definite integral from a to b equals F(b) − F(a). This theorem bridges the concept of differentiation and integration.

若 f(x) 在 [a, b] 上连续，且 F(x) 是 f(x) 的一个原函数，则从 a 到 b 的定积分等于 F(b) − F(a)。该定理架起了微分与积分之间的桥梁。

The evaluation process is written as:

∫ₐᵇ f(x)dx = [F(x)]ₐᵇ = F(b) − F(a)

In A-Level exams, always remember to substitute the upper limit first, then subtract the value at the lower limit. Never omit the evaluation step.

在 A-Level 考试中，务必先代入上限，再减去下限处的值。切勿省略代入求值的步骤。

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2. Standard Definite Integral Formulas | 标准定积分公式

You must memorise these fundamental integrals. They appear in nearly every exam paper, whether in pure mathematics or in applied contexts.

以下是必须熟记的基础积分公式。它们在几乎每份试卷中都会出现，无论是在纯数学还是应用数学中。

Integral | 积分	Result | 结果
∫xⁿ dx (n ≠ −1)	xⁿ⁺¹/(n+1) + C
∫eˣ dx	eˣ + C
∫1/x dx	ln|x| + C
∫sin x dx	−cos x + C
∫cos x dx	sin x + C
∫sec² x dx	tan x + C

For example, evaluating ∫₀¹ x² dx gives [x³/3]₀¹ = 1/3 − 0 = 1/3. The constant C cancels out in definite integrals, which is why we omit it during evaluation.

例如，计算 ∫₀¹ x² dx 得到 [x³/3]₀¹ = 1/3 − 0 = 1/3。常数 C 在定积分中相互抵消，因此代入求值时无需写出。

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3. Properties of Definite Integrals | 定积分的性质

Understanding the algebraic properties of definite integrals helps you simplify complex expressions quickly and spot shortcut opportunities in exams.

理解定积分的代数性质，有助于快速简化复杂表达式，并在考试中发现捷径。

• Reversal of limits | 交换上下限：∫ₐᵇ f(x)dx = −∫ᵦₐ f(x)dx. Swapping the limits changes the sign.
• Zero interval | 零区间：∫ₐᵃ f(x)dx = 0. Integrating over zero width always yields zero.
• Linear combination | 线性组合：∫ₐᵇ [cf(x) + dg(x)]dx = c∫ₐᵇ f(x)dx + d∫ₐᵇ g(x)dx.
• Interval splitting | 区间分割：∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx + ∫ᶜᵇ f(x)dx, where c is any point in [a, b].

These properties are especially useful when dealing with piecewise functions or when a function has a discontinuity that must be handled separately.

这些性质在处理分段函数或函数存在间断点而必须分段处理时尤为有用。

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4. Definite Integrals of Trigonometric Functions | 三角函数的定积分

Trigonometric integrals appear frequently in both pure and applied mathematics. One key fact to remember is symmetry over the interval [0, 2π] or [−π, π].

三角函数积分在纯数学和应用数学中经常出现。一个关键要点是记住函数在 [0, 2π] 或 [−π, π] 上的对称性。

∫₀^{π/2} sin x dx = [−cos x]₀^{π/2} = 0 − (−1) = 1

Similarly, ∫₀^{π/2} cos x dx = [sin x]₀^{π/2} = 1 − 0 = 1. Both results are equal because sin and cos are phase shifts of each other.

同样地，∫₀^{π/2} cos x dx = [sin x]₀^{π/2} = 1 − 0 = 1。两者结果相同，因为 sin 和 cos 互为相位平移。

For higher powers, use the double-angle identities: sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. These reduce the integrand to a form you can integrate directly.

对于高次幂，使用二倍角公式：sin²x = (1 − cos 2x)/2 和 cos²x = (1 + cos 2x)/2。这些公式将被积函数化为可直接积分的形式。

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5. Integration by Substitution in Definite Integrals | 定积分的换元积分法

When using substitution in a definite integral, you must also transform the limits of integration. This is a common source of student error.

在定积分中使用换元法时，必须同时变换积分的上下限。这是学生常犯错误的一个常见来源。

Suppose we evaluate ∫₀¹ 2x·√(x² + 1) dx. Let u = x² + 1, so du = 2x dx. The new limits are: when x = 0, u = 1; when x = 1, u = 2.

我们计算 ∫₀¹ 2x·√(x² + 1) dx。设 u = x² + 1，则 du = 2x dx。新限为：当 x = 0 时，u = 1；当 x = 1 时，u = 2。

∫₀¹ 2x·√(x² + 1) dx = ∫₁² u^{1/2} du = [2u^{3/2}/3]₁² = 2(2√2 − 1)/3

Always change the limits alongside the substitution — never substitute back into x at the end when working with definite integrals.

始终在换元时同步变换上下限——处理定积分时，切勿在最后重新代回 x。

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6. Integration by Parts for Definite Integrals | 定积分的分部积分法

The formula for integration by parts is derived from the product rule. For definite integrals, we apply the limits to each term after integration.

分部积分公式由乘法法则推导而来。对于定积分，我们在积分完成后对各部分代入上下限。

∫ₐᵇ u·(dv/dx) dx = [u·v]ₐᵇ − ∫ₐᵇ v·(du/dx) dx

As a worked example, evaluate ∫₀¹ x·eˣ dx. Choose u = x and dv = eˣ dx, so du = dx and v = eˣ.

作为示例，计算 ∫₀¹ x·eˣ dx。选取 u = x，dv = eˣ dx，则 du = dx，v = eˣ。

∫₀¹ x·eˣ dx = [x·eˣ]₀¹ − ∫₀¹ eˣ dx = (e − 0) − (e − 1) = 1

The result is exactly 1. The LIATE rule (Log, Inverse trig, Algebraic, Trig, Exponential) helps you choose u strategically: pick the function that simplifies when differentiated.

结果恰为 1。LIATE 法则（对数、反三角、代数、三角、指数）帮助你合理选择 u：优先选择求导后能简化的函数作为 u。

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7. Area Under a Curve | 曲线下方的面积

The most direct application of definite integrals is calculating the area between a curve y = f(x) and the x-axis from x = a to x = b.

定积分最直接的应用是计算曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的面积。

Area = ∫ₐᵇ |f(x)| dx

You must take the absolute value because area is always positive. If f(x) is negative on part of the interval, the integral alone would give a negative contribution.

必须取绝对值，因为面积始终为正。如果 f(x) 在区间的一部分为负，仅靠积分会得到负贡献。

For example, the area between y = sin x and the x-axis from 0 to 2π is:

例如，y = sin x 与 x 轴从 0 到 2π 之间的面积为：

∫₀^π sin x dx + |∫_π^{2π} sin x dx| = 2 + 2 = 4

Each hump has area 2. Without the absolute value, the second integral would incorrectly cancel the first, giving zero.

每个拱形的面积均为 2。若不取绝对值，第二个积分会错误地抵消第一个，得到零。

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8. Area Between Two Curves | 两曲线之间的面积

To find the area enclosed between two curves y = f(x) and y = g(x), integrate the difference of the upper curve minus the lower curve.

求两条曲线 y = f(x) 和 y = g(x) 所围成的面积，只需积分上方曲线减下方曲线的差。

Area = ∫ₐᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x) on [a, b]

As an example, find the area between y = x² and y = x. First solve x² = x to get x = 0 or x = 1. On [0, 1], y = x is the upper curve.

例如，求 y = x² 与 y = x 之间的面积。先解 x² = x 得 x = 0 或 x = 1。在 [0, 1] 上，y = x 是上方曲线。

Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6

Always sketch the graphs first to identify which curve is on top. Marking the intersection points is essential for determining the integration limits.

始终先画草图以确定哪条曲线在上方。标出交点对于确定积分限至关重要。

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9. Volumes of Revolution | 旋转体体积

When a curve y = f(x) is rotated through 360° about the x-axis between x = a and x = b, the resulting volume is given by a definite integral.

当曲线 y = f(x) 绕 x 轴从 x = a 到 x = b 旋转 360° 时，所得旋转体体积由定积分给出。

V = π∫ₐᵇ [f(x)]² dx

For rotation about the y-axis, the formula becomes V = π∫_c^d [f⁻¹(y)]² dy, using the inverse function and limits in y.

绕 y 轴旋转时，公式变为 V = π∫_c^d [f⁻¹(y)]² dy，使用反函数和 y 方向的限。

Consider the curve y = √x rotated about the x-axis from x = 0 to x = 1:

考虑曲线 y = √x 绕 x 轴从 x = 0 到 x = 1 旋转：

V = π∫₀¹ x dx = π[x²/2]₀¹ = π/2

The radius of each disc is y = √x, so the cross-sectional area is πy² = πx. Integrating this over the interval gives the total volume.

每个圆盘的半径是 y = √x，因此截面积为 πy² = πx。在区间上积分即可得总体积。

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10. Definite Integrals of Even and Odd Functions | 偶函数与奇函数的定积分

Symmetry can dramatically simplify definite integrals over symmetric intervals like [−a, a]. Recognising even and odd functions saves valuable exam time.

对称性可以极大简化在 [−a, a] 等对称区间上的定积分。识别偶函数和奇函数能节省宝贵的考试时间。

• Even function | 偶函数：f(−x) = f(x), so ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx. Examples: x², cos x.
• Odd function | 奇函数：f(−x) = −f(x), so ∫₋ₐᵃ f(x)dx = 0. Examples: x³, sin x.

For instance, ∫₋₁¹ (x³ + x⁵) dx = 0 immediately, because both terms are odd functions. No integration is needed.

例如，∫₋₁¹ (x³ + x⁵) dx 直接等于 0，因为两项都是奇函数。无需进行积分计算。

Be cautious: a function like x + x² is neither even nor odd overall. You must split it and apply the rules term by term.

请注意：像 x + x² 这样的函数整体上既非偶也非奇。你必须将其拆分，逐项应用这些规则。

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11. Numerical Integration: The Trapezium Rule | 数值积分：梯形法则

When a function cannot be integrated analytically, numerical methods provide an approximation. The trapezium rule is the most common one in A-Level.

当函数无法解析积分时，数值方法可提供近似值。梯形法则是 A-Level 中最常用的一种。

∫ₐᵇ f(x)dx ≈ (h/2)[y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]

Here h = (b − a)/n, where n is the number of strips, and yᵢ = f(xᵢ) are the function values at equally spaced points. The accuracy improves as n increases.

其中 h = (b − a)/n，n 是条带数量，yᵢ = f(xᵢ) 是等距点上的函数值。n 越大，精度越高。

For example, approximate ∫₀¹ x² dx using 4 strips. With h = 0.25, we get y₀ = 0, y₁ = 0.0625, y₂ = 0.25, y₃ = 0.5625, y₄ = 1.

例如，用 4 个条带近似 ∫₀¹ x² dx。取 h = 0.25，得到 y₀ = 0，y₁ = 0.0625，y₂ = 0.25，y₃ = 0.5625，y₄ = 1。

(0.25/2)[0 + 1 + 2(0.0625 + 0.25 + 0.5625)] = 0.125 × 2.75 = 0.34375

The exact value is 1/3 ≈ 0.3333, so the approximation is reasonably close. Increasing n to 10 would give a better estimate.

精确值为 1/3 ≈ 0.3333，因此近似值相当接近。将 n 增加到 10 会得到更好的估计。

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12. Real-World Applications and Exam Tips | 实际应用与考试技巧

Definite integrals model average values, total distance from velocity, work done in physics, and consumer surplus in economics. The average value of f(x) over [a, b] is:

定积分可用于建模平均值、由速度求总路程、物理学中的做功以及经济学中的消费者剩余。f(x) 在 [a, b] 上的平均值为：

Average value = (1/(b − a))∫ₐᵇ f(x) dx

For exam success, always show your working step by step. Write the antiderivative, substitute the limits separately, and simplify fully. Make sure your calculator is in the correct angle mode when dealing with trigonometric integrals.

要在考试中取得成功，务必逐步展示运算过程。写出原函数，分别代入上下限，并完整化简。处理三角函数积分时，确保计算器处于正确的角度模式。

Finally, always check the sign: if an area should be positive but your integral is negative, you have likely forgotten the absolute value or swapped the limits.

最后，始终检查符号：如果面积应为正而积分为负，你很可能忘记了绝对值或交换了上下限。

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