A-Level数学 分部积分 换元积分 定积分

A-Level数学 分部积分 换元积分 定积分

引言 Introduction

积分是A-Level数学中最重要的技能之一。在纯数学2和3（P2/P3）中，你需要掌握多种积分技巧来解决复杂的函数积分问题。本文聚焦三个核心方法：分部积分法（Integration by Parts）、换元积分法（Integration by Substitution）以及定积分的应用。这些方法不仅出现在纯数学考试中，也是力学和统计学的计算基础。

Integration is one of the most important skills in A-Level Mathematics. In Pure Mathematics 2 and 3 (P2/P3), you need to master multiple integration techniques to solve complex function integrals. This article focuses on three core methods: Integration by Parts, Integration by Substitution, and the application of Definite Integrals. These methods appear not only in pure mathematics exams but also serve as the computational foundation for mechanics and statistics.

分部积分法 Integration by Parts

分部积分法源于微分的乘法法则。如果u和v都是x的函数，那么乘积法则告诉我们：(uv)’ = u’v + uv’。重新排列得到分部积分公式：∫ u dv = uv – ∫ v du。这个公式将复杂的积分转化为可能更简单的形式。

Integration by Parts derives from the product rule of differentiation. If u and v are both functions of x, the product rule tells us that (uv)’ = u’v + uv’. Rearranging gives the Integration by Parts formula: ∫ u dv = uv – ∫ v du. This formula transforms a difficult integral into a potentially simpler form.

选择u和dv是关键步骤。一个常用的记忆法是”LIATE”规则：对数（Logarithmic）、反三角（Inverse trig）、代数（Algebraic）、三角（Trigonometric）、指数（Exponential）。排在LIATE前面的函数通常设为u。例如，对于 ∫ x e^x dx，x是代数的，e^x是指数的，根据LIATE，设u = x，dv = e^x dx。

Choosing u and dv is the critical step. A commonly used mnemonic is the “LIATE” rule: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Functions appearing earlier in LIATE are typically set as u. For example, for ∫ x e^x dx, x is algebraic and e^x is exponential; according to LIATE, set u = x and dv = e^x dx.

让我们看一个经典例子。计算 ∫ x sin x dx。设 u = x, dv = sin x dx。那么 du = dx, v = -cos x。代入公式：∫ x sin x dx = x(-cos x) – ∫ (-cos x) dx = -x cos x + sin x + C。验证：对结果求导应该得到 x sin x。

Let us look at a classic example. Evaluate ∫ x sin x dx. Set u = x, dv = sin x dx. Then du = dx, v = -cos x. Substituting into the formula: ∫ x sin x dx = x(-cos x) – ∫ (-cos x) dx = -x cos x + sin x + C. Verification: differentiating the result should yield x sin x.

有时需要重复使用分部积分。例如 ∫ x^2 e^x dx。第一次分部积分后得到 x^2 e^x – ∫ 2x e^x dx，然后对 ∫ x e^x dx 再次使用分部积分。这种”表格法”（Tabular Method）可以高效处理多项式乘以指数或三角函数的积分。

Sometimes Integration by Parts must be applied repeatedly. For example, ∫ x^2 e^x dx. After the first integration by parts we get x^2 e^x – ∫ 2x e^x dx, then apply integration by parts again to ∫ x e^x dx. The “Tabular Method” can efficiently handle integrals of polynomials multiplied by exponentials or trigonometric functions.

还有一个重要的技巧：循环分部积分。当 ∫ e^x sin x dx 出现时，两次分部积分后会回到原积分形式，从而解出方程。设 I = ∫ e^x sin x dx，经过两次分部积分得到 I = e^x sin x – e^x cos x – I，所以 2I = e^x(sin x – cos x) + C，即 I = (1/2)e^x(sin x – cos x) + C。

There is also an important technique: cyclic integration by parts. When ∫ e^x sin x dx appears, after two rounds of integration by parts you return to the original integral, allowing you to solve an equation. Let I = ∫ e^x sin x dx; after two applications we get I = e^x sin x – e^x cos x – I, so 2I = e^x(sin x – cos x) + C, giving I = (1/2)e^x(sin x – cos x) + C.

对数函数的积分也需要分部积分法，因为 ln x 本身不是基本积分公式中的一部分。计算 ∫ ln x dx 时，设 u = ln x，dv = dx，那么 du = (1/x)dx，v = x。代入公式得到 ∫ ln x dx = x ln x – ∫ x·(1/x) dx = x ln x – x + C。这个结果值得记住。

Integrating logarithmic functions also requires Integration by Parts, since ln x does not appear in the basic integration formula list. To evaluate ∫ ln x dx, set u = ln x, dv = dx; then du = (1/x)dx, v = x. Substituting yields ∫ ln x dx = x ln x – ∫ x·(1/x) dx = x ln x – x + C. This result is worth memorising.

换元积分法 Integration by Substitution

换元积分法是链式法则的反向操作。如果被积函数包含复合函数的形式 f(g(x))g'(x)，那么设 u = g(x)，积分变为 ∫ f(u) du。在A-Level考试中，题目通常会直接给出代换：例如”Use the substitution u = √x”。

Integration by Substitution is the reverse of the chain rule. If the integrand contains a composite function of the form f(g(x))g'(x), then setting u = g(x) transforms the integral into ∫ f(u) du. In A-Level exams, the substitution is usually given directly: for example, “Use the substitution u = √x”.

标准步骤是：设 u = 某个关于x的表达式，计算 du/dx，将 dx 替换为 du ÷ (du/dx)，将所有x替换为u的表达式，求出关于u的积分，最后代回x。对于定积分，还需要同时变换积分限。

The standard procedure is: set u = some expression in x, compute du/dx, replace dx with du ÷ (du/dx), replace all x with expressions in u, evaluate the integral in terms of u, and finally substitute back to x. For definite integrals, you must also transform the limits of integration.

例题：使用代换 u = 3x + 2 计算 ∫ (3x + 2)^5 dx。du/dx = 3，所以 dx = du/3。积分变为 ∫ u^5 · (du/3) = (1/3) ∫ u^5 du = (1/3)(u^6/6) + C = u^6/18 + C = (3x+2)^6/18 + C。

Example: Use the substitution u = 3x + 2 to evaluate ∫ (3x + 2)^5 dx. du/dx = 3, so dx = du/3. The integral becomes ∫ u^5 · (du/3) = (1/3) ∫ u^5 du = (1/3)(u^6/6) + C = u^6/18 + C = (3x+2)^6/18 + C.

三角代换是换元积分法的重要子类。当被积函数包含 √(a^2-x^2)、√(a^2+x^2) 或 √(x^2-a^2) 时，分别使用 x = a sin θ、x = a tan θ 或 x = a sec θ。例如，∫ dx/√(1-x^2)，设 x = sin θ，dx = cos θ dθ，积分变为 ∫ cos θ dθ/√(1-sin^2θ) = ∫ cos θ/cos θ dθ = ∫ dθ = θ + C = arcsin x + C。

Trigonometric substitution is an important subclass of the substitution method. When the integrand contains √(a^2-x^2), √(a^2+x^2), or √(x^2-a^2), use x = a sin θ, x = a tan θ, or x = a sec θ respectively. For example, ∫ dx/√(1-x^2), set x = sin θ, dx = cos θ dθ; the integral becomes ∫ cos θ dθ/√(1-sin^2θ) = ∫ cos θ/cos θ dθ = ∫ dθ = θ + C = arcsin x + C.

A-Level考试中最常见的换元类型是线性代换（u = ax + b）和根式代换（u = √x）。在处理定积分的换元时，不要忘记更改积分上下限。如果原积分限是 x = a 到 x = b，新积分限应该是 u(a) 到 u(b)。

The most common substitution types in A-Level exams are linear substitution (u = ax + b) and radical substitution (u = √x). When handling definite integrals with substitution, do not forget to change the limits of integration. If the original limits are x = a to x = b, the new limits should be u(a) to u(b).

一个更具挑战的例子：使用代换 u = 2x – 1 计算 ∫[1,5] x√(2x-1) dx。当 x = 1 时 u = 1，当 x = 5 时 u = 9。由 u = 2x – 1 得 x = (u+1)/2，dx = du/2。积分变为 ∫[1,9] ((u+1)/2)·√u·(du/2) = (1/4)∫[1,9] (u^(3/2) + u^(1/2)) du = (1/4)[(2/5)u^(5/2) + (2/3)u^(3/2)]₁⁹。代入计算得最终数值结果。

A more challenging example: use the substitution u = 2x – 1 to evaluate ∫[1,5] x√(2x-1) dx. When x = 1, u = 1; when x = 5, u = 9. From u = 2x – 1 we get x = (u+1)/2, dx = du/2. The integral becomes ∫[1,9] ((u+1)/2)·√u·(du/2) = (1/4)∫[1,9] (u^(3/2) + u^(1/2)) du = (1/4)[(2/5)u^(5/2) + (2/3)u^(3/2)]₁⁹. Substituting the limits yields the final numerical result.

定积分应用 Definite Integral Applications

定积分在A-Level中有三个核心应用：计算曲线下面积、旋转体积和参数方程下的面积。曲线 y = f(x) 从 x = a 到 x = b 下的面积由 ∫[a,b] f(x) dx 给出。如果曲线在x轴下方，积分为负值，需要取绝对值。

Definite integrals have three core applications in A-Level: calculating area under a curve, volume of revolution, and area under parametric equations. The area under the curve y = f(x) from x = a to x = b is given by ∫[a,b] f(x) dx. If the curve lies below the x-axis, the integral is negative and you must take the absolute value.

两条曲线之间的面积需要格外小心。若 y = f(x) 在 y = g(x) 上方，从 x = a 到 x = b 之间的面积为 ∫[a,b] [f(x) – g(x)] dx。必须先画出草图，确定哪条曲线在上方、交点的x坐标，然后分段积分。

The area between two curves requires extra care. If y = f(x) is above y = g(x), the area between them from x = a to x = b is ∫[a,b] [f(x) – g(x)] dx. You must first sketch the curves, determine which is above, find the x-coordinates of intersection points, then integrate by sections.

旋转体积公式是A-Level的重点考点。曲线 y = f(x) 绕x轴旋转360度，从 x = a 到 x = b，形成的体积为 V = π ∫[a,b] [f(x)]^2 dx。注意：必须先平方再积分，顺序不能颠倒。如果是两条曲线之间的区域绕x轴旋转，体积为 π ∫[a,b] ([f(x)]^2 – [g(x)]^2) dx。

The volume of revolution formula is a key exam topic in A-Level. When the curve y = f(x) is rotated 360 degrees about the x-axis from x = a to x = b, the volume generated is V = π ∫[a,b] [f(x)]^2 dx. Note: you must square first and then integrate; the order cannot be reversed. If the region between two curves is rotated about the x-axis, the volume is π ∫[a,b] ([f(x)]^2 – [g(x)]^2) dx.

参数方程下的积分需要将积分变量从x转换为参数t。如果 x = f(t), y = g(t)，那么 ∫ y dx = ∫ g(t) f'(t) dt，积分限相应转换为t值。例如，参数曲线 x = t^2, y = 2t，在 t = 0 到 t = 2 之间，dx/dt = 2t，面积 = ∫ y dx = ∫[0,2] 2t·2t dt = ∫[0,2] 4t^2 dt = [4t^3/3]₀² = 32/3。

Integration under parametric equations requires converting the integration variable from x to the parameter t. If x = f(t), y = g(t), then ∫ y dx = ∫ g(t) f'(t) dt, with the limits converted to t values accordingly. For example, the parametric curve x = t^2, y = 2t, from t = 0 to t = 2: dx/dt = 2t, area = ∫ y dx = ∫[0,2] 2t·2t dt = ∫[0,2] 4t^2 dt = [4t^3/3]₀² = 32/3.

旋转体积的完整计算示例：曲线 y = √x 从 x = 0 到 x = 4 绕x轴旋转。V = π ∫[0,4] (√x)^2 dx = π ∫[0,4] x dx = π [x^2/2]₀⁴ = π · 8 = 8π 立方单位。若绕y轴旋转，需先将函数改写为 x = y^2，然后使用公式 V = π ∫[c,d] x^2 dy。

A complete volume of revolution example: the curve y = √x from x = 0 to x = 4, rotated about the x-axis. V = π ∫[0,4] (√x)^2 dx = π ∫[0,4] x dx = π [x^2/2]₀⁴ = π · 8 = 8π cubic units. If rotated about the y-axis, first rewrite the function as x = y^2, then use V = π ∫[c,d] x^2 dy.

考试技巧 Exam Tips

在A-Level考试中，积分题通常占纯数学试卷的15-20%。常见陷阱包括：忘记在不定积分后加积分常数C、换元定积分时忘记变换积分限、旋转体积计算中忘记乘以π。另一点是要仔细检查被积函数中的负号：这是最容易出错的地方。

In A-Level exams, integration questions typically account for 15-20% of pure mathematics papers. Common pitfalls include: forgetting to add the constant of integration C after indefinite integration, forgetting to transform limits when using substitution on definite integrals, and forgetting to multiply by π in volume of revolution calculations. Another point is to carefully check negative signs in the integrand ： this is where mistakes occur most easily.

建议的复习策略：先掌握基本积分公式（多项式、三角、指数、对数），然后练习分部积分（至少20道不同题型），接着是换元积分（特别注意三角代换），最后才是定积分的综合应用题。每道题完成后都要用微分验证结果。

Recommended revision strategy: first master the basic integration formulas (polynomial, trigonometric, exponential, logarithmic), then practice integration by parts (at least 20 different question types), followed by integration by substitution (pay special attention to trigonometric substitution), and finally comprehensive application problems of definite integrals. Verify every result by differentiation after completing each question.

最后要特别注意题目的表述方式。如果题目说”Hence evaluate”（由此计算），说明题目要求你使用前面部分的结果，通常会节省大量计算时间。如果题目说”Find the exact value”（求精确值），答案必须保留根号和π等精确形式，不能用小数近似。区分”prove that”和”show that”两种问法：前者从已知条件出发推导，后者可以代入验证。

Finally, pay special attention to how questions are phrased. If a question says “Hence evaluate”, it means you must use the result from the previous part, which usually saves substantial calculation time. If it says “Find the exact value”, the answer must be kept in exact form with radicals and π, not decimal approximations. Distinguish between “prove that” and “show that”: the former requires deriving from known conditions, while the latter allows verification by substitution.