pdfjoiner_(4)-173 | Edexcel A-Level 积分方法全览及题173详解

📚 pdfjoiner_(4)-173 | Edexcel A-Level 积分方法全览及题173详解

Welcome to this focused revision guide on integration techniques for Edexcel A-Level Mathematics. In many past paper compilations, such as the one labelled ‘pdfjoiner_(4)-173’, students encounter a wide range of integral problems that require a solid command of methods from basic formulas to advanced substitutions. This article will systematically review key integration strategies and conclude with a step-by-step solution to Question 173 – a definite integral that brings together substitution and algebraic simplification.

欢迎阅读本Edexcel A-Level数学积分技巧专项复习指南。在许多如“pdfjoiner_(4)-173”这样的历年真题汇编中，学生们会遇到各种各样的积分问题，需要从基本公式到高级代换的扎实掌握。本文系统回顾关键的积分方法，并以问题173的逐步解答收尾——这是一个结合了代换与代数化简的定积分问题。

1. Integration in the Edexcel P4 Syllabus | Edexcel P4 大纲中的积分

Integration forms a cornerstone of the Edexcel Pure Mathematics 4 (P4) syllabus. Candidates are expected to integrate a wide variety of functions, apply definite integration to find areas under curves and volumes of revolution, and solve first-order differential equations using an integrating factor. Mastery of integration techniques is directly tested in both structured and unstructured problems across the paper.

积分是Edexcel纯数学P4大纲的基石。考生需要能够对各类函数进行积分，运用定积分求曲线下面积和旋转体体积，并利用积分因子求解一阶微分方程。对积分技巧的掌握在试卷中的结构化和非结构化问题中都会直接考查。

2. Standard Integration Formulas | 标准积分公式

The following standard integrals must be memorised and instantly recognised. They serve as building blocks for more complex problems.

以下标准积分公式必须熟记并能立即辨认，它们是构建更复杂问题的基础。

∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, n ≠ -1

∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C，n ≠ -1

∫ 1/x dx = ln|x| + C

∫ eˣ dx = eˣ + C ∫ aˣ dx = aˣ/ln(a) + C

∫ sin x dx = -cos x + C ∫ cos x dx = sin x + C

∫ sec² x dx = tan x + C ∫ cosec² x dx = -cot x + C

3. Integration by Substitution | 代换积分法

Substitution is used when an integral contains a composite function. The idea is to replace a complicated inner expression by a single variable u, transform the differential dx, and evaluate the resulting simpler integral.

当被积函数含有复合函数时，可使用代换法。其思路是将复杂的内层表达式用单变量u代替，转换微分dx，并计算所得简单积分。

For example, consider ∫ 2x√(x²+1) dx. Let u = x²+1, then du = 2x dx. The integral becomes ∫ √u du = (2/3)u^(3/2) + C = (2/3)(x²+1)^(3/2) + C. Always remember to revert to the original variable in indefinite integrals.

例如，考虑 ∫ 2x√(x²+1) dx。令 u = x²+1，则 du = 2x dx。积分化为 ∫ √u du = (2/3)u^(3/2) + C = (2/3)(x²+1)^(3/2) + C。在不定积分中，一定要记得换回原变量。

For definite integrals, change the limits as well. If ∫₁² … the limits in u must be calculated from the substitution equation. This avoids the need to return to x at the end.

对于定积分，需要同时转换上下限。如果 ∫₁² … 则必须根据代换方程计算u的上下限，这样最终无需换回x。

4. Integration by Parts | 分部积分法

Integration by parts is derived from the product rule for differentiation. The formula is:

分部积分法源自乘法微分法则。公式如下：

∫ u dv = uv – ∫ v du

Typically, you choose u to be a function that becomes simpler upon differentiation (such as ln x, xⁿ, or inverse trig functions), and dv is the remaining part. A classic example is ∫ x eˣ dx. Let u = x, dv = eˣ dx; then du = dx, v = eˣ. The integral becomes x eˣ – ∫ eˣ dx = x eˣ – eˣ + C.

通常，选择微分后变得简单的函数为u（如 ln x, xⁿ 或反三角函数），dv为剩余部分。经典示例：∫ x eˣ dx。令 u = x, dv = eˣ dx，则 du = dx, v = eˣ。积分化为 x eˣ – ∫ eˣ dx = x eˣ – eˣ + C。

For repeated integration by parts, such as ∫ x² sin x dx, apply the method twice, carefully tracking signs and coefficients.

对于需要多次分部积分的情况，如 ∫ x² sin x dx，需连续两次应用该方法，并小心处理符号和系数。

5. Using Partial Fractions | 部分分式积分

Rational functions where the denominator factorises are often integrated by expressing them as a sum of simpler fractions. For instance, to integrate (3x+5)/(x² – x – 2), factorise the denominator as (x-2)(x+1) and write:

当分母可分解因式的有理函数，常将其表示为简单分式之和进行积分。例如，积分 (3x+5)/(x² – x – 2)，将分母因式分解为 (x-2)(x+1) 并写出：

(3x+5)/((x-2)(x+1)) = A/(x-2) + B/(x+1)

Solving for A and B yields A=11/3, B=-2/3. The integral becomes (11/3)ln|x-2| – (2/3)ln|x+1| + C. This technique is frequently examined when combined with substitution or after polynomial division.

解得 A=11/3, B=-2/3。积分化为 (11/3)ln|x-2| – (2/3)ln|x+1| + C。该技巧常与代换或多项式除法结合考查。

6. Trigonometric Integrals | 三角函数的积分

Integrating powers of sine and cosine often requires the use of double-angle identities. For example, sin²x can be integrated by writing it as (1 – cos 2x)/2, which gives ∫ sin²x dx = (1/2)x – (1/4)sin 2x + C.

对正弦和余弦的次幂积分常需使用倍角公式。例如，sin²x 可改写为 (1 – cos 2x)/2，从而 ∫ sin²x dx = (1/2)x – (1/4)sin 2x + C。

Another useful identity is cos²x = (1 + cos 2x)/2. For products like sin³x, separate one power and use the identity sin²x = 1 – cos²x, then substitute u = cos x.

另一个有用公式是 cos²x = (1 + cos 2x)/2。对于 sin³x 这类乘积，可分离出一个幂次，利用 sin²x = 1 – cos²x，再代换 u = cos x。

7. Definite Integrals and Area Problems | 定积分与面积问题

The definite integral ∫ₐᵇ f(x) dx gives the signed area between the curve y = f(x) and the x-axis from x = a to x = b. When the curve crosses the axis, you must split the integral where f(x) changes sign to find the total area.

定积分 ∫ₐᵇ f(x) dx 给出曲线 y = f(x) 与 x 轴之间从 x=a 到 x=b 的符号面积。当曲线穿越 x 轴时，必须在 f(x) 变号处拆分积分以求总面积。

Volumes of revolution about the x-axis are computed as V = π ∫ₐᵇ [f(x)]² dx. About the y-axis, the formula becomes V = π ∫ᵤᵥ [g(y)]² dy after rearranging the function. These applications rely heavily on accurate integration techniques.

绕 x 轴旋转的体积计算公式为 V = π ∫ₐᵇ [f(x)]² dx。绕 y 轴旋转时，需将函数整理为 x=g(y)，应用 V = π ∫ᵤᵥ [g(y)]² dy。这些应用高度依赖精确的积分技巧。

8. Step-by-Step Solution to Question 173 | 题173 逐步详解

Question 173 (from ‘pdfjoiner_(4)-173’): Evaluate the definite integral ∫₀¹ √x/(1+√x) dx using the substitution u = 1 + √x.

问题173（选自“pdfjoiner_(4)-173”）：用代换 u = 1 + √x 计算定积分 ∫₀¹ √x/(1+√x) dx。

Step 1: Express everything in terms of u. From u = 1 + √x, we get √x = u – 1, x = (u – 1)², and dx = 2(u – 1) du. Change the limits: when x = 0, u = 1; when x = 1, u = 2.

步骤1：将所有变量用u表示。由 u = 1 + √x 得 √x = u – 1，x = (u – 1)²，dx = 2(u – 1) du。转换积分限：x = 0 → u = 1；x = 1 → u = 2。

Step 2: Substitute into the integral. The integrand √x/(1+√x) becomes (u – 1)/u. The integral transforms to ∫₁² (u – 1)/u * 2(u – 1) du = 2 ∫₁² (u – 1)² / u du.

步骤2：代入积分。被积函数 √x/(1+√x) 变为 (u – 1)/u。积分化为 ∫₁² (u – 1)/u * 2(u – 1) du = 2 ∫₁² (u – 1)² / u du。

Step 3: Expand and simplify. (u – 1)² = u² – 2u + 1, so the integrand is (u² – 2u + 1)/u = u – 2 + 1/u. The integral becomes 2 ∫₁² (u – 2 + 1/u) du.

步骤3：展开并化简。(u – 1)² = u² – 2u + 1，故被积函数为 (u² – 2u + 1)/u = u – 2 + 1/u。积分变为 2 ∫₁² (u – 2 + 1/u) du。

Step 4: Integrate term by term. ∫ u du = (1/2)u², ∫ -2 du = -2u, ∫ 1/u du = ln|u|. Hence the antiderivative is 2[(1/2)u² – 2u + ln|u|] evaluated from 1 to 2.

步骤4：逐项积分。∫ u du = (1/2)u²，∫ -2 du = -2u，∫ 1/u du = ln|u|。因此原函数为 2[(1/2)u² – 2u + ln|u|]，上下限从1到2。

Step 5: Apply the limits. At u = 2: (1/2)(4) – 4 + ln 2 = 2 – 4 + ln 2 = -2 + ln 2. At u = 1: (1/2)(1) – 2 + ln 1 = 0.5 – 2 + 0 = -1.5. Subtract and multiply by 2: 2[(-2 + ln 2) – (-1.5)] = 2[ln 2 – 0.5] = 2 ln 2 – 1.

步骤5：代入上下限。u=2时：(1/2)(4) – 4 + ln 2 = 2 – 4 + ln 2 = -2

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pdfjoiner_(4)-173 | Edexcel A-Level 积分方法全览及题173详解