A-Level Mathematics: Core Integration Concepts & Problem-Solving Methods | A-Level 数学：积分核心考点与解题方法

📚 A-Level Mathematics: Core Integration Concepts & Problem-Solving Methods | A-Level 数学：积分核心考点与解题方法

Integration is one of the most important topics in A-Level Mathematics. It is the reverse of differentiation and is used to calculate areas, volumes, and solve differential equations. To score well in CIE exams, you must understand both the techniques and the underlying principles.

积分是 A-Level 数学中最重要的内容之一。它是微分的逆运算，用于计算面积、体积以及求解微分方程。要在 CIE 考试中取得高分，你既需要掌握各种积分技巧，也需要理解其背后的原理。

---

1. What Is Integration? | 积分是什么

Integration is the process of finding a function when its derivative is known. The result of an indefinite integral is a family of functions that differ by a constant. For example, the derivative of x² is 2x, so the integral of 2x is x² + C, where C is the constant of integration.

积分是在已知导数的情况下求原函数的过程。不定积分的结果是一族相差一个常数的函数。例如，x² 的导数是 2x，所以 2x 的积分是 x² + C，其中 C 是积分常数。

• The symbol ∫ means "the integral of", and dx tells us the variable we are integrating with respect to.

  符号 ∫ 表示“对……积分”，dx 表示我们针对哪个变量进行积分。

• If F'(x) = f(x), then ∫ f(x) dx = F(x) + C.

  若 F'(x) = f(x)，则 ∫ f(x) dx = F(x) + C。

---

2. Indefinite Integration: Reverse of Differentiation | 不定积分：微分的逆运算

The power rule for integration states that for any real number n ≠ -1:

积分幂法则指出，对任意实数 n ≠ -1：

∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C

Always remember to add the constant of integration C when you perform an indefinite integral. This constant represents all possible vertical shifts of the original function.

进行不定积分时，务必加上积分常数 C。这个常数表示原函数所有可能的垂直平移。

• For example, ∫ x³ dx = x⁴/4 + C.

  例如，∫ x³ dx = x⁴/4 + C。

• Constant multiples factor out: ∫ 3x² dx = 3 ∫ x² dx = x³ + C.

  常数倍数可以提出：∫ 3x² dx = 3 ∫ x² dx = x³ + C。

• For n = -1, the rule does not apply. Instead, ∫ x⁻¹ dx = ∫ 1/x dx = ln|x| + C.

  当 n = -1 时，该法则不适用。此时 ∫ x⁻¹ dx = ∫ 1/x dx = ln|x| + C。

---

3. Definite Integrals & The Fundamental Theorem | 定积分与微积分基本定理

erentiation: to evaluate ∫ₐᵇ f(x) dx, find an antiderivative F(x) and compute F(b) - F(a).

定积分带有上限和下限，其结果是一个数值而非函数。微积分基本定理将积分与微分联系起来：求 ∫ₐᵇ f(x) dx 时，先找原函数 F(x)，再计算 F(b) - F(a)。

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) - F(a)

• Example: ∫₁³ x² dx = [x³/3]₁³ = (27/3) - (1/3) = 26/3.

  例如：∫₁³ x² dx = [x³/3]₁³ = 27/3 - 1/3 = 26/3。

• If the upper and lower limits are equal, the integral is zero.

  若上下限相等，则定积分值为零。

• Reversing the limits changes the sign: ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx.

  交换上下限会改变符号：∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx。

---

4. Integrating Composite Functions (Reverse Chain Rule) | 复合函数积分（反向链式法则）

Many functions are of the form f' When you see a function multiplied by its own derivative, you can integrate by inspection.

许多函数具有 f'(x) × g(f(x)) 的形式。例如，(x² + 1)⁵ 的导数是 5(x² + 1)⁴ × 2x。当题目中出现一个函数乘以其自身的导数时，可以直接“观察”积分。

• If ∫ f'(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ / (n+1) + C.

  若 ∫ f'(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ / (n+1) + C。

• For example, ∫ 2x(x² + 1)³ dx = (x² + 1)⁴ / 4 + C.

  例如，∫ 2x(x² + 1)³ dx = (x² + 1)⁴ / 4 + C。

• For trigonometric forms: ∫ f'(x) sin(f(x)) dx = -cos(f(x)) + C, and ∫ f'(x) cos(f(x)) dx = sin(f(x)) + C.

  对于三角函数形式：∫ f'(x) sin(f(x)) dx = -cos(f(x)) + C，∫ f'(x) cos(f(x)) dx = sin(f(x)) + C。

---

5. Integration by Substitution | 换元积分法

ome function of x, differentiate to find du, and rewrite the whole integral in terms of u.

换元积分法通过更改变量来化简积分。你选择 u = x 的某个函数，求微分得到 du，然后将整个积分改写为关于 u 的形式。

∫ f(g(x)) g'(x) dx = ∫ f(u) du, where u = g(x)

• en you substitute. For example, if u = x² and x goes from 0 to 2, then u goes from 0 to 4.

  对于定积分，换元时也要更换上下限。例如，若 u = x²，且 x 从 0 到 2，则 u 从 0 到 4。

• Example: ∫ 2x√x² + 1 dx. Let u = x² + 1, du = 2x dx. Then the integral becomes ∫ √u du = (2/3)u³ᐟ² + C = (2/3)(x² + 1)³ᐟ² + C.

  例如：∫ 2x√x² + 1 dx。令 u = x² + 1，du = 2x dx，则积分变为 ∫ √u du = (2/3)u³ᐟ² + C = (2/3)(x² + 1)³ᐟ² + C。

• When you see x × (linear in x²), try substitution x² + a.

  当看到 x 乘以关于 x² 的一次式时，可尝试令 u = x² + a。

---

6. Integration by Parts | 分部积分法

Integration by parts is used when the integrand is a product of two different types of functions, such as x ln x or x eˣ. The formula comes from the product rule for differentiation.

分部积分法适用于被积函数是两类不同函数乘积的情况，例如 x ln x 或 x eˣ。该公式来源于乘法求导法则。

∫ u dv = uv - ∫ v du

• Choose u to be the function that becomes simpler when differentiated, such as x or ln x. Choose dv to be the remaining part, which you can integrate easily.

  一般选择求导后变得更简单的函数作为 u，例如 x 或 ln x；选择容易积分的部分作为 dv。

• Example: ∫ x eˣ dx. Let u = x and dv = eˣ dx. Then du = dx and v = eˣ. So ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C.

  例如：∫ x eˣ dx。令 u = x，dv = eˣ dx，则 du = dx，v = eˣ。因此 ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C。

• For ln x, write it as 1 × ln x, then integrate by parts.

  对 ln x 积分时，可把被积函数写成 1 × ln x，再用分部积分法。

---

7. Integrating Rational Functions & Partial Fractions | 有理函数与部分分式积分

Rational functions are fractions of polynomials. To integrate them, rewrite the fraction as partial fractions, then integrate each term using standard rules. This technique is frequently tested in CIE A-Level papers.

有理函数是多项式的比值。积分时先把该分式改写为部分分式，再用标准法则逐项积分。这是 CIE A-Level 考试中常见的考点。

Form of denominator 分母形式	Partial fraction form 部分分式形式
(ax + b)(cx + d)	A/(ax + b) + B/(cx + d)
(ax + b)²	A/(ax + b) + B/(ax + b)²

• After decomposition, use ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C.

  分解后，使用 ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C。

• For a repeated linear factor, include a term with the squared denominator in the denominator.

  若存在重复线性因式，部分分式中要包含以平方因式为分母的项。

• Always check whether the degree of the numerator is less than the degree of the denominator before applying partial fractions. If not, perform polynomial long division first.

  在应用部分分式前，务必检查分子的次数是否低于分母的次数。若不是，需要先进行多项式长除法。

---

广告 / Sponsored

8. Area Between Curves | 曲线间面积

e area between two curves, integrate the difference of the functions, with the upper function minus the lower function.

定积分 ∫ₐᵇ f(x) dx 给出曲线 y = f(x) 与 x 轴之间的有向面积。求两条曲线之间的面积时，需要积分两函数之差，即上方曲线减下方曲线。

Area = ∫ₐᵇ [f(x) - g(x)] dx

• First find the x-coordinates of the intersection points of the two curves. These become the limits a and b.

  首先要找出两条曲线交点的 x 坐标，作为积分上下限 a 和 b。

• If the curve crosses the x-axis, split the integral into separate parts so that each part has a constant sign.

  若曲线与 x 轴相交，应分段积分，确保每一段内的函数值符号不变。

• For areas between y = f(x) and the y-axis, integrate with respect to y, or use x = f⁻¹(y).

  若求曲线 y = f(x) 与 y 轴之间的面积，可对 y 积分，或使用反函数 x = f⁻¹(y)。

---

9. Volumes of Revolution | 旋转体体积

he x-axis, the volume of the solid formed is found by integrating πy² with respect to x. Similarly, rotation about the y-axis uses πx² with respect to y.

当曲线 y = f(x) 绕 x 轴旋转时，所形成立体的体积可通过 πy² 对 x 积分得到。类似地，绕 y 轴旋转时则使用 πx² 对 y 积分。

V = ∫ₐᵇ π y² dx (about x-axis)

• Always square the whole function before multiplying by π. A common mistake is to integrate πf(x) rather than π[f(x)]².

  计算时一定要先对完整函数平方，再乘以 π。常见错误是积分 πf(x) 而不是 π[f(x)]²。

• xis, rewrite y = f(x) as x = g(y), then integrate πx² dy between the appropriate y-limits.

  若区域绕 y 轴旋转，先将 y = f(x) 改写为 x = g(y)，再在相应的 y 上下限之间对 πx² dy 积分。

• ves is rotated, subtract the inner volume from the outer volume: V = ∫π[f(x)² - g(x)²] dx.

  若两曲线之间的区域旋转，则用外体积减去内体积：V = ∫π[f(x)² - g(x)²] dx。

---

10. Trapezium Rule (Numerical Integration) | 梯形法则（数值积分）

Some functions cannot be integrated exactly using standard A-Level methods. In such cases, the trapezium rule gives an approximate value of a definite integral by dividing the area under the curve into strips and approximating each strip by a trapezium.

有些函数无法用 A-Level 的标准方法精确积分。此时，梯形法则通过将曲线下的面积分成若干窄条，并用梯形近似每个窄条，从而给出定积分的近似值。

∫ₐᵇ y dx ≈ (h/2)[y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)], where h = (b - a)/n

• The larger the number of strips n, the more accurate the approximation.

  窄条数量 n 越大，近似值越精确。

• You should be able to list the y-values from the equation or from a given table.

  你应该能够从方程或给定表格中列出所有 y 值。

• Do not forget that the first and last y-values have coefficient 1, while all interior y-values have coefficient 2.

  不要忘记首尾两个 y 值的系数是 1，而中间所有 y 值的系数是 2。

---

11. Differential Equations: Separation of Variables | 微分方程：变量分离

A differential equation contains derivatives such as dy/dx. In CIE A-Level, you often solve these by separating the variables: move all y terms to one side and all x terms to the other, then integrate both sides.

微分方程包含 dy/dx 等导数项。在 CIE A-Level 中，通常通过变量分离来求解：把所有含 y 的项移到一边，所有含 x 的项移到另一边，然后两边积分。

If dy/dx = f(x)g(y), then ∫ 1/g(y) dy = ∫ f(x) dx

• Include the constant of integration after integrating the right-hand side.

  对右边积分后需要加上积分常数。

• If initial conditions are given, substitute them to find the value of C. This produces a particular solution.

  若给出初值条件，代入即可求出常数 C，从而得到特解。

• Be careful with absolute values: ln|y| often appears, but the modulus sign may be omitted if the context ensures y is positive.

  注意绝对值：结果中常出现 ln|y|，若题目条件保证 y 为正数，则可省略绝对值符号。

---

12. Common Exam Pitfalls & Final Tips | 常见考点陷阱与最终建议

Integration questions in CIE A-Level are designed to test accuracy and careful reasoning. The most common errors are forgetting the constant, misapplying the power rule, and ignoring absolute values in logarithmic answers.

CIE A-Level 的积分题旨在考查准确性和严谨推理。最常见的错误包括忘记积分常数、误用幂法则，以及对数答案中忽略绝对值。

Common mistake 常见错误	Correct approach 正确做法
∫ 1/x dx = x⁰/0 + C	∫ 1/x dx = ln|x| + C
∫ e²ˣ dx = e²ˣ/2 + C (forgetting the inner derivative)	∫ e²ˣ dx = e²ˣ/2 + C
Omitting C in indefinite integrals	Always write + C

• ng (ax + b)ⁿ, divide by the coefficient of x: ∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n+1)] + C.

  积分 (ax + b)ⁿ 时，需要除以 x 的系数：∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n+1)] + C。

• Check the result by differentiating your answer. The derivative of your answer should be the original integrand.

  通过求导来检验结果。你的答案的导数应该等于原被积函数。

• Practice many past-paper questions, especially those mixing integration with trigonometry, logarithms, and exponentials.

  多练习历年真题，尤其是将积分与三角函数、对数函数和指数函数结合起来的题目。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）