Integration for GCSE CCEA Mathematics | GCSE CCEA 数学：积分 考点精讲

📚 Integration for GCSE CCEA Mathematics | GCSE CCEA 数学：积分 考点精讲

Integration is one of the two cornerstones of calculus, the reverse process of differentiation. In the CCEA GCSE Mathematics Higher Tier, you will learn to find indefinite integrals of polynomial functions, use definite integration to calculate areas under curves and between curves, solve problems involving velocity and acceleration, and approximate areas using the trapezoidal rule. A clear grasp of integration not only boosts your exam performance but also prepares you for advanced study in maths and physics.

积分是微积分的两大基石之一，是微分的逆运算。在 CCEA GCSE 数学高等级考试中，你将学习如何求多项式函数的不定积分，利用定积分计算曲线下方以及曲线之间的面积，解决涉及速度与加速度的应用题，并利用梯形法则进行面积近似。扎实掌握积分不但能提升考试成绩，还为更高层次的数学和物理学习铺平道路。

1. What Is Integration? | 什么是积分？

Integration can be thought of as finding the ‘whole’ from its rate of change. If differentiation gives the gradient function f'(x), integration recovers the original function f(x). In geometry, a definite integral represents the signed area between a curve y = f(x) and the x‑axis over an interval [a, b].

积分可以理解为从变化率中找出“整体”。如果微分得到的是梯度函数 f'(x)，那么积分就能恢复出原来的函数 f(x)。在几何中，定积分表示区间 [a, b] 上曲线 y = f(x) 与 x 轴之间的有向面积。

The symbol for integration is ∫, introduced by Leibniz. An indefinite integral has no limits and includes a constant of integration +C. A definite integral has lower and upper limits a and b, and yields a numerical value.

积分符号为 ∫，由莱布尼茨引入。不定积分没有上下限，并带有积分常数 +C。定积分有下上限 a 与 b，结果是一个数值。

2. Power Rule for Integration | 幂函数的积分公式

The most important integration rule for GCSE is the power rule. For any real number n ≠ –1,

对 GCSE 而言，最重要的积分法则就是幂函数法则。对于任意 n ≠ –1 的实数，

∫ xⁿ dx = (xⁿ⁺¹)/(n + 1) + C

Simply increase the exponent by 1 and divide by the new exponent. Remember to add the constant of integration C for an indefinite integral. This rule works for negative and fractional powers as well, provided n ≠ –1.

只需将指数增加 1，再除以新的指数。不定积分一定要记得在末尾加上积分常数 C。这个法则对负指数和分数指数同样成立，只要指数不等于 –1。

Example:

例子：

• ∫ x³ dx = x⁴/4 + C
• ∫ √x dx = ∫ x^½ dx = x^(3/2) / (3/2) + C = (2/3) x^(3/2) + C
• ∫ 1/x² dx = ∫ x⁻² dx = x⁻¹/(-1) + C = –1/x + C

3. Integrating Constant and Multiple Terms | 常数与多项式的积分

The integral of a constant k with respect to x is kx + C, because the derivative of kx is k. For a sum or difference of terms, integrate each term separately and combine the constants into a single +C at the end.

常数 k 关于 x 的积分是 kx + C，因为 kx 的导数恰好为 k。对于多项式的和或差，可以逐项积分，最后将所有常数合并成一个 +C。

∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx

Example: ∫ (4x³ – 6x + 5) dx = x⁴ – 3x² + 5x + C.

示例：∫ (4x³ – 6x + 5) dx = x⁴ – 3x² + 5x + C.

Do not forget that constant multipliers can be taken outside the integral sign: ∫ k·f(x) dx = k ∫ f(x) dx.

不要忘记，常数因子可以提到积分号外面：∫ k·f(x) dx = k ∫ f(x) dx。

4. The Constant of Integration and Finding the Original Function | 积分常数与求原函数

When we integrate a derivative f'(x), we obtain a family of curves f(x) + C. A specific curve can be identified if an extra condition is given, for example a point (x₀, y₀) lying on the curve. Substitute the coordinates to find the value of C. This is a very common CCEA exam question.

对导数 f'(x) 积分，我们得到一族曲线 f(x) + C。如果题目给出了额外条件，比如曲线经过某点 (x₀, y₀)，就可以代入坐标求出 C 的值。这是 CCEA 考试中十分常见的题型。

Worked example: Given f'(x) = 6x² – 2x + 3 and the curve passes through (1, 6), find f(x).
Integrate: f(x) = 2x³ – x² + 3x + C.
Substitute x = 1, y = 6: 2(1) – 1 + 3 + C = 6 ⇒ 4 + C = 6 ⇒ C = 2.
Hence f(x) = 2x³ – x² + 3x + 2.

解题示例：已知 f'(x) = 6x² – 2x + 3，且曲线过点 (1, 6)，求 f(x)。
积分得：f(x) = 2x³ – x² + 3x + C。
代入 x = 1, y = 6：2(1) – 1 + 3 + C = 6 ⇒ 4 + C = 6 ⇒ C = 2。
因此 f(x) = 2x³ – x² + 3x + 2.

5. Definite Integration and the Area Under a Curve | 定积分与曲线下方面积

A definite integral evaluates the net area between the curve y = f(x), the x‑axis, and the vertical lines x = a and x = b. The formula is called the Newton–Leibniz theorem:

定积分计算的是曲线 y = f(x)、x 轴以及直线 x = a、x = b 围成的净面积，其公式被称为牛顿—莱布尼茨公式：

∫ₐᵇ f(x) dx = F(b) – F(a)

where F(x) is any antiderivative of f(x). The steps: find the indefinite integral (ignore +C), substitute the upper limit, subtract the value when the lower limit is substituted. Always use square brackets to show the antiderivative with limits.

其中 F(x) 是 f(x) 的任意一个原函数。步骤为：先求出不定积分（忽略 +C），代入上限计算值，再减去下限代入后的值。解题时务必用方括号写出带上下限的原函数。

Example: Find the area under y = 2x + 3 from x = 1 to x = 4.
∫₁⁴ (2x + 3) dx = [x² + 3x]₁⁴ = (16 + 12) – (1 + 3) = 28 – 4 = 24 square units.

示例：求 y = 2x + 3 在 x = 1 到 x = 4 之间与 x 轴围成的面积。
∫₁⁴ (2x + 3) dx = [x² + 3x]₁⁴ = (16 + 12) – (1 + 3) = 28 – 4 = 24 平方单位。

6. Areas Below the x‑Axis | x 轴下方的面积

If the curve lies below the x‑axis on an interval, the definite integral gives a negative value. The actual area is the absolute value of that integral. Always check where the graph crosses the x‑axis and split the calculation into separate positive and negative parts, then add the absolute areas.

如果曲线在某个区间内位于 x 轴的下方，定积分的结果为负值。此时真正的面积是积分的绝对值。务必先确定曲线与 x 轴的交点，将积分区间拆分为正值部分和负值部分，最后将各部分的绝对值相加。

For instance, to find the total area enclosed by y = x² – 4 between x = 1 and x = 3, note that the curve crosses the x‑axis at x = 2. Calculate ∫₁² (x² – 4) dx (negative) and ∫₂³ (x² – 4) dx (positive). The total area = |first part| + second part.

例如，求 y = x² – 4 在 x = 1 与 x = 3 之间包围的总面积。注意曲线在 x = 2 处穿过 x 轴。分别计算 ∫₁² (x² – 4) dx（负值）和 ∫₂³ (x² – 4) dx（正值）。总面积 = |第一部分| + 第二部分。

7. Area Between Two Curves | 两条曲线之间的面积

To find the area enclosed between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b, use the formula:

求上方曲线 y = f(x) 与下方曲线 y = g(x) 在 x = a 到 x = b 之间围成的面积时，可使用公式：

Area = ∫ₐᵇ [f(x) – g(x)] dx

Always identify which function is on top by selecting a test point within the interval. If the curves intersect between a and b, split the interval at the intersection points and calculate each sub‑area separately.

可通过在区间内取测试点来判断哪条曲线在上方。如果两条曲线在 a 与 b 之间有交点，则需在交点处拆分区间，分段计算每一部分的面积。

Example: Find the area enclosed by y = x + 3 and y = (x – 1)² from x = 0 to x = 2.
Identify upper function: at x = 1, x+3=4, (x-1)²=0, so upper is y = x+3.
Area = ∫₀² [(x + 3) – (x² – 2x + 1)] dx = ∫₀² (–x² + 3x + 2) dx = [–x³/3 + (3/2)x² + 2x]₀² = (–8/3 + 6 + 4) – 0 = 22/3 square units.

示例：求 y = x + 3 与 y = (x – 1)² 在 x = 0 到 x = 2 之间围成的面积。
判断上方函数：取 x = 1，x+3=4，(x-1)²=0，因此上方为 y = x+3。
面积 = ∫₀² [(x + 3) – (x² – 2x + 1)] dx = ∫₀² (–x² + 3x + 2) dx = [–x³/3 + (3/2)x² + 2x]₀² = (–8/3 + 6 + 4) – 0 = 22/3 平方单位。

8. Kinematics: Velocity and Acceleration | 运动学：速度与加速度

In CCEA GCSE, integration is often applied to motion problems. If you are given the velocity v(t) of a particle, integration gives the displacement s(t). If acceleration a(t) is given, the first integration gives velocity, and the second gives displacement. Initial conditions s(0) and v(0) are used to find constants.

在 CCEA GCSE 中，积分经常用于运动问题。若给出了质点的速度 v(t)，积分可得到位移 s(t)。若给了加速度 a(t)，一次积分得到速度，再次积分得到位移。利用初始条件 s(0) 和 v(0) 可求出积分常数。

Key relationships: v = ds/dt, a = dv/dt = d²s/dt². Hence, s = ∫ v dt and v = ∫ a dt.

关键关系：v = ds/dt，a = dv/dt = d²s/dt²。因此，s = ∫ v dt，v = ∫ a dt。

Worked example: A particle moves with acceleration a = 6t – 2. Its initial velocity is v(0) = 3 m/s and initial displacement s(0) = 1 m. Find s(t).
v(t) = ∫ (6t – 2) dt = 3t² – 2t + C. Using v(0) = 3 ⇒ C = 3, so v(t) = 3t² – 2t + 3.
s(t) = ∫ (3t² – 2t + 3) dt = t³ – t² + 3t + D. Using s(0) = 1 ⇒ D = 1, so s(t) = t³ – t² + 3t + 1.

解题示例：质点加速度为 a = 6t – 2，初始速度 v(0) = 3 m/s，初始位移 s(0) = 1 m。求 s(t)。
v(t) = ∫ (6t – 2) dt = 3t² – 2t + C。代入 v(0) = 3 ⇒ C = 3，得 v(t) = 3t² – 2t + 3。
s(t) = ∫ (3t² – 2t + 3) dt = t³ – t² + 3t + D。代入 s(0) = 1 ⇒ D = 1，故 s(t) = t³ – t² + 3t + 1。

9. The Trapezoidal Rule for Approximating Areas | 梯形法则与面积的近似计算

When the function cannot be integrated easily, or only a table of values is given, the trapezoidal rule provides an estimate for a definite integral. The rule divides the area into n strips of equal width h = (b – a)/n and sums the areas of the trapezia.

当函数难以积分，或题目只给出了数值表格时，梯形法则可以为定积分提供一个近似值。该法则将区间分成 n 个等宽的小段，宽度 h = (b – a)/n，然后把各个梯形的面积相加。

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

where yᵢ = f(xᵢ) and x₀ = a, xₙ = b. The more strips used, the better the approximation generally becomes. CCEA questions often ask you to calculate using 4 or 5 ordinates and perhaps compare with the exact value.

其中 yᵢ = f(xᵢ)，且 x₀ = a，xₙ = b。分段越多，近似值通常越精确。CCEA 考题经常要求使用 4 条或 5 条纵坐标进行计算，甚至与精确值作比较。

Example: Approximate ∫₀² (x² + 1) dx using 4 strips.
h = (2–0)/4 = 0.5. x: 0, 0.5, 1.0, 1.5, 2.0. y: 1, 1.25, 2, 3.25, 5.
Area ≈ (0.5/2)[1 + 2(1.25 + 2 + 3.25) + 5] = 0.25[1 + 13 + 5] = 4.75. (Exact area = 14/3 ≈ 4.667)

示例：用 4 个梯形近似计算 ∫₀² (x² + 1) dx。
h = (2–0)/4 = 0.5。x 值：0, 0.5, 1.0, 1.5, 2.0。y 值：1, 1.25, 2, 3.25, 5。
面积 ≈ (0.5/2)[1 + 2(1.25 + 2 + 3.25) + 5] = 0.25[1 + 13 + 5] = 4.75。（精确面积为 14/3 ≈ 4.667）

10. Common Mistakes and Tips for the Exam | 常见错误与应考技巧

Even strong candidates lose marks on integration through avoidable errors. Watch out for these traps:

即使是实力强的考生，也常因一些可避免的错误在积分题上丢分。请留意以下陷阱：

• Forgetting the constant +C in indefinite integration. If the question asks for “the integral of”, always include +C unless it’s a definite integral.
• 忘记在不定积分中加常数 +C。如果题目要求“求积分”，除非是定积分，否则一律要写 +C。
• Misapplying the power rule for n = –1. ∫ x⁻¹ dx = ln|x| + C, but this is beyond GCSE; CCEA will not ask you to integrate 1/x unless in a context where other methods are given.
• 错误地对 n = –1 使用幂法则。∫ x⁻¹ dx = ln|x| + C，但这超出了 GCSE 范围；CCEA 不会直接要求你积分 1/x，除非给出了其他方法。
• Sign errors with limits. Always do ‘upper minus lower’. When the lower limit is larger, the result may be negative, which can be correct for net area.
• 上下限代入时符号错误。一定要“上限减下限”。当下限大于上限时，结果可能为负，这对净面积来说是正确的。
• Area between curves: always subtract lower curve from upper curve over the entire interval. If they cross, split the integral.
• 曲线之间的面积：整个区间内始终用上方曲线减去下方曲线。如果有交叉，务必拆分积分。
• Units: include units for area or displacement where applicable, and use square brackets notation for evaluation.
• 单位：在涉及面积或位移时要注意写出单位，并用方括号表示代入上下限的过程。

Exam tip: Show all steps clearly. In CCEA mark schemes, method marks are awarded for correct antiderivative and correct handling of limits, even if a numerical slip occurs.

考试建议：清晰展示每一步。在 CCEA 的评分方案中，即使计算过程中出现小错，只要原函数正确、上下限代入方法正确，仍能拿到方法分。

11. Exam‑Style Worked Examples | 考试风格例题精解

Let’s consolidate the concepts with a range of typical CCEA questions.

让我们通过一系列典型的 CCEA 试题来巩固这些概念。

Example 1 – Indefinite integral with condition:
Find the equation of the curve whose derivative is dy/dx = 3x² – 4 and which passes through (–1, 2).
y = ∫ (3x² – 4) dx = x³ – 4x + C.
Substitute (–1, 2): (–1)³ – 4(–1) + C = 2 → –1 + 4 + C = 2 → C = –1.
Hence y = x³ – 4x – 1.

例题 1 – 带条件的不定积分：
已知某曲线的导数为 dy/dx = 3x² – 4，且曲线过点 (–1, 2)，求曲线方程。
y = ∫ (3x² – 4) dx = x³ – 4x + C。
代入 (–1, 2)：(–1)³ – 4(–1) + C = 2 → –1 + 4 + C = 2 → C = –1。
因此 y = x³ – 4x – 1。

Example 2 – Definite integral and area:
Find the area bounded by y = 4x – x² and the x‑axis. First find where the curve meets the x‑axis: 4x – x² = 0 → x(4 – x) = 0 → x = 0 and x = 4.
Area = ∫₀⁴ (4x – x²) dx = [2x² – x³/3]₀⁴ = (32 – 64/3) – 0 = (96/3 – 64/3) = 32/3 square units.

例题 2 – 定积分与面积：
求曲线 y = 4x – x² 与 x 轴围成的面积。首先求曲线与 x 轴的交点：4x – x² = 0 → x(4 – x) = 0 → x = 0 与 x = 4。
面积 = ∫₀⁴ (4x – x²) dx = [2x² – x³/3]₀⁴ = (32 – 64/3) – 0 = (96/3 – 64/3) = 32/3 平方单位。

Example 3 – Kinematics:
A particle moves along a line with velocity v(t) = 2t – 5 m/s. Find the displacement between t = 1 s and t = 4 s, and the total distance travelled in that interval.
Displacement = ∫₁⁴ (2t – 5) dt = [t² – 5t]₁⁴ = (16 – 20) – (1 – 5) = –4 – (–4) = 0 m.
For distance, note v = 0 at t = 2.5 s. Distance = |∫₁².⁵ (2t – 5) dt| + ∫₂.₅⁴ (2t – 5) dt.
First: [t² – 5t]₁²·⁵ = (6.25 – 12.5) – (–4) = –6.25 + 4 = –2.25; absolute = 2.25 m.
Second: [t² – 5t]₂.₅⁴ = (–4) – (–6.25) = 2.25 m. Total distance = 2.25 + 2.25 = 4.5 m.

例题 3 – 运动学：
一质点沿直线运动，速度为 v(t) = 2t – 5 m/s。求 t = 1 s 到 t = 4 s 间的位移以及该时段内行驶的总路程。
位移 = ∫₁⁴ (2t – 5) dt = [t² – 5t]₁⁴ = (16 – 20) – (1 – 5) = –4 – (–4) = 0 m。
对于路程，注意在 t = 2.5 s 时 v = 0。路程 = |∫₁².⁵ (2t – 5) dt| + ∫₂.₅⁴ (2t – 5) dt。
第一部分：[t² – 5t]₁²·⁵ = (6.25 – 12.5) – (–4) = –6.25 + 4 = –2.25，绝对值为 2.25 m。
第二部分：[t² – 5t]₂.₅⁴ = (–4) – (–6.25) = 2.25 m。总路程 = 2.25 + 2.25 = 4.5 m。

12. Summary and Final Advice | 总结与最后建议

Integration in CCEA GCSE Mathematics rotates around the power rule, the constant +C, definite integration for net and total areas, area between curves, and kinematic applications. The trapezoidal rule gives a numerical backup when exact integration is not possible. Mastering these ideas requires practice with a variety of functions, including fractional and negative exponents.

CCEA GCSE 数学的积分内容围绕幂法则、常数 +C、定积分求净面积与总面积、曲线间面积以及运动学应用展开。当无法精确积分时，梯形法则提供了数值计算的备选方案。要掌握这些概念，需要针对包括分数指数和负指数在内的各种函数进行充分练习。

Always sketch the curve when tackling area problems, label x‑intercepts, and show every substitution step. Double‑check whether the question asks for net area or total area, and whether a kinematic question wants displacement or distance. With methodical working, you will secure full marks on integration questions.

在解决面积问题时，务必画出曲线草图，标出与 x 轴的交点，并展示每一步代入过程。仔细审题，分清问题是求净面积还是总面积，运动题要求位移还是路程。有条理地书写过程，你就能在积分题上拿下满分。

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