📚 IB Maths Exercise 7G.1: Definite Integration – Techniques and Applications | IB数学练习7G.1：定积分技巧与应用
In Exercise 7G.1 of the IB Mathematics course, students are challenged to evaluate definite integrals and apply them to geometric and physical contexts. This revision guide unpacks the core skills tested—from recognising the Fundamental Theorem of Calculus to using substitution, finding areas between curves, and computing volumes of revolution. Whether you are preparing for the Analysis & Approaches or Applications & Interpretation routes, mastering these techniques will sharpen your problem‑solving fluency and deepen your understanding of how integration connects accumulation, geometry, and motion.
在IB数学课程的练习7G.1中，学生需要计算定积分并将其应用于几何和物理场景。本篇复习指南梳理了该练习所考察的核心技能——从识别微积分基本定理，到使用换元法、计算曲线间的面积以及旋转体的体积。无论你学习的是分析与方法路径还是应用与解释路径，掌握这些技巧都将提升你的解题熟练度，并加深你对积分如何联系累积、几何与运动的理解。
1. Understanding Definite Integrals | 理解定积分
A definite integral ∫ₐᵇ f(x) dx ed area under the curve y = f(x) between x = a and x = b. If the function lies above the x‑ as Δx → 0, underpins this idea, but in practice the evaluation relies on antiderivatives.
定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x) 与 x = a、x = b 之间所围成的有向面积。若函数位于 x 轴上方，积分给出正面积；若位于下方，则积分为负值。在练习7G.1中，你常常需要根据图像或给定情境先建立积分式，然后再求值。用黎曼和 Σ f(xᵢ) Δx（当 Δx → 0）给出的极限定义是这一概念的基础，但实际计算则依赖于原函数。
2. The Fundamental Theorem of Calculus | 微积分基本定理
The Fundamental Theorem of Calculus (FTC) bridges differentiation and integration. Part 1 states that if F is an antiderivative of f on [a, b], then ∫ₐᵇ f(x) dx = F(b) − F(a). Part 2 concerns the derivative of an accumulation function: d/dx ∫ₐˣ ys remember to include the constant of integration when finding F(x), but it cancels out when evaluating the definite integral.
微积分基本定理（FTC）为微分与积分建立了桥梁。第一部分指出，若 F 是 f 在区间 [a, b] 上的一个原函数，则 ∫ₐᵇ f(x) dx = F(b) − F(a)。第二部分涉及累积函数的导数：d/dx ∫ₐˣ f(t) dt = f(x)。练习7G.1经常会要求你应用第一部分：找到原函数，代入上限和下限，然后相减。在求 F(x) 时请记住要添加积分常数，但计算定积分时该常数会相互抵消。
3. Basic Integration Rules and Properties | 基本积分法则与性质
dx = k ∫ₐᵇ f(x) dx; ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx; and ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜ These are used constantly in Exercise 7G.1 to break complex integrals into simpler pieces.
在着手解题之前，回顾标准的积分公式及能拆分定积分的性质很有帮助。下表总结了一些基本的原函数。此外，请牢记这些关键性质：∫ₐᵇ k·f(x) dx = k ∫ₐᵇ f(x) dx；∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx；以及对于 a 与 b 之间的任意 c，有 ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx。在练习7G.1中，这些性质被不断用于将复杂积分拆分为更简单的形式。
| Function f(x) | Indefinite Integral ∫ f(x) dx |
|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹/(n+1) + C |
| 1/x | ln|x| + C |
| eˣ | eˣ + C |
| sin x | −cos x + C |
| cos x | sin x + C |
| sec² x | tan x + C |
| 1/√1−x² | arcsin x + C |
4. Integration by Substitution for Definite Integrals | 定积分的换元积分法
When an integral contains a composite function, substitution (u‑substitution) simplifies the evaluation. For a definite integral ∫ₐᵇ f(g(x))·g'(x) dx, set u = g(x) so that du = g' imits must also change: when x = a, u = g(a); when x = b, u = g(b). The integral becomes ∫g(a)g(b) f(u) du. Exercise 7G.1 often includes problems such as ∫₀¹ 2x·ex² Always check that the substitution fully replaces the original variable and its differential.
当积分中包含复合函数时，换元法（u‑代换）可以简化计算。对于定积分 ∫ₐᵇ f(g(x))·g'(x) dx，令 u = g(x)，则 du = g'(x) dx。积分限也必须随之改变：当 x = a 时，u = g(a)；当 x = b 时，u = g(b)。积分式变为 ∫g(a)g(b) f(u) du。练习7G.1常包含诸如 ∫₀¹ 2x·ex² dx 的问题，令 u = x² 则 du = 2x dx，积分限由 x=0 → u=0、x=1 → u=1，得到 ∫₀¹ eᵘ du = e − 1。请务必检查代换是否完整地替换了原变量及其微分。
∫ₐᵇ f(g(x))·g'(x) dx = ∫g(a)g(b) f(u) du
5. Calculating Area Between Curves | 计算曲线间的面积
e intersection points by solving f(x) = g(x) to determine the limits a and b. If the upper‑ r regions bounded by functions of y, switch the roles: area = ∫ₚᵠ [right(x) − left(x)] dy.
练习7G.1的经典题型是求两条曲线围成的面积。若在区间 [a, b] 上 y = f(x) 为上曲线，y = g(x) 为下曲线，则面积为 ∫ₐᵇ [f(x) − g(x)] dx。你必须先通过解方程 f(x) = g(x) 确定交点，以得到积分限 a 和 b。如果上‑下关系发生交换，应在交点处拆分积分，并使用绝对值或分别积分。对于由 y 的函数围成的区域，可交换角色：面积 = ∫ₚᵠ [右函数 − 左函数] dy。
6. Volume of Revolution: Disk Method | 旋转体体积：圆盘法
When a region is rotated about the x‑ nd y = g(x) with f(x) ≥ g(x) ≥ 0, the volume of the solid formed by revolving around the x‑axis is π ∫ₐᵇ ([f(x)]² − [g(x)]²) dx (washer method). For rotation about the y‑ k you to set up such integrals from given diagrams or descriptions before evaluating them.
当区域绕 x 轴旋转时，可采用圆盘法求体积：V = π ∫ₐᵇ [f(x)]² dx。若区域由两条曲线 y = f(x) 与 y = g(x) 围成且 f(x) ≥ g(x) ≥ 0，绕 x 轴旋转所得立体的体积为 π ∫ₐᵇ ([f(x)]² − [g(x)]²) dx（垫圈法）。若绕 y 轴旋转，则将 x 表示为 y 的函数并对 y 积分：V = π ∫ₚᵠ [x(y)]² dy。练习7G.1可能会要求你根据所给图形或描述先建立积分式，再求值。
V = π ∫ₐᵇ [f(x)]² dx (disk about x‑axis)
7. Definite Integrals in Kinematics | 运动学中的定积分
are given a velocity function v(t), the displacement over the time interval [t₁, t₂] is ∫t₁t₂ v(t) dt. The total distance travelled, however, is ∫t₁t₂ |v(t)| milarly, the integral of acceleration a(t) gives the change in velocity: v(t₂) − v(t₁) = ∫t₁t₂ a(t) dt. In Exercise 7G.1 you might be asked to interpret the physical meaning of a definite integral in context, such as finding the distance travelled by a particle during a specific time window or determining its final position given an initial condition.
IB数学常将定积分与运动学联系起来。若给定速度函数 v(t)，在时间区间 [t₁, t₂] 上的位移即为 ∫t₁t₂ v(t) dt。然而，总路程为 ∫t₁t₂ |v(t)| dt，因为速率是速度的绝对值。同理，对加速度 a(t) 积分则得到速度的变化量：v(t₂) − v(t₁) = ∫t₁t₂ a(t) dt。在练习7G.1中，你可能需要解释一个定积分在具体情境中的物理含义，例如求一个质点在特定时段内行驶的路程，或根据初始条件确定其最终位置。
8. Common Mistakes and Tips for Exercise 7G.1 | 练习7G.1常见错误与技巧
Even confident students can slip up on definite integrals. Here are key pitfalls to avoid: (1) Forgetting to change limits when using substitution—always map x‑limits to u‑limits. (2) Mishandling negative areas: if the graph crosses the x‑axis, split the integral at intercepts and use absolute values for total area. (3) Mixing up the order in the Fundamental Theorem: remember it is F(b) − F(a), not F(a) − F(b). (4) Ignoring the need for exact forms: IB often expects answers in terms of π, e, or ln. (5) Misapplying the volume formula: for rotation about the x‑axis the integrand is [f(x)]², not f(x²). Before submitting your solutions, double‑check that you have correctly identified the upper and lower functions, the limits, and any simplifications.
即便是自信的学生也可能在定积分上失误。以下是一些需要避免的关键陷阱：(1) 使用换元时忘记改变积分限——必须将 x‑限映射为 u‑限。(2) 错误处理负面积：若图形穿过 x 轴，应在截点处拆分积分并用绝对值求总面积。(3) 混淆基本定理中的次序：记住是 F(b) − F(a)，而非 F(a) − F(b)。(4) 忽视精确形式的要求：IB常要求答案用 π、e 或 ln 表示。(5) 误用体积公式：绕 x 轴旋转时被积函数为 [f(x)]²，而非 f(x²)。在提交解答之前，请仔细核对是否已正确识别上下函数、积分限以及所有化简步骤。
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