Review Set 13C: Integration Techniques and Applications | IB数学复习题集13C：积分技巧与应用

📚 Review Set 13C: Integration Techniques and Applications | IB数学复习题集13C：积分技巧与应用

Review Set 13C in the IB Mathematics: Analysis and Approaches course is designed to consolidate your understanding of integral calculus, from basic antiderivatives to applications such as area, volume, and kinematics. This article walks you through the key concepts and typical problems presented in that set, offering bilingual explanations to sharpen your skills and exam technique.

IB数学：分析与方法的复习题集13C旨在巩固你对积分微积分的理解，从基本反导函数到面积、体积和运动学等应用。本文带你梳理该题集中的核心概念和典型问题，提供双语讲解，助你提升技巧与应试能力。

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1. Overview of the Review Set | 复习题集概览

Review Set 13C contains a mixture of indefinite and definite integrals, substitution, area between curves, volumes of revolution, kinematic problems, and an introduction to improper integrals. The questions are structured to gradually increase in difficulty, mirroring the IB exam style.

复习题集13C包含不定积分和定积分、换元法、曲线间面积、旋转体体积、运动学问题及反常积分入门，题目难度逐步递增，模拟IB考试风格。

You should treat this set as a diagnostic tool. Identify which types of integrals cause you trouble and go back to the relevant sections in your textbook before attempting the next set. A solid grasp of differentiation rules is a prerequisite, as integration often requires you to recognize derivatives in reverse.

你应将该题集视为诊断工具。找出困扰你的积分类型，在尝试下一题集之前回归课本相关章节。扎实掌握微分规则是前提，因为积分常需要逆向识别导数。

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2. Basic Antiderivatives | 基本反导函数

Mastering basic antiderivatives is essential for the first few questions in Review Set 13C. The table below summarises the most common indefinite integrals you must know by heart.

掌握基本反导函数是应对复习题集13C前几道题的关键。下表总结了你必须牢记的最常见不定积分。

Function f(x)	Indefinite Integral ∫ f(x) dx
xⁿ (n ≠ -1)	xⁿ⁺¹/(n+1) + C
1/x	ln|x| + C
eˣ	eˣ + C
aˣ (a > 0)	aˣ/ln a + C
sin x	-cos x + C
cos x	sin x + C
sec² x	tan x + C
1/√1 - x²	arcsin x + C
1/(1 + x²)	arctan x + C

Always remember to add the constant of integration C for indefinite integrals. In the context of Review Set 13C, forgetting C can cost you marks even when the rest of the work is correct.

始终记住对不定积分加上积分常数C。在复习题集13C中，即使其余部分正确，遗漏C也可能导致失分。

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3. Integration by Substitution | 换元积分法

Substitution is the most common technique required in this set. The idea is to let u equal an inner function so that du appears in the integrand, transforming a complicated integral into a standard form.

换元法是该题集最常见的方法，其思想是令u等于内层函数，使被积函数中出现du，从而将复杂积分化为标准形式。

Consider the integral ∫ 2x · e^x² dx. Let u = x², then du/dx = 2x, so du = 2x dx. The integral becomes ∫ e^u du = e^u + C, and substituting back gives e^x² + C.

考虑积分∫ 2x · e^x² dx。令u = x²，则du/dx = 2x，即du = 2x dx。积分化为∫ e^u du = e^u + C，代回得e^x² + C。

iginal limits are x = 0 to x = 1, the new limits are u = 0 to u = 1, so you evaluate ∫₀¹ e^u du directly.

对于定积分，换元时记住要改变积分限。若u = x²，原积分限为x=0到x=1，则新积分限为u=0到u=1，可直接计算∫₀¹ e^u du。

is not perfectly present, requiring you to adjust by a constant factor. For example, ∫ x·√2x²+1 dx needs u = 2x²+1, du = 4x dx, so you insert a factor of 1/4.

复习题集13C包含多个换元题，其中内层函数导数并不完全匹配，需通过常数因子调整。例如∫ x·√2x²+1 dx需令u = 2x²+1, du = 4x dx，因此补上因子1/4。

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4. Definite Integrals and Area | 定积分与面积

n Review Set 13C, you are often asked to find the exact area, so you must evaluate the integral precisely and apply the fundamental theorem of calculus.

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的带符号面积。在复习题集13C中常要求精确面积，需准确计算积分并应用微积分基本定理。

ed to evaluate ∫₋₁¹ (x² - 1) dx and ∫₁² (x² - 1) dx separately and sum their absolute values.

若曲线位于x轴下方，定积分得负值；求实际面积时需取绝对值或将积分按函数变号区域拆分。例如求y = x² - 1与x轴从x=-1到x=2围成的面积，需分别计算∫₋₁¹ (x² - 1) dx 和∫₁² (x² - 1) dx，并将绝对值相加。

Typical questions in the set ask you to set up the integral for an area given a sketch and then compute it, often requiring substitution or trigonometric identities. Practice these until you can confidently determine limits and signs.

题集中典型问题要求根据草图建立积分表达式并计算，往往需要换元或三角恒等式。反复练习直至能自信确定积分限和符号。

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5. Area Between Curves | 曲线间的面积

unded by y = f(x) (top) and y = g(x) (bottom) from x = a to x = b is ∫ₐᵇ [f(x) - g(x)] dx.

当两条曲线相交时，曲线间的面积通过对函数之差积分求得。由y = f(x)（上方）和y = g(x)（下方）在x=a到x=b所围面积公式为∫ₐᵇ [f(x) - g(x)] dx。

on is the upper one on the interval; a quick test with a sample x-value can prevent sign errors.

在复习题集13C中，你会遇到需先解方程f(x)=g(x)求交点的问题，然后判断区间上哪条是上方函数；用一个样本x值快速检验可避免符号错误。

s above y = x², so the area is ∫₋₁² [(x + 2) - x²] dx. Evaluating yields 9/2 square units.

经典例题是求y = x²与y = x + 2所围面积。解x² = x + 2得x = -1, 2。在(-1,2)上直线y = x + 2位于y = x²之上，因此面积为∫₋₁² [(x + 2) - x²] dx，计算得9/2平方单位。

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6. Volumes of Revolution: Disk Method | 旋转体体积：圆盘法

dx, provided the solid has no hollow part. This is a direct application of integration that appears regularly in Review Set 13C.

当平面区域绕x轴旋转时，若生成的立体无空心部分，可用圆盘法计算体积：V = π ∫ₐᵇ [f(x)]² dx。这是复习题集13C中常出现的积分直接应用。

For example, to find the volume when the region bounded by y = √x, the x-axis, and x = 4 is rotated about the x-axis, compute V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π cubic units.

例如求由y = √x、x轴和x=4所围区域绕x轴旋转所得体积，计算V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π 立方单位。

Always square the function before integrating, and do not forget the π factor. Many students in the set lose points by omitting π or squaring incorrectly when the function is negative—the square makes it positive anyway.

始终先平方函数再积分，且勿遗漏π因子。题集中不少学生因遗漏π或对负函数平方处理不当而失分——平方后终为正值。

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7. Volumes of Revolution: Washer Method | 旋转体体积：垫圈法

e is V = π ∫ₐᵇ ([f(x)]² - [g(x)]²) dx, where f is the outer radius and g the inner radius.

若旋转区域未延伸至旋转轴，生成的立体带孔洞，需用垫圈法。体积为V = π ∫ₐᵇ ([f(x)]² - [g(x)]²) dx，其中f为外半径，g为内半径。

inner radius is x, giving V = π ∫₀¹ [(x²+1)² - x²] dx. Expand and integrate term by term.

复习题集13C中典型任务是绕x轴旋转两曲线间区域。对y = x² + 1与y = x在x=0至x=1间区域，外半径为x²+1，内半径为x，得V = π ∫₀¹ [(x²+1)² - x²] dx，展开后逐项积分。

ust adjust the radii by subtracting c from each function. The same logic applies to vertical revolutions using functions of y.

若旋转轴为水平线但不是x轴，例如y = c，需用每个函数减去c来调整半径。同理，绕竖直轴旋转时使用y的函数。

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8. Kinematics and Integration | 运动学与积分

Integration connects acceleration, velocity, and displacement. In Review Set 13C, kinematic problems usually provide an acceleration function a(t) and initial conditions; you integrate to find v(t) and integrate again to find s(t).

积分联系着加速度、速度和位移。复习题集13C的运动学问题通常给出加速度函数a(t)和初始条件；积分得v(t)，再积分得s(t)。

3t² - 2t + 3. Then s(t) = ∫ (3t² - 2t + 3) dt = t³ - t² + 3t + D; s(0)=0 gives D=0, so s(2

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