Review Set 17C | IB 数学复习题集 17C

📚 Review Set 17C | IB 数学复习题集 17C

This review set is designed to consolidate core calculus skills required for the IB Mathematics: Analysis and Approaches and Applications and Interpretation courses. The problems mix differentiation, integration, kinematics, area calculations, and optimisation. By working through these examples, you will strengthen your ability to apply rules accurately and link concepts across different function types. Each step is explained with standard IB terminology, and solutions are paired in English and Chinese to support bilingual learning.

本复习题集旨在巩固 IB 数学（分析与方法、应用与解释）所需的核心微积分技能。题目涵盖求导、积分、运动学、面积计算和优化。通过这些例题，你将提升准确应用法则的能力，并能将不同函数类型的概念串联起来。每一步解答都使用了标准的 IB 术语，并以英语和中文双语配对解释，帮助双语学习。

1. Chain Rule Mastery | 链式法则掌握

Problem: Differentiate y = (5x² – 3x + 1)⁴ with respect to x.

题目：对 y = (5x² – 3x + 1)⁴ 关于 x 求导。

Solution strategy: Identify the outer function u⁴ and the inner function u = 5x² – 3x + 1. The chain rule states that dy/dx = dy/du × du/dx. First, dy/du = 4u³ = 4(5x² – 3x + 1)³. Then, du/dx = 10x – 3.

解题策略：识别外层函数 u⁴ 和内层函数 u = 5x² – 3x + 1。链式法则表明 dy/dx = dy/du × du/dx。首先，dy/du = 4u³ = 4(5x² – 3x + 1)³。然后，du/dx = 10x – 3。

Multiply the two derivatives: dy/dx = 4(5x² – 3x + 1)³ × (10x – 3) = 4(10x – 3)(5x² – 3x + 1)³. Simplify the constant if needed.

将两个导数相乘：dy/dx = 4(5x² – 3x + 1)³ × (10x – 3) = 4(10x – 3)(5x² – 3x + 1)³。如有需要可简化常数。

2. Product and Quotient Rules | 乘积与商法则

Problem a (Product Rule): Find dy/dx if y = x³ ln(x).

题目 a（乘积法则）：若 y = x³ ln(x)，求 dy/dx。

Recall the product rule: (uv)’ = u’v + uv’. Here u = x³, v = ln(x). Then u’ = 3x², v’ = 1/x. Therefore dy/dx = 3x² ln(x) + x³ × (1/x) = 3x² ln(x) + x².

回顾乘积法则：(uv)’ = u’v + uv’。此处 u = x³, v = ln(x)。那么 u’ = 3x², v’ = 1/x。因此 dy/dx = 3x² ln(x) + x³ × (1/x) = 3x² ln(x) + x²。

Problem b (Quotient Rule): Differentiate f(x) = eˣ / (x + 1).

题目 b（商法则）：求 f(x) = eˣ / (x + 1) 的导数。

Using the quotient rule: (u/v)’ = (u’v – uv’)/v². Let u = eˣ ⇒ u’ = eˣ; v = x + 1 ⇒ v’ = 1. Then f ‘(x) = [eˣ(x + 1) – eˣ·1] / (x + 1)² = [eˣ(x + 1 – 1)] / (x + 1)² = x eˣ / (x + 1)².

使用商法则：(u/v)’ = (u’v – uv’)/v²。设 u = eˣ ⇒ u’ = eˣ；v = x + 1 ⇒ v’ = 1。则 f ‘(x) = [eˣ(x + 1) – eˣ·1] / (x + 1)² = [eˣ(x + 1 – 1)] / (x + 1)² = x eˣ / (x + 1)²。

3. Basic Integration | 基本积分

Problem: Evaluate ∫ (4x³ – 6x + 2/√x) dx.

题目：计算 ∫ (4x³ – 6x + 2/√x) dx。

Rewrite the integrand using powers: 4x³ – 6x + 2x⁻¹/². Integrate term by term: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ –1). For 4x³: ∫ 4x³ dx = 4 × x⁴/4 = x⁴. For –6x: ∫ –6x dx = –6 × x²/2 = –3x². For 2x⁻¹/²: ∫ 2x⁻¹/² dx = 2 × x^(½) / (½) = 4x^(½) = 4√x.

用幂函数重写被积函数：4x³ – 6x + 2x⁻¹/²。逐项积分：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C（n ≠ –1）。对于 4x³：∫ 4x³ dx = 4 × x⁴/4 = x⁴。对于 –6x：∫ –6x dx = –6 × x²/2 = –3x²。对于 2x⁻¹/²：∫ 2x⁻¹/² dx = 2 × x^(½) / (½) = 4x^(½) = 4√x。

Combining the terms, the general antiderivative is: x⁴ – 3x² + 4√x + C. Always include the constant of integration in indefinite integrals.

合并各项，原函数为：x⁴ – 3x² + 4√x + C。不定积分中务必写上积分常数。

4. Definite Integrals and Signed Area | 定积分与带符号面积

Problem: Calculate the total area enclosed between the curve y = x² – 4x and the x-axis from x = 0 to x = 4.

题目：计算曲线 y = x² – 4x 与 x 轴在 x = 0 到 x = 4 之间围成的总面积。

First find intercepts: x² – 4x = 0 ⇒ x(x – 4) = 0, so roots at x = 0 and x = 4. The curve is a parabola opening upwards, lying below the x-axis for 0 < x < 4. Thus the signed integral ∫₀⁴ (x² – 4x) dx is negative, but area is absolute value.

首先求交点：x² – 4x = 0 ⇒ x(x – 4) = 0，因此在 x = 0 和 x = 4 处有根。该曲线为开口向上的抛物线，在 0 < x < 4 区间位于 x 轴下方。因此带符号积分 ∫₀⁴ (x² – 4x) dx 为负，但面积取绝对值。

Compute the definite integral: ∫₀⁴ (x² – 4x) dx = [x³/3 – 2x²]₀⁴ = (64/3 – 32) – (0) = 64/3 – 96/3 = –32/3. The signed area is –32/3, so the total bounded area is |–32/3| = 32/3 square units.

计算定积分：∫₀⁴ (x² – 4x) dx = [x³/3 – 2x²]₀⁴ = (64/3 – 32) – (0) = 64/3 – 96/3 = –32/3。带符号面积为 –32/3，因此总包围面积为 |–32/3| = 32/3 平方单位。

5. Kinematics with Calculus | 微积分在运动学中的应用

Problem: A particle moves along a straight line such that its displacement s(t) = t³ – 9t² + 24t + 5, t ≥ 0. Find its velocity and acceleration at time t = 2 seconds.

题目：一质点沿直线运动，其位移 s(t) = t³ – 9t² + 24t + 5，t ≥ 0。求 t = 2 秒时的速度和加速度。

Velocity v(t) = ds/dt = 3t² – 18t + 24. Acceleration a(t) = dv/dt = 6t – 18. Evaluate at t = 2: v(2) = 3(4) – 18(2) + 24 = 12 – 36 + 24 = 0 m/s. a(2) = 6(2) – 18 = 12 – 18 = –6 m/s².

速度 v(t) = ds/dt = 3t² – 18t + 24。加速度 a(t) = dv/dt = 6t – 18。代入 t = 2：v(2) = 3(4) – 18(2) + 24 = 12 – 36 + 24 = 0 m/s。a(2) = 6(2) – 18 = 12 – 18 = –6 m/s²。

Interpretation: at the instant t = 2, the particle is momentarily at rest but its acceleration is negative, which means it is about to move in the negative direction (decelerating from positive direction).

解释：在 t = 2 瞬间，质点瞬时静止，但加速度为负，意味着它即将向负方向运动（从正方向减速）。

6. Trigonometric Integrals | 三角函数积分

Problem: Find ∫ sin²θ dθ.

题目：求 ∫ sin²θ dθ。

Use the power-reduction identity: sin²θ = (1 – cos 2θ)/2. Then ∫ sin²θ dθ = ½ ∫ (1 – cos 2θ) dθ = ½ [θ – ½ sin 2θ] + C = θ/2 – (sin 2θ)/4 + C.

使用降幂恒等式：sin²θ = (1 – cos 2θ)/2。于是 ∫ sin²θ dθ = ½ ∫ (1 – cos 2θ) dθ = ½ [θ – ½ sin 2θ] + C = θ/2 – (sin 2θ)/4 + C。

This technique frequently appears in IB calculus problems involving periodic functions and area calculations. Always check the final expression by differentiating to verify it returns sin²θ.

这一技巧在涉及周期函数和面积计算的 IB 微积分题目中经常出现。务必通过求导验证最终表达式是否回到 sin²θ。

7. Integration by Substitution | 换元积分法

Problem: Evaluate ∫ 6x² √(x³ + 1) dx using the substitution u = x³ + 1.

题目：使用换元 u = x³ + 1 计算 ∫ 6x² √(x³ + 1) dx。

If u = x³ + 1, then du/dx = 3x², so du = 3x² dx. The integral contains 6x² dx, which is exactly 2 du. So ∫ 6x² √(x³ + 1) dx = ∫ √u × 2 du = 2 ∫ u^(½) du.

设 u = x³ + 1，则 du/dx = 3x²，因此 du = 3x² dx。积分中含有 6x² dx，恰好等于 2 du。于是 ∫ 6x² √(x³ + 1) dx = ∫ √u × 2 du = 2 ∫ u^(½) du。

Integrate: 2 × [u^(3/2) / (3/2)] + C = 2 × (⅔) u^(3/2) + C = (4/3) u^(3/2) + C. Substitute back u = x³ + 1 to obtain final answer: (4/3)(x³ + 1)^(3/2) + C.

积分：2 × [u^(3/2) / (3/2)] + C = 2 × (⅔) u^(3/2) + C = (4/3) u^(3/2) + C。回代 u = x³ + 1 得最终结果：(4/3)(x³ + 1)^(3/2) + C。

8. Rational Functions and Logarithms | 有理函数与对数积分

Problem: Determine ∫ (2x + 3)/(x² + 3x – 4) dx.

题目：求 ∫ (2x + 3)/(x² + 3x – 4) dx。

Inspect the numerator: it is exactly the derivative of the denominator since d/dx(x² + 3x – 4) = 2x + 3. Hence the integral is of the form ∫ (f ‘(x))/f(x) dx, which integrates to ln|f(x)| + C.

观察分子：它恰好是分母的导数，因为 d/dx(x² + 3x – 4) = 2x + 3。因此该积分形如 ∫ (f ‘(x))/f(x) dx，其积分为 ln|f(x)| + C。

Thus ∫ (2x + 3)/(x² + 3x – 4) dx = ln|x² + 3x – 4| + C. If the numerator were not an exact derivative, partial fraction decomposition would be required, which is another important skill in IB.

因此 ∫ (2x + 3)/(x² + 3x – 4) dx = ln|x² + 3x – 4| + C。若分子并非精准的导数形式，则需使用部分分式分解，这也是 IB 课程中的另一项重要技能。

9. Area Between Two Curves | 曲线间的面积

Problem: Find the area of the region bounded by the curves y = x² – 2x and y = 4 – x².

题目：求由曲线 y = x² – 2x 和 y = 4 – x² 所围区域的面积。

Find intersection points: set x² – 2x = 4 – x² ⇒ 2x² – 2x – 4 = 0 ⇒ x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0. The curves intersect at x = –1 and x = 2.

求交点：令 x² – 2x

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Review Set 17C | IB 数学复习题集 17C