📚 AP Calculus: Integral Definition and Calculation Practice Questions | AP 微积分：积分学定义与计算真题精讲
In AP Calculus, the concept of integration is fundamental. Mastering its definition and computational techniques is crucial for success on the exam. This article provides a detailed review of integral definitions, essential integration methods, and step-by-step solutions to exam-style questions, helping you build a solid foundation.
在 AP 微积分中，积分概念是基础。掌握其定义和计算技巧对考试成功至关重要。本文详细回顾了积分的定义、基本积分方法，并逐步讲解真题风格的题目，助你打下坚实基础。
1. Definition of Indefinite Integral | 不定积分的定义
An indefinite integral represents the antiderivative of a function. If F'(x) = f(x), then ∫ f(x) dx = F(x) + C, where C is the constant of integration. The notation is read ‘the integral of f of x dx’ and yields a family of functions.
不定积分表示函数的反导数。若 F'(x) = f(x)，则 ∫ f(x) dx = F(x) + C，其中 C 为积分常数。符号 ∫ f(x) dx 读作“f 关于 x 的积分”，得到一族函数。
For example, because d/dx (x³) = 3x², we have ∫ 3x² dx = x³ + C.
例如，由于 d/dx (x³) = 3x²，我们有 ∫ 3x² dx = x³ + C。
2. Definition of Definite Integral and Riemann Sums | 定积分与黎曼和的定义
The definite integral ∫ₐᵇ f(x) dx is defined as the limit of Riemann sums: limₙ→∞ Σᵢ₌₁ⁿ f(xᵢ*) Δx, where Δx = (b-a)/n and xᵢ* is a sample point in the i-th subinterval. It computes the net area between the curve and the x-axis from a to b.
定积分 ∫ₐᵇ f(x) dx 定义为黎曼和的极限：limₙ→∞ Σᵢ₌₁ⁿ f(xᵢ*) Δx，其中 Δx = (b-a)/n，xᵢ* 是第 i 个子区间内的样本点。它计算了从 a 到 b 曲线与 x 轴之间的净面积。
When f(x) ≥ 0 on [a, b], the definite integral gives the exact area under the graph. The Riemann sum approximates this area using rectangles, and the limit makes it exact.
当在 [a, b] 上 f(x) ≥ 0 时，定积分给出曲线下方精确的面积。黎曼和通过矩形近似该面积，取极限后精确化。
3. Fundamental Theorem of Calculus | 微积分基本定理
The Fundamental Theorem of Calculus (FTC) links differentiation and integration. Part 1: If F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). Part 2: ∫ₐᵇ f(x) dx = F(b) – F(a), where F is any antiderivative of f.
微积分基本定理（FTC）将微分与积分联系起来。第一部分：若 F(x) = ∫ₐˣ f(t) dt，则 F'(x) = f(x)。第二部分：∫ₐᵇ f(x) dx = F(b) – F(a)，其中 F 是 f 的任意一个反导数。
In practice, to evaluate a definite integral like ∫₁³ 2x dx, we find an antiderivative F(x)=x², then compute F(3)-F(1)=9-1=8.
实际计算定积分如 ∫₁³ 2x dx 时，我们先求出反导数 F(x)=x²，然后计算 F(3)-F(1)=9-1=8。
4. Basic Integration Formulas | 基本积分公式
Memorizing the following fundamental antiderivatives is essential for the AP exam. Each formula can be verified by differentiation.
熟记以下基本反导数公式对 AP 考试至关重要，每一条都可通过求导验证。
| 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, a≠1) | 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 |
Linearity rules also apply: ∫ [k·f(x) ± g(x)] dx = k ∫ f(x) dx ± ∫ g(x) dx.
线性法则同样适用：∫ [k·f(x) ± g(x)] dx = k ∫ f(x) dx ± ∫ g(x) dx。
5. Integration by Substitution | 换元积分法
Substitution (u-substitution) reverses the chain rule. Choose u = g(x), then compute du = g'(x) dx and rewrite the integral entirely in terms of u.
换元法（u-代换）是链式法则的逆运算。令 u = g(x)，计算 du = g'(x) dx，并将积分完全用 u 表示。
Example (indefinite): Evaluate ∫ 2x cos(x²) dx. Let u = x², so du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x²) + C.
不定积分示例：计算 ∫ 2x cos(x²) dx。令 u = x²，则 du = 2x dx。积分化为 ∫ cos u du = sin u + C = sin(x²) + C。
Definite integral version: ∫₀^√π x sin(x²) dx. Set u = x², du = 2x dx → x dx = du/2. Limits: x=0 → u=0; x=√π → u=π. Integral = ½ ∫₀^π sin u du = ½ [-cos u]₀^π = ½ ( -cos π + cos 0 ) = ½ (1+1) = 1.
定积分版本：∫₀^√π x sin(x²) dx。令 u = x²，du = 2x dx → x dx = du/2。积分限：x=0 → u=0；x=√π → u=π。积分 = ½ ∫₀^π sin u du = ½ [-cos u]₀^π = ½ ( -cos π + cos 0 ) = ½ (1+1) = 1。
6. Integration by Parts | 分部积分法
This technique follows from the product rule: ∫ u dv = uv – ∫ v du. Choose u and dv so that the new integral ∫ v du is simpler. The LIATE rule (Log, Inverse trig, Algebraic, Trig, Exponential) helps select u.
此法源于乘法法则：∫ u dv = uv – ∫ v du。选择 u 和 dv 时应使新积分 ∫ v du 更简单。LIATE 法则（对数、反三角、代数、三角、指数）可辅助选定 u。
Example: ∫ x eˣ dx. Let u = x, dv = eˣ dx → du = dx, v = eˣ. Then ∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C.
示例：∫ x eˣ dx。令 u = x，dv = eˣ dx → du = dx，v = eˣ。故 ∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C。
For definite integrals, apply the limits to the fully integrated expression: ∫₀¹ x eˣ dx = [x eˣ – eˣ]₀¹ = (e – e) – (0 – 1) = 1.
对于定积分，将上下限代入完整积分表达式：∫₀¹ x eˣ dx = [x eˣ – eˣ]₀¹ = (e – e) – (0 – 1) = 1。
7. Integration Using Partial Fractions | 部分分式积分法
When integrating a rational function P(x)/Q(x) where degree(P) < degree(Q), factor Q(x) and decompose into simpler fractions. Then integrate each term individually.
当对有理函数 P(x)/Q(x) 积分且 deg(P) < deg(Q) 时，先分解 Q(x) 为部分分式，再逐项积分。
Example: ∫ 1/(x² – 1) dx. Factor denominator: x² – 1 = (x-1)(x+1). Decompose: 1/(x²-1) = ½ [1/(x-1) – 1/(x+1)]. Then integrate: ½ (∫ 1/(x-1) dx – ∫ 1/(x+1) dx) = ½ (ln|x-1| – ln|x+1|) + C = ½ ln| (x-1)/(x+1) | + C.
示例：∫ 1/(x² – 1) dx。分解分母：x² – 1 = (x-1)(x+1)。拆项：1/(x²-1) = ½ [1/(x-1) – 1/(x+1)]。积分：½ (∫ 1/(x-1) dx – ∫ 1/(x+1) dx) = ½ (ln|x-1| – ln|x+1|) + C = ½ ln| (x-1)/(x+1) | + C。
8. Definite Integral Calculation Examples (AP Style) | 定积分计算真题示例
Below are typical AP Calculus questions that test fundamental skills. Work through each one step by step.
以下为典型的 AP 微积分题目，考查基本技能。逐步解答每一道。
Example 1: Evaluate ∫₁⁴ (2√x – 1/x) dx.
示例 1：计算 ∫₁⁴ (2√x – 1/x) dx。
Rewrite √x as x^(½). Antiderivative: 2 * (x^(³/₂) / (3/2)) – ln|x| = (4/3) x^(³/₂) – ln x. Evaluate from 1 to 4: [(4/3)·4^(³/₂) – ln 4] – [(4/3)·1^(³/₂) – ln 1] = (4/3)·8 – ln 4 – 4/3 = 32/3 – 4/3 – ln 4 = 28/3 – ln 4.
将 √x 改写为 x^(½)。反导数：2·(x^(³/₂) / (3/2)) – ln|x| = (4/3) x^(³/₂) – ln x。代入 1 到 4：[(4/3)·4^(³/₂) – ln 4] – [(4/3)·1^(³/₂) – ln 1] = (4/3)·8 – ln 4 – 4/3 = 32/3 – 4/3 – ln 4 = 28/3 – ln 4。
Example 2: If ∫₀³ f(x) dx = 5 and ∫₀³ g(x) dx = 2, find ∫₀³ (2f(x) – 3g(x)) dx.
示例 2：已知 ∫₀³ f(x) dx = 5，∫₀³ g(x) dx = 2，求 ∫₀³ (2f(x) – 3g(x)) dx。
Using linearity: ∫₀³ 2f(x) dx – ∫₀³ 3g(x) dx = 2 ∫₀³ f(x) dx – 3 ∫₀³ g(x) dx = 2(5) – 3(2) = 10 – 6 = 4.
利用线性性质：∫₀³ 2f(x) dx – ∫₀³ 3g(x) dx = 2 ∫₀³ f(x) dx – 3 ∫₀³ g(x) dx = 2×5 – 3×2 = 10 – 6 = 4。
Example 3 (FTC Part 1): Let F(x) = ∫₂ˣ ln(t) dt. Find F'(x).
示例 3（FTC 第一部分）：设 F(x) = ∫₂ˣ ln(t) dt。求 F'(x)。
By FTC Part 1, F'(x) = ln(x). If the upper limit were x², then by the chain rule: d/dx ∫₂ˣ^² ln(t) dt = ln(x²) · 2x.
根据 FTC 第一部分，F'(x) = ln(x)。若上限为 x²，则由链式法则：d/dx ∫₂ˣ^² ln(t) dt = ln(x²) · 2x。
9. Application: Area Between Curves | 应用：曲线间面积
The area between y = f(x) and y = g(x) from a to b is given by ∫ₐᵇ |f(x) – g(x)| dx. It is often necessary to find intersection points to set up the integral correctly.
从 a 到 b 由 y = f(x) 与 y = g(x) 所围区域面积为 ∫ₐᵇ |f(x) – g(x)| dx。通常需要先求出交点以正确设立积分。
Example: Find the area bounded by y = x² and y = 2x. Intersections: x² = 2x → x=0, x=2. On [0,2], 2x ≥ x², so Area = ∫₀² (2x – x²) dx = [x² – x³/3]₀² = (4 – 8/3) – 0 = 4/3.
示例：求 y = x² 与 y = 2x 所围面积。交点：x² = 2x → x=0, x=2。在 [0,2] 上 2x ≥ x²，故面积 = ∫₀² (2x – x²) dx = [x² – x³/3]₀² = (4 – 8/3) – 0 = 4/3。
If the curves switch positions, split the interval accordingly and sum the absolute values.
若曲线上下位置互换，需分割区间并求和各段的绝对值。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
Always include ‘+ C’ for indefinite integral answers; omitting it loses points on the AP exam. For definite integrals using substitution, change the limits to u-values to avoid back-substitution mistakes.
不定积分答案务必加上“+ C”，AP 考试漏写会扣分。用换元法计算定积分时，务必将积分限换成 u 值，避免回代错误。
Check if the integrand has symmetry on symmetric limits. If f is
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