Problem Solving by Integration | 通过积分解决问题

📚 Problem Solving by Integration | 通过积分解决问题

Integration is a cornerstone of calculus that empowers us to solve a vast array of problems — from geometric quantities like areas and volumes to dynamic situations involving motion and growth. In the IB Mathematics curriculum, both Analysis & Approaches (AA) and Applications & Interpretation (AI) rely heavily on integration as a problem-solving tool. This article walks you through the most common types of integration problems, highlighting techniques, typical exam questions, and strategies to avoid pitfalls.

积分是微积分的基石，它使我们能够解决大量问题——从面积、体积等几何量到涉及运动和增长的动态情况。在 IB 数学课程中，无论是分析与方法（AA）还是应用与解释（AI），都高度依赖积分作为解决问题的工具。本文带你梳理最常见的积分问题类型，重点介绍技巧、典型考题以及避免陷阱的策略。

1. Areas Between Curves | 曲线之间的面积

The most direct application of definite integration is finding the area enclosed by curves. If two functions f(x) and g(x) satisfy f(x) ≥ g(x) on an interval [a, b], the area between them is given by ∫ₐᵇ (f(x) − g(x)) dx. When curves intersect, you must split the integral at intersection points to ensure the integrand is always non-negative.

定积分最直接的应用是求曲线围成的面积。如果两个函数 f(x) 和 g(x) 在区间 [a, b] 上满足 f(x) ≥ g(x)，则它们之间的面积为 ∫ₐᵇ (f(x) − g(x)) dx。当曲线相交时，必须在交点处拆分积分，以确保被积函数始终非负。

Always sketch the region first, find intersection points by solving f(x) = g(x), and check which function is above the other on each subinterval. A common mistake is forgetting to subtract the lower curve, leading to a negative area.

务必先画出区域的草图，通过解 f(x) = g(x) 求交点，并检查在每个子区间上哪条曲线在上方。常见错误是忘记减去下方的曲线，导致面积为负。

Area = ∫ₐᵇ [upper function − lower function] dx

面积 = ∫ₐᵇ [上方函数 − 下方函数] dx

For areas bounded by functions of y, i.e., x = f(y) and x = g(y), the formula becomes ∫cᵈ (right − left) dy. This is especially useful when the region is more easily described horizontally.

对于用 y 的函数表示的区域，即 x = f(y) 和 x = g(y)，公式变为 ∫cᵈ (右 − 左) dy。当区域更适合水平描述时，这尤其有用。

2. Volumes of Revolution | 旋转体体积

When a region is revolved around an axis, integration calculates the volume of the resulting solid. For rotation about the x-axis, the disc method gives V = π ∫ₐᵇ [f(x)]² dx. If the region is between two curves, the washer method yields V = π ∫ₐᵇ [ (f(x))² − (g(x))² ] dx.

当一个区域绕轴旋转时，积分可以计算所生成立体的体积。绕 x 轴旋转时，圆盘法给出 V = π ∫ₐᵇ [f(x)]² dx。如果区域位于两条曲线之间，垫圈法给出 V = π ∫ₐᵇ [ (f(x))² − (g(x))² ] dx。

For revolution about the y-axis, rewrite the functions as x = f(y) and integrate with respect to y. The key is to correctly identify the radius (or outer and inner radii) and the limits of integration. Always check whether the solid is fully formed or has a hollow centre.

对于绕 y 轴旋转，将函数改写为 x = f(y)，并对 y 积分。关键在于正确识别半径（或外半径与内半径）以及积分限。务必检查立体是完全实心还是中空的。

V = π ∫ₐᵇ [R(x)]² dx (disc method | 圆盘法)

V = π ∫ₐᵇ [R(x)]² − [r(x)]² dx (washer method | 垫圈法)

A typical IB question might ask for the volume generated when the region bounded by y = √x, the x-axis, and x = 4 is revolved about the x-axis. The volume is π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [½x²]₀⁴ = 8π.

一个典型的 IB 考题可能要求求由 y = √x、x 轴和 x = 4 围成的区域绕 x 轴旋转所生成的体积。体积为 π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [½x²]₀⁴ = 8π。

3. Kinematics: Displacement, Velocity, and Acceleration | 运动学：位移、速度与加速度

Integration links displacement s(t), velocity v(t), and acceleration a(t). Since v(t) = ds/dt and a(t) = dv/dt, we recover displacement by s(t) = ∫ v(t) dt + C and velocity by v(t) = ∫ a(t) dt + D. Definite integrals give the total change over a time interval.

积分将位移 s(t)、速度 v(t) 和加速度 a(t) 联系起来。由于 v(t) = ds/dt 且 a(t) = dv/dt，我们可以通过 s(t) = ∫ v(t) dt + C 得到位移，通过 v(t) = ∫ a(t) dt + D 得到速度。定积分给出一个时间段内的总变化量。

Displacement (net change in position) from t₁ to t₂ is ∫ₜ₁ᵗ² v(t) dt. Total distance travelled, however, requires integrating the absolute value of velocity, |v(t)|, because the particle may change direction. Always break the interval where v(t) = 0 to handle sign changes.

从 t₁ 到 t₂ 的位移（位置的净变化）为 ∫ₜ₁ᵗ² v(t) dt。然而，总路程需要积分速度的绝对值 |v(t)|，因为质点可能改变运动方向。务必在 v(t) = 0 处拆分区间，以处理符号变化。

Displacement = ∫ₜ₁ᵗ² v(t) dt Total distance = ∫ₜ₁ᵗ² |v(t)| dt

位移 = ∫ₜ₁ᵗ² v(t) dt 总路程 = ∫ₜ₁ᵗ² |v(t)| dt

For example, if v(t) = t² − 4t + 3, find the distance travelled in the first 4 seconds. First find when v(t) = 0 ⇒ t=1, 3. Then integrate piecewise: |v(t)| from 0 to 1, 1 to 3, and 3 to 4, summing the positive results.

例如，若 v(t) = t² − 4t + 3，求前 4 秒内经过的路程。首先求 v(t) = 0 ⇒ t=1, 3。然后分段积分：从 0 到 1、1 到 3、3 到 4 分别积分 |v(t)|，再将正值相加。

4. Solving Separable Differential Equations | 解可分离变量的微分方程

Many real-world models involve differential equations of the form dy/dx = g(x) h(y). These can be solved by separating variables: (1/h(y)) dy = g(x) dx, then integrating both sides. An initial condition allows us to find the particular solution.

许多现实世界模型涉及形如 dy/dx = g(x) h(y) 的微分方程。这些可通过分离变量来求解：(1/h(y)) dy = g(x) dx，然后两边积分。已知初始条件可帮助我们求出特解。

Always check that h(y) ≠ 0 before separating. After integration, the constant C appears; use the given point to solve for C. The final answer should be expressed as y = f(x) or an implicit equation if the explicit form is too cumbersome.

在分离变量前，务必检查 h(y) ≠ 0。积分后会出现常数 C；利用给定点解出 C。最终答案应表示为 y = f(x)，或者如果显式形式过于繁琐，也可用隐式方程表示。

Separate: ∫ (1/h(y)) dy = ∫ g(x) dx + C

分离：∫ (1/h(y)) dy = ∫ g(x) dx + C

Classic examples include exponential growth dy/dx = ky, where separation leads to y = Aeᵏˣ, and logistic growth models. In exams, you may be asked to verify a given solution or to solve an initial-value problem step by step.

经典例子包括指数增长 dy/dx = ky，分离后得 y = Aeᵏˣ，以及逻辑斯蒂增长模型。考试中，可能会要求验证一个给定解，或者逐步求解一个初值问题。

5. Accumulated Change from a Rate | 从变化率求累积变化

If a quantity Q changes at a known rate r(t), the net change in Q over the interval [a, b] is the definite integral of the rate: ΔQ = ∫ₐᵇ r(t) dt. This concept underpins problems involving water flow, population growth rate, temperature change, and more.

如果一个量 Q 以已知的速率 r(t) 变化，那么在区间 [a, b] 上 Q 的净变化就是该速率的定积分：ΔQ = ∫ₐᵇ r(t) dt。这一概念是涉及水流、种群增长率、温度变化等问题的基础。

Be careful to distinguish between the rate and the total amount. The integral adds up the instantaneous changes; adding the initial amount Q(a) gives the final amount: Q(b) = Q(a) + ∫ₐᵇ r(t) dt. This is a direct application of the Fundamental Theorem of Calculus.

注意区分速率与总量。积分将瞬时变化累积起来；加上初始量 Q(a) 便得到最终量：Q(b) = Q(a) + ∫ₐᵇ r(t) dt。这是微积分基本定理的直接应用。

Total change = ∫ₐᵇ (rate) dt Final amount = Initial + ∫ₐᵇ (rate) dt

总变化 = ∫ₐᵇ (速率) dt 最终量 = 初始量 + ∫ₐᵇ (速率) dt

For instance, if water flows into a tank at r(t) = 4 − e⁻ᵗ litres per minute, the total volume added in the first 5 minutes is ∫₀⁵ (4 − e⁻ᵗ) dt. Use technology or integration by parts if required.

例如，若水以 r(t) = 4 − e⁻ᵗ 升/分钟的速度流入水箱，则前 5 分钟内加入的总水量为 ∫₀⁵ (4 − e⁻ᵗ) dt。如有需要可使用计算器或分部积分。

6. Average Value of a Function | 函数的平均值

The average value of a continuous function f over [a, b] is defined as f_avg = (1/(b−a)) ∫ₐᵇ f(x) dx. Geometrically, this is the height of a rectangle with base [a, b] whose area equals the area under the curve.

连续函数 f 在 [a, b] 上的平均值定义为 f_avg = (1/(b−a)) ∫ₐᵇ f(x) dx。从几何上看，这是一个以 [a, b] 为底、面积等于曲线下面积的矩形的高。

This concept appears in contexts like average temperature, average velocity (as distinct from average speed), or mean concentration. It is also the continuous analogue of the arithmetic mean.

这一概念出现在平均温度、平均速度（有别于平均速率）或平均浓度等情境中。它也是算术平均值的连续类比。

Average = 1/(b−a) ∫ₐᵇ f(x) dx

平均值 = 1/(b−a) ∫ₐᵇ f(x) dx

A typical question might ask: Find the average value of f(x) = sin x on [0, π]. The integral ∫₀ᵖⁱ sin x dx = 2, so the average is 2/π. Note that the exact answer is left as 2/π.

典型考题可能要求：求 f(x) = sin x 在 [0, π] 上的平均值。积分 ∫₀ᵖⁱ sin x dx = 2，因此平均值为 2/π。注意精确答案保持为 2/π。

7. Arc Length (AA HL) | 弧长（仅 AA HL）

For students in Analysis & Approaches HL, the arc length of a curve y = f(x) from x = a to x = b is given by L = ∫ₐᵇ √(1 + [f'(x)]²) dx. This formula arises from summing infinitesimal hypotenuses along the curve.

对于分析与方法 HL 的学生，曲线 y = f(x) 从 x = a 到 x = b 的弧长由 L = ∫ₐᵇ √(1 + [f'(x)]²) dx 给出。该公式源于将沿曲线的无穷小斜边累加起来。

Arc length problems often involve functions whose derivatives produce simple expressions under the square root. For example, find the length of y = (2/3)(x² + 1)^(3/2) from x = 0 to x = 1. First compute dy/dx = 2x√(x²+1), then 1+(dy/dx)² = 1+4x²(x²+1) = (2x²+1)², so the integrand simplifies to 2x²+1, making the integration straightforward.

弧长问题通常涉及导数能在根号下产生简单表达式的函数。例如，求 y = (2/3)(x² + 1)^(3/2) 从 x = 0 到 x = 1 的长度。首先计算 dy/dx = 2x√(x²+1)，然后 1+(dy/dx)² = 1+4x²(x²+1) = (2x²+1)²，因此被积函数简化为 2x²+1，积分变得简单直接。

L = ∫ₐᵇ √(1 + (dy/dx)²) dx or L = ∫ₜ₁ᵗ² √( (dx/dt)² + (dy/dt)² ) dt

L = ∫ₐᵇ √(1 + (dy/dx)²) dx 或 L = ∫ₜ₁ᵗ² √( (dx/dt)² + (dy/dt)² ) dt

For parametrically defined curves, use the second form. Arc length is also a prime example where a graphics calculator (GDC) is often used to evaluate the definite integral when an antiderivative is not elementary.

对于参数定义的曲线，使用第二种形式。弧长也是当原函数非初等时通常使用图形计算器（GDC）计算定积分的典型例子。

8. Probability Density Functions | 概率密度函数

For continuous random variables, the probability density function (pdf) f(x) must satisfy f(x) ≥ 0 and ∫_all f(x) dx = 1. Probabilities are found by integrating the pdf over an interval: P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx.

对于连续随机变量，概率密度函数 f(x) 必须满足 f(x) ≥ 0 且 ∫_全域 f(x) dx = 1。概率通过在一个区间上积分 pdf 求得：P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx。

The median m is the value such that ∫_−∞ᵐ f(x) dx = 0.5. The mean (expected value) is μ = ∫_all x f(x) dx. Integration is also used to find variance via E(X²) − μ².

中位数 m 是满足 ∫_−∞ᵐ f(x) dx = 0.5 的值。均值（期望值）为 μ = ∫_全域 x f(x) dx。积分还可用于通过 E(X²) − μ² 求方差。

P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx E(X) = ∫ x f(x) dx

A common IB problem gives a piecewise pdf, such as f(x) = kx(4−x) for 0 ≤ x ≤ 4, and asks to find k, the probability that X > 2, or the median. First use the total area condition ∫₀⁴ kx(4−x) dx = 1 to solve for k, then proceed with the required calculations.

常见的 IB 题会给出一个分段 pdf，例如 f(x) = kx(4−x) 在 0 ≤ x ≤ 4 上，要求求 k、X > 2 的概率或中位数。首先利用总面积为 1 的条件 ∫₀⁴ kx(4−x) dx = 1 解出 k，然后进行所需的计算。

9. Work and Other Physical Applications | 功与其他物理应用

In physics, the work done by a variable force F(x) along a straight line from x = a to x = b is W = ∫ₐᵇ F(x) dx. This is a direct extension of the constant force formula W = F·d. Similarly, pumping water out of a tank or stretching a spring uses integration to sum infinitesimal work contributions.

在物理学中，变力 F(x) 沿直线从 x = a 到 x = b 所做的功为 W = ∫ₐᵇ F(x) dx。这是恒力公式 W = F·d 的直接推广。类似地，从水箱中抽水或拉伸弹簧均使用积分来累加无穷小的功。

Hooke’s Law, F(x) = kx, for a spring leads to W = ∫₀ᵈ kx dx = ½kd². This is a classic application that also appears in some IB questions, particularly in the options or as a modelling scenario.

胡克定律 F(x) = kx 对弹簧做功给出 W = ∫₀ᵈ kx dx = ½kd²。这是一个经典应用，也出现在一些 IB 题目中，尤其是在选修部分或作为建模场景。

W = ∫ₐᵇ F(x) dx (work done by a variable force | 变力做功)

W = ∫ₐᵇ F(x) dx (变力做功)

Consumer and producer surplus in economics are other practical integrals. If a demand curve is p = D(q), consumer surplus is ∫₀ᵠ* [D(q) − p*] dq. These problems test the ability to read real-world contexts and set up the correct integral.

经济学中的消费者剩余和生产者剩余是另一类实用积分。如果需求曲线为 p = D(q)，则消费者剩余为 ∫₀ᵠ* [D(q) − p*] dq。这类问题测试学生阅读现实情境并建立正确积分的能力。

10. Common Mistakes and Tips | 常见错误与技巧

Integration problems often trip up students when they neglect to adjust limits after substitution or forget the constant of integration in indefinite integrals. Always double-check the limits — if you use u-substitution, change the limits to u-values, or substitute back before evaluating.

积分问题常令学生栽跟头的地方包括：换元后忘记调整积分限，或者在不定积分中遗漏积分常数。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Problem Solving by Integration | 通过积分解决问题