📚 Integration Applications: Common Question Types in IB Maths | IB数学：积分应用常见题型解析
Integration is one of the most versatile tools in the IB Mathematics curriculum, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI), at Standard Level and Higher Level. Beyond the mechanical skill of finding antiderivatives, examiners are deeply interested in whether students can apply integration to real-world and geometric contexts. This article breaks down the most frequently tested question types involving integration applications, with a clear focus on what IB examiners expect in Paper 2 and Paper 3.
积分是IB数学课程中最具多功能性的工具之一，既出现在分析与方法（AA）中，也出现在应用与解释（AI）中，覆盖标准级别和高级级别。除了求原函数的机械技能之外，考官更关注学生能否将积分应用于真实情境和几何情境。本文将拆解积分应用中最常考的题型，并明确聚焦IB考官在Paper 2和Paper 3中的评分期待。
1. Area Under a Curve | 曲线下方面积
The most fundamental application of definite integration is computing the area bounded by a curve and the x-axis over an interval [a, b]. The key subtlety that separates top-scoring students from the rest is the treatment of regions below the x-axis. Since area is a positive quantity, a definite integral that evaluates to a negative value must be interpreted carefully: the geometric area is the absolute value of that integral. If a function crosses the x-axis within the interval, the total area must be split at the roots and computed as the sum of absolute values.
定积分最基本的应用是计算曲线与x轴在区间[a, b]上围成的面积。能够区分高分考生与其他考生的一个关键细节是：如何处理x轴下方的区域。由于面积是正量，当定积分结果为负时，必须仔细解读：几何面积为该积分的绝对值。如果函数在区间内穿过x轴，则总面积必须在根处分段，并将各段绝对值相加。
ate the roots (x = 0 and x = 3). The function is negative on (0, 3), so the total area is:
例如，要求 f(x) = x² − 3x 与x轴在 x = 0 到 x = 3 之间围成的总面积，首先找出根（x = 0 和 x = 3）。函数在 (0, 3) 上为负，因此总面积为：
Total Area = |∫₀³ (x² − 3x) dx| = |[x³/3 − 3x²/2]₀³| = |(9 − 13.5)| = 4.5
Always sketch the graph or identify roots before computing area.
在计算面积之前，务必画草图或先求根。
Use the absolute value of the integral for each sub-interval where the sign of f(x) is constant.
在f(x)符号恒定的每个子区间上，使用积分的绝对值。
In exam marking schemes, failing to split the integral at roots is a common cause of lost marks.
在考试评分标准中，未在根处分段积分是常见失分原因。
2. Area Between Two Curves | 两曲线之间的面积
o curves intersect within the interval, it is crucial to determine all intersection points and to split the integral accordingly, because the "upper" function may switch between f and g.
面积问题的自然延伸是求两条函数曲线（例如 f(x) 与 g(x)）在某个区间内围成的区域面积。标准做法是对上函数与下函数之差积分：∫ (上 − 下) dx。当两条曲线在区间内相交时，必须求出所有交点并相应分段积分，因为"上方"函数可能在f与g之间切换。
cos x from x = 0 to x = π: the curves intersect at x = π/4. Thus, the area is computed as:
例如，求 y = sin x 与 y = cos x 在 x = 0 到 x = π 之间的面积：两曲线在 x = π/4 处相交。因此面积为：
Area = ∫₀π/4 (cos x − sin x) dx + ∫π/4π (sin x − cos x) dx
Use a graphing calculator or sketch to identify which function is on top in each sub-interval.
使用图形计算器或草图判断各子区间上哪个函数在上方。
Set f(x) = g(x) to find intersection points — these are the limits of integration.
令 f(x) = g(x) 求交点——这些交点就是积分限。
Do not integrate |f − g| directly without considering where the curves cross.
不要在不考虑曲线交点的情况下直接对 |f − g| 积分。
3. Volume of Revolution: Disk Method | 旋转体体积：圆盘法
ᵇ [f(x)]² dx. This is known as the disk method. The key idea is that each infinitesimally thin vertical strip of the region sweeps out a circular disk. Students must square the function — forgetting the square is one of the most common algebraic mistakes in this topic.
当曲线 y = f(x) 下方的区域绕x轴旋转时，所得三维立体的体积为 V = π∫ₐᵇ [f(x)]² dx。这称为圆盘法。核心思想是：区域中每一条无限薄的竖直条带扫出一个圆盘。学生必须将函数平方——忘记平方是本主题中最常见的代数错误之一。
V = π ∫ₐᵇ y² dx (rotation about the x-axis) | V = π ∫ₐᵇ y² dx （绕x轴旋转）
For rotation about the y-axis using the disk method, the formula is V = π ∫꜀ᵈ x² dy, where x is expressed as a function of y and c, d are the y-limits.
若使用圆盘法绕y轴旋转，则公式为 V = π ∫꜀ᵈ x² dy，其中x表示为y的函数，c、d为y方向上的积分限。
Identify the axis of rotation first — this determines whether you integrate with respect to x or y.
首先确定旋转轴——这决定了你是对x积分还是对y积分。
tion about the x-axis, express the curve as y = f(x); for the y-axis, rewrite as x = g(y).
绕x轴旋转时，将曲线写成 y = f(x)；绕y轴旋转时，改写为 x = g(y)。
Check the limits of integration carefully — they refer to the axis of rotation.
仔细核对积分限——它们对应于旋转轴方向。
4. Volume of Revolution: Washer Method | 旋转体体积：垫圈法
When the region between two curves is revolved about an axis, the cross-section perpendicular to the axis is not a solid disk but a washer (a disk with a hole). The volume is obtained by subtracting the inner volume from the outer volume:
当两条曲线之间的区域绕某轴旋转时，垂直于轴的截面不是实心圆盘，而是垫圈（带孔的圆盘）。体积等于外体积减去内体积：
V = π ∫ₐᵇ ([f(x)]² − [g(x)]²) dx
classic IB example is finding the volume generated by rotating the region enclosed by y = √x and y = x² about the x-axis.
其中 f(x) 为外半径，g(x) 为内半径。一个经典的IB例题是：求由 y = √x 与 y = x² 围成的区域绕x轴旋转所得体积。
The outer radius is always the function farther from the axis of rotation.
外半径始终是离旋转轴更远的函数。
Never compute π∫(f − g)² dx — this is a common and serious error.
切勿计算 π∫(f − g)² dx——这是一个常见且严重的错误。
Find intersection points of f and g to determine the integration limits.
求f与g的交点以确定积分限。
5. Kinematics: Displacement and Distance | 运动学：位移与路程
) dt. However, displacement can be negative if the particle moves backwards. The total distance travelled is the integral of speed, |v(t)|:
在运动学中，若 v(t) 表示质点沿直线运动的速度，则速度的定积分给出时间区间内的位移：s = ∫ₐᵇ v(t) dt。然而，如果质点向后运动，位移可能为负。质点经过的总路程是速率 |v(t)| 的积分：
Displacement = ∫ₐᵇ v(t) dt | Distance = ∫ₐᵇ |v(t)| dt
in the given interval. These times partition the interval into sub-intervals on which the sign of v(t) is constant. The distance is then the sum of the absolute values of displacement on each sub-interval.
要计算总路程，学生必须首先找出给定时间区间内 v(t) = 0 的所有时刻。这些时刻将区间划分为若干子区间，在每个子区间上v(t)的符号恒定。总路程即为各子区间位移绝对值之和。
Displacement is a signed quantity; distance is always non-negative.
位移是有符号的量；路程始终非负。
If v(t) does not change sign, then distance = |displacement|.
若v(t)不变号，则路程 = |位移|。
Acceleration is the derivative of velocity: a(t) = v′(t). Thus Δv = ∫ₐᵇ a(t) dt.
加速度是速度的导数：a(t) = v′(t)。因此 Δv = ∫ₐᵇ a(t) dt。
6. Average Value of a Function | 函数的平均值
Another common application of definite integrals is finding the average value of a function f(x) over an interval [a, b]. The formula is:
定积分的另一个常见应用是求函数 f(x) 在区间 [a, b] 上的平均值。公式为：
Average Value = (1/(b − a)) ∫ₐᵇ f(x) dx
This is not the same as the average of the endpoints f(a) and f(b). It represents the height of a rectangle with width (b − a) whose area equals the area under the curve. In IB questions, this often appears in physics contexts such as average temperature or average current over a time interval.
这与端点平均值 f(a) 和 f(b) 的算术平均不同。它表示一个宽度为 (b − a) 且面积等于曲线下面积的矩形的高度。在IB题目中，这常出现在物理情境中，例如某时间段内的平均温度或平均电流。
Divide by the interval length (b − a), not by the number of points.
除以区间长度 (b − a)，而不是除以点的个数。
The Mean Value Theorem for Integrals guarantees that this average value is attained by f at least once on [a, b] if f is continuous.
积分中值定理保证：若f连续，则这个平均值在 [a, b] 上至少被f取到一次。
Memorise the formula in the form (1/(b − a)) ∫, not ∫/(b − a) — the placement of the factor is crucial.
记住公式的形式是 (1/(b − a)) ∫，而不是 ∫/(b − a)——因子位置至关重要。
7. Differential Equations: Separation of Variables | 微分方程：变量分离法
ion of variables involves rearranging to put all y-terms on one side and all x-terms on the other, then integrating both sides:
许多积分应用题通过一阶微分方程的形式设定：dy/dx = f(x)g(y)。变量分离法是将所有含y的项移到一边、所有含x的项移到另一边，然后对两边积分：
∫ (1/g(y)) dy = ∫ f(x) dx
After integrating, a constant of integration appears, which is then determined by an initial condition or boundary condition given in the problem. IB exam questions often embed this in growth/decay models, Newton's law of cooling, or logistic growth scenarios.
积分之后会出现积分常数，该常数由题目所给的初始条件或边界条件确定。IB考试题常将其嵌入增长/衰减模型、牛顿冷却定律或逻辑斯谛增长情境中。
State the constant of integration before applying the initial condition — examiners reward this.
在代入初始条件之前先写出积分常数——考官对此给分。
main: if g(y) = 0 for some value of y, that value may be a singular solution worth noting.
检查定义域：若存在某个y值使 g(y) = 0，该值可能是一个值得注意的奇解。
Always verify your solution by differentiating it back into the original differential equation.
务必通过求导将你的解代回原微分方程来验证。
8. Kinematics from Acceleration: Integrating Twice | 由加速度求运动：两次积分
When a problem provides the acceleration function a(t) and initial conditions for both velocity and displacement, two successive integrations are required. First, integrate a(t) to obtain v(t), using the initial velocity v(0) to find the constant. Then integrate v(t) to obtain the displacement or position function, using the initial position to determine the second constant.
当题目给出加速度函数 a(t) 以及速度和位移的初始条件时，需要进行两次连续积分。首先对 a(t) 积分得到 v(t)，利用初始速度 v(0) 求出积分常数；然后对 v(t) 积分得到位移或位置函数，利用初始位置确定第二个常数。
For example, if a(t) = 6t − 4, v(0) = 2, and s(0) = 1, then:
例如，若 a(t) = 6t − 4，v(0) = 2，s(0) = 1，则：
v(t) = ∫(6t − 4) dt = 3t² − 4t + C, C = 2 → v(t) = 3t² − 4t + 2
s(t) = ∫(3t² − 4t + 2) dt = t³ − 2t² + 2t + C, C = 1 → s(t) = t³ − 2t² + 2t + 1
Organise your constants carefully — label them C₁ and C₂.
仔细整理积分常数——用C₁和C₂区分标注。
Be careful with units: acceleration in m/s², velocity in m/s, displacement in metres.
注意单位：加速度单位为m/s²，速度单位为m/s，位移单位为米。
Often the question asks for maximum height, which occurs when v(t) = 0.
题目常要求求最大高度，此时 v(t) = 0。
9. Improper Integrals | 反常积分
Improper integrals appear when either the interval of integration is infinite or the integrand has an infinite discontinuity within the interval. The technique is to replace the problematic limit with a variable and take the limit as that variable approaches the boundary. If the limit exists and is finite, the improper integral converges; otherwise it diverges.
反常积分出现在积分区间无穷或被积函数在区间内有无穷间断点时。处理技巧是将有问题的边界替换为一个变量，并取该变量趋近边界时的极限。若极限存在且有限，则反常积分收敛；否则发散。
∫₁∞ (1/x²) dx = lima→∞ ∫₁ᵃ x⁻² dx = lima→∞ [−x⁻¹]₁ᵃ = lima→∞ (1 − 1/a) = 1 (converges)
In IB, improper integrals often arise in probability density functions — re domain equals 1, or computing expected values via limits. Students must recognise when an integral is improper even if it is not written with infinity signs.
在IB中，反常积分常出现在概率密度函数中——验证整个定义域上 ∫f(x)dx 等于1，或通过极限计算期望值。学生必须能够识别一个积分是反常的，即使它没有以无穷号写出。
Check the domain of the integrand for vertical asymptotes before integrating.
在积分前检查被积函数定义域中的垂直渐近线。
Use a dummy variable (such as a or R) and take the limit explicitly in your working.
使用一个哑变量（例如a或R）并在解题过程中明确写极限。
If either limit fails to exist, state that the integral diverges.
若任一极限不存在，需说明该积分发散。
10. Numerical Integration: Trapezoidal Rule | 数值积分：梯形法则
When an antiderivative cannot be found easily, or when the function is known only as discrete data points, numerical integration provides an approximation. The trapezoidal rule approximates the area under a curve by summing the areas of trapezoids formed over n sub-intervals:
当原函数不易求出，或函数仅以离散数据点形式给出时，数值积分提供了一种近似方法。梯形法则通过将曲线下的区域划分为n个子区间并累加梯形面积来近似积分：
∫ₐᵇ f(x) dx ≈ (h/2) × (y₀ + 2y₁ + 2y₂ + … + 2yn−1 + yₙ), h = (b − a)/n
IB requires students to compute such approximations by hand, to interpret the approximation as an over- or under-estimate depending on the concavity of f, and to express answers to a sensible degree of accuracy. The trapezoidal rule underestimates when f is concave down and overestimates when f is concave up.
IB要求学生手算这类近似值，根据函数的凹凸性判断近似值是高估还是低估，并以合理精度给出答案。当f下凹时梯形法则低估，当f上凸时梯形法则高估。
The trapezoidal rule is not exact; it is an approximation that improves as n increases.
梯形法则不是精确的；它是一种近似，随着n增大而更精确。
If f is concave up on [a, b], the trapezoidal estimate is an overestimate.
若f在 [a, b] 上上凸，梯形估计为高估。
Keep at least one extra decimal place in intermediate calculations to avoid rounding errors in the final answer.
在中间计算中至少保留一位额外小数，以避免最终答案出现舍入误差。
11. Areas with Parametric Curves | 参数曲线的面积
) must be carefully considered: if x decreases as t increases, the integral may produce a negative value, and the area should be taken as the absolute value.
对于由参数方程 x = x(t)、y = y(t) 定义的曲线，从 t = t₁ 到 t = t₂ 的曲线下面积为 A = ∫ y dx = ∫ y(t) x′(t) dt。必须仔细考虑 x′(t) 的符号：若x随t增大而减小，积分可能为负，面积应取其绝对值。
or 0 ≤ t ≤ 2π is often computed using symmetry so that negative contributions are avoided:
例如，圆 x = a cos t、y = a sin t 在 0 ≤ t ≤ 2π 内的面积通常利用对称性计算，以避免负贡献：
A = 4∫₀π/2 y(t) x′(t) dt = 4∫₀π/2 (a sin t)(−a sin t) dt = 4a² ∫₀π/2 (−sin²t) dt = πa²
Always consider the orientation of the parameter direction.
始终考虑参数方向的方向性。
Using symmetry can dramatically simplify the computation, but justify each symmetry step.
利用对称性可以大幅简化计算，但需说明每一个对称性步骤的理由。
Double-check that the limits t₁ and t₂ correspond to the correct endpoints of the curve.
核对t₁和t₂这两个积分限是否对应于曲线的正确端点。
12. Modelling: Accumulated Change | 建模：累积变化量
For example, if water flows into a tank at a rate r(t) = 5 + 2t litres per minute, the total volume added from t = 0 to t = 10 is:
IB中最复杂的积分应用是利用定积分对累积变化量进行建模，例如人口增长、资源消耗、流速和能量消耗。若已知变化率 r(t)，则总变化量就是 r(t) 的积分。例如，若水以 r(t) = 5 + 2t 升/分钟的速率流入水箱，则从 t = 0 到 t = 10 的总加入水量为：
Total Volume = ∫₀¹⁰ (5 + 2t) dt = [5t + t²]₀¹⁰ = 50 + 100 = 150 litres
Such problems require translating the word problem into an integral setup, identifying the appropriate limits, and interpreting the result in the context of the original question. Examiners often award one mark for setting up the correct integral and one for evaluating it; communicate your reasoning clearly.
这类问题要求将文字题转化为积分结构、确定合适的积分限，并在原问题情境中解读结果。考官通常给1分用于写出正确的积分式，1分用于求值；请清晰表达你的推理过程。
Identify the rate function and the variable of integration (usually time).
确定速率函数和积分变量（通常是时间）。
Pay attention to initial conditions: if the tank already contains water, add the initial amount.
注意初始条件：若水箱中原已有水，需加上初始量。
Write your final answer as a sentence with the correct units to demonstrate full understanding.
用带有正确单位的完整句子写出最终答案，以展示你的完整理解。
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