IB Maths: Integration by Parts for Definite Integrals | IB数学：分部积分法求解定积分

📚 IB Maths: Integration by Parts for Definite Integrals | IB数学：分部积分法求解定积分

Integration by parts is one of the most heavily examined techniques in the IB Mathematics: Analysis and Approaches HL calculus syllabus, and it almost always appears as a definite integral with limits attached. Students who can set up the formula correctly still lose marks because of one small detail – the boundary term. This guide rebuilds the technique from the product rule, shows you how to choose u and dv with confidence, and works through the exact question types that examiners recycle every session.

分部积分是IB数学分析与方法（AA）HL 微积分部分考查频率最高的技巧之一，而且几乎总是以带上下限的定积分形式出现。很多学生公式背得没错，却因为一个细节丢分——边界项。本文从乘积法则出发重建整个方法，教你如何有把握地选择 u 与 dv，并逐题讲透考官每个考季都在重复使用的题型。

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1. Why Integration by Parts Is a Guaranteed Exam Topic | 为什么分部积分是必考题型

In the current IB guide, integration by parts sits in the HL calculus content of Analysis and Approaches. It is examined in both Paper 1 (no calculator) and Paper 2 (calculator allowed), and it rarely appears alone: it is usually combined with exponential, logarithmic, trigonometric or inverse trigonometric functions, or with a follow-up question about an area or a volume of revolution.

在现行 IB 大纲中，分部积分属于 AA HL 的微积分内容，Paper 1（不可用计算器）和 Paper 2（可用计算器）都会考。它很少单独出现，通常与指数函数、对数函数、三角函数或反三角函数结合，后面还会接一问求面积或旋转体体积。

The typical command terms are “Find”, “Show that” and “Hence”. A “show that” part is your best friend because it tells you the target answer – if your value disagrees, you know immediately that a sign or a boundary term has gone wrong.

常见的指令词是 Find、Show that 和 Hence。其中 show that 是最友好的，因为它直接给出目标答案；你的结果一旦对不上，就能立刻判断出是符号还是边界项出了问题。

On other IB routes the scope of calculus differs, so always confirm with your teacher whether integration by parts is examinable for your particular course before drilling it.

其他 IB 数学路线的微积分范围不同，因此在集中训练前，请先向老师确认分部积分是否属于你所在课程的考查内容。

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2. From the Product Rule to the Integration by Parts Formula | 从乘积法则到分部积分公式

Everything starts with the product rule for differentiating two functions u and v of x:

一切都从两个关于 x 的函数 u 与 v 的乘积求导法则开始：

d(uv)/dx = u (dv/dx) + v (du/dx)

Integrating both sides with respect to x gives uv = ∫ u dv + ∫ v du, because the integral of a derivative simply returns the function. Rearranging produces the standard indefinite form.

两边对 x 积分，由于导数的积分还原为原函数，得到 uv = ∫ u dv + ∫ v du。移项即得标准的不定积分形式。

∫ u dv = u v − ∫ v du

The logic of the method is simple: you trade a difficult integral for an easier one. You differentiate one part (u becomes du) and integrate the other (dv becomes v), then hope that the new integral is friendlier. That trade is the whole idea, and judging whether the trade is a good one is the skill being tested.

方法的逻辑很简单：用一个较难的积分换一个较易的积分。你对一部分求导（u 变 du），对另一部分积分（dv 变 v），再希望新的积分更好算。这个”交换”就是全部思想，而判断交换是否划算，正是被考查的能力。

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3. The Definite Integral Version: Never Drop the Boundary Term | 定积分版本：绝不能漏掉边界项

For a definite integral from x = a to x = b, the formula gains one extra piece: the product uv must itself be evaluated between the limits. Written in words rather than symbols, the rule is:

对于从 x = a 到 x = b 的定积分，公式多出一部分：乘积 uv 本身也必须代入上下限求值。用文字表述，规则是：

∫ (from a to b) u dv = [u v] (from a to b) − ∫ (from a to b) v du

The bracket [uv] evaluated from a to b is called the boundary term. In most markschemes it carries its own mark, which means writing it explicitly is not optional. Students who skip straight to the second integral usually lose two marks even when their final number happens to be right.

方括号 [uv] 代入上下限的结果称为边界项。在多数评分标准中它单独占一分，也就是说不写出来不行。直接跳到第二个积分的学生，即使最后数字碰巧对了，通常也会丢两分。

Notice also that the limits do not change. In a definite integral there is no “+ C”, and there is no need to convert limits back to x after integrating, because you never leave the variable x in the first place.

还要注意上下限保持不变。定积分中没有 “+ C”，也不需要在积分后把变量换回 x，因为整个过程从未改变变量。

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4. Choosing u and dv: The LIATE Strategy | 选择 u 与 dv：LIATE 策略

Every integration by parts question begins with a decision: which factor becomes u, and which becomes dv? A reliable heuristic is LIATE, which ranks the candidates for u in order of priority.

每一道分部积分题都始于一个决定：哪个因子作 u，哪个作 dv？一个可靠的启发式规则是 LIATE，它按优先顺序排列 u 的候选对象。

Letter	Type	Example	Why it becomes u
L	Logarithmic	ln x, log₂ x	Differentiating simplifies it to 1/x
I	Inverse trig	sin⁻¹x, tan⁻¹x	Derivative is algebraic, no inverse left
A	Algebraic	x, x², 3x + 1	Repeated differentiation kills it
T	Trigonometric	sin x, cos 2x	Cycles rather than disappears
E	Exponential	eˣ, e⁻²ˣ	Almost never choose this as u

Work down the list: whichever type appears earlier should be your u. So for x eˣ, x is algebraic (A) and eˣ is exponential (E), hence u = x. For x ln x, ln x wins, so u = ln x.

从上往下看：出现得更靠前的类型就作 u。因此对 x eˣ，x 是代数函数（A）、eˣ 是指数函数（E），所以 u = x。对 x ln x，ln x 优先，所以 u = ln x。

LIATE is a heuristic, not a theorem. When both choices look equally workable, test the trade: if differentiating u does not simplify things and integrating dv does not make things worse, you have probably picked correctly.

LIATE 是启发式规则，不是定理。当两种选择看起来都可行时，就检验这次交换：如果 u 求导后没有变简单、而 dv 积分后也没有变复杂，那你多半选对了。

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5. Worked Example 1: A Polynomial Times an Exponential | 例1：多项式乘指数函数

Evaluate ∫₀¹ x eˣ dx. By LIATE, u = x and dv = eˣ dx, so du = dx and v = eˣ.

计算 ∫₀¹ x eˣ dx。按 LIATE，取 u = x、dv = eˣ dx，于是 du = dx、v = eˣ。

Substituting into the definite formula gives [x eˣ]₀¹ − ∫₀¹ eˣ dx. The boundary term is (1 · e¹) − (0 · e⁰) = e.

代入定积分公式得 [x eˣ]₀¹ − ∫₀¹ eˣ dx。边界项为 (1 · e¹) − (0 · e⁰) = e。

The remaining integral is ∫₀¹ eˣ dx = [eˣ]₀¹ = e − 1. Therefore the answer is e − (e − 1) = 1.

剩余积分为 ∫₀¹ eˣ dx = [eˣ]₀¹ = e − 1。因此答案为 e − (e − 1) = 1。

Notice how neat the cancellation is. This is the signature of a well-chosen split, and it is exactly what the examiner wants to see: a boundary term that is easy to evaluate and a leftover integral that is elementary.

注意抵消得非常漂亮。这正是选对拆分的标志，也是考官想看到的：边界项易于求值，剩下的积分是基本积分。

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6. Worked Example 2: The dv = dx Trick for ln x | 例2：ln x 的 dv = dx 技巧

Evaluate ∫₁^e ln x dx. There appears to be only one factor, but every function can be written as the product of itself and 1, so take u = ln x and dv = dx. Then du = (1/x) dx and v = x.

计算 ∫₁^e ln x dx。看上去只有一个因子，但任何函数都可以写成自身与 1 的乘积，所以取 u = ln x、dv = dx，于是 du = (1/x) dx、v = x。

Applying the formula: [x ln x] from 1 to e − ∫₁^e x · (1/x) dx. The boundary term is (e · ln e) − (1 · ln 1) = e − 0 = e, and the integral simplifies to ∫₁^e 1 dx = e − 1.

套用公式：[x ln x] 从 1 到 e 求值 − ∫₁^e x · (1/x) dx。边界项为 (e · ln e) − (1 · ln 1) = e − 0 = e，而积分化简为 ∫₁^e 1 dx = e − 1。

The final answer is e − (e − 1) = 1. The same trick works for the inverse trigonometric functions: write u = sin⁻¹x with dv = dx, and the derivative 1/√(1 − x²) turns the problem into an algebraic integral.

最终答案为 e − (e − 1) = 1。同样的技巧也适用于反三角函数：取 u = sin⁻¹x、

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