📚 Integration by Parts: Definite Integrals | 分部积分法：定积分
Mastering integration by parts for definite integrals is a cornerstone of IB Mathematics. It transforms complex products into manageable forms and, with careful handling of limits, yields exact numerical values. This article walks you through the formula, strategic choices, detailed examples, and common pitfalls to ensure you can evaluate definite integrals confidently and accurately.
掌握定积分的分部积分法是 IB 数学的基石。它将复杂的乘积转化为易于处理的形式，并通过谨慎处理上下限得出精确的数值结果。本文将带你深入学习公式、策略选择、详细示例和常见陷阱，确保你能自信、准确地计算定积分。
1. The Definite Integration by Parts Formula | 定积分的分部积分公式
tion. If u(x) and v(x) are differentiable, the derivative of their product is d(uv)/dx = u' v + u v'. Integrating both sides from a to b yields the fundamental identity: ∫ab u dv = [u v]ab – ∫ab v du. Here, dv denotes v' dx and du denotes u' dx. The beauty of this formula is that the first term on the right-hand side is simply evaluated at the upper and lower limits, while the remaining integral may be easier to solve.
定积分的分部积分公式直接源于微分的乘积法则。若 u(x) 与 v(x) 可导，其乘积的导数为 d(uv)/dx = u' v + u v'。在区间 [a,b] 上对等式两边积分，即得基本恒等式：∫ab u dv = [u v]ab – ∫ab v du。其中 dv 代表 v' dx，du 代表 u' dx。此公式的精妙之处在于，右侧第一项只需在上下限处求值，而余下的积分往往更易计算。
Unlike indefinite integration by parts, where you must add a constant of integration C, the definite version directly yields a number. The key is to apply the limits to the product u v immediately, before focusing on the new integral. Many mistakes occur when students forget to evaluate [u v] at both limits or mishandle signs when substituting.
与不定积分不同，定积分版本无需添加积分常数 C，而是直接得出数值。关键在于立即将上下限代入乘积 u v，然后再集中处理新积分。许多错误源于学生在代入时忘记计算 [u v] 在两个限处的值，或处理符号时出现偏差。
2. Choosing u and dv: The LIATE Rule | 选择 u 与 dv：LIATE 法则
emainder (including dx). This ordering helps reduce the complexity of the integral ∫ v du.
分部积分法的成功取决于对 u 与 dv 的明智选择。LIATE 法则给出了优先级：对数函数、反三角函数、代数函数（多项式）、三角函数、指数函数。通常，u 应从列表中靠前的函数类型中选取，dv 则包含余下的部分（包括 dx）。这一顺序有助于降低新积分 ∫ v du 的复杂程度。
For example, in ∫ x ex dx, algebraic x precedes exponential, so you set u = x and dv = ex When two functions fall in the same category, often experience or the goal of simplifying the integrand guides the choice. In definite integrals, the same rule applies; you only need to keep track of the limits throughout.
例如，在 ∫ x ex dx 中，代数函数 x 优先级高于指数函数，因此设 u = x，dv = ex dx。在 ∫ x ln x dx 中，对数函数 ln x 优先于代数函数 x，所以 u = ln x，dv = x dx。当两个函数属于同一类别时，通常要靠经验或化简被积函数的目标来做出选择。对于定积分，同样适用该法则；你只需全程留意上下限即可。
3. Example 1: Polynomial × Exponential | 示例1：多项式乘以指数函数
Evaluate ∫01 x ex dx. Let u = x, so du = dx. Then dv = ex dx, giving v = ex. Apply the formula: ∫01 x ex dx = [x ex]01 – ∫01 ex dx. Evaluate the boundary term: (1·e1 – 0·e0) = e. The remaining integral is ∫01 ex dx = [ex]01 = e – 1. Thus, the result is e – (e – 1) = 1.
计算 ∫01 x ex dx。设 u = x，则 du = dx。同时 dv = ex dx，得 v = ex。应用公式：∫01 x ex dx = [x ex]01 – ∫01 ex dx。计算边界项：(1·e1 – 0·e0) = e。余下的积分为 ∫01 ex dx = [ex]01 = e – 1。因此最终结果为 e – (e – 1) = 1。
This example illustrates how a single application reduces the integral to a basic one. Notice that the algebraic factor x disappears after differentiation, leaving a pure exponential to integrate. The definite limits are applied both to the u v term and to the antiderivative of v du, ensuring no constant of integration is needed.
本例说明如何通过一次分部积分将原积分化为基本积分。注意到代数因子 x 经微分后消失，留下纯指数函数进行积分。定限不仅应用于 u v 项，也应用于 v du 的原函数，因此无需积分常数。
4. Example 2: Polynomial × Trigonometric | 示例2：多项式乘以三角函数
Consider ∫0π/2 x dx. Set u = x (algebraic), dv = cos x dx, so du = dx and v = sin x. Using the formula: ∫0π/2 x cos x dx = [x sin x]0π/2 – ∫0π/2 sin x dx. The boundary term gives (π/2)·sin(π/2) – 0·sin 0 = π/2. The new integral is ∫0π/2 sin x dx = [–cos x]0π/2 = (–0) – (–1) = 1. Hence, the answer is π/2 – 1.
考虑 ∫0π/2 x cos x dx。设 u = x（代数函数），dv = cos x dx，则 du = dx，v = sin x。代入公式：∫0π/2 x cos x dx = [x sin x]0π/2 – ∫0π/2 sin x dx。边界项得 (π/2)·sin(π/2) – 0·sin 0 = π/2。新积分为 ∫0π/2 sin x dx = [–cos x]0π/2 = (–0) – (–1) = 1。因此结果为 π/2 – 1。
When working with trigonometric functions, pay careful attention to the signs of antiderivatives. A common slip is forgetting that the integral of sin x is –cos x. Checking the evaluation at both limits systematically prevents sign errors. In an exam setting, it is wise to factor out constants early and double-check the evaluation of the [u v] term.
处理三角函数时，要密切注意原函数的符号。一个常见失误是忘记 sin x 的积分是 –cos x。系统检查两处限值的计算可以防止符号错误。在考试环境下，及早提取常数并复查 [u v] 项的计算是明智之举。
5. Example 3: Integrating ln x | 示例3：对数函数的积分
Integrating a lone logarithm on a definite interval is a classic case. Evaluate ∫1e ts u = ln x (logarithmic) and dv = 1 dx. Then du = (1/x) dx and v = x. The formula gives ∫1e ln x dx = [x ln x]1e – ∫1e x·(1/x) dx = [x ln x]1e – ∫1e 1 dx.
在定区间上积分单独的对数函数是经典案例。计算 ∫1e ln x dx。此处我们将被积函数视为 (1)·ln x。根据 LIATE 设 u = ln x（对数函数），dv = 1 dx，则 du = (1/x) dx，v = x。代入公式得 ∫1e ln x dx = [x ln x]1e – ∫1e x·(1/x) dx = [x ln x]1e – ∫1e 1 dx。
Now [x ln x]1e = e·ln e – 1·ln 1 = e·1 – 1·0 = e. The remaining integral is ∫1e 1 dx = [x]1e = e – 1. Subtract to obtain e – (e – 1) = 1. Remarkably, the area under y = ln x from 1 to e is exactly 1 square unit.
现在 [x ln x]1e = e·ln e – 1·ln 1 = e·1 – 1·0 = e。余下的积分为 ∫1e 1 dx = [x]1e = e – 1。相减得 e – (e – 1) = 1。值得注意的是，y = ln x 在 1 到 e 之间的面积恰好为 1 平方单位。
rctan x dx, let u be the transcendental function and dv be dx. The differentiation of the logarithm removes it from the new integrand, leaving a simple rational or constant function.
本例突出了一个策略技巧：每当面对 ∫ ln x dx 或 ∫ arctan x dx 时，设 u 为超越函数，dv 为 dx。对数的微分会将其从新被积函数中移除，留下简单的有理函数或常数。
6. Example 4: Inverse Trigonometric Function | 示例4：反三角函数的积分
Find ∫01 dx. Again, use u = arctan x, dv = dx. Then du = 1/(1 + x²) dx, v = x. The formula yields ∫01 arctan x dx = [x arctan x]01 – ∫01 x/(1 + x²) dx. The boundary term: 1·arctan 1 – new integral, use a simple substitution w = 1 + x², dw = 2x dx, so ∫ x/(1 + x²) dx = ½ ln|1 + x²|. Evaluating from 0 to 1: ½ [ln(2) – ln(1)] = ½ ln 2. The final answer is π/4 – ½ ln 2.
求 ∫01 arctan x dx。同样的，设 u = arctan x，dv = dx，则 du = 1/(1 + x²) dx，v = x。代入公式得 ∫01 arctan x dx = [x arctan x]01 – ∫01 x/(1 + x²) dx。边界项：1·arctan 1 – 0·arctan 0 = π/4。对于新积分，使用简单换元 w = 1 + x²，dw = 2x dx，得 ∫ x/(1 + x²) dx = ½ ln|1 + x²|。在 0 到 1 上求值：½ [ln(2) – ln(1)] = ½ ln 2。最终答案为 π/4 – ½ ln 2。
Integration by parts converts an integrand without a product into one that involves a rational function. Combining it with substitution is a powerful pattern. Always look for opportunities where the derivative of u simplifies the integrand dramatically, as the derivative of arctan x is a rational expression.
分部积分法将原本无乘积形式的被积函数转化为包含有理函数的积分。将其与换元法结合是一种强大的模式。始终要寻找 u 的导数能显著简化被积函数的机会，正如 arctan x 的导数为有理表达式那样。
7. Example 5: Cyclic Integral with ex sin x | 示例5：循环积分 ex sin x
pplications of integration by parts, leading back to the original integral. Evaluate I = ∫0π ex sin x dx. Let u = sin x, dv = ex dx, then du = cos x dx, v = ex. First application: I = [ex sin x]0π – ∫0π ex cos x dx. The boundary term is (eπ sin π – e0 sin 0) = 0. So I = – ∫0π ex cos x dx.
某些定积分需要两次应用分部积分法，并最终回到原积分。计算 I = ∫0π ex sin x dx。设 u = sin x，dv = ex dx，则 du = cos x dx，v = ex。第一次应用：I = [ex sin x]0π – ∫0π ex cos x dx。边界项为 (eπ sin π – e0 sin 0) = 0。故 I = – ∫0π ex cos x dx。
Now apply integration by parts to J = ∫0π ex cos x dx. Set u = cos x, dv = ex dx, yielding du = –sin x dx, v = ex. Then J = [ex cos x]0π – ∫0π ex (–sin x) dx = (eπ cos π – e0 cos 0) + ∫0π ex sin x dx = (–eπ – 1) + I. Substitute back into the expression for I: I = –[(–eπ – 1) + I] = eπ + 1 – I. Solving for I gives 2I = eπ + 1, so I = ½(eπ + 1).
现在对 J = ∫0π ex cos x dx 应用分部积分法。设 u = cos x，dv = ex dx，则 du = –sin x dx，v = ex。于是 J = [ex cos x]0π – ∫0π ex (–sin x) dx = (eπ cos π – e0 cos 0) + ∫0π ex sin x dx = (–eπ – 1) + I。代回 I 的表达式：I = –[(–eπ – 1) + I] = eπ + 1 – I。求解 I 得 2I = eπ + 1，故 I = ½(eπ + 1)。
stitutions consistent; if you initially chose u as the trig function, maintain that choice in the second application to avoid undoing progress. Always check that the boundary terms are correctly simplified—here sin π = 0 and cos π = –1 were essential.
循环积分在 IB 考试中频繁出现，尤其是指数函数与三角函数的乘积。关键在于保持替换的一致性；如果你最初选择三角函数作为 u，在第二次应用时也应维持该选择，以避免前功尽弃。务必检查边界项是否已正确化简——此处 sin π = 0 和 cos π = –1 至关重要。
8. Common Pitfalls and How to Avoid Them | 常见陷阱与对策
One recurring mistake is forgetting to evaluate the [u v] term at both limits, especially when one limit is zero and the product vanishes. Write out the full expression [u(x)v(x)]ab explicitly as u(b)v(b) – u(a)v(a) to avoid dropping a negative sign. Another pitfall is mixing up u and dv, which can make the new integral more complicated rather than simpler.
一个反复出现的错误是忘记在上下限处计算 [u v] 项，尤其是当其中一个限为零且乘积为零时。明确写出完整表达式 [u(x)v(x)]ab = u(b)v(b) – u(a)v(a)，以避免丢失负号。另一个陷阱是混淆 u 与 dv，这可能使新积分变得更复杂而不是更简单。
g trigonometric or exponential antiderivatives are also common. For instance, ∫ sin x dx = – , carefully track negative signs through each application. Using brackets liberally helps.
积分三角函数或指数函数的原函数时，符号错误也很常见。例如，∫ sin x dx = –cos x + C，遗漏负号会导致最终答案出错。当积分为 ∫ f(x)g(x) dx 且需要两次分部积分时，应谨慎追踪每一步的负号。大量使用括号有助于避免错误。
Students sometimes attempt to apply the indefinite integration by parts first, find an antiderivative, and then evaluate limits. While this works, it introduces extra steps and the risk of mishandling the C. The direct definite formula is faster and less error-prone. Finally, always simplify the integrand before choosing u and dv—sometimes algebraic manipulation or substitution makes the problem trivial.
学生有时会尝试先求不定积分，找到原函数后再代入上下限。虽然可行，但引入了额外步骤和误处理常数 C 的风险。直接使用定积分公式更快且更少出错。最后，在选择 u 和 dv 之前，始终先化简被积函数——有时代数变形或换元可使问题迎刃而解。
9. Practice Tips for IB Exams | IB 考试的练习建议
When preparing for the IB Mathematics Analysis & Approaches or Applications & Interpretation papers, consistent practice with diverse functions is vital. Start with straightforward polynomials times ex, sin x, or cos x, then progress to logarithms, inverse trig, and cyclic forms. Time yourself to build speed while maintaining accuracy; many exam questions award marks for correct setup even if the final evaluation has a minor slip.
在准备 IB 数学分析与方法或应用与解释的考试时，针对不同函数进行持续练习至关重要。从简单的多项式乘 ex、sin x 或 cos x 开始，逐步过渡到对数、反三角函数和循环形式。计时练习以提升速度，同时保持准确性；许多考题即使最终计算有小失误，正确的列式仍能获得分数。
Always present your working in a structured way: clearly state u, du, dv, v, the formula with limits, and the evaluation of boundary terms. In questions asking for exact values, leave answers in terms of e, π, or ln 2 rather than decimal approximations. Familiarise yourself with the formula booklet provided—it includes the integration by parts formula, so you can quickly reference it under exam pressure.
始终以结构化方式呈现解题步骤：明确写出 u、du、dv、v、带限值的公式以及边界项的求值。对于要求精确值的问题，以 e、π 或 ln 2 等形式给出答案，而非小数近似。熟悉提供的公式手册——其中包含分部积分公式，因此可以在考试压力下快速查阅。
arts. Review past IB papers to identify common patterns, such as integrals of the form ∫ xn ln x dx or those where the boundary term vanishes identically. The more you expose yourself to variations, the more instinctive your choice of u and dv becomes.
用需要结合多种技巧的积分来挑战自己：例如，先通过换元化简被积函数，再使用分部积分法的定积分。复习历年 IB 试卷，识别常见模式，例如 ∫ xn ln x dx 形式的积分，或那些边界项恒为零的情形。接触的变体越多，你对 u 和 dv 的选择就越本能。
10. Summary and Key Takeaways | 总结与核心要点
Integration by parts for definite integrals is a methodical process: identify u and dv using LIATE, differentiate u and integrate dv, apply the formula with limits, and evaluate the resulting expressions. Remember that the bracket term [u v] is always evaluated at the upper and lower limits before tackling the new integral. Steadfast attention to signs, clean notation, and systematic substitution ensure you reach the correct exact value.
定积分的分部积分法是一个系统过程：利用 LIATE 确定 u 和 dv，对 u 微分并对 dv 积分，应用带限值的公式，并计算得出的表达式。请记住，在处理新积分之前，总是要先计算括号项 [u v] 在上限和下限处的值。坚定关注符号、清晰的记号以及系统的代入，可确保你得出正确的精确值。
From simplifying logarithms and inverse trig to handling cyclic integrals and combined techniques, this method opens up a vast class of integrals that are otherwise intractable. Build a strong foundation through repetitive practice, and you will find that what once seemed challenging becomes a reliable source of marks in your IB Mathematics journey.
从化简对数和反三角函数，到处理循环积分和组合技巧，该方法打开了原本难以求解的一大类积分的大门。通过反复练习打下坚实基础，你将发现，曾经看似困难的内容会成为 IB 数学之旅中一个可靠的得分来源。
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