📚 Exercise 16F.1: Integrating x eˣ Using Integration by Parts | 习题16F.1：利用分部积分法求解 x eˣ 的积分
Exercise 16F.1 is a classic IB Mathematics: Analysis and Approaches HL problem that introduces the essential technique of integration by parts. Mastering this problem not only strengthens your calculus skills but also prepares you for more complex integrands involving products of algebraic and exponential functions. In this article, we will walk through the full solution, extend the concept, and discuss common pitfalls to ensure you are fully prepared for any exam variation.
习题16F.1 是 IB 数学分析与方法 HL 课程中一道经典的分部积分法入门题。掌握这道题不仅能夯实你的微积分基本功，还能为你应对更复杂的代数与指数函数乘积积分打下基础。本文将带你完整求解，拓展概念，并梳理常见错误，确保你为考试中的任何变式做好充分准备。
1. Understanding the Problem | 理解问题
Exercise 16F.1 typically asks you to evaluate the indefinite integral ∫ x eˣ dx. Unlike simpler integrals such as ∫ eˣ dx, here we have a product of two functions: a polynomial term x and an exponential term eˣ ethod of integration by parts, which is derived from the product rule for differentiation.
习题16F.1 通常要求计算不定积分 ∫ x eˣ dx。与简单的 ∫ eˣ dx 不同，这里被积函数是两个函数的乘积：多项式项 x 与指数项 eˣ。直接求原函数行不通，因此必须使用由微分乘法法则推导而来的分部积分法。
2. The Integration by Parts Formula | 分部积分公式
chosen from the original integrand such that dv is easily integrable and du simplifies the overall expression. The formula essentially exchanges one integral for another, which should be easier to evaluate.
分部积分公式为：∫ u dv = u v − ∫ v du。其中 u 和 dv 由原被积函数选出，要求 dv 易于积分，且 du 能使整体表达式简化。该公式实质上将一个积分转化为另一个，而后者应更易于计算。
3. Choosing u and dv | 选择 u 和 dv
For ∫ x eˣ unctions (like x) are chosen as u over exponential functions. So we set: u = x, and dv = eˣ dx. Then we compute du = 1 dx, and v = ∫ eˣ dx = eˣ.
对于 ∫ x eˣ dx，实用的LIATE法则（对数、反三角、代数、三角、指数）可以帮助选择 u。根据LIATE，代数函数（如 x）应优先于指数函数被选作 u。因此我们设：u = x，dv = eˣ dx。进而得到 du = 1 dx，v = ∫ eˣ dx = eˣ。
4. Step-by-Step Solution for ∫ x eˣ dx | 逐步求解 ∫ x eˣ dx
Plug these into the formula ∫ u dv = u v − ∫ v du: ∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx. Now the remaining integral is just ∫ eˣ dx = eˣ. Therefore, ∫ x eˣ dx = x eˣ − eˣ + C, where C is the constant of integration. Factoring gives the compact answer eˣ(x − 1) + C.
代入公式：∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx。剩下的积分正是 ∫ eˣ dx = eˣ。因此，∫ x eˣ dx = x eˣ − eˣ + C，其中 C 为积分常数。因式分解后可得简洁结果 eˣ(x − 1) + C。
5. Verifying by Differentiation | 通过求导验证
To validate our result, differentiate F(x) = eˣ(x − 1) + C. Using the product rule: d/dx [eˣ(x − 1)] = eˣ·1 + eˣ(x − 1) = eˣ + eˣ x − eˣ = x eˣ. The derivative matches the original integrand, confirming our answer.
为验证结果，对 F(x) = eˣ(x − 1) + C 求导。利用乘法法则：d/dx [eˣ(x − 1)] = eˣ·1 + eˣ(x − 1) = eˣ + eˣ x − eˣ = x eˣ。导数与原被积函数一致，确认答案无误。
6. Extension: Integrating xⁿ eˣ | 拓展：积分 xⁿ eˣ
Now consider the general form Iₙ = ∫ xⁿ eˣ dx for n ∈ ℕ. Applying integration by parts with u = xⁿ, dv = eˣ dx gives du = n xⁿ⁻¹ dx, v = eˣ. Then Iₙ = xⁿ eˣ − n ∫ xⁿ⁻¹ eˣ dx. Notice the recurrence relation Iₙ = xⁿ eˣ − n Iₙ₋₁. This formula reduces the power of x step by step.
现在考虑一般形式 Iₙ = ∫ xⁿ eˣ dx，其中 n 为自然数。应用分部积分法，设 u = xⁿ，dv = eˣ dx，则 du = n xⁿ⁻¹ dx，v = eˣ。于是 Iₙ = xⁿ eˣ − n ∫ xⁿ⁻¹ eˣ dx。注意到递推关系 Iₙ = xⁿ eˣ − n Iₙ₋₁。该公式可逐步降低 x 的幂次。
7. Reduction Formula for Iₙ = ∫ xⁿ eˣ dx | 递推公式 Iₙ
The reduction formula Iₙ = xⁿ eˣ − n Iₙ₋₁ is extremely powerful. Starting from the base case I₀ = ∫ eˣ dx = eˣ, you can compute I₁ = ∫ x eˣ dx = x eˣ − I₀ = x eˣ − eˣ, which matches our earlier result. For I₂ = ∫ x² eˣ dx, apply the formula: I₂ = x² eˣ − 2 I₁ = x² eˣ − 2(x eˣ − eˣ) = eˣ ing a structure that is extremely useful for IB exam questions requiring integrals of xⁿ eˣ for small n.
递推公式 Iₙ = xⁿ eˣ − n Iₙ₋₁ 功能强大。从基础情形 I₀ = ∫ eˣ dx = eˣ 出发，可计算 I₁ = ∫ x eˣ dx = x eˣ − I₀ = x eˣ − eˣ，与之前结果吻合。对于 I₂ = ∫ x² eˣ dx，应用公式：I₂ = x² eˣ − 2 I₁ = x² eˣ − 2(x eˣ − eˣ) = eˣ(x² − 2x + 2) + C。这一规律持续下去，揭示的结构对 IB 考试中涉及低次幂的 xⁿ eˣ 积分极为有用。
8. Applying the Reduction Formula | 应用递推公式
Let's demonstrate with n = 3: I₃ = ∫ x³ eˣ dx = x³ eˣ − 3 I₂. Using I₂ = eˣ(x² − 2x + 2), we get I₃ = x³ eˣ − 3eˣ(x² − 2x + 2) = eˣ ce the alternating signs and factorial-like coefficients. In general, the integral of xⁿ eˣ yields eˣ ts involving factorials. Being familiar with this pattern can save valuable time in exams.
以 n = 3 为例：I₃ = ∫ x³ eˣ dx = x³ eˣ − 3 I₂。代入 I₂ = eˣ(x² − 2x + 2)，得 I₃ = x³ eˣ − 3eˣ(x² − 2x + 2) = eˣ(x³ − 3x² + 6x − 6) + C。注意其中交替的符号与阶乘式系数。一般而言，∫ xⁿ eˣ dx 的结果为 eˣ 乘以一个 n 次多项式，系数与阶乘有关。熟悉这一模式可在考试中节省宝贵时间。
9. Common Mistakes to Avoid | 常见错误
switching u and dv, which makes the new integral more complex. For example, choosing u = eˣ, dv = x dx leads to du = eˣ dx, v = x²/2, giving ∫ x eˣ dx = (x²/2)eˣ − ∫ (x²/2)eˣ e careful with signs when subtracting ∫ v du; a missing negative sign is a frequent error.
许多学生忘记添加积分常数 +C。另一些则误用 LIATE 法则，颠倒 u 和 dv，导致新积分更为复杂。例如设 u = eˣ，dv = x dx，则 du = eˣ dx，v = x²/2，得到 ∫ x eˣ dx = (x²/2)eˣ − ∫ (x²/2)eˣ dx，x 的幂次不降反升。请始终以降低多项式次数为目标。同时，在减去 ∫ v du 时要小心符号；遗漏负号是常见错误。
10. Practice Variations | 变式练习
Once you are confident with ∫ x eˣ dx, try related integrals: ∫ (2x + 3)eˣ dx, ∫ x e²ˣ dx, and ∫ x² e⁻ˣ o a familiar form after adjusting signs. Also, practice definite integrals such as ∫₀¹ x eˣ ou apply the antiderivative and evaluate limits. This solidifies both technique and speed.
熟练掌握 ∫ x eˣ dx 后，不妨尝试相关练习题：∫ (2x + 3)eˣ dx，∫ x e²ˣ dx，以及 ∫ x² e⁻ˣ dx。对于后者，可令 w = −x 转化为熟悉的形式并调整符号。同时，练习定积分如 ∫₀¹ x eˣ dx，应用原函数并代入上下限。这将同时强化解题技巧与速度。
11. IB Exam Tips | IB考试技巧
In IB Mathematics AA HL Paper 1 and Paper 2, integration by parts often appears in Section B questions linked with area, volume, or differential equations. Always show the u and dv assignments clearly. If the question asks for an exact value, ensure your constant of integration is correctly handled when evaluating definite integrals. The reduction formula is sometimes given in the problem statement; if not, you may need to derive it – practicing the derivation is essential.
在 IB 数学 AA HL 试卷一和试卷二中，分部积分法常出现在 Section B 的题目中，与面积、体积或微分方程相结合。务必清晰地写出 u 和 dv 的设定。若题目要求精确值，计算定积分时确保积分常数处理正确。递推公式有时会在题目中给出；若未给出，则需自行推导——练习推导过程至关重要。
12. Conclusion | 总结
Exercise 16F.1 is more than a simple drill; it lays the groundwork for mastering integration by parts and recognizing reduction patterns. By understanding the choice of u and dv, verifying through differentiation, and extending to higher powers, you build a robust calculus toolkit. Approach every similar integral with the confidence that a systematic application of the integration by parts formula will lead you to success.
习题16F.1 不仅是简单的练习；它为你掌握分部积分法、识别递推模式奠定了基础。通过理解 u 和 dv 的选择、用求导验证结果，并推广至高阶幂次，你将构建起强大的微积分工具库。带着对分部积分公式系统化应用的信心去面对每一道同类积分，成功必将水到渠成。
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