📚 Integration Techniques and Marking Points | 积分运算的方法与得分要点
Integration is one of the most powerful tools in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI). This article systematically covers the core techniques, common pitfalls, and practical scoring strategies to help you earn full marks.
积分是IB数学中最强大的工具之一，在分析与方法（AA）和应用与解释（AI）两个方向中都会出现。本文系统梳理核心方法、常见陷阱和实用得分策略，助你轻松拿满分。
1. Understanding Integration as Anti-Differentiation | 理解积分作为微分的逆运算
Before attempting any integral, recognise that integration reverses differentiation. The fundamental rule states that the derivative of xⁿ is nxⁿ⁻¹, so the integral of xⁿ is xⁿ⁺¹/(n+1), provided n ≠ -1.
在尝试任何积分之前，要认识到积分是微分的逆运算。基本法则表明xⁿ的导数是nxⁿ⁻¹，因此xⁿ的不定积分是xⁿ⁺¹/(n+1)，前提是n ≠ -1。
/4 + C. Always include the constant of integration C when evaluating indefinite integrals.
例如，∫ x³ dx = x⁴/4 + C。计算不定积分时务必加上积分常数C。
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1
Scoring key: This simplest rule is easy marks. However, the most common error is forgetting C. Examiners deduct marks for missing C in indefinite integrals. Always verify your answer by differentiating it.
得分要点：这一最简单的法则属于容易得分点。但最常见的错误是忘记C。考官会在不定积分中因缺少C而扣分。务必通过对结果求导来验证答案。
2. Expanding Before Integrating | 先展开再积分
tance, ∫ (x + 1)(x - 2) dx should be expanded to ∫ (x² - x - 2) dx = x³/3 - x²/2 - 2x + C.
许多函数在代数展开后更容易积分。例如，∫ (x + 1)(x - 2) dx 应先展开为 ∫ (x² - x - 2) dx = x³/3 - x²/2 - 2x + C。
Similarly, rewrite radicals and reciprocals in index form: √x = x¹ᐟ² and 1/x² = x⁻². Then apply the power rule.
类似地，将根式和倒数改写为指数形式：√x = x¹ᐟ²，1/x² = x⁻²，然后使用幂法则。
identities such as sin²x = (1 - cos2x)/2 to convert products into sums before integrating.
对三角表达式，先使用恒等式如 sin²x = (1 - cos2x)/2 将乘积转化为和的形式再积分。
3. The Substitution Method (u-Substitution) | 换元法（u-代换）
ite function, use substitution. The key steps are: (1) choose u = g(x), (2) compute du = g'dx, (3) rewrite the entire integral in terms of u, (4) integrate, and (5) substitute back.
当被积函数包含复合函数时，使用换元法。关键步骤是：(1) 令 u = g(x)，(2) 计算 du = g'(x)dx，(3) 将整个积分改写为u的表达式，(4) 积分，(5) 代回。
Example: ∫ 2x·eˣ² dx. Let u = x², then du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = eˣ² + C.
例：∫ 2x·eˣ² dx。令 u = x²，则 du = 2x dx。积分变为 ∫ eᵘ du = eᵘ + C = eˣ² + C。
Scoring key: In IB exams, substitution often carries 4-6 marks. You receive marks for: choosing the correct substitution, correctly computing du, completely eliminating x from the integral, evaluating the integral in u, and converting back to x.
得分要点：IB考试中，换元法通常占4-6分。得分点包括：选择正确的代换、正确计算du、完全消去积分中的x、对u求积分、以及转换回x的表达式。
4. Definite Integrals with Substitution | 定积分的换元法
For definite integrals, when applying substitution, you have two options: (1) change the limits to u-values, or (2) keep limits in x and substitute back before evaluating.
对于定积分，应用换元法时有两个选择：(1) 将上下限改为对应的u值；(2) 保持上下限为x值，代回后再代入求值。
= 1, u = 2. The integral equals (1/2)∫₁² u³ du = (1/2)[u⁴/4]₁² = (1/2)(16/4 - 1/4) = 15/8.
例：∫₀¹ x·(x² + 1)³ dx。令 u = x² + 1，du = 2x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 2。积分等于 (1/2)∫₁² u³ du = (1/2)[u⁴/4]₁² = (1/2)(16/4 - 1/4) = 15/8。
Scoring key: Marks are awarded for changing limits correctly. Failure to change limits is the single largest source of lost marks in IB definite integrals. Alternatively, if you keep original limits, you must clearly show the substitution of x-values after anti-differentiation.
得分要点：正确变换上下限是得分关键。未变换上下限是IB定积分中最主要的失分原因。如果你选择保留原上下限，则必须在反微分后清楚展示代入x值的步骤。
5. Integration by Parts | 分部积分法
ctions, use integration by parts: ∫ u dv = uv - ∫ v du. The choice of u and dv is crucial.
当被积函数是两个无关函数的乘积时，使用分部积分法：∫ u dv = uv - ∫ v du。u和dv的选择至关重要。
The acronym LIATE helps prioritise u: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u based on this order, and set dv as the remaining factor.
首字母缩写LIATE帮助你优先选择u：对数函数、反三角函数、代数函数、三角函数、指数函数。按照此顺序选择u，将剩余因子设为dv。
Example: ∫ x·eˣ dx. Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Result: x·eˣ - ∫ eˣ dx = x·eˣ - eˣ + C = eˣ(x - 1) + C.
例：∫ x·eˣ dx。令 u = x，dv = eˣ dx。则 du = dx，v = eˣ。结果为 x·eˣ - ∫ eˣ dx = x·eˣ - eˣ + C = eˣ(x - 1) + C。
ting u, integrating dv, and substituting into the formula. For circular integrals like ∫ eˣ·sinx dx, expect to apply parts twice and solve algebraically.
得分要点：IB按步骤给分——正确设置u和dv、对u求导、对dv积分、代入公式各得一分。对于循环积分如 ∫ eˣ·sinx dx，需要两次分部积分并代数求解。
6. Partial Fractions | 部分分式法
Rational functions of the form P(x)/Q(x) can be decomposed into simpler fractions. First check that the degree of the numerator is less than the degree of the denominator; if not, perform polynomial division first.
形如 P(x)/Q(x) 的有理函数可以分解为更简单的分式。首先确认分子的次数低于分母的次数；若不是，先进行多项式除法。
- 1) + B/(x + 2). Solving gives A = 1/3, B = -1/3. Thus ∫ 1/[(x - 1)(x + 2)] dx = (1/3)ln|x - 1| - (1/3)ln|x + 2| + C.
例：1/[(x - 1)(x + 2)] = A/(x - 1) + B/(x + 2)。解得 A = 1/3，B = -1/3。因此 ∫ 1/[(x - 1)(x + 2)] dx = (1/3)ln|x - 1| - (1/3)ln|x + 2| + C。
For repeated linear factors, include terms up to the multiplicity: A/(x - a) + B/(x - a)². For irreducible quadratic factors, use (Ax + B)/(x² + px + q).
对于重线性因子，需包含直到重数为止的各项：A/(x - a) + B/(x - a)²。对于不可约二次因子，使用 (Ax + B)/(x² + px + q)。
7. Definite Integrals and the Fundamental Theorem | 定积分与微积分基本定理
e F is any anti-derivative of f. This theorem connects differentiation and integration and is central to evaluating areas.
微积分基本定理表明：∫ₐᵇ f(x) dx = F(b) - F(a)，其中F是f的任意反导数。该定理连接了微分和积分，是计算面积的核心。
Scoring key: When evaluating definite integrals, always perform the following in order: (1) find the anti-derivative, (2) substitute the upper limit, (3) substitute the lower limit, (4) subtract, and (5) simplify. IB examiners award marks for visible substitution and subtraction steps — do not skip them.
得分要点：求定积分时务必按顺序完成以下步骤：(1) 求反导数，(2) 代入上限，(3) 代入下限，(4) 相减，(5) 化简。IB考官对可见的代入和相减步骤给分——不要跳过这些步骤。
Be careful with negative areas. If the function is below the x-axis within the interval, the integral gives a negative value. For area, integrate |f(x)| or split at the roots.
注意负面积。若函数在区间内位于x轴下方，积分值为负。求面积时应积分 |f(x)| 或在根处分段处理。
8. Area Between Curves | 曲线之间的面积
urves cross within the interval, find intersection points and integrate each segment separately.
两条曲线 y = f(x) 和 y = g(x) 之间的面积，在 f(x) ≥ g(x) 于 [a, b] 成立时，为 ∫ₐᵇ [f(x) - g(x)] dx。若曲线在区间内相交，需找到交点并分段积分。
IB marking scheme typically awards: 1 mark for finding limits (roots or intersections), 1-2 marks for setting up the correct integrand with top-minus-bottom, 1 mark for integration, 1 mark for substitution, and 1 mark for the final answer.
IB评分方案通常为：求上下限（零点或交点）得1分，正确建立上减下的被积函数得1-2分，积分得1分，代值得1分，最终答案得1分。
e curves intersect, do not attempt to integrate across the intersection without splitting.
得分要点：明确标注哪条曲线是上函数、哪条是下函数。将积分写成 ∫ (上 - 下) dx 的形式。若曲线相交，切勿不分段就跨交点积分。
9. Common Trigonometrical Integrals | 常见三角函数积分
x dx = sinx + C, ∫ sec²x dx = tanx + C, ∫ csc²x dx = -cotx + C, ∫ secx·tanx dx = secx + C.
牢记标准结果：∫ sinx dx = -cosx + C，∫ cosx dx = sinx + C，∫ sec²x dx = tanx + C，∫ csc²x dx = -cotx + C，∫ secx·tanx dx = secx + C。
x = (1 - cos2x)/2, cos²x = (1 + cos2x)/2. For ∫ sin²x dx, the result is x/2 - sin2x/4 + C.
对高次幂，使用恒等式：sin²x = (1 - cos2x)/2，cos²x = (1 + cos2x)/2。∫ sin²x dx 的结果为 x/2 - sin2x/4 + C。
ith different angles, use product-to-sum formulas: sinA·cosB = (1/2)[sin(A+B) + sin(A-B)].
对于不同角度的正弦余弦乘积，使用积化和差公式：sinA·cosB = (1/2)[sin(A+B) + sin(A-B)]。
10. Applications: Displacement, Velocity and Area | 应用：位移、速度与面积
velocity. For a particle moving along a line, displacement = ∫ v(t) dt, while distance = ∫ |v(t)| dt.
在运动学中，对速度积分得到位移，对加速度积分得到速度。对于沿直线运动的粒子，位移 = ∫ v(t) dt，而路程 = ∫ |v(t)| dt。
al distance travelled over [0, T]. The procedure is: (1) find when v(t) = 0, (2) integrate |v(t)| over the interval, splitting at the zeros.
IB题目常结合这些概念：已知 v(t)，求在 [0, T] 内的总路程。步骤是：(1) 求 v(t) = 0 的时刻，(2) 在区间内对 |v(t)| 积分，在零点处分段。
Scoring key: The most common error is confusing displacement and distance. Displacement allows cancellation (negative and positive areas cancel), while distance adds absolute values. Clearly state which quantity you are calculating.
得分要点：最常见的错误是混淆位移和路程。位移允许抵消（正负面积相消），而路程需加绝对值。明确说明你在计算哪个量。
11. Review Checklist and Marking Strategy | 复习清单与答题策略
Before the exam, verify you can: (1) integrate all standard forms instantly, (2) recognise when to use each method, (3) handle definite integrals with correct limits, and (4) apply integration to area and kinematics problems.
考试前，请确认你能：(1) 快速写出所有标准形式的积分，(2) 识别每种方法的使用时机，(3) 正确处理定积分的上下限，(4) 将积分应用于面积和运动学问题。
During the exam, follow this strategy:
考试中，遵循以下策略：
Read the question carefully — identify whether it asks for an indefinite integral, definite integral, area, or displacement.
仔细读题——判断题目要求不定积分、定积分、面积还是位移。
Write every step explicitly. IB partial credit rewards visible working.
写出每一步。IB按步骤给分，展示完整的解题过程。
Check the constant C for indefinite integrals; check units for applied problems.
不定积分检查常数C；应用题检查单位。
Verify by differentiation when time permits — this is the fastest self-checking method.
时间允许时通过求导验证——这是最快的自查方法。
Finally, practise past paper questions under timed conditions. Consistent practice with the marking scheme is the most reliable way to maximise your score.
最后，在限时条件下练习历年真题。结合评分方案进行系统练习是提分最可靠的方法。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图