📚 Integrating Vectors | 向量积分
In Edexcel A-Level Mathematics, especially in mechanics, integrating vectors is the reverse process of differentiating vector functions. It is used to move from acceleration to velocity and from velocity to displacement when acceleration or velocity varies with time. Because vectors have components, the integration is performed separately on each component.
在 Edexcel A-Level 数学（尤其是力学）中，向量积分是向量函数求导的逆过程。当加速度或速度随时间变化时，它被用来从加速度求速度、从速度求位移。由于向量具有分量，积分要对每个分量分别进行。
1. What Does Integrating a Vector Mean? | 向量积分的含义
A vector function with respect to time t can be written as r(t) = x(t)i + y(t)j, where x(t) and y(t) are scalar functions and i, j are constant unit vectors. Since integration is linear, the integral of the vector is the sum of the integrals of its components.
关于时间 t 的向量函数可以写成 r(t) = x(t)i + y(t)j，其中 x(t) 和 y(t) 是标量函数，i、j 是恒定的单位向量。由于积分是线性的，向量的积分等于其各分量积分之和。
∫ r(t) dt = (∫ x(t) dt) i + (∫ y(t) dt) j + C
Here C is a vector constant of integration, usually expressed as c₁ i + c₂ j. In kinematics, this constant is found from initial conditions such as initial velocity or initial position.
这里 C 是积分常向量，通常表示为 c₁ i + c₂ j。在运动学中，该常数由初始速度或初始位置等初始条件确定。
2. Component-by-Component Integration | 逐分量积分
To integrate a vector, reverse the differentiation rules separately for each component. For example, if a(t) = 6t i + 4 j, then integrating with respect to t gives v(t) = 3t² i + 4t j + C. The constant C must be a vector, not a scalar.
对向量积分时，分别对每个分量反向使用求导规则。例如，若 a(t) = 6t i + 4 j，则对 t 积分得到 v(t) = 3t² i + 4t j + C。常数 C 必须是向量，而不是标量。
| Vector function | Indefinite integral |
|---|---|
| 2t i + 3 j | (t² + c₁) i + (3t + c₂) j |
| 6t² i − 2 j | (2t³ + c₁) i + (−2t + c₂) j |
| −sin t i + cos t j | (cos t + c₁) i + (sin t + c₂) j |
Each component uses the same power, trigonometric or exponential rules as ordinary scalar integration. The i and j directions are independent, so mistakes in one component do not affect the other.
每个分量使用与普通标量积分相同的幂函数、三角函数或指数函数规则。i 和 j 方向相互独立，因此一个分量的错误不会影响另一个分量。
3. Indefinite Integrals and Constant Vectors | 不定积分与常向量
When evaluating an indefinite integral, the added constant is C = c₁ i + c₂ j. This reflects two independent scalar constants, one for each component. In kinematics, these constants are determined by the initial velocity or initial position.
计算不定积分时，所加的常数是 C = c₁ i + c₂ j。它反映了两个独立的标量常数，每个分量一个。在运动学中，这些常数由初始速度或初始位置确定。
For instance, if v(0) = 2i − 3j, then C = 2i − 3j. You cannot simply write “+ C” and then substitute a single scalar value; both components must be matched separately.
例如，若 v(0) = 2i − 3j，则 C = 2i − 3j。你不能只写 ‘+ C’ 然后代入一个标量值；必须分别匹配两个分量。
4. Definite Integrals of Vector Functions | 向量函数的定积分
A definite integral of a vector function from t = a to t = b is computed component by component. It gives the net change in the vector quantity over that interval. In mechanics, the definite integral of v(t) from a to b equals displacement, not distance.
向量函数从 t = a 到 t = b 的定积分要逐分量计算。它给出该向量量在该区间内的净变化。在力学中，速度 v(t) 从 a 到 b 的定积分等于位移，而不是距离。
∫ₐᵇ v(t) dt = (x(b) − x(a)) i + (y(b) − y(a)) j
This follows from the fundamental theorem of calculus applied to each scalar component. If v = dr/dt, then the left-hand side is exactly r(b) − r(a), the displacement vector.
这由对每个标量分量应用微积分基本定理得出。如果 v = dr/dt，那么左边正是 r(b) − r(a)，即位移向量。
5. Kinematics: From Acceleration to Velocity | 运动学：从加速度到速度
In kinematics, acceleration is the derivative of velocity: a = dv/dt. Therefore velocity is found by integrating acceleration with respect to time. If an exam gives a(t) and an initial velocity, integrate and substitute t = 0 to find C.
在运动学中，加速度是速度的导数：a = dv/dt。因此，速度可以通过对时间积分加速度来求得。如果考题给出 a(t) 和初始速度，先积分再代入 t = 0 求出 C。
Example: Given a = (6t − 4)i + 8j and v(0) = 2i − j, integrate to get v = (3t² − 4t)i + 8t j + C. Substituting t = 0 gives C = 2i − j, so v = (3t² − 4t + 2)i + (8t − 1)j.
例题：已知 a = (6t − 4)i + 8j 且 v(0) = 2i − j，积分得 v = (3t² − 4t)i + 8t j + C。代入 t = 0 得 C = 2i − j，因此 v = (3t² − 4t + 2)i + (8t − 1)j。
6. Kinematics: From Velocity to Position | 运动学：从速度到位移
Similarly, velocity is the derivative of position: v = dr/dt. Integrating velocity gives the position vector r(t). The constant vector here represents initial position r(0). Always write r as x i + y j.
类似地，速度是位置的导数：v = dr/dt。对速度积分可得到位置向量 r(t)。这里的常向量表示初始位置 r(0)。始终将 r 写成 x i + y j 的形式。
Example: If v = (2t + 1)i + 3t² j and r(0) = i + 2j, then r = (t² + t)i + t³ j + D. Substituting t = 0 gives D = i + 2j, so r = (t² + t + 1)i + (t³ + 2)j.
例题：若 v = (2t + 1)i + 3t² j 且 r(0) = i + 2j，则 r = (t² + t)i + t³ j + D。代入 t = 0 得 D = i + 2j，因此 r = (t² + t + 1)i + (t³ + 2)j。
7. Using Initial Conditions | 利用初始条件
Problems often require two successive integrations. First, integrate acceleration to get velocity and use the initial velocity. Then integrate velocity to get position and use the initial position. Phrases such as ‘initially at rest’ mean v(0) = 0; ‘initially at the origin’ means r(0) = 0.
题目通常需要两次连续积分。首先对加速度积分得到速度，并利用初始速度；然后对速度积分得到位置，并利用初始位置。’初始静止’ 表示 v(0) = 0；’初始位于原点’ 表示 r(0) = 0。
- v(0) fixes the constant after the first integration. | 初始速度 v(0) 确定第一次积分后的常数。
- r(0) fixes the constant after the second integration. | 初始位置 r(0) 确定第二次积分后的常数。
- Always check whether the question gives ‘at rest’, ‘at the origin’ or another specific vector value. | 始终检查题目是否给出 ‘静止’、’在原点’ 或其他具体向量值。
8. Worked Example: Variable Acceleration | 例题：变加速度
A particle P moves with acceleration a = (6t − 6)i + 4j m s⁻². At t = 0, P is at rest at the origin. Find v(t), r(t) and the speed when t = 2.
质点 P 以加速度 a = (6t − 6)i + 4j m s⁻² 运动。t = 0 时，P 静止于原点。求 v(t)、r(t) 以及 t = 2 时的速率。
First integrate acceleration to get velocity:
首先对加速度积分得到速度：
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