📚 Differentiation and Integration: Mixed Practice for Edexcel A Level | 微分与积分：Edexcel A-Level混合练习
This revision article reviews the core differentiation and integration techniques that appear in Edexcel A Level Mathematics Paper 1 and Paper 2. The worked examples and commentary focus on the mixed problem style that often combines implicit differentiation, parametric differentiation, integration by substitution, and applications to area and volume.
本文复习 Edexcel A-Level 数学 Paper 1 和 Paper 2 中出现的核心微分与积分技巧。例题与解析围绕混合问题类型展开，涵盖隐函数微分、参数方程微分、换元积分以及面积与体积的应用。
1. Understanding the Mixed Problem | 理解混合问题
Mixed questions in Edexcel often ask you to differentiate an implicit relation, then use the derivative to find a tangent or normal, and finally integrate a related expression to find an area or volume. This structure tests whether you can switch confidently between differentiation and integration.
Edexcel 的混合题通常要求先对隐函数关系求导，然后利用导数求切线或法线，最后对相关表达式积分以求出面积或体积。这种结构考查你是否能够在微分与积分之间自如切换。
You should be comfortable with the chain rule, product rule, implicit differentiation, parametric differentiation, integration by substitution, integration by parts, and definite integrals. A common exam command is “Find dy/dx and hence evaluate the area bounded by the curve.”
你应当熟练掌握链式法则、乘法法则、隐函数微分、参数方程微分、换元积分、分部积分以及定积分。常见的考试指令是“求 dy/dx，并由此计算曲线围成的面积”。
2. Implicit Differentiation | 隐函数微分
When an equation gives y implicitly as a function of x, differentiate both sides with respect to x and multiply every derivative of a y-term by dy/dx. For example, given x² + y² = 25, differentiating gives:
当方程以隐函数形式给出 y 与 x 的关系时，对等式两边关于 x 求导，并对每一个含 y 的项乘以 dy/dx。例如，给定 x² + y² = 25，求导可得：
2x + 2y(dy/dx) = 0
Rearranging gives dy/dx = −x/y. This derivative can then be used to find gradients of tangents and normals at specific points on the curve.
整理后得到 dy/dx = −x/y。这个导数可用于求曲线上具体点处切线和法线的斜率。
Always remember that differentiating a term such as y² produces 2y(dy/dx), not just 2y. Forgetting the dy/dx factor is one of the most common errors in implicit differentiation.
务必牢记，对 y² 这样的项求导会产生 2y(dy/dx)，而不仅仅是 2y。忘记 dy/dx 因子是隐函数微分中最常见的错误之一。
3. Parametric Differentiation | 参数方程微分
For a curve defined by x = f(t) and y = g(t), the derivative dy/dx is found by dividing dy/dt by dx/dt:
对于由 x = f(t) 和 y = g(t) 定义的曲线，导数 dy/dx 等于 dy/dt 除以 dx/dt：
dy/dx = (dy/dt) ÷ (dx/dt)
For example, if x = t² and y = 2t, then dx/dt = 2t and dy/dt = 2, so dy/dx = 2 ÷ 2t = 1/t. Parametric equations often appear in questions that ask you to find the equation of a tangent at a given value of t.
例如，若 x = t² 且 y = 2t，则 dx/dt = 2t 且 dy/dt = 2，因此 dy/dx = 2 ÷ 2t = 1/t。参数方程经常出现在要求你在给定 t 值处求切线方程的题目中。
Do not confuse dy/dx with dx/dy, and make sure you substitute the original value of t into both x and y to obtain the point before writing the tangent equation.
不要混淆 dy/dx 与 dx/dy，并且在写切线方程前，务必将原始的 t 值代入 x 和 y 以得到点的坐标。
4. Integration by Substitution | 换元积分法
Integration by substitution reverses the chain rule. If an integrand is of the form f(g(x))g′(x), set u = g(x) so that du = g′(x) dx. The integral becomes ∫ f(u) du, which is often easier to evaluate.
换元积分法是链式法则的逆运算。如果被积函数形如 f(g(x))g′(x)，设 u = g(x)，则 du = g′(x) dx。积分变为 ∫ f(u) du，通常更容易计算。
For example, to integrate ∫ 2x(x² + 1)³ dx, let u = x² + 1, then du = 2x dx. The integral becomes ∫ u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C.
例如，要计算 ∫ 2x(x² + 1)³ dx，设 u = x² + 1，则 du = 2x dx。积分变为 ∫ u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C。
When the limits are given, remember to change them to u-values or substitute back to x before applying the original limits. Both methods are acceptable, but changing limits is often quicker.
当给定积分上下限时，记住将上下限转换为 u 的值，或者在代入原上下限之前先把变量换回 x。两种方法都可以，但转换上下限通常更快。
5. Integration by Parts | 分部积分法
Integration by parts is used for products of functions, such as x eˣ or ln x. The formula is:
分部积分法用于函数乘积的积分，例如 x eˣ 或 ln x。公式为：
∫ u dv = uv − ∫ v du
For ∫ x eˣ dx, choose u = x and dv = eˣ dx. Then du = dx and v = eˣ, so the integral equals x eˣ − ∫ eˣ dx = x eˣ − eˣ + C.
对于 ∫ x eˣ dx，选择 u = x 且 dv = eˣ dx。则 du = dx 且 v = eˣ，因此积分等于 x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。
When choosing u, use the mnemonic LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. The function appearing earlier in this list is usually chosen as u.
选择 u 时，可使用助记词 LIATE：对数函数、反三角函数、代数函数、三角函数、指数函数。出现在该列表中较前的函数通常选作 u。
6. Definite Integrals and Area | 定积分与面积
The area between a curve y = f(x) and the x-axis from x = a to x = b is given by A = ∫ₐᵇ y dx, provided y ≥ 0 on [a, b]. If the curve lies below the x-axis, the integral is negative, so take its absolute value or integrate −y.
曲线 y = f(x) 与 x 轴在 x = a 到 x = b 之间的面积为 A = ∫ₐᵇ y dx，前提是在 [a, b] 上 y ≥ 0。如果曲线位于 x 轴下方，积分为负，因此应取其绝对值或对 −y 积分。
For example, the area under y = x² from x = 1 to x = 3 is ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3.
例如，y = x² 在 x = 1 到 x = 3 之间的面积为 ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3。
If the curve crosses the x-axis within the interval, split the integral at the crossing point and add the absolute values of the separate areas. This prevents cancellation of positive and negative regions.
如果曲线在区间内穿过 x 轴，应在交点处拆分积分，并分别加上各区域面积的绝对值。这样可以避免正值和负值区域相互抵消。
7. Volumes of Revolution | 旋转体体积
When a region under a curve is rotated 360° about the x-axis, the volume of the solid formed is:
当曲线下方的区域绕 x 轴旋转 360° 时，所形成立体的体积为：
V = π ∫ₐᵇ y² dx
For example, rotating y = √x from x = 0 to x = 4 about the x-axis gives V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π.
例如，将 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转，得到 V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π。
Volumes about the y-axis use V = π ∫ₐᵇ x² dy, where the integrand is expressed in terms of y and the limits are y-values. Always sketch the region to avoid using the wrong axis limits.
绕 y 轴旋转的体积使用 V = π ∫ₐᵇ x² dy，其中被积函数用 y 表示，上下限为 y 值。务必先画出区域草图，以避免使用错误的轴上下限。
8. Differential Equations | 微分方程
Edexcel A Level also tests first-order separable differential equations of the form dy/dx = g(x)h(y). Separate the variables to obtain ∫ 1/h(y) dy = ∫ g(x) dx and integrate both sides.
Edexcel A-Level 还会考查一阶可分离变量微分方程，其形式为 dy/dx = g(x)h(y)。分离变量得到 ∫ 1/h(y) dy = ∫ g(x) dx，然后对两边积分。
For example, solve dy/dx = xy. Separating gives ∫ 1/y dy = ∫ x dx, so ln|y| = x²/2 + C. Exponentiating gives y = Ae^(x²/2), where A = ±e^C.
例如，求解 dy/dx = xy。分离变量得到 ∫ 1/y dy = ∫ x dx，因此 ln|y| = x²/2 + C。两边取指数得到 y = Ae^(x²/2)，其中 A = ±e^C。
If an initial condition is given, substitute it to find the constant A. Do not forget the arbitrary constant, as omitting it can cost a method mark even if the rest of the integration is correct.
如果给出初始条件，代入即可求出常数 A。不要忘记任意常数，因为即使其余积分正确，漏掉常数也可能丢掉方法分。
9. Worked Example in Edexcel Style | Edexcel 风格例题
A curve is defined implicitly by x² + 3xy + y² = 11. Find dy/dx, and hence find the equation of the tangent to the curve at the point (2, 1).
一条曲线由隐函数 x² + 3xy + y² = 11 定义。求 dy/dx，并由此求曲线在点 (2, 1) 处的切线方程。
Differentiate both sides with respect to x. The derivative of x² is 2x, the derivative of 3xy is 3(y + x dy/dx) by the product rule, and the derivative of y² is 2y(dy/dx). This gives:
对等式两边关于 x 求导。x² 的导数为 2x，3xy 的导数由乘法法则得 3(y + x dy/dx)，y² 的导数为 2y(dy/dx)。得到：
2x + 3(y + x dy/dx) + 2y(dy/dx) = 0
Substitute x = 2 and y = 1: 4 + 3(1 + 2 dy/dx) + 2 dy/dx = 0. This simplifies to 7 + 8 dy/dx = 0, so dy/dx = −7/8.
代入 x = 2 和 y = 1：4 + 3(1 + 2 dy/dx) + 2 dy/dx = 0。化简得 7 + 8 dy/dx = 0，因此 dy/dx = −7/8。
The tangent gradient is m = −7/8. Using the point-slope form, y − 1 = −7/8(x − 2). Multiplying by 8 and rearranging gives 7x + 8y − 22 = 0.
切线斜率为 m = −7/8。利用点斜式方程，y − 1 = −7/8(x − 2)。两边乘以 8 并整理得 7x + 8y − 22 = 0。
10. Common Mistakes and Checking | 常见错误与检查
When differentiating implicit equations, always multiply the derivative of any y-term by dy/dx. In parametrics, divide dy/dt by dx/dt, not the other way round. In substitution, match du exactly to part of the integrand before integrating.
对隐函数方程求导时，任何含 y 项的导数都要乘以 dy/dx。在参数方程中，用 dy/dt 除以 dx/dt，而不是反过来。在换元积分中，先将被积函数中的一部分与 du 精确匹配，再进行积分。
For definite integrals, watch for sign changes when a curve crosses the x-axis. For volumes of revolution, square y before integrating, not after. For area, integrate y, but for volume, integrate πy².
对于定积分，当曲线穿过 x 轴时要注意符号变化。对于旋转体体积，先对 y 平方再积分，而不是积分后再平方。对于面积，对被积函数 y 积分；对于体积，对被积函数 πy² 积分。
Always leave enough time to check your final tangent equation by substituting the original point. If the point satisfies your equation, it is a good sign that the derivative and rearranging are correct.
务必留出足够时间检查最终的切线方程，将原点代入。如果该点满足你的方程，就说明导数与整理过程很可能正确。
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