Using Integration: Area, Volume and Differential Equations | 积分的应用：面积、旋转体体积与微分方程

📚 Using Integration: Area, Volume and Differential Equations | 积分的应用：面积、旋转体体积与微分方程

Integration is one of the central pillars of A-Level Mathematics, extending far beyond simply ‘opposite of differentiation’. In Edexcel modules, you are expected to use integration to compute areas bounded by curves, volumes of revolution, and to solve simple differential equations that model real-world situations. This article revisits the core integration techniques and then shows how to apply them systematically, ensuring you are fully prepared for exam-style questions that test both routine and problem-solving skills.

积分是A-Level数学的核心支柱之一，其应用远不止“微分的逆运算”。在Edexcel考试中，你需要运用积分计算曲线围成的面积、旋转体体积，并求解描述现实世界的简单微分方程。本文回顾基本积分技巧，然后系统展示如何应用它们，帮助你全面准备考试中既考常规操作又考问题解决能力的题目。

1. Review of Basic Integration Techniques | 基本积分方法复习

Before tackling applications, you must be fluent with indefinite integrals. The power rule states ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ –1. For trigonometric functions, ∫ sin x dx = –cos x + C and ∫ cos x dx = sin x + C. The exponential function eˣ integrates to itself: ∫ eˣ dx = eˣ + C. The reciprocal function integrates to a natural logarithm: ∫ (1/x) dx = ln|x| + C. Remember also that constant multiples and sums can be integrated term by term.

在解决应用问题之前，你必须熟练掌握不定积分。幂法则指出：当 n ≠ –1 时，∫ xⁿ dx = xⁿ⁺¹/(n+1) + C。对于三角函数，∫ sin x dx = –cos x + C，∫ cos x dx = sin x + C。指数函数 eˣ 积分后仍是自身：∫ eˣ dx = eˣ + C。倒数函数积分得到自然对数：∫ (1/x) dx = ln|x| + C。还要记住常数倍和和差可以逐项积分。

2. Definite vs Indefinite Integrals | 定积分与不定积分

An indefinite integral produces a family of functions with a constant of integration C. A definite integral ∫_a^b f(x) dx evaluates to a number, representing the net area between the curve and the x-axis from a to b. It is computed using the Fundamental Theorem of Calculus: ∫_a^b f(x) dx = F(b) – F(a), where F is any antiderivative of f. Always use square brackets notation to show the evaluation step, e.g. [½ x²]₁³ = 4.

不定积分产生一族带有积分常数 C 的函数。定积分 ∫_a^b f(x) dx 则计算出一个数值，表示从 a 到 b 曲线上方与 x 轴之间的净值面积。定积分通过微积分基本定理计算：∫_a^b f(x) dx = F(b) – F(a)，其中 F 是 f 的任意一个原函数。总是用方括号表示代入步骤，例如 [½ x²]₁³ = 4。

3. Finding the Constant of Integration | 求积分常数

Many problems give a point through which the curve passes, allowing you to determine the +C. For instance, given dy/dx = 6x and (2, 10) lying on the curve, integrate to get y = 3x² + C. Substituting x = 2, y = 10 gives 10 = 3(4) + C → C = –2. Hence the particular solution is y = 3x² – 2. Always check your constant by plugging the point back into the final equation.

许多题目会给出曲线经过的某一点，从而确定常数 C。例如，给定 dy/dx = 6x 并且点 (2, 10) 在曲线上，积分得 y = 3x² + C。代入 x = 2, y = 10 得 10 = 3(4) + C → C = –2。因此特解为 y = 3x² – 2。记得将点代入最终方程检验所得常数。

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4. Area Under a Curve | 曲线下的面积

When y = f(x) ≥ 0 on [a, b], the area between the curve, the x-axis, and the lines x = a and x = b is exactly A = ∫_a^b f(x) dx. If the function dips below the x-axis, the integral yields a negative contribution. To find the total enclosed area, you must split the interval where f(x) changes sign and take absolute values: Area = ∫_a^c f(x) dx + |∫_c^b f(x) dx|. On a graph, always shade the region before setting up the integral.

当 y = f(x) 在 [a, b] 上非负时，曲线、x 轴以及直线 x = a 和 x = b 之间的面积正好等于 A = ∫_a^b f(x) dx。如果函数下降到 x 轴下方，积分会产生负值。为求得总的包围面积，必须在 f(x) 变号处拆分区间并取绝对值：面积 = ∫_a^c f(x) dx + |∫_c^b f(x) dx|。做题时，先画出图形并将区域涂上阴影，再建立积分式。

Example: Find the area bounded by y = 4 – x² and the x-axis. The curve cuts the x-axis at x = –2 and x = 2. The area is ∫_–2² (4 – x²) dx = [4x – ⅓ x³]_–2² = (8 – 8/3) – (–8 + 8/3) = 32/3.

示例：求 y = 4 – x² 与 x 轴围成的面积。曲线与 x 轴交于 x = –2 和 x = 2。面积为 ∫_–2² (4 – x²) dx = [4x – ⅓ x³]_–2² = (8 – 8/3) – (–8 + 8/3) = 32/3。

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5. Area Between Two Curves | 两条曲线之间的面积

To find the area enclosed between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b, use A = ∫_a^b [f(x) – g(x)] dx. Always identify the intersections to determine a and b. If the upper/lower relationship swaps, split the interval accordingly. A common mistake is to integrate f – g without checking which is above; sketching a quick graph helps.

若要计算两条曲线 y = f(x)（上曲线）和 y = g(x)（下曲线）在 x = a 到 x = b 之间包围的面积，使用 A = ∫_a^b [f(x) – g(x)] dx。总是先求出交点以确定上下限 a 和 b。如果上下曲线关系发生反转，务必分段处理。常见错误是在没判断谁上谁下的情况下直接积分 f – g；快速画一下草图可以有效避免。

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6. Parametric Integration | 参数方程下的积分

When a curve is given parametrically as x = x(t), y = y(t), the area under the curve between t = t₁ and t = t₂ is found by A = ∫_t₁^t₂ y · (dx/dt) dt. Note that you replace y and dx with expressions in t, and the limits are t-values. This is particularly useful for loops and asymmetric shapes where Cartesian integration is messy.

当曲线以参数方程 x = x(t), y = y(t) 给出时，位于 t = t₁ 到 t = t₂ 之间的曲线下方面积由 A = ∫_t₁^t₂ y · (dx/dt) dt 计算。注意用含 t 的表达式替换 y 和 dx，并且上下限是 t 的值。该方法对于环状或不对称图形尤其有用，因为此时普通直角坐标积分会很繁琐。

For example, the curve given by x = 2 cos t, y = sin t from t = 0 to π gives area ∫₀^π (sin t)(–2 sin t) dt = –2 ∫ sin² t dt. Use cos 2t identity to integrate.

例如，曲线 x = 2 cos t，y = sin t 从 t = 0 到 π 的面积为 ∫₀^π (sin t)(–2 sin t) dt = –2 ∫ sin² t dt。利用 cos 2t 公式进行积分。

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7. Volumes of Revolution: x-axis | 绕x轴旋转的体积

When a region bounded by y = f(x), the x-axis, x = a and x = b is rotated 360° about the x-axis, the volume of the solid generated is V = π ∫_a^b [f(x)]² dx. This formula sweeps out disks of radius y and thickness dx. Always square the function before integrating. If the region is between two curves, the volume is V = π ∫ [f(x)² – g(x)²] dx.

将由 y = f(x)、x 轴、x = a 和 x = b 围成的区域绕 x 轴旋转 360°，所得立体的体积为 V = π ∫_a^b [f(x)]² dx。该公式源自半径为 y、厚度为 dx 的圆盘叠加。务必在积分前先对函数平方。如果区域位于两条曲线之间，体积公式为 V = π ∫ [f(x)² – g(x)²] dx。

Example: Rotate y = √x from x = 0 to x = 4 about the x-axis. Volume = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [½ x²]₀⁴ = 8π.

示例：将 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转，体积 = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [½ x²]₀⁴ = 8π。

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8. Volumes of Revolution: y-axis | 绕y轴旋转的体积

For rotation about the y-axis, rearrange the curve into the form x = g(y). The volume is V = π ∫_y=c^y=d [g(y)]² dy. The limits must be y-values. Sometimes you need to subtract an inner cylinder when the region is not adjacent to the axis. Be careful with square units: the integration variable is y, not x.

对于绕 y 轴旋转，需将曲线改写为 x = g(y) 的形式。体积为 V = π ∫_y=c^y=d [g(y)]² dy，其中上下限应为 y 的取值。若区域不是紧邻坐标轴，可能还需要减去内部圆柱的体积。注意积分变量是 y 而非 x。

For instance, the region bounded by y = x², y = 1 and x = 0 rotated about the y-axis: x = √y, limits y = 0 to 1, volume = π ∫₀¹ (√y)² dy = π [½ y²]₀¹ = ½π.

例如，将 y = x²、y = 1 以及 x = 0 围成的区域绕 y 轴旋转：x = √y，上下限为 y = 0 到 1，体积 = π ∫₀¹ (√y)² dy = π [½ y²]₀¹ = ½π。

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9. Solving Differential Equations by Separation of Variables | 分离变量法解微分方程

Many modelling questions lead to first-order separable differential equations of the form dy/dx = f(x)g(y). Rearrange to bring all y terms to one side: ∫ 1/g(y) dy = ∫ f(x) dx. Integrate both sides, add the constant of integration on one side (usually the right), and use any given initial condition to find the particular solution. Always express the final answer in a simple form, e.g. y = … or an implicit equation if required.

许多建模题会引出一阶可分离变量的微分方程 dy/dx = f(x)g(y)。移项使所有含 y 的项集中到一边：∫ 1/g(y) dy = ∫ f(x) dx。两边积分，在一边（通常是右边）加上积分常数，并利用给定的初始条件求出特解。最终答案应表示为显函数 y = … 的形式，或题目要求的隐式方程。

Typical example: dy/dx = 2xy with y(0) = 3. Separate: ∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C. Then y = e^{x² + C} = A e^{x²}. Using y(0)=3 gives A=3, so y = 3 e^{x²}.

典型例子：dy/dx = 2xy，且 y(0)=3。分离变量：∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C。于是 y = e^{x² + C} = A e^{x²}。代入 y(0)=3 得到 A=3，所以 y = 3 e^{x²}。

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10. Applications in Kinematics | 在运动学中的应用

In mechanics, velocity v(t) is the derivative of displacement s(t); thus s(t) = ∫ v(t) dt. Similarly, acceleration a(t) = dv/dt, so v(t) = ∫ a(t) dt. Definite integrals give the change in displacement over a time interval, while integrating the absolute value of velocity yields the total distance travelled. When initial conditions such as “from rest” or “at t=0, s=0” are given, you can determine the constants of integration.

在力学中，速度 v(t) 是位移 s(t) 的导数，因此 s(t) = ∫ v(t) dt。同样，加速度 a(t) = dv/dt，故 v(t) = ∫ a(t) dt。定积分给出一个时间段内位移的变化量，而对速度的绝对值进行积分则得到总路程。当题目给出“从静止开始”或“在 t=0 时 s=0”等初始条件时，即可确定积分常数。

For example, a particle moves with v(t) = 3t² – 4, and at t=0 its displacement is 5. Then s(t) = ∫ (3t² – 4) dt = t³ – 4t + C. Using s(0)=5 gives C=5, so s = t³ – 4t + 5.

例如，某质点以 v(t) = 3t² – 4 运动，且在 t=0 时位移为 5。则 s(t) = ∫ (3t² – 4) dt = t³ – 4t + C。利用 s(0)=5 得出 C=5，因此 s = t³ – 4t + 5。

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11. Key Exam Tips | 关键考试技巧

Always write down the formula you are using before substituting. Check if an area needs to be split into pieces because the curve crosses the x‑axis or the two curves intersect. For volumes of revolution, remember to square first – forgetting this is a classic mistake. In differential equations, show the separation step clearly and include the constant of integration before applying initial conditions. Finally, double-check your limits and signs; a negative area or volume is a red flag.

在代入数值之前，始终先写出所使用的公式。检查面积是否需要分段计算，因为曲线可能穿过 x 轴或两条曲线相交。在计算旋转体体积时，务忘先平方——忘记平方是经典错误。在解微分方程时，清晰地展示变量分离步骤，并在应用初始条件之前加上积分常数。最后，再次检查积分上下限和符号；出现负的面积或体积就是警示信号。

Manage your time by practising past Edexcel papers; many ‘using integration’ questions combine two or three of these topics into one multi-part problem. With solid technique and careful reading, integration can become one of your highest-scoring topics.

通过练习Edexcel历年真题来管理时间；许多“积分应用”题会结合本文中的两至三个知识点构成一道多部分题目。扎实的方法与仔细审题会让你把积分变成得分最高的板块之一。

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