Area Under Curves and Applications of Definite Integrals | 曲线下面积与定积分的应用

📚 Area Under Curves and Applications of Definite Integrals | 曲线下面积与定积分的应用

The definite integral is one of the most powerful tools in A-Level Mathematics. It allows us to calculate the exact area enclosed by a curve and the x-axis, the area between two curves, and even quantities in physics such as displacement and work. This article explains the theory, worked methods, and exam-focused techniques for applying definite integrals to area problems.

定积分是 A-Level 数学中最强大的工具之一。它使我们能够精确计算曲线与 x 轴围成的面积、两条曲线之间的面积，甚至物理中的位移和功等物理量。本文将讲解定积分应用于面积问题的理论、方法以及考试重点技巧。

1. The Definite Integral as a Limit of Sums | 定积分作为和的极限

Before we use the notation ∫ₐᵇ f(x) dx, we must understand what it means. The area under a curve from x = a to x = b can be approximated by dividing the region into n thin rectangles, each of width Δx = (b − a)/n, and summing their areas. As n approaches infinity, this sum approaches the true area.

在我们将符号 ∫ₐᵇ f(x) dx 投入使用之前，必须先理解它的含义。曲线从 x = a 到 x = b 下方的面积可以通过将该区域分割成 n 个细长矩形来近似，每个矩形的宽度为 Δx = (b − a)/n，然后将它们的面积相加。当 n 趋向于无穷大时，这个和趋近于真实面积。

Area = limₙ→∞ Σᵢ₌₁ⁿ f(xᵢ) Δx = ∫ₐᵇ f(x) dx

This idea is called a Riemann sum, and it is the foundation of integral calculus. For a typical A-Level question, you will not be asked to compute this limit directly; instead, you will use the Fundamental Theorem of Calculus to evaluate the integral exactly.

这一思想被称为黎曼和，它是积分学的基础。对于典型的 A-Level 考题，你不会被要求直接计算该极限；而是运用微积分基本定理来精确计算积分值。

2. The Fundamental Theorem of Calculus | 微积分基本定理

The Fundamental Theorem of Calculus connects differentiation and integration. It states that if F(x) is an antiderivative of f(x), then:

微积分基本定理将微分与积分联系起来。它指出：如果 F(x) 是 f(x) 的一个原函数，那么：

∫ₐᵇ f(x) dx = F(b) − F(a)

This theorem allows us to evaluate definite integrals without calculating limits. For example, ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3.

该定理使我们无需计算极限即可求定积分。例如，∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3。

In exams, always write the antiderivative in square brackets with the limits before substituting, to avoid arithmetic errors. Also remember: the constant of integration cancels out when evaluating F(b) − F(a), so you may omit it in definite integrals.

在考试中，代入上下限之前，务必将原函数写在方括号内并标出上下限，以避免算术错误。还要记得：在计算 F(b) − F(a) 时，积分常数会相互抵消，因此在定积分中可以不写常数项。

3. Area Between a Curve and the x-Axis | 曲线与 x 轴之间的面积

When f(x) ≥ 0 on the interval [a, b], the area between the curve, the x-axis, and the lines x = a and x = b is simply ∫ₐᵇ f(x) dx.

当 f(x) ≥ 0 在区间 [a, b] 上成立时，曲线、x 轴以及直线 x = a 和 x = b 所围成的面积就是 ∫ₐᵇ f(x) dx。

Worked Example. Find the area enclosed by the curve y = 4 − x² and the x-axis.

实例分析。求曲线 y = 4 − x² 与 x 轴所围成的面积。

The curve crosses the x-axis where 4 − x² = 0, so x = ±2. Since the curve is symmetric and positive between −2 and 2, the area is:

曲线与 x 轴相交于 4 − x² = 0，即 x = ±2。由于曲线关于 y 轴对称，并且在 −2 和 2 之间取正值，因此面积为：

Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16/3 + 16/3 = 32/3

Notice that we did not use the symmetry shortcut in the calculation; whether you use symmetry or compute directly, the result is the same. Symmetry is a useful verification tool but should be applied carefully.

注意在计算中我们并未使用对称性的技巧；无论用对称性还是直接计算，结果都一样。对称性是一种有效的检验工具，但使用时需谨慎。

4. Negative Areas and the Absolute Value | 负面积与绝对值

If f(x) < 0 on part of the interval, the definite integral returns a negative value, which represents "negative area." The actual geometric area must be positive. Therefore, when the curve dips below the x-axis, you must split the integral at the roots and take absolute values.

如果 f(x) 在区间的某一部分上取负值，定积分返回负值，这表示“负面积”。实际几何面积必须为正。因此，当曲线下降到 x 轴以下时，必须在根处拆分积分并取绝对值。

Example. Find the total area bounded by y = x³ − x and the x-axis from x = −1 to x = 1.

实例。求 y = x³ − x 与 x 轴从 x = −1 到 x = 1 所围成的总面积。

First factor: x(x² − 1) = x(x − 1)(x + 1), so roots are x = −1, 0, 1. On (−1, 0), the function is positive; on (0, 1), it is negative. Thus:

先因式分解：x(x² − 1) = x(x − 1)(x + 1)，所以根为 x = −1、0、1。在 (−1, 0) 上函数取正值；在 (0, 1) 上取负值。因此：

Total Area = ∫₋₁⁰ (x³ − x) dx + |∫₀¹ (x³ − x) dx|

= [x⁴/4 − x²/2]₋₁⁰ + |[x⁴/4 − x²/2]₀¹|

= (0 − (1/4 − 1/2)) + |(1/4 − 1/2) − 0| = 1/4 + 1/4 = 1/2

Had we simply evaluated ∫₋₁¹ (x³ − x) dx, we would have obtained 0 — an incorrect geometric answer. Recognizing when to split the integral is a crucial exam skill.

如果我们直接计算 ∫₋₁¹ (x³ − x) dx，结果会是 0——这是一个错误的几何答案。判断何时需要拆分积分是一项关键的考试技能。

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

To find the area enclosed by two curves y = f(x) and y = g(x), first find their intersection points. On an interval where f(x) ≥ g(x), the area is given by ∫ₐᵇ [f(x) − g(x)] dx.

要求两条曲线 y = f(x) 和 y = g(x) 所围成的面积，首先找出它们的交点。在满足 f(x) ≥ g(x) 的区间上，面积为 ∫ₐᵇ [f(x) − g(x)] dx。

Worked Example. Find the area enclosed by y = x² and y = x + 2.

实例分析。求曲线 y = x² 和 y = x + 2 所围成的面积。

Intersection: x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0, so x = −1 or x = 2. On this interval, x + 2 ≥ x². Therefore:

交点：x² = x + 2，即 x² − x − 2 = 0，即 (x − 2)(x + 1) = 0，所以 x = −1 或 x = 2。在该区间上，x + 2 ≥ x²。因此：

Area = ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁²

= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = (6 − 8/3) − (−3/2 + 1/3) = 10/3 + 7/6 = 9/2

Notice how we consistently subtract the lower curve from the upper curve. If you reverse the order, you will get a negative result; taking its absolute value works only if the upper curve stays on top throughout the interval.

注意我们始终是用上方曲线减去下方曲线。如果顺序颠倒，结果将为负；只有当上方曲线在整个区间内都保持在上方时，取绝对值才成立。

6. Splitting the Interval at Intersection Points | 在交点处拆分区间

When two curves cross each other multiple times, the upper and lower functions can swap. In such cases, you must split the integral at every intersection point and apply the appropriate sign.

当两条曲线多次相交时，上方与下方的函数可能会互换。在这种情况下，必须在每个交点处拆分积分，并应用相应的正负号。

For instance, consider y = sin x and y = cos x between x = 0 and x = π. They intersect at x = π/4. From 0 to π/4, cos x ≥ sin x; from π/4 to π, sin x ≥ cos x. Hence:

例如，考虑 y = sin x 和 y = cos x 在 x = 0 到 x = π 之间的情形。它们在 x = π/4 处相交。从 0 到 π/4，cos x ≥ sin x；从 π/4 到 π，sin x ≥ cos x。因此：

Area = ∫₀^{π/4} (cos x − sin x) dx + ∫_{π/4}^{π} (sin x − cos x) dx

Failing to split the integral at an intersection will produce cancellation between positive and negative contributions, leading to a wrong answer. Always sketch the graphs — or at least determine sign patterns — before integrating.

在交点处不拆分积分会导致正负贡献相互抵消，从而得出错误答案。在积分之前，务必先画出草图——或者至少确定符号模式——然后再积分。

7. Area Enclosed by a Curve and the y-Axis | 曲线与 y 轴围成的面积

Sometimes the area is most naturally calculated with respect to the y-axis. If x = h(y) and h(y) ≥ 0 on c ≤ y ≤ d, the area between the curve, the y-axis, and the horizontal lines y = c and y = d is ∫꜀ᵈ h(y) dy.

有时，面积最自然的计算方式是相对于 y 轴进行的。如果 x = h(y) 在 c ≤ y ≤ d 上满足 h(y) ≥ 0，那么曲线、y 轴以及水平线 y = c 和 y = d 之间的面积为 ∫꜀ᵈ h(y) dy。

Example. Find the area bounded by y = √x (i.e. x = y²), the y-axis, and the line y = 2.

实例。求由 y = √x（即 x = y²）、y 轴和直线 y = 2 所围成的面积。

Area = ∫₀² y² dy = [y³/3]₀² = 8/3

When you integrate along the y-axis, the upper and lower limits are y-values rather than x-values. Ensure that your function is correctly written as x in terms of y before integrating.

当沿 y 轴积分时，上下限是 y 值而非 x 值。在积分之前，务必先将函数正确地写成用 y 表示 x 的形式。

8. The Trapezium Rule for Approximate Integration | 梯形法则近似积分

When a function cannot be integrated analytically, or when only discrete data are available, we use numerical methods. The trapezium rule approximates the area by straight line segments replacing the curve:

当函数无法解析积分，或仅有离散数据可用时，我们使用数值方法。梯形法则用直线段替代曲线以近似面积：

Area ≈ (h/2) × [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]

where h = (b − a)/n and yᵢ = f(xᵢ). The trapezium rule is an overestimate for convex functions and an underestimate for concave functions. In A-Level exams, you may be asked to estimate an integral using n strips and then compare it to the exact value.

其中 h = (b − a)/n，yᵢ = f(xᵢ)。梯形法则对凸函数高估，对凹函数低估。在 A-Level 考试中，你可能需要先用 n 条分割估算积分，再与精确值进行比较。

Example. Approximate ∫₁² (1/x) dx using 4 strips. Here h = 0.25, and y-values at x = 1, 1.25, 1.5, 1.75, 2 are 1, 0.8, 0.6667, 0.5714, 0.5. Thus:

实例。用 4 条分割近似 ∫₁² (1/x) dx。此处 h = 0.25，在 x = 1、1.25、1.5、1.75、2 处的 y 值分别为 1、0.8、0.6667、0.5714、0.5。因此：

Area ≈ (0.25/2) × [1 + 0.5 + 2(0.8 + 0.6667 + 0.5714)] = 0.125 × [1.5 + 4.0762] = 0.6970

The exact value is ln 2 ≈ 0.6931, so our approximation is very close. The accuracy improves as n increases.

精确值为 ln 2 ≈ 0.6931，因此我们的近似值非常接近。随着 n 增大，精度会进一步提高。

9. Definite Integrals in Kinematics | 运动学中的定积分

In kinematics, the velocity v(t) of a particle is the derivative of displacement s(t). Therefore, the displacement between t = t₁ and t = t₂ is:

在运动学中，粒子的速度 v(t) 是位移 s(t) 的导数。因此，在 t = t₁ 和 t = t₂ 之间的位移为：

Displacement = ∫ₜ₁ᵗ₂ v(t) dt

However, the total distance travelled is the integral of the speed |v(t)|:

然而，总路程是速率 |v(t)| 的积分：

Distance = ∫ₜ₁ᵗ₂ |v(t)| dt

Example. A particle moves with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 5. Find the displacement and the total distance travelled.

实例。一质点以速度 v(t) = t² − 4t + 3 在 0 ≤ t ≤ 5 内运动。求位移和总路程。

Displacement = ∫₀⁵ (t² − 4t + 3) dt = [t³/3 − 2t² + 3t]₀⁵ = 125/3 − 50 + 15 = 20/3 ≈ 6.67 m. For distance, v(t) = (t − 1)(t − 3), which is negative on (1, 3). Hence:

位移 = ∫₀⁵ (t² − 4t + 3) dt = [t³/3 − 2t² + 3t]₀⁵ = 125/3 − 50 + 15 = 20/3 ≈ 6.67 m。路程需注意 v(t) = (t − 1)(t − 3)，在 (1, 3) 上为负。因此：

Distance = ∫₀¹ v dt + |∫₁³ v dt| + ∫₃⁵ v dt

This example perfectly illustrates why “distance” and “displacement” are not the same, and why the algebraic sign of the integral must be treated with care.

这个例子完美地说明了“路程”和“位移”并不相同，以及积分符号必须谨慎处理的道理。

10. Volume of Revolution | 旋转体体积

Another major application of definite integrals in A-Level is the volume of a solid formed by rotating a curve around an axis. Rotating y = f(x) around the x-axis between a and b gives:

定积分在 A-Level 中的另一大应用是曲线绕坐标轴旋转所得旋转体的体积。将 y = f(x) 绕 x 轴从 a 到 b 旋转一周，所得体积为：

Volume = π ∫ₐᵇ [f(x)]² dx

Example. The region under y = √(9 − x²) from x = −3 to x = 3 is rotated about the x-axis. Find the volume.

实例。将 y = √(9 − x²) 从 x = −3 到 x = 3 的曲线下方的区域绕 x 轴旋转。求体积。

V = π ∫₋₃³ (9 − x²) dx = π [9x − x³/3]₋₃³ = π(27 − 9) − π(−27 + 9) = 36π

This solid is a sphere of radius 3, whose volume is indeed (4/3)π(3³) = 36π, confirming our result. Similarly, rotating around the y-axis requires x² = h(y) and uses π ∫꜀ᵈ [h(y)]² dy.

该旋转体是半径为 3 的球体，体积确实为 (4/3)π(3³) = 36π，验证了我们的结果。类似地，绕 y 轴旋转则需要 x² = h(y)，并使用 π ∫꜀ᵈ [h(y)]² dy。

11. Common Pitfalls in Exam Questions | 考试题目中的常见陷阱

Across examination boards (Edexcel, AQA, OCR, CAIE), several recurring pitfalls appear:

在各考试局（Edexcel、AQA、OCR、CAIE）中，有几个反复出现的坑点：

Forgetting to find roots first. Always determine where f(x) = 0 before integrating, because the sign of f(x) decides the splitting scheme.

忘记先找根。在积分之前应总是先确定 f(x) = 0 的位置，因为 f(x) 的正负决定了拆分方案。
Using incorrect limits. The limits of integration are x-values at intersections or given boundaries, not arbitrary values.

使用错误的上下限。积分上下限是交点或给定边界处对应的 x 值，而不是任意数字。
Subtracting in the wrong order. For area between curves, always integrate (upper − lower); for volume, it is (outer² − inner²) when a washer is formed.

相减顺序错误。对于两曲线之间的面积，应始终积分（上方 − 下方）；对于旋转体，如果形成垫圈形状，则需用（外半径² − 内半径²）。
Confusing displacement and distance. In kinematics problems, use |v(t)| for distance, not v(t).

混淆位移与路程。在运动学问题中，路程应使用 |v(t)| 积分，而不是 v(t)。

It is also essential to leave your answer in exact form where required, such as 16/3 or 5π, and only give decimals when the question explicitly asks for them.

在需要精确值的地方，务必保留分数或 π 等精确形式，例如 16/3 或 5π；只有在题目明确要求时再给出小数。

12. Summary and Final Advice | 总结与终级建议

The definite integral transforms a geometric measurement — the area under a curve — into a simple arithmetic computation of antiderivative values. Mastery of this topic requires: (1) finding intersection points carefully; (2) identifying which function is on top; (3) splitting integrals where necessary to respect sign; and (4) knowing which application formula to use.

定积分将几何测量——曲线下方面积——转化为原函数值的简单算术运算。掌握这一主题需要：(1) 小心求解交点；(2) 确定哪条函数在上方；(3) 在必要时拆分积分以处理正负号；(4) 知道运用哪个应用公式。

Finally, practise drawing a quick sketch for every area problem. A correct diagram leads to correct limits and correct order of subtraction. The time you invest in a 10-second sketch will save you from losing 5 or more marks in an exam.

最后，对于每一个面积问题都练习画一个快速草图。正确的图形能带来正确的上下限和正确的相减顺序。花 10 秒钟画图所投入的时间，将帮你避免在考试中丢掉 5 分甚至更多。

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Area Under Curves and Applications of Definite Integrals | 曲线下面积与定积分的应用