Finding Areas by Integration | 用积分求面积

📚 Finding Areas by Integration | 用积分求面积

In Edexcel A-Level Mathematics, finding areas is one of the most important applications of integration. You will meet questions that ask for the area under a curve, the area between two curves, areas from parametric equations, and approximate areas using the trapezium rule.

在 Edexcel A-Level 数学中，求面积是积分最重要的应用之一。考试中会遇到求曲线下方面积、两条曲线之间的面积、参数方程对应的面积，以及使用梯形法则求近似面积等问题。

1. The Core Idea: Area as a Limit of Sums | 核心思想：面积作为和的极限

The area under a curve y = f(x) between x = a and x = b can be approximated by splitting the region into n thin rectangles. The width of each rectangle is Δx, and the height is f(xᵢ). As n becomes very large and Δx tends to zero, the sum of the rectangle areas approaches the exact area.

曲线 y = f(x) 在 x = a 与 x = b 之间的下方区域，可以通过将其分割成 n 个细长矩形来近似。每个矩形的宽度为 Δx，高度为 f(xᵢ)。当 n 非常大且 Δx 趋于零时，矩形面积之和就趋近于精确面积。

A ≈ Σ f(xᵢ) Δx, and as Δx → 0, A = ∫ₐᵇ f(x) dx

A ≈ Σ f(xᵢ) Δx，当 Δx → 0 时，A = ∫ₐᵇ f(x) dx

This limiting process is the foundation of the definite integral. In exam work, you will normally use the Fundamental Theorem of Calculus rather than calculating limits of sums by hand.

这一极限过程是定积分的基础。在考试中，你通常会使用微积分基本定理，而不是手动计算和的极限。

2. Definite Integral Notation and the Fundamental Theorem | 定积分符号与微积分基本定理

The definite integral ∫ₐᵇ f(x) dx represents the signed area between the curve y = f(x) and the x-axis from x = a to x = b. The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x), then:

定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x) 与 x 轴在 x = a 到 x = b 之间的带符号面积。微积分基本定理指出，如果 F(x) 是 f(x) 的一个原函数，那么：

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) – F(a)

Notice that no constant of integration is needed in a definite integral because the constants cancel when subtracting F(a) from F(b).

注意，在定积分中不需要加积分常数，因为计算 F(b) 减去 F(a) 时常数会相互抵消。

Function	Integral	中文
xⁿ, n ≠ -1	xⁿ⁺¹/(n+1) + C	xⁿ 的积分
1/x	ln\|x\| + C	1/x 的积分
eˣ	eˣ + C	eˣ 的积分
sin x	-cos x + C	sin x 的积分
cos x	sin x + C	cos x 的积分

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

If f(x) is positive on the interval [a, b], then the area between the curve and the x-axis is simply the definite integral ∫ₐᵇ f(x) dx. No absolute value is needed because the integral is already positive.

如果 f(x) 在区间 [a, b] 上为正，那么曲线与 x 轴之间的面积就是定积分 ∫ₐᵇ f(x) dx。此时不需要取绝对值，因为积分值本来就是正的。

For example, find the area under y = x² from x = 0 to x = 3. Since x² ≥ 0 on this interval, the area is:

例如，求 y = x² 在 x = 0 到 x = 3 之间的下方面积。由于在该区间上 x² ≥ 0，面积为：

A = ∫₀³ x² dx = [x³/3]₀³ = 27/3 – 0 = 9

So the shaded area is 9 square units. Always write the correct primitive and substitute the limits carefully.

因此阴影面积为 9 平方单位。一定要写出正确的原函数，并仔细代入上下限。

4. Handling Areas Below the x-axis | 处理 x 轴下方的面积

If f(x) is negative on [a, b], the definite integral ∫ₐᵇ f(x) dx will be negative. This is because the integral gives a signed area. To obtain the actual area, you must take the absolute value or place a minus sign in front of the integral.

如果 f(x) 在 [a, b] 上为负，那么定积分 ∫ₐᵇ f(x) dx 会是负值。这是因为积分给出的是带符号的面积。要得到实际面积，你必须取绝对值，或在积分前面加上负号。

Area = -∫ₐᵇ f(x) dx when f(x) ≤ 0

当 f(x) ≤ 0 时，面积 = -∫ₐᵇ f(x) dx

For example, find the area between y = x² – 4 and the x-axis from x = 0 to x = 2. Here the curve is below the x-axis, so the integral will be negative.

例如，求 y = x² – 4 与 x 轴在 x = 0 到 x = 2 之间的面积。这里曲线位于 x 轴下方，因此积分值会是负数。

∫₀² (x² – 4) dx = [x³/3 – 4x]₀² = (8/3 – 8) – 0 = -16/3

The area is therefore | -16/3 | = 16/3 square units. A common exam error is to give the negative value as the final answer.

因此面积为 | -16/3 | = 16/3 平方单位。考试中常见的错误是把负值直接作为最终答案。

5. Splitting the Integral at Sign Changes | 在符号变化处拆分积分

When a curve crosses the x-axis within the interval, the signed integral may partly cancel. You must find the roots and split the integral into separate intervals where the sign of f(x) is constant.

当曲线在区间内穿过 x 轴时，带符号积分可能会部分抵消。你必须求出曲线与 x 轴的交点，并将积分拆分成 f(x) 符号恒定的若干区间。

For example, find the total area enclosed by y = x² – 1 and the x-axis from x = 0 to x = 2. The curve crosses the x-axis at x = 1, so split at x = 1.

例如，求 y = x² – 1 与 x 轴在 x = 0 到 x = 2 之间围成的总面积。曲线在 x = 1 处穿过 x 轴，因此在 x = 1 处拆分。

From 0 to 1: ∫₀¹ (x² – 1) dx = [x³/3 – x]₀¹ = 1/3 – 1 = -2/3

从 0 到 1：∫₀¹ (x² – 1) dx = [x³/3 – x]₀¹ = 1/3 – 1 = -2/3

From 1 to 2: ∫₁² (x² – 1) dx = [x³/3 – x]₁² = 4/3

从 1 到 2：∫₁² (x² – 1) dx = [x³/3 – x]₁² = 4/3

The total area is therefore 2/3 + 4/3 = 2 square units. If you had integrated straight from 0 to 2, you would obtain 2/3, which is incorrect for the actual area.

因此总面积为 2/3 + 4/3 = 2 平方单位。如果你直接从 0 到 2 积分，会得到 2/3，这对实际面积来说是错误的。

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

To find the area between two curves y = f(x) and y = g(x), first identify which curve is on top. If f(x) ≥ g(x) on [a, b], then the area is:

要求两条曲线 y = f(x) 与 y = g(x) 之间的面积，首先要确定哪条曲线在上方。如果 f(x) ≥ g(x) 在 [a, b] 上成立，那么面积为：

A = ∫ₐᵇ [f(x) – g(x)] dx

For example, find the area enclosed by y = x + 2 and y = x². First solve x + 2 = x², giving x² – x – 2 = 0, so x = -1 and x = 2. The line y = x + 2 is above the parabola y = x² on this interval.

例如，求 y = x + 2 与 y = x² 围成的面积。首先解 x + 2 = x²，得到 x² – x – 2 = 0，因此 x = -1 和 x = 2。在该区间上，直线 y = x + 2 在抛物线 y = x² 上方。

A = ∫₋₁² [(x + 2) – x²] dx = [x²/2 + 2x – x³/3]₋₁² = 27/6 = 9/2

Always sketch the graphs or test a value between the intersection points to confirm which function is on top.

始终画出草图，或在交点之间取一个测试值，以确认哪个函数在上方。

7. Integrating with Respect to y | 对 y 积分求面积

Sometimes it is easier to treat x as a function of y. If a region is bounded by x = f(y), the y-axis, and horizontal lines y = c and y = d, then the area is:

有时把 x 看作 y 的函数会更简单。如果区域由 x = f(y)、y 轴以及水平线 y = c 和 y = d 围成，那么面积为：

A = ∫꜀ᵈ f(y) dy

For example, find the area bounded by x = y², the y-axis, y = 0 and y = 3. This region lies to the right of the y-axis, so the area is positive.

例如，求由 x = y²、y 轴、y = 0 和 y = 3 围成的面积。该区域位于 y 轴右侧，因此面积为正。

A = ∫₀³ y² dy = [y³/3]₀³ = 27/3 = 9

This is often quicker than rewriting as y = √x and integrating with respect to x.

这通常比重写为 y = √x 并对 x 积分更快。

8. Parametric Curves and Areas | 参数曲线与面积

If a curve is given parametrically by x = x(t) and y = y(t), the area under the curve from t = t₁ to t = t₂ is found using:

如果曲线由参数方程 x = x(t) 和 y = y(t) 给出，则曲线下方从 t = t₁ 到 t = t₂ 的面积由下式求得：

A = ∫ₜ₁ᵗ² y (dx/dt) dt

This formula comes from replacing dx in the ordinary integral ∫ y dx by (dx/dt) dt. You must use the t-limits that correspond to the required x-values.

这个公式来自将普通积分 ∫ y dx 中的 dx 替换为 (dx/dt) dt。你必须使用与所需 x 值对应的 t 参数上下限。

For example, a curve has x = t² and y = t for t from 0 to 2. Since dx/dt = 2t, the area is:

例如，一条曲线为 x = t²，y = t，t 从 0 到 2。由于 dx/dt = 2t，面积为：

A = ∫₀² (t)(2t) dt = ∫₀² 2t² dt = [2t³/3]₀² = 16/3

Always convert both the integrand and the limits into the parameter t before integrating.

在积分之前，一定要将被积函数和上下限都转换为参数 t 的形式。

9. Trapezium Rule for Approximate Areas | 用梯形法则求近似面积

When an integral cannot be found exactly, the trapezium rule gives a numerical approximation. Split [a, b] into n equal strips of width h = (b – a)/n. The ordinates are y₀, y₁, …, yₙ. Then:

当积分无法精确求出时，梯形法则可以提供数值近似。将 [a, b] 分成 n 个等宽条带，宽度为 h = (b – a)/n。对应的纵坐标为 y₀, y₁, …, yₙ。那么：

∫ₐᵇ f(x) dx ≈ h/2 [ y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ ]

For example, approximate ∫

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Finding Areas by Integration | 用积分求面积