📚 A-Level Mathematics: Finding Areas Enclosed by Curves and Lines | A-Level数学：曲线与直线围成面积的求法
Integration is a powerful tool in A-Level Mathematics for calculating the area enclosed by curves and straight lines. This guide explains the underlying principles and demonstrates the steps through clear examples.
在A-Level数学中，积分是计算曲线与直线所围区域面积的强大工具。本文将讲解其基本原理，并通过清晰的例题展示解题步骤。
1. Definite Integrals and Signed Area | 定积分与有向面积
The definite integral of a function f(x) from x = a to x = b is written as ∫ab f(x) dx. This integral represents the signed area between the curve and the x-axis over the interval [a, b].
函数 f(x) 从 x = a 到 x = b 的定积分记作 ∫ab f(x) dx。该积分表示曲线与 x 轴在区间 [a, b] 上围成的有向面积。
If f(x) ≥ 0, the integral gives the actual area. If f(x) ≤ 0, the integral is negative, so we take its absolute value for the area.
若 f(x) ≥ 0，积分值即为实际面积；若 f(x) ≤ 0，积分值为负，需要取绝对值得到面积。
∫ab f(x) dx = F(b) − F(a)
By the Fundamental Theorem of Calculus, F is an antiderivative of f.
根据微积分基本定理，F 是 f 的一个原函数。
2. Area Under a Curve (Above x-axis) | 曲线与x轴围成的面积（曲线上方）
To find the area enclosed by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b, where f(x) ≥ 0 on [a, b], simply evaluate the definite integral.
要求由曲线 y = f(x)、x 轴以及两条竖直线 x = a 与 x = b 围成的面积，且 f(x) 在 [a, b] 上非负，直接计算定积分即可。
Area = ∫ab f(x) dx
For example, the area under y = x² from x = 0 to x = 1 is ∫01 x² dx = 1/3.
例如，y = x² 从 x = 0 到 x = 1 下方的面积为 ∫01 x² dx = 1/3。
3. Area Below x-axis | 曲线在x轴下方的面积
When the curve lies entirely below the x-axis on [a, b], the definite integral is negative. The area is therefore its absolute value:
当曲线在 [a, b] 上完全位于 x 轴下方时，定积分值为负。因此面积为绝对值：
Area = −∫ab f(x) dx
Consider y = x³ − x. On [0, 1] the curve is below the axis, so the enclosed area is −∫01 (x³ − x) dx = 1/4.
例如，y = x³ − x。在 [0, 1] 上曲线位于 x 轴下方，因此围成面积为 −∫01 (x³ − x) dx = 1/4。
4. Mixed Regions and the x-axis | 曲线与x轴的混合区域面积
If a curve crosses the x-axis within the interval, the integral over the whole interval would cancel positive and negative areas. Instead, split the interval at the roots and add the absolute values of the integrals.
若曲线在区间内穿越 x 轴，直接对整个区间积分会使正负面积相互抵消。应先求出根，并在每个子区间上分别积分后取绝对值相加。
- Find roots of f(x) = 0 in [a, b]
- For each subinterval, evaluate ∫ f(x) dx
- Add absolute values
- 求出 f(x) = 0 在 [a, b] 内的根
- 在每个子区间上计算 ∫ f(x) dx
- 取绝对值后相加
For y = x² − 1 from x = 0 to x = 2, the root is x = 1. Area = |∫01 (x² − 1) dx| + |∫12 (x² − 1) dx| = 2/3 + 4/3 = 2.
例如，y = x² − 1 从 x = 0 到 x = 2，根为 x = 1。面积 = |∫01 (x² − 1) dx| + |∫12 (x² − 1) dx| = 2/3 + 4/3 = 2。
5. Area Between a Curve and a Line | 曲线与直线围成的面积
The area enclosed by a curve y = f(x) and a line y = g(x) between their intersection points x = a and x = b is found by integrating the difference of the functions. If the line lies above the curve on the interval:
由曲线 y = f(x) 与直线 y = g(x) 在其交点 x = a 与 x = b 之间围成的面积，可通过积分两函数之差求得。若直线在曲线上方：
Area = ∫ab (g(x) − f(x)) dx
If the curve is above the line, swap the order. Always sketch the graphs or test a point to determine which function is larger.
若曲线在直线上方，则交换相减顺序。务必先画草图或在区间内取点判断哪个函数更大。
6. Finding Limits of Integration | 通过联立方程求积分上下限
To find the x-coordinates of the intersection points, set the two equations equal: f(x) = g(x). Solve for x.
要求交点横坐标，令两方程相等：f(x) = g(x)，然后解出 x。
For a line and a quadratic, this gives a quadratic equation, which may be solved by factorisation or the quadratic formula.
对于直线与二次曲线，这会得到一个一元二次方程，可通过因式分解或求根公式求解。
For y = x + 2 and y = x², set x + 2 = x² → x² − x − 2 = 0 → (x − 2)(x + 1) = 0, so x = −1 and x = 2.
例如，对 y = x + 2 与 y = x²，令 x + 2 = x² → x² − x − 2 = 0 → (x − 2)(x + 1) = 0，所以 x = −1 与 x = 2。
7. Worked Example 1: Line and Parabola | 例题1：直线与抛物线
We find the area enclosed by the parabola y = x² and the line y = x + 2.
求抛物线 y = x² 与直线 y = x + 2 围成的面积。
The intersection points are at x = −1 and x = 2. On the interval [−1, 2], the line is above the parabola. Therefore:
交点横坐标为 x = −1 与 x = 2。在区间 [−1, 2] 上，直线位于抛物线上方。因此：
Area = ∫−12 (x + 2 − x²) dx
The antiderivative is x²/2 + 2x − x³/3. Evaluating from −1 to 2 gives (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 − (−7/6) = 9/2.
原函数为 x²/2 + 2x − x³/3。从 −1 到 2 计算得 (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 − (−7/6) = 9/2。
Therefore the enclosed area is 9/2 square units.
因此围成面积为 9/2 平方单位。
8. Worked Example 2: Two Parabolas | 例题2：两条抛物线
Find the area enclosed by y = x² and y = 2 − x².
求 y = x² 与 y = 2 − x² 所围成的面积。
Set x² = 2 − x², so 2x² = 2, giving x = ±1. On [−1, 1], the curve y = 2 − x² is above y
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