📚 Areas under the x-axis | x轴下方的面积
In A-Level mathematics, calculating the area bounded by a curve and the x-axis often involves definite integration. However, when a curve lies below the x-axis, the definite integral yields a negative value. To find the actual geometric area, students must carefully handle these negative contributions by using absolute values or splitting the integral at the points where the curve crosses the axis.
在A-Level数学中，计算由曲线与x轴围成的面积常常用到定积分。然而，当曲线位于x轴下方时，定积分会得出负值。要得到实际的几何面积，学生必须通过取绝对值或在曲线穿过x轴的点处分割积分，谨慎处理这些负贡献。
1. Definite Integrals and Geometric Area | 定积分与几何面积
The definite integral ∫ab to x = b. If f(x) ≥ 0 on [a, b], the integral directly equals the area under the curve.
定积分 ∫ab f(x) dx 给出曲线 y = f(x) 与 x 轴在 x = a 到 x = b 之间的有向面积。如果 f(x) ≥ 0 在 [a, b] 上，积分直接等于曲线下方的面积。
2. What Happens When the Curve Dips Below the Axis? | 曲线降到轴下方时会发生什么？
If f(x) ≤ 0 on an interval, the definite integral produces a negative number. For example, ∫01 actual area of the region is 1/3. This occurs because integration sums the signed heights of rectangles: negative heights contribute negative 'area'.
如果 f(x) ≤ 0 在一个区间上，定积分会得出一个负数。例如，∫01 (−x²) dx = −1/3，但该区域的实际面积是 1/3。这是因为积分是对矩形有向高度的累加：负高度贡献负的“面积”。
3. The Signed Area vs Actual Area | 有向面积与实际面积
Distinguish between signed area (which can be positive or negative) and actual geometric area (always non‑negative). In exam questions, when asked for 'the area bounded by the curve and the x‑axis', you must provide the total geometric area, summing the absolute values of the signed areas.
区分有向面积（可以是正或负）和实际几何面积（始终非负）。在考题中，如果要求“曲线与x轴所围成的面积”，你必须给出总的几何面积，即将各段有向面积的绝对值相加。
4. Using the Absolute Value: ∫ |f(x)| dx | 使用绝对值：∫ |f(x)| dx
One method is to integrate the absolute value of the function:
Area = ∫ab |f(x)| dx
However, integrating |f(x)| is rarely possible without splitting the integral. On an interval where f(x) is negative, |f(x)| = −f(x), so you integrate the negative of the function.
一种方法是积分函数的绝对值。如上式所示。然而，如果不分割积分，直接积分 |f(x)| 通常不可能。在 f(x) 为负的区间上，|f(x)| = −f(x)，所以你需要积分函数的相反数。
5. Splitting the Integral at x‑intercepts | 在 x 截距处分割积分
Find the points where the curve crosses the x‑ ivide the integration range. For each subinterval, determine the sign of f(x). If positive, integrate f(x); if negative, integrate −f(x). Add the results to get total area.
通过解 f(x) = 0 找到曲线与 x 轴的交点。这些根将积分区间分割。对每个子区间，判断 f(x) 的符号。若为正，积分 f(x)；若为负，积分 −f(x)。将结果相加得到总面积。
6. Worked Example: A Simple Negative Region | 例题：简单负区域
Find the area between y = x² − 4 and the x‑axis from x = 0 to x = 3.
求曲线 y = x² − 4 与 x 轴在 x = 0 到 x = 3 之间的面积。
ve crosses at x = 2. For 0 ≤ x < 2, f(x) is negative; for 2 < x ≤ 3, positive.
首先求截距：x² − 4 = 0 ⇒ x = 2 或 x = −2。在 [0,3] 内，曲线在 x = 2 处穿过 x 轴。对于 0 ≤ x < 2，f(x) 为负；对于 2 < x ≤ 3，f(x) 为正。
Area = ∫02 −(x² − 4) dx + ∫23 (x² − 4) dx.
Compute:
∫02 (−x² + 4) dx = [−x³/3 + 4x]02 = (−8/3 + 8) − 0 = 16/3.
∫23 (x² − 4) dx = [x³/3 − 4x]23 = (27/3 − 12) − (8/3 − 8) = (9 − 12) − (−16/3) = 7/3.
Total area = 16/3 + 7/3 = 23/3 square units.
面积 = ∫02 −(x² − 4) dx + ∫23 (x² − 4) dx。计算得出总面积 = 23/3 平方单位。
7. Curve That Crosses Multiple Times | 多次穿过 x 轴的曲线
When a curve crosses the x‑axis several times, divide the interval into segments separated by the roots. For each segment, take the absolute value of the definite integral (or integrate the absolute function). This ensures that all regions count positively.
当曲线多次穿过 x 轴时，用根将区间分割成段。对每一段，取定积分的绝对值（或积分绝对值函数）。这确保所有区域都以正数计入。
8. Area Between a Curve and the x‑axis: Step‑by‑Step Strategy | 曲线与 x 轴之间的面积：分步策略
Here is a reliable method:
- 1. Sketch the graph or identify the shape of f(x).
- 2. Solve f(x) = 0 to find all x‑intercepts in the given interval.
- 3. List the subintervals determined by these intercepts.
- 4. For each subinterval, decide whether f(x) ≥ 0 or f(x) ≤ 0 (test a point or use knowledge of the function).
- 5. If f(x) ≥ 0, compute ∫ f(x) dx; if f(x) ≤ 0, compute ∫ −f(x) dx.
- 6. Sum the contributions.
Alternatively, compute ∫ f(x) dx for each subinterval and then add the absolute values of these individual integrals.
这里有一个可靠的方法：1. 画函数草图或确定 f(x) 的形状。2. 解 f(x) = 0 找出给定区间内所有 x 截距。3. 列出这些截距划分的子区间。4. 对每个子区间，判断 f(x) ≥ 0 还是 f(x) ≤ 0（测试一个点或利用函数知识）。5. 若 f(x) ≥ 0，计算 ∫ f(x) dx；若 f(x) ≤ 0，计算 ∫ −f(x) dx。6. 相加各贡献。或者，计算每个子区间的 ∫ f(x) dx，然后把各积分的绝对值相加。
9. Using Symmetry to Reduce Work | 利用对称性减少计算量
If the curve is symmetric about the y‑axis or the origin, you might integrate over half the interval and double the result. However, be cautious when the curve crosses the axis in the symmetric region. Always check that the area you double corresponds to a region entirely above or entirely below the axis. If the region straddles the axis, symmetry might apply if you consider absolute area.
如果曲线关于 y 轴或原点对称，你可以对一半区间积分然后乘2。但要注意当曲线在对称区域内穿过轴的情况。始终检查你所加倍的区域是否完全在轴上方或完全在轴下方。如果区域跨轴，若考虑绝对面积，对称性可能仍然适用。
10. Common Pitfalls & How to Avoid Them | 常见陷阱及避免方法
Pitfall 1: Forgetting to take the absolute value and simply integrating f(x) over an interval where it is negative, giving a negative answer for area. Area is always non‑negative.
Pitfall 2: Adding signed integrals without converting negatives, leading to cancellation. For example, ∫02 (x−1) dx = 0, but the area is 1.
itfall 3: Not checking for roots within the integration limits. Always solve f(x)=0 first.
Pitfall 4: Using the wrong sign when removing the absolute value. If f(x) < 0, |f(x)| = −f(x). Confirm by testing a point.
陷阱1：忘记取绝对值，直接在 f(x) 为负的区间上积分 f(x)，导致面积为负。面积始终为非负。
陷阱2：不将负值转正，将有向积分相加，导致抵消。例如 ∫02 (x−1) dx = 0，但面积是1。
陷阱3：未检查积分限内的根。始终先解 f(x)=0。
陷阱4：去掉绝对值时符号错误。如果 f(x) < 0，|f(x)| = −f(x)。通过测试点来确认。
11. Practice Problem with Full Solution | 练习题及完整解答
Find the total area enclosed between the curve y = x³; − 4x and the x‑axis.
求曲线 y = x³; − 4x 与 x 轴之间所围成的总面积。
First, find x‑ + 2) = 0 ⇒ x = −2, 0, 2. The region is bounded between x = −2 and x = 2.
首先求 x 截距：x³ − 4x = x(x² − 4) = x(x − 2)(x + 2) = 0 ⇒ x = −2, 0, 2。区域范围在 x = −2 到 x = 2 之间。
Determine the sign on each subinterval:
r x ∈ (−2, 0), test x = −1 ⇒ f(−1) = 3 > 0, so f(x) ≥ 0.
For x ∈ (0, 2), test x = 1 ⇒ f(1) = −3 < 0, so f(x) ≤ 0.
判断每个子区间的符号：x ∈ (−2,0) 时 f(x) ≥ 0；x ∈ (0,2) 时 f(x) ≤ 0。
Thus, Area = ∫−20 (x³ − 4x) dx + ∫02 −(x³ − 4x) dx.
Compute:
∫ (x³ − 4x) dx = x⁴/4 − 2x².
From −2 to 0: [0] − [((−2)⁴/4 − 2(−2)²)] = 0 − (16/4 − 8) = −(−4) = 4.
From 0 to 2 for the absolute part: −[x⁴/4 − 2x²]02 = −( (16/4 − 8) − 0 ) = −(−4) = 4.
Total area = 4 + 4 = 8 square units.
计算：∫ (x³ − 4x) dx = x⁴/4 − 2x²。从 −2 到 0 得 4；从 0 到 2 的绝对值部分得 4。总面积 = 8 平方单位。
12. Key Takeaways for Exam Success | 考试成功的关键要点
quantity. The definite integral gives signed area. To find the geometric area under the x‑ of f(x) = 0 to split the range, and never assume the whole integral can be computed as a single ∫ f(x) dx. Practice with functions like quadratics, cubics, and trigonometric graphs to master this technique.
永远记住：面积是正量。定积分给出有向面积。要求 x 轴下方的几何面积，要么在负区间积分 −f(x)，要么对每个子区间的定积分取绝对值。找出 f(x) = 0 的根来分割区间，决不假设整个积分可以作为一个
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