A-Level Maths: Calculating Areas Below the x-axis | A-Level数学：x轴下方区域的面积计算

📚 A-Level Maths: Calculating Areas Below the x-axis | A-Level数学：x轴下方区域的面积计算

When calculating the area between a curve and the x-axis, one of the most common mistakes in A-Level Mathematics is to simply evaluate a definite integral and assume the result is the area. However, when the curve lies below the x-axis, the integral gives a negative value, because the signed area is negative. To find the actual geometric area, you must take the absolute value of the integral, or split the interval at the x-intercepts.

在A-Level数学中，计算曲线与x轴之间的面积时，最常见的错误之一就是直接计算定积分并认为结果就是面积。然而，当曲线位于x轴下方时，定积分得到的是负值，因为“有向面积”是负的。要找到真正的几何面积，你必须对积分取绝对值，或者在x轴交点处拆分积分区间。

1. Definite Integrals and Signed Area | 定积分与有向面积

A definite integral \(\int_a^b f(x)\,dx\) represents the signed area between the curve \(y = f(x)\) and the x-axis. If \(f(x) \ge 0\) on \([a,b]\), the integral is positive. If \(f(x) \le 0\), the integral is negative. If the curve crosses the x-axis, the integral combines positive and negative contributions, which can cancel out.

定积分 \(\int_a^b f(x)\,dx\) 表示曲线 \(y = f(x)\) 与x轴之间的有向面积。若在 \([a,b]\) 上 \(f(x) \ge 0\)，积分值为正；若 \(f(x) \le 0\)，积分值为负；若曲线穿过x轴，积分会将正负贡献相加，可能相互抵消。

The integral gives signed area, not geometric area.
积分给出的是有向面积，不是几何面积。
For area, use \(|f(x)|\) as the integrand.
求面积时，应对 \(|f(x)|\) 进行积分。

2. The Classic Trap: A Negative Integral | 经典陷阱：积分为负

Consider \(f(x) = x\) from \(x = -2\) to \(x = 0\). The integral is \(\int_{-2}^{0} x\,dx = \left[\frac{x^2}{2}\right]_{-2}^{0} = 0 – 2 = -2\). The actual geometric area of the triangle below the x-axis is 2 square units, not -2. The negative sign simply tells us the region is below the x-axis.

考虑 \(f(x) = x\) 从 \(x = -2\) 到 \(x = 0\)。积分 \(\int_{-2}^{0} x\,dx = \left[\frac{x^2}{2}\right]_{-2}^{0} = 0 – 2 = -2\)。实际几何面积是x轴下方三角形的面积，为2平方单位，而不是-2。负号只说明区域位于x轴下方。

Geometric Area = |∫ₐᵇ f(x) dx| when f(x) ≤ 0 on [a,b]

几何面积 = |∫ₐᵇ f(x) dx|，当 f(x) ≤ 0 在 [a,b] 上时

3. Splitting at x-intercepts: The General Method | 在x轴交点处拆分：通用方法

When a curve crosses the x-axis within the interval of integration, the signed areas of the portions above and below the axis have opposite signs. To compute the total geometric area, you must split the integral at each x-intercept, integrate each segment separately, take the absolute value of each result, and then sum them.

当曲线在积分区间内穿过x轴时，x轴上方和下方部分的有向面积符号相反。要计算总几何面积，必须在每个x轴交点处拆分积分，分别对每一段求积分，对每个结果取绝对值，然后相加。

Example	Method	Area
∫₋₁² x dx	Split at x=0: ∫₋₁⁰ x dx + ∫₀² x dx	\|−½\| + \|2\| = 2.5
∫₀³ (x²−4) dx	Split at x=2: ∫₀² (x²−4) dx + ∫₂³ (x²−4) dx	\|−16/3\| + \|7/3\| = 23/3

Always start by solving \(f(x)=0\) to find the x-intercepts within your interval.

务必先解 \(f(x)=0\)，找出区间内的x轴交点。

4. Worked Example: Quadratic Below the x-axis | 例题：x轴下方的二次函数

Find the area bounded by \(y = x^2 – 4\), the x-axis, and the lines \(x = 0\) and \(x = 3\).

求由 \(y = x^2 – 4\)、x轴以及直线 \(x = 0\) 和 \(x = 3\) 所围成的面积。

First, solve \(x^2 – 4 = 0\), giving \(x = \pm 2\). Only \(x = 2\) lies in \([0,3]\). On \([0,2]\), the curve is below the x-axis; on \([2,3]\), it is above.

首先解 \(x^2 – 4 = 0\)，得 \(x = \pm 2\)。在 \([0,3]\) 内只有 \(x = 2\)。在 \([0,2]\) 上曲线位于x轴下方；在 \([2,3]\) 上位于x轴上方。

Area = |∫₀² (x²−4) dx| + ∫₂³ (x²−4) dx

Compute the first integral: \(\int_0^2 (x^2-4)\,dx = \left[\frac{x^3}{3} – 4x\right]_0^2 = \frac{8}{3} – 8 = -\frac{16}{3}\). Its absolute value is \(\frac{16}{3}\).

计算第一个积分：\(\int_0^2 (x^2-4)\,dx = \left[\frac{x^3}{3} – 4x\right]_0^2 = \frac{8}{3} – 8 = -\frac{16}{3}\)。其绝对值为 \(\frac{16}{3}\)。

Compute the second integral: \(\int_2^3 (x^2-4)\,dx = \left[\frac{x^3}{3} – 4x\right]_2^3 = (9-12) – (\frac{8}{3}-8) = -3 – (-\frac{16}{3}) = \frac{7}{3}\).

计算第二个积分：\(\int_2^3 (x^2-4)\,dx = \left[\frac{x^3}{3} – 4x\right]_2^3 = (9-12) – (\frac{8}{3}-8) = -3 – (-\frac{16}{3}) = \frac{7}{3}\)。

Total Area = 16/3 + 7/3 = 23/3 ≈ 7.67 square units

总面积 = 16/3 + 7/3 = 23/3 ≈ 7.67 平方单位

5. Using Absolute Value Inside the Integral | 在积分内使用绝对值

An alternative method is to integrate the absolute value function \(\int_a^b |f(x)|\,dx\). This automatically handles the sign change. However, in practice, you still need to find the x-intercepts to write \(|f(x)|\) as a piecewise function.

另一种方法是对绝对值函数积分 \(\int_a^b |f(x)|\,dx\)。这会自动处理符号变化。但在实际操作中，你仍然需要找出x轴交点，才能将 \(|f(x)|\) 写成分段函数。

Find roots of \(f(x)=0\).
求出 \(f(x)=0\) 的根。
Split the integral at the roots.
在根处拆分积分。
Replace each segment with its absolute value.
将每一段替换为其绝对值。

This method is equivalent to splitting and taking absolute values, but it may be easier to set up in the exam when multiple roots exist.

这种方法与拆分后取绝对值等价，但在有多个根时可能更容易书写。

6. Worked Example: Cubic Crossing the Axis | 例题：穿过x轴的三次函数

Find the total area between \(y = x^3 – 4x\) and the x-axis from \(x = -2\) to \(x = 2\).

求 \(y = x^3 – 4x\) 与x轴从 \(x = -2\) 到 \(x = 2\) 之间的总面积。

Factor: \(x^3 – 4x = x(x^2-4) = x(x-2)(x+2)\). Roots are \(x = -2, 0, 2\). The interval \([-2,2]\) includes the root \(x=0\) in its interior. Check the sign of the function in each subinterval:

因式分解：\(x^3 – 4x = x(x^2-4) = x(x-2)(x+2)\)。根为 \(x = -2, 0, 2\)。区间 \([-2,2]\) 的内部包含根 \(x=0\)。在每个子区间检查函数符号：

On \([-2,0]\), test \(x=-1\): \((-1)^3 – 4(-1) = -1 + 4 = 3 > 0\).
在 \([-2,0]\) 上，测试 \(x=-1\)：\((-1)^3 – 4(-1) = -1 + 4 = 3 > 0\)。
On \([0,2]\), test \(x=1\): \(1 – 4 = -3 < 0\).
在 \([0,2]\) 上，测试 \(x=1\)：\(1 – 4 = -3 < 0\)。

Therefore the curve is above the axis on \([-2,0]\) and below on \([0,2]\).

因此曲线在 \([-2,0]\) 上位于x轴上方，在 \([0,2]\) 上位于x轴下方。

Compute the integrals:

计算积分：

∫₋₂⁰ (x³−4x) dx = [x⁴/4 − 2x²]₋₂⁰ = 0 − (4 − 8) = 4

∫₀² (x³−4x) dx = [x⁴/4 − 2x²]₀² = (4 − 8) − 0 = −4

Taking absolute values: \(|4| + |-4| = 8\) square units.

取绝对值：\(|4| + |-4| = 8\) 平方单位。

Notice that the single integral \(\int_{-2}^{2} (x^3-4x)\,dx = 0\), which would incorrectly suggest zero area. This illustrates why splitting is essential.

注意单个积分 \(\int_{-2}^{2} (x^3-4x)\,dx = 0\)，这会错误地暗示面积为零。这说明拆分是必要的。

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

In Edexcel A-Level Mathematics, you may also be asked to find the area between a curve and the y-axis. In that case, integrate with respect to \(y\): \(\int_c^d x\,dy\), where \(x = g(y)\). The same principle applies: if the curve lies to the left of the y-axis, the integral is negative, and you take the absolute value or split at the y-intercepts.

在Edexcel A-Level数学中，你也可能会被要求求曲线与y轴之间的面积。此时应对 \(y\) 积分：\(\int_c^d x\,dy\)，其中 \(x = g(y)\)。同样的原则适用：若曲线位于y轴左侧，则积分为负，需取绝对值或在y轴交点处拆分。

Rearrange the curve equation to make \(x\) the subject.
将曲线方程变形，使 \(x\) 成为因变量。
Find the y-coordinates of the boundary points.
求出边界点的y坐标。
Split at y-intercepts if necessary.
如有必要，在y轴交点处拆分。

8. Common Exam Mistakes and Tips | 常见考试错误与提示

Many students lose marks because they treat every definite integral as an area. Here are the most frequent pitfalls and how to avoid them.

许多学生因为把每个定积分都当作面积而失分。以下是最常见的陷阱以及如何避免它们。

Mistake	Correction
Ignoring negative values	Always check signs on each subinterval
Splitting at every root outside the interval	Only split at roots inside \([a,b]\)
Forgetting the absolute value	Write \|∫\| for below-axis regions
Using one integral when signs change	Split or integrate \(\|f(x)\|\)

Also, remember to include the correct units in your final answer when asked for a geometrical interpretation.

此外，若题目要求几何解释，记得在最终答案中包含正确的单位。

9. Practice Questions | 练习题目

Try these questions to test your understanding. Solutions are provided below.

尝试以下题目以测试你的理解。解答附后。

Find the area between \(y = -x^2 + 1\) and the x-axis from \(x = 0\) to \(x = 2\).
求 \(y = -x^2 + 1\) 与x轴从 \(x = 0\) 到 \(x = 2\) 之间的面积。
Find the total area between \(y = \sin x\) and the x-axis from \(x = 0\) to \(x = 2\pi\).
求 \(y = \sin x\) 与x轴从 \(x = 0\) 到 \(x = 2\pi\) 之间的总面积。
Find the area between \(y = x^2 – 3x + 2\) and the x-axis from \(x = 0\) to \(x = 3\).
求 \(y = x^2 – 3x + 2\) 与x轴从 \(x = 0\) 到 \(x = 3\) 之间的面积。

10. Solutions to Practice Questions | 练习题目解答

Question 1: \(y = -x^2 + 1\). Roots: \(x = \pm 1\). Only \(x=1\) is in \([0,2]\). On \([0,1]\), \(y \ge 0\); on \([1,2]\), \(y \le 0\).

第1题： \(y = -x^2 + 1\)。根：\(x = \pm 1\)。在 \([0,2]\) 内只有 \(x=1\)。在 \([0,1]\) 上 \(y \ge 0\)；在 \([1,2]\) 上 \(y \le 0\)。

Area = ∫₀¹ (1−x²) dx + |∫₁² (1−x²) dx| = 2/3 + |−4/3| = 2/3 + 4/3 = 2

面积 = ∫₀¹ (1−x²) dx + |∫₁² (1−x²) dx| = 2/3 + |−4/3| = 2/3 + 4/3 = 2

Question 2: \(y = \sin x\). Roots in \([0,2\pi]\): \(x = 0, \pi, 2\pi\). On \([0,\pi]\), \(\sin x \ge 0\); on \([\pi,2\pi]\), \(\sin x \le 0\).

第2题： \(y = \sin x\)。在 \([0,2\pi]\) 内的根：\(x = 0, \pi, 2\pi\)。在 \([0,\pi]\) 上 \(\sin x \ge 0\)；在 \([\pi,2\pi]\) 上 \(\sin x \le 0\)。

Area = ∫₀^π sin x dx + |∫_π^{2π} sin x dx| = 2 + |−2| = 4

面积 = ∫₀^π sin x dx + |∫_π^{2π} sin x dx| = 2 + |−2| = 4

Question 3: \(y = x^2 – 3x + 2 = (x-1)(x-2)\). Roots \(x=1,2\), both inside \([0,3]\). Test intervals: \([0,1]\), \([1,2]\), \([2,3]\). Signs: positive, negative, positive.

第3题： \(y = x^2 – 3x + 2 = (x-1)(x-2)\)。根 \(x=1,2\) 都在 \([0,3]\) 内。测试区间：\([0,1]\)、\([1,2]\)、\([2,3]\)。符号：正、负、正。

Compute each segment:

计算每一段：

\(\int_0^1 (x^2-3x+2)\,dx = [x^3/3 – 3x^2/2 + 2x]_0^1 = 1/3 – 3/2 + 2 = 5/6\)

\(\int_1^2 (x^2-3x+2)\,dx = [x^3/3 – 3x^2/2 + 2x]_1^2 = (8/3 – 6 + 4) – (1/3 – 3/2 + 2) = 2/3 – 5/6 = -1/6\)

\(\int_2^3 (x^2-3x+2)\,dx = [x^3/3 – 3x^2/2 + 2x]_2^3 = (9 – 27/2 + 6) – (8/3 – 6 + 4) = 3/2 – 2/3 = 5/6\)

Total Area = 5/6 + |−1/6| + 5/6 = 11/6 ≈ 1.833

总面积 = 5/6 + |−1/6| + 5/6 = 11/6 ≈ 1.833

11. Integration by Symmetry and Special Cases | 对称性与特殊情况下的积分

Sometimes symmetry can simplify area calculations. For an odd function like \(y = x^3\), the areas above and below the x-axis on a symmetric interval cancel when integrated, but they are equal in magnitude. For an even function, the areas on symmetric intervals are equal, so you can double one side.

有时对称性可以简化面积计算。对于奇函数如 \(y = x^3\)，在对称区间上，x轴上下方的面积在积分时会抵消，但它们的几何大小相等。对于偶函数，对称区间上的面积相等，因此可以将一侧的结果翻倍。

Odd function: \(f(-x) = -f(x)\). On \([-a,a]\), \(\int_{-a}^{a} f(x)\,dx = 0\), but geometric area = \(2\int_0^a |f(x)|\,dx\).
奇函数：\(f(-x) = -f(x)\)。在 \([-a,a]\) 上，\(\int_{-a}^{a} f(x)\,dx = 0\)，但几何面积 = \(2\int_0^a |f(x)|\,dx\)。
Even function: \(f(-x) = f(x)\). On \([-a,a]\), area = \(2\int_0^a f(x)\,dx\) (if \(f(x) \ge 0\)).
偶函数：\(f(-x) = f(x)\)。在 \([-a,a]\) 上，面积 = \(2\int_0^a f(x)\,dx\)（若 \(f(x) \ge 0\)）。

This is particularly useful in trigonometric integration and in questions involving polynomial curves.

这在三角积分和涉及多项式曲线的问题中尤其有用。

12. Final Advice for Exams | 考试最终建议

Always sketch the curve or at least determine the sign of \(f(x)\) in each subinterval before integrating. This prevents sign errors and ensures you split the integral correctly. Write your working clearly, showing the absolute value brackets where needed.

在积分之前，务必先画出曲线草图，或至少确定每个子区间上 \(f(x)\) 的符号。这可以防止符号错误，并确保你正确拆分积分。书写步骤要清晰，在需要的地方标出绝对值符号。

Remember: for Edexcel A-Level Mathematics, most area questions are worth 4–6 marks, and the marks are often awarded for the method of splitting, the integration, and the final absolute value. A correct sketch alone can earn you a method mark.

记住：在Edexcel A-Level数学中，大多数面积题的分值为4–6分，分数通常授予拆分方法、积分计算和最终的绝对值处理。仅画对草图就可以获得方法分。

Area below x-axis = |∫ₐᵇ f(x) dx| (no sign change)

x轴下方的面积 = |∫ₐᵇ f(x) dx|（无符号变化时）

For intervals with roots, split first, integrate each, take absolute values, then add.

对于含有根的区间，先拆分，再逐段积分，取绝对值，最后相加。

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A-Level Maths: Calculating Areas Below the x-axis | A-Level数学：x轴下方区域的面积计算