A-Level Maths: Calculating Areas Below the x-axis by Integration | A-Level 数学：x轴下方面积的积分计算

📚 A-Level Maths: Calculating Areas Below the x-axis by Integration | A-Level 数学：x轴下方面积的积分计算

When we use definite integration to find the area between a curve and the x-axis, we must remember that a definite integral gives the signed area. If the curve lies entirely below the x-axis, the integral produces a negative value, even though a physical area is always positive.

当我们用定积分求曲线与x轴之间的面积时，必须记住定积分给出的是有向面积（有符号面积）。如果曲线完全位于x轴下方，积分结果是负数，而实际面积始终为正。

1. Signed Area vs Geometric Area | 有向面积与几何面积

For a function f(x) ≥ 0 on [a, b], the integral ∫_a^b f(x) dx equals the ordinary area between y = f(x) and the x-axis. For f(x) ≤ 0, the integral equals the negative of that area.

对于在区间 [a, b] 上满足 f(x) ≥ 0 的函数，∫_a^b f(x) dx 等于曲线 y = f(x) 与x轴之间的普通面积。当 f(x) ≤ 0 时，积分等于该面积的相反数。

This signed aspect means that if a curve goes above and below the x-axis, simply integrating from start to finish will cancel positive and negative contributions. To find the total geometric area, we must split the interval at every x-intercept and take the absolute value of each sub-integral.

这种有向性质意味着，如果曲线在x轴上下起伏，直接从起点积分到终点会抵消正、负贡献。为了求出总的几何面积，我们必须在每一个与x轴的交点处分割区间，并对每一段积分取绝对值。

2. Why Is the Integral Negative Below the x-axis? | 为什么x轴下方的积分为负？

Consider a constant negative function f(x) = -k (k > 0). Then ∫_a^b (-k) dx = -k(b – a) < 0. The rectangle has height k and width (b - a), so its area is k(b - a). The integral has the opposite sign because it measures “oriented” area relative to the positive direction of the y-axis.

考虑常数负函数 f(x) = -k（k > 0）。则 ∫_a^b (-k) dx = -k(b – a) < 0。矩形高度为 k，宽度为 (b - a)，其面积为 k(b - a)。积分的符号相反，因为它是相对于y轴正方向测量的“有向”面积。

In physical terms, if the velocity is negative (moving left), the displacement (integral of velocity over time) becomes negative. Similarly, a negative integrand represents a negative contribution to the total signed area.

用物理语言来说，如果速度为负（向左运动），位移（速度对时间的积分）就是负的。类似地，负的被积函数代表对总的有向面积产生负贡献。

3. Strategy: Splitting at the x-axis | 策略：在x轴处分割区间

To compute the geometric area between a curve and the x-axis over [a, b]:

计算曲线与x轴在区间 [a, b] 上围成的几何面积，步骤如下：

Find all solutions of f(x) = 0 inside (a, b), say c₁, c₂, …
找到 f(x) = 0 在 (a, b) 内的所有解，记为 c₁, c₂, …
Split [a, b] into sub-intervals [a, c₁], [c₁, c₂], …
将 [a, b] 分成子区间 [a, c₁], [c₁, c₂], …
On each sub-interval, determine whether f(x) is positive or negative.
在每个子区间上判断 f(x) 的正负。
Compute the absolute value of the definite integral on each sub-interval, then sum them.
计算每个子区间上定积分的绝对值，然后求和。

The total area is therefore

因此总面积为

A = ∫_a^c₁ |f(x)| dx + ∫_c₁^c₂ |f(x)| dx + ⋯ + ∫_cₙ^b |f(x)| dx

where we evaluate each integral separately by checking the sign of f(x) on that interval.

其中我们分别计算每一段积分，并判断该区间上 f(x) 的符号。

4. Worked Example 1: Parabola Below the x-axis | 例题1：x轴下方的抛物线

Find the area enclosed by y = x² – 4, the x-axis, and the lines x = -2 and x = 2.

求由 y = x² – 4、x轴以及直线 x = -2 和 x = 2 所围成的面积。

First solve x² – 4 = 0 ⇒ x = ±2. These are the interval endpoints, so on (-2, 2) the function is always negative (e.g. at x = 0, f(0) = -4). Thus we need the absolute value of the integral:

首先解 x² – 4 = 0，得 x = ±2。这两个点正是区间端点，因此在 (-2, 2) 内函数恒为负（例如 x = 0 时 f(0) = -4）。所以我们需求积分绝对值为：

| ∫_-2² (x² – 4) dx |

Compute:

计算：

∫_-2² (x² – 4) dx = [ x³/3 – 4x ]_-2² = (8/3 – 8) – (-8/3 + 8) = -16/3 – 16/3 = -32/3

Therefore the area is 32/3 square units.

因此面积为 32/3 平方单位。

5. Worked Example 2: Cubic with Mixed Intervals | 例题2：正负区间混合的三次函数

Find the total area between y = x³ – 3x and the x-axis from x = 0 to x = 2.

求曲线 y = x³ – 3x 与x轴从 x = 0 到 x = 2 之间围成的总面积。

Factor: x³ – 3x = x(x² – 3) = x(x – √3)(x + √3). Within [0, 2], the roots are x = 0 and x = √3 (since √3 ≈ 1.732). Check signs:

因式分解：x³ – 3x = x(x² – 3) = x(x – √3)(x + √3)。在 [0, 2] 内，根为 x = 0 和 x = √3（因为 √3 ≈ 1.732）。判断符号：

For 0 < x < √3, choose x = 1: f(1) = 1 - 3 = -2 (negative)
对于 0 < x < √3，取 x = 1：f(1) = 1 - 3 = -2（负）
For √3 < x < 2, choose x = 1.9: f(1.9) = 1.9³ - 5.7 ≈ 6.859 - 5.7 = 1.159 (positive)
对于 √3 < x < 2，取 x = 1.9：f(1.9) = 1.9³ - 5.7 ≈ 6.859 - 5.7 = 1.159（正）

So the area is

所以面积为

A = | ∫₀^√3 (x³ – 3x) dx | + ∫_√3² (x³ – 3x) dx

First integral:

第一段积分：

∫₀^√3 (x³ – 3x) dx = [ x⁴/4 – 3x²/2 ]₀^√3 = (9/4 – 9/2) – 0 = 9/4 – 18/4 = -9/4

Second integral:

第二段积分：

∫_√3² (x³ – 3x) dx = [ x⁴/4 – 3x²/2 ]_√3² = (16/4 – 12/2) – (9/4 – 9/2) = (4 – 6) – (9/4 – 18/4) = -2 – (-9/4) = -2 + 9/4 = 1/4

Total area = | -9/4 | + 1/4 = 9/4 + 1/4 = 10/4 = 5/2 square units.

总面积 = | -9/4 | + 1/4 = 9/4 + 1/4 = 10/4 = 5/2 平方单位。

6. Worked Example 3: Sine Curve over One Period | 例题3：正弦曲线一个周期

Find the total area between y = sin x and the x-axis on [0, 2π].

求曲线 y = sin x 与x轴在 [0, 2π] 上围成的总面积。

The sine function is positive on (0, π) and negative on (π, 2π). So we compute:

正弦函数在 (0, π) 为正，在 (π, 2π) 为负。因此我们计算：

A = ∫₀^π sin x dx + | ∫_π^2π sin x dx |

Evaluate:

计算：

∫₀^π sin x dx = [ -cos x ]₀^π = (-cos π) – (-cos 0) = 1 – (-1) = 2

∫_π^2π sin x dx = [ -cos x ]_π^2π = (-cos 2π) – (-cos π) = (-1) – (1) = -2

Therefore A = 2 + | -2 | = 4 square units.

因此 A = 2 + | -2 | = 4 平方单位。

Notice that the single integral ∫₀^2π sin x dx = 0 because the two contributions cancel. This is exactly why splitting is essential.

注意直接计算 ∫₀^2π sin x dx = 0，因为两段贡献相互抵消。这正是必须分割区间的原因。

7. Using Symmetry to Simplify | 利用对称性简化计算

When a curve is symmetrical about the y-axis or origin, we can often avoid unnecessary work. For instance, because sin x is an odd function, the absolute area over [-a, a] is twice the area over [0, a].

当曲线关于y轴或原点对称时，常常可以避免不必要的计算。例如，因为 sin x 是奇函数，在 [-a, a] 上的绝对面积等于 [0, a] 上面积的两倍。

For even functions like x² – 4, the area below the axis from -a to a equals 2 times the area from 0 to a, provided the function stays negative throughout. Always check whether the symmetry actually applies to the sign of the function on each sub-interval.

对于偶函数如 x² – 4，在 -a 到 a 上位于轴下方的面积等于 0 到 a 上面积的两倍，前提是函数在整个区间上保持为负。务必检查对称性是否适用于每个子区间上函数的符号。

8. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Mistake 1: Taking the absolute value of the whole integral when the curve crosses the x-axis. This does not give the total area if positive and negative parts exist.
错误1：当曲线穿过x轴时，对整个积分取绝对值。如果正负部分都存在，这样得不到总面积。
Mistake 2: Forgetting to find all roots in the interval. A missing intercept leads to an incorrect area.
错误2：忘记找出区间内的所有根。漏掉一个交点会导致面积错误。
Mistake 3: Assuming the integral from a to b always represents a geometric area, even when the function is negative.
错误3：认为从 a 到 b 的积分总是代表几何面积，即使函数为负。

To avoid these, always sketch or mentally check the sign of f(x) on each sub-interval before integrating.

为避免这些错误，在积分前请画出草图或在脑海中检查 f(x) 在每个子区间上的符号。

9. Worked Example 4: Two Roots and Mixed Signs | 例题4：两个交点与正负混合

Find the total area between y = x(x – 1)(x – 3) and the x-axis from x = 0 to x = 2.

求曲线 y = x(x – 1)(x – 3) 与x轴从 x = 0 到 x = 2 之间围成的总面积。

The roots are x = 0, 1, 3. Only x = 1 lies inside [0, 2]. Test signs:

根为 x = 0, 1, 3。只有 x = 1 在 [0, 2] 内。测试符号：

On (0, 1), take x = 0.5: f(0.5) = 0.5 × (-0.5) × (-2.5) = 0.625 > 0
在 (0, 1) 上，取 x = 0.5：f(0.5) = 0.5 × (-0.5) × (-2.5) = 0.625 > 0
On (1, 2), take x = 1.5: f(1.5) = 1.5 × 0.5 × (-1.5) = -1.125 < 0
在 (1, 2) 上，取 x = 1.5：f(1.5) = 1.5 × 0.5 × (-1.5) = -1.125 < 0

Area = ∫₀¹ f(x) dx + | ∫₁² f(x) dx |.

面积 = ∫₀¹ f(x) dx + | ∫₁² f(x) dx |。

Expand f(x) = x³ – 4x² + 3x. Then:

展开 f(x) = x³ – 4x² + 3x。则：

∫₀¹ (x³ – 4x² + 3x) dx = [ x⁴/4 – 4x³/3 + 3x²/2 ]₀¹ = 1/4 – 4/3 + 3/2 = 3/12 – 16/12 + 18/12 = 5/12

∫₁² (x³ – 4x² + 3x) dx = [ x⁴/4 – 4x³/3 + 3x²/2 ]₁² = (16/4 – 32/3 + 12/2) – (5/12) = (4 – 32/3 + 6) – 5/12 = (10 – 32/3) – 5/12 = (30/3 – 32/3) – 5/12 = -2/3 – 5/12 = -8/12 – 5/12 = -13/12

Total area = 5/12 + | -13/12 | = 18/12 = 3/2 square units.

总面积 = 5/12 + | -13/12 | = 18/12 = 3/2 平方单位。

10. Area with Curves Crossing the x-axis Multiple Times | 曲线多次穿过x轴时的面积

If there are many roots, the principle remains the same: split at every root, determine the sign on each interval, and sum the absolute values of the integrals. Always list the roots in increasing order to keep the integration limits correct.

如果有很多根，原理仍然相同：在每个根处分割，判断每个区间上的符号，并将各积分绝对值求和。务必按递增顺序列出根，以保证积分限正确。

A graphic calculator or a quick sketch of the graph helps you identify the roots accurately and avoids missing a sign change. However, in an exam you must be able to show the algebraic working without relying on technology.

图形计算器或快速画图可以帮助你准确找出根，避免漏掉符号变化。但在考试中，你必须展示代数过程，不能依赖技术。

11. Summary of Steps | 步骤总结

To find the geometric area between a curve and the x-axis on [a, b]:

求曲线与x轴在 [a, b] 上围成的几何面积：

Solve f(x) = 0 for all roots in (a, b).
解 f(x) = 0 得到 (a, b) 内的所有根。
Split [a, b] into intervals using these roots.
用这些根将 [a, b] 分割成若干区间。
Determine the sign of f(x) on each sub-interval.
判断每个子区间上 f(x) 的符号。
Integrate on each sub-interval and take the absolute value for any negative result.
在每个子区间上求积分，若结果为负则取绝对值。
Add all the absolute values together.
将所有绝对值相加。

Mastering this process is essential for A-Level maths questions involving area, whether the function is a polynomial, trigonometric, exponential, or a combination.

掌握这一过程对于A-Level数学中涉及面积的题目至关重要，无论函数是多项式、三角函数、指数函数还是组合函数。

12. Practice Question | 练习题

Find the total area enclosed by the curve y = cos x and the x-axis from x = 0 to x = π.

求曲线 y = cos x 与x轴从 x = 0 到 x = π 围成的总面积。

Hint: cos x = 0 at x = π/2. The curve is positive on (0, π/2) and negative on (π/2, π).

提示：cos x = 0 在 x = π/2 处。曲线在 (0, π/2) 为正，在 (π/2, π) 为负。

Answer: ∫₀^π/2 cos x dx + | ∫_π/2^π cos x dx | = 1 + | -1 | = 2 square units.

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A-Level Maths: Calculating Areas Below the x-axis by Integration | A-Level 数学：x轴下方面积的积分计算