Area Between Two Curves | 两曲线之间的面积

📚 Area Between Two Curves | 两曲线之间的面积

In IB Mathematics, the definite integral is a powerful tool for calculating the area of a region bounded by curves. When we need to find the area enclosed between two functions y = f(x) and y = g(x) over an interval, we integrate the difference of the functions. This topic appears in both Analysis & Approaches and Applications & Interpretation, and mastering it is essential for success in calculus-based exam questions. Understanding how to set up the integral correctly, handle intersections, and interpret the geometry is key.

在IB数学中，定积分是计算曲线所围区域面积的有力工具。当需要求两个函数 y = f(x) 与 y = g(x) 在某一区间之间的面积时，我们对两函数之差进行积分。该主题同时出现在分析与方法和应用与解释课程中，掌握它对解答微积分考题至关重要。正确建立积分式、处理交点以及理解几何意义是关键。

1. Introduction to Area Between Curves | 曲线间面积简介

The basic idea is simple: if f(x) ≥ g(x) for all x in [a, b], then the area between the curves from a to b is given by ∫ₐᵇ [f(x) − g(x)] dx. Visually, we are summing up the infinitesimally thin vertical strips whose height is the difference between the top function and the bottom function. This extends the familiar concept of area under a single curve by subtracting the lower boundary.

基本思路很简单：如果在区间 [a, b] 上恒有 f(x) ≥ g(x)，那么从 a 到 b 两曲线之间的面积就是 ∫ₐᵇ [f(x) − g(x)] dx。直观上，我们是在累加无数个竖直狭窄长条，每个长条的高度等于上方函数减去下方函数。这通过减掉下侧边界，将熟悉的单一曲线下的面积概念进行了推广。

---

2. The Definite Integral as Area | 定积分表示面积

Recall that for a non‑negative function y = f(x), the definite integral ∫ₐᵇ f(x) dx represents the area between the curve and the x‑axis. When a second function g(x) is introduced, we are effectively finding the area between f(x) and g(x) by considering the integral of the difference. Even if one or both functions dip below the x‑axis, the vertical distance f(x) − g(x) still measures the height of the strip, provided we maintain the correct order (upper minus lower).

回顾一下，对于非负函数 y = f(x)，定积分 ∫ₐᵇ f(x) dx 表示曲线与 x 轴之间的面积。当引入第二个函数 g(x) 后，我们实际上是利用差的积分来求 f(x) 与 g(x) 之间的面积。即使其中一个或两个函数落到 x 轴下方，只要保持正确的顺序（上方减下方），竖直距离 f(x) − g(x) 依然能度量长条的高度。

---

3. Setting Up the Integral: Upper Minus Lower | 建立积分：上减下

The most critical step is identifying which function is the upper one and which is the lower over the interval of interest. A quick way is to test a point between the intersection points: plug a convenient x‑value into both functions; the larger output belongs to the upper curve. Once established, the area formula is simply:

A = ∫ₐᵇ [f_upper(x) − f_lower(x)] dx

Always draw a rough sketch or use your GDC to confirm the relative positions of the curves. A mistake in the order will yield a negative result, which you can correct by taking the absolute value, but it is better to set up the integral correctly from the start.

最关键的一步是确定在所考虑的区间内哪个函数是上方的、哪个是下方的。一个快速的方法是测试交点之间的一个点：将某个方便的 x 值代入两个函数，输出较大的那个属于上方曲线。一旦确定了，面积公式就是：

A = ∫ₐᵇ [f_upper(x) − f_lower(x)] dx

始终画个草图或使用图形计算器来确认两条曲线的相对位置。顺序出错会得到负值，你可以取绝对值来纠正，但最好是从一开始就正确建立积分式。

---

4. Finding Intersection Points | 求交点

To determine the limits of integration a and b, solve the equation f(x) = g(x). In many IB questions, the intersections can be found algebraically; sometimes the calculator’s solve function is needed. The solutions give the x‑coordinates where the curves meet. If there are more than two intersections, the region may consist of multiple parts, and you will need to split the area accordingly.

为了确定积分下限 a 和上限 b，需要解方程 f(x) = g(x)。在许多 IB 考题中，交点可以用代数方法求出；有时则需要使用计算器的求解功能。方程的解就是曲线相交处的 x 坐标。如果有多于两个交点，该区域可能由多个部分组成，你就需要相应地将面积分割计算。

Always check whether the intersection points are within the domain you are considering. Extraneous solutions outside the interval must be disregarded. Label the intersections clearly and use them as the boundaries for your integrals.

始终要检查交点是否在你所考虑的区间内。位于区间之外的增根必须舍去。清晰地标出交点，并将它们作为积分的界限。

---

5. When Curves Cross: Splitting the Integral | 曲线交叉时：分割积分

In many problems, one function is not consistently above the other throughout the whole interval. At an intersection point, the roles of upper and lower may swap. In such cases you must split the integral at every intersection point and set up a separate integral for each sub‑interval, always taking upper minus lower for that section.

在许多问题中，一条函数并不始终在整个区间内位于另一条之上。在交点处，上下角色可能会互换。这时，你必须在每个交点处分割积分，为每个子区间分别建立积分，始终对那段取上方减下方。

For example, if f(x) and g(x) intersect at x = c within [a, b] and f(x) ≥ g(x) on [a, c] while g(x) ≥ f(x) on [c, b], the total area becomes ∫ₐᶜ [f(x) − g(x)] dx + ∫ᶜᵇ [g(x) − f(x)] dx. A sign‑based mistake that adds the integrals without swapping the order is a common pitfall.

举个例子，如果 f(x) 和 g(x) 在 [a, b] 内的 x = c 处相交，且在 [a, c] 上 f(x) ≥ g(x)，而在 [c, b] 上 g(x) ≥ f(x)，那么总面积为 ∫ₐᶜ [f(x) − g(x)] dx + ∫ᶜᵇ [g(x) − f(x)] dx。不交换顺序就把积分相加，这种基于符号的错误是一个非常常见的陷阱。

---

6. Area Between Curves with Respect to y | 关于y轴的曲线间面积

Sometimes it is more convenient to treat x as a function of y. When the curves are more naturally expressed as x = f(y) and x = g(y), the area between them from y = c to y = d is given by ∫ᶜᵈ [f_right(y) − f_left(y)] dy. The principle is the same: right minus left, integrated along the y‑axis.

有时把 x 视为 y 的函数更方便。当曲线较自然地表为 x = f(y) 和 x = g(y) 时，它们之间从 y = c 到 y = d 的面积由 ∫ᶜᵈ [f_right(y) − f_left(y)] dy 给出。原理是一样的：右减左，沿 y 轴积分。

This approach is particularly useful when one or both curves are not functions of x (e.g., sideways parabolas) or when the region’s boundaries are horizontal. Always determine the upper and lower y‑bounds by solving f(y) = g(y), then identify the rightmost and leftmost curves on that y‑interval.

当一条或两条曲线不是 x 的函数（比如侧向抛物线），或者区域的边界是水平线时，这种方法尤其有用。总是通过解 f(y) = g(y) 来确定 y 的上限和下限，然后找出在该 y 区间上最右和最左的曲线。

---

7. Using Absolute Value Approach | 使用绝对值方法

A compact way to express the area between two curves without worrying about which function is on top is A = ∫ₐᵇ |f(x) − g(x)| dx. By using the absolute value, you automatically account for any crossing of the curves. Many graphing calculators can directly evaluate such an integral numerically, but you must still enter the absolute value correctly.

一种无需操心哪个函数在上方的紧凑表达方式是 A = ∫ₐᵇ |f(x) − g(x)| dx。通过使用绝对值，你自动处理了曲线的任何交叉。许多图形计算器都能直接对这种积分进行数值计算，但你必须正确输入绝对值。

However, for exact (analytic) answers, IB exam questions often expect you to split the integral or at least recognise that the absolute value forces a piecewise integration. Understanding both the algebraic split and the absolute‑value shortcut strengthens your problem‑solving toolkit.

然而，若要得到精确（解析）答案，IB 考题常会要求你分割积分，或至少认识到绝对值会迫使用分段积分。既要懂得代数分割，又要掌握绝对值捷径，这能增强你解题的工具箱。

---

8. Worked Example: Parabola and Line | 例题：抛物线与直线

Find the area enclosed by the curves y = x² and y = x + 2.

求由曲线 y = x² 与 y = x + 2 所围成的面积。

Step 1: Intersections. Solve x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0, so x = −1 and x = 2.

第1步：求交点。 解 x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0，得 x = −1 和 x = 2。

Step 2: Upper/lower test. Try x = 0: 0² = 0 and 0+2 = 2, so y = x+2 is above y = x² on [−1, 2].

第2步：上下测试。 试 x = 0：0² = 0 而 0+2 = 2，因此在 [−1, 2] 上 y = x+2 在 y = x² 上方。

Step 3: Set up integral. A = ∫₋₁² [(x + 2) − x²] dx.

第3步：建立积分。 A = ∫₋₁² [(x + 2) − x²] dx.

Step 4: Integrate. ∫(x+2−x²) dx = ½x² + 2x − ⅓x³. Evaluate from −1 to 2:

第4步：积分。 ∫(x+2−x²) dx = ½x² + 2x − ⅓x³. 从 −1 到 2 计算：

At x = 2: ½(4) + 4 − ⅓(8) = 2 + 4 − 8/3 = 10/3.

在 x = 2 处：½(4) + 4 − ⅓(8) = 2 + 4 − 8/3 = 10/3.

At x = −1: ½(1) − 2 − ⅓(−1) = ½ − 2 + ⅓ = −3/2 + 1/3 = −7/6.

在 x = −1 处：½(1) − 2 − ⅓(−1) = ½ − 2 + ⅓ = −3/2 + 1/3 = −7/6.

Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 = 4.5 square units.

面积 = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 = 4.5 平方单位。

---

9. Tips for Using a Graphing Calculator | 使用图形计算器提示

IB students can often use a graphing calculator (GDC) to find the area between two curves directly. After graphing both functions, use the ‘Analyze Graph’ or ‘Integral’ function, select the interval between intersections, and the calculator will compute the numerical or exact area. However, always show the setup of the integral in your written solution; the calculator is a tool, not a replacement for reasoning.

IB 学生常常可以使用图形计算器（GDC）直接求两曲线之间的面积。在画出两个函数的图像后，使用“图像分析”或“积分”功能，选择交点之间的区间，计算器就会计算出数值面积或精确面积。然而，在书写的解答中始终要展示积分的建立；计算器是工具，不能替代推理过程。

When using the absolute value method, enter |f(x)−g(x)| for the integrand. Double‑check that you have captured all intersections; missing a cross‑over point can lead to an incorrect total area.

使用绝对值方法时，输入 |f(x)−g(x)| 作为被积函数。再次检查你是否已捕捉到所有交点；漏掉一个交叉点会导致总面积错误。

---

10. Common Mistakes to Avoid | 常见错误避免

(1) Forgetting to find and use all intersection points as integration limits. Students sometimes guess the interval or only use the ones visible on a default window, missing negative or far‑off solutions.

（1）忘记求出并使用所有交点作为积分限。学生有时会猜测区间，或者只用默认窗口里可见的交点，而漏掉负解或远处的解。

(2) Wrong order of subtraction: always upper minus lower. Writing f(x) − g(x) without checking which is larger can give a negative area; while you can take the absolute value afterwards, it is much safer to confirm the order first.

（2）减法顺序错误：总是上方减下方。不检查哪个更大就写出 f(x) − g(x) 可能会得到负面积；虽然后面可以取绝对值，但先确认顺序要稳妥得多。

(3) Mixing up the variable: when working with respect to y, students often still write dx instead of dy, or use the wrong bounds. Clear labelling of axes prevents this.

（3）混淆变量：当关于 y 积分时，学生常常仍写成 dx 而不是 dy，或用错界限。清楚地标出坐标轴可以防止这一点。

(4) Ignoring the need to split the integral when the curves cross, leading to cancellation of areas and an incorrect result.

（4）当曲线交叉时忽略分割积分的必要性，这会导致面积相互抵消，得到错误的结果。

---

11. Practice Problem Walkthrough | 练习题解析

Determine the area bounded by y = sin x and y = cos x between x = 0 and x = π/2.

求由 y = sin x 和 y = cos x 在 x = 0 到 x = π/2 之间所围成的面积。

Intersections: sin x = cos x ⇒ tan x = 1 ⇒ x = π/4 within the interval. So we split at x = π/4. On [0, π/4], cos x ≥ sin x; on [π/4, π/2], sin x ≥ cos x.

交点： sin x = cos x ⇒ tan x = 1 ⇒ 在区间内 x = π/4。因此我们在 x = π/4 处分割。在 [0, π/4] 上，cos x ≥ sin x；在 [π/4, π/2] 上，sin x ≥ cos x。

Area = ∫₀^{π/4} (cos x − sin x) dx + ∫_{π/4}^{π/2} (sin x − cos x) dx = [sin x + cos x]₀^{π/4} + [−cos x − sin x]_{π/4}^{π/2}. Evaluating gives (√2 − 1) + (√2 − 1) = 2√2 − 2. This is one of the classic IB area questions and highlights the split integral technique clearly.

面积 = ∫₀^{π/4} (cos x − sin x) dx + ∫_{π/4}^{π/2} (sin x − cos x) dx = [sin x + cos x]₀^{π/4} + [−cos x − sin x]_{π/4}^{π/2}。计算得到 (√2 − 1) + (√2 − 1) = 2√2 − 2。这是经典的 IB 面积题之一，清楚地凸显了分割积分的技巧。

---

12. Summary | 总结

To succeed with area‑between‑curves problems, follow a structured approach: (i) find all intersection points, (ii) determine upper/lower (or right/left) on each subinterval, (iii) set up one integral per subinterval using the correct difference, and (iv) evaluate accurately. When in doubt, a quick sketch or a table of values helps confirm the relative positions. The absolute value shortcut is a handy check, but be prepared to split integrals for exact analytic answers. With systematic practice, you will handle any IB calculus question involving two functions with confidence.

要成功解决曲线间面积问题，请遵循结构化的步骤：（i）求出所有交点，（ii）确定每个子区间上的上方/下方（或右侧/左侧），（iii）在每个子区间上用正确的差建立一个积分，（iv）准确计算。当有疑惑时，快速画个草图或列个数值表有助于确认相对位置。绝对值捷径是一个方便的检验方法，但要准备好分割积分以得到精确的解析答案。通过系统的练习，你将能自信地应对任何涉及两个函数的IB微积分问题。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）