Iterated Integrals with Constant Limits | IB数学：常数限下的累次积分

📚 Iterated Integrals with Constant Limits | IB数学：常数限下的累次积分

In IB Mathematics Higher Level, one of the essential techniques in multivariate calculus is the iterated integral. When both limits of integration are constants, the process becomes particularly systematic and elegant. This article provides a comprehensive guide to understanding, evaluating, and applying iterated integrals with constant limits.

在IB数学高级水平课程中，累次积分是多元微积分中的核心技巧之一。当两个积分限均为常数时，积分过程尤其具有系统性和简洁性。本文将为读者全面讲解常数限下累次积分的理解、求值与应用。

1. What is an Iterated Integral? | 累次积分简介

An iterated integral is the result of applying integration twice, once for each variable in a two-variable function. For a function f(x, y), we write the iterated integral as ∫ₐᵇ ∫ₐᵈ f(x, y) dy dx, which means we first integrate with respect to y, then integrate the result with respect to x.

累次积分是对二元函数依次进行两次积分运算的结果。对于函数 f(x, y)，我们将其累次积分记为 ∫ₐᵇ ∫ₐᵈ f(x, y) dy dx，这表示先对 y 进行积分，然后再对所得结果关于 x 积分。

The notation ∫ₐᵇ ∫ₐᵈ f(x, y) dy dx contains two integral signs, two sets of limits, and two differentials. The order of the differentials tells us which integration to perform first — the differential closest to the function is always integrated first. Here, dy appears closest, so y is integrated first.

记号 ∫ₐᵇ ∫ₐᵈ f(x, y) dy dx 包含两个积分号、两组积分限和两个微分。微分的顺序决定了积分的先后次序——紧靠着函数的微分先积分。在这个例子中，dy 紧靠着函数，因此我们先对 y 积分。

2. Fubini’s Theorem | 富比尼定理

Fubini’s Theorem is the mathematical foundation that justifies evaluating a double integral as an iterated integral. It states that if f(x, y) is continuous on the rectangular region R = [a, b] × [c, d], then the double integral over R equals the iterated integral in either order.

富比尼定理是双重积分能够转化为累次积分的数学理论基础。该定理指出：若 f(x, y) 在矩形区域 R = [a, b] × [c, d] 上连续，则 R 上的双重积分等于按任意顺序计算的累次积分。

∬ᵣ f(x, y) dA = ∫ₐᵇ ∫ₐᵈ f(x, y) dy dx = ∫ₐᵈ ∫ₐᵇ f(x, y) dx dy

This theorem gives us tremendous flexibility: we may choose the order of integration that makes the calculation easier. With constant limits, both orders are always valid for continuous functions, and both will produce the same numerical answer.

这一定理为我们提供了极大的灵活性：我们可以选择使计算更为简便的积分顺序。对于常数限，只要函数连续，两种顺序都有效，并且计算结果完全相同。

3. Evaluating Iterated Integrals Step-by-Step | 逐步求值累次积分

The procedure for evaluating an iterated integral requires careful attention to which variable is being integrated at each stage. The key principle is to treat all variables other than the integration variable as constants.

求值累次积分的过程需要特别注意每一步正在对哪个变量进行积分。核心原则是：除当前积分变量外的所有变量均视为常数。

Let us evaluate ∫₀¹ ∫₀² xy dy dx in stages:

我们以 ∫₀¹ ∫₀² xy dy dx 为例，分步骤求值：

Step 1 — Inner integral with respect to y:

第一步——内层积分（对 y）：

∫₀² xy dy = x ∫₀² y dy = x [y² / 2]₀² = x (4 / 2 − 0) = 2x

Step 2 — Outer integral with respect to x:

第二步——外层积分（对 x）：

∫₀¹ 2x dx = [x²]₀¹ = 1² − 0² = 1

Thus, the value of the iterated integral is 1. Notice that in Step 1, x was treated as a constant factor; in Step 2, we integrated a function of x alone.

因此，该累次积分的值为 1。注意在第一步中，x 被当作常数因子；在第二步中，我们仅对 x 的函数进行积分。

4. The Order of Integration | 积分的顺序

When the limits are constants, either order of integration can be used. Let us verify this with the example ∫₀² ∫₁³ (xy + 2x) dx dy, evaluated in both ways.

当积分限为常数时，两种积分顺序都可以使用。下面我们以 ∫₀² ∫₁³ (xy + 2x) dx dy 为例，分别用两种顺序验证。

Order 1 — integrate with respect to x first:

顺序一——先对 x 积分：

∫₁³ (xy + 2x) dx = y ∫₁³ x dx + 2 ∫₁³ x dx = y [x² / 2]₁³ + 2 [x² / 2]₁³

= y (9/2 − 1/2) + 2 (9/2 − 1/2) = 4y + 8

Then ∫₀² (4y + 8) dy = [2y² + 8y]₀² = 8 + 16 = 24

然后 ∫₀² (4y + 8) dy = [2y² + 8y]₀² = 8 + 16 = 24

Order 2 — integrate with respect to y first:

顺序二——先对 y 积分：

∫₀² (xy + 2x) dy = x ∫₀² y dy + 2x ∫₀² dy = x [y² / 2]₀² + 2x [y]₀² = 2x + 4x = 6x

Then ∫₁³ 6x dx = [3x²]₁³ = 27 − 3 = 24

然后 ∫₁³ 6x dx = [3x²]₁³ = 27 − 3 = 24

Both orders yield 24, confirming Fubini’s Theorem. When choosing an order, look for the arrangement that produces a simpler inner integral — for instance, if one variable appears linearly and the other quadratically, integrating the linear one first often simplifies the computation.

两种顺序都得到 24，验证了富比尼定理。在选择顺序时，应寻找能使内层积分更简单的安排——例如，若一个变量以一次形式出现而另一个以二次形式出现，通常先对一次形式的变量积分会使计算更简便。

5. Geometric Interpretation | 几何解释

An iterated integral with constant limits computes the volume under the surface z = f(x, y) above the rectangular region [a, b] × [c, d] in the xy-plane. Each infinitesimal element dA = dx dy represents a small rectangle, and f(x, y) dA represents the volume of a thin vertical column above that rectangle.

具有常数限的累次积分计算的是曲面 z = f(x, y) 之下、xy 平面上矩形区域 [a, b] × [c, d] 之上的体积。每一个微小面积元 dA = dx dy 代表一个小矩形，而 f(x, y) dA 表示该小矩形上方薄柱体的体积。

For example, ∫₀¹ ∫₀² (x + y) dy dx gives the volume of the solid whose base is the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 and whose top surface is the plane z = x + y. Computing this integral:

例如，∫₀¹ ∫₀² (x + y) dy dx 表示底面为矩形 0 ≤ x ≤ 1, 0 ≤ y ≤ 2、顶面为平面 z = x + y 的立体体积。计算得：

∫₀² (x + y) dy = [xy + y² / 2]₀² = 2x + 2, then ∫₀¹ (2x + 2) dx = [x² + 2x]₀¹ = 3

∫₀² (x + y) dy = [xy + y² / 2]₀² = 2x + 2，然后 ∫₀¹ (2x + 2) dx = [x² + 2x]₀¹ = 3

Geometrically, this solid is a wedge. Its volume 3 can also be calculated using the formula for the volume of a prism: base area 2 multiplied by the average height (0 + 1 + 2) / 3 = 1, giving 2 × 1.5 = 3. This connection between integration and geometry reinforces the power of the iterated integral as a computational tool.

从几何上看，该立体是一个楔形体。其体积 3 也可以通过棱柱体积公式验证：底面积 2 乘以平均高度 (0 + 1 + 2) / 3 = 1，得到 2 × 1.5 = 3。这种积分与几何之间的联系进一步体现了累次积分作为计算工具的强大之处。

6. Worked Examples | 详细例题

Below are additional worked examples covering typical IB examination questions on iterated integrals with constant limits.

以下例题涵盖了IB考试中典型的常数限累次积分问题。

Example 1: Evaluate ∫₀¹ ∫₀² (x² + y²) dy dx.

例1：求 ∫₀¹ ∫₀² (x² + y²) dy dx。

Inner integral: ∫₀² (x² + y²) dy = [x²y + y³ / 3]₀² = 2x² + 8/3

内层积分：∫₀² (x² + y²) dy = [x²y + y³ / 3]₀² = 2x² + 8/3

Outer integral: ∫₀¹ (2x² + 8/3) dx = [2x³ / 3 + 8x / 3]₀¹ = 2/3 + 8/3 = 10/3

外层积分：∫₀¹ (2x² + 8/3) dx = [2x³ / 3 + 8x / 3]₀¹ = 2/3 + 8/3 = 10/3

Example 2: Evaluate ∫₀^π/2 ∫₀^π/2 sin x cos y dx dy.

例2：求 ∫₀^π/2 ∫₀^π/2 sin x cos y dx dy。

Since the integrand separates into a product of a function of x and a function of y, the iterated integral factorises:

由于被积函数可分离为 x 的函数与 y 的函数的乘积，累次积分可以进行因式分解：

∫₀^π/2 sin x dx × ∫₀^π/2 cos y dy = [−cos x]₀^π/2 × [sin y]₀^π/2 = 1 × 1 = 1

This product property holds whenever f(x, y) = g(x) · h(y), saving significant computation time in examinations.

当 f(x, y) = g(x) · h(y) 时，这一乘积性质恒成立，可以大幅节省考试中的计算时间。

Example 3: Evaluate ∫₀¹ ∫₀¹ (3x²y + 4xy³) dy dx.

例3：求 ∫₀¹ ∫₀¹ (3x²y + 4xy³) dy dx。

Inner: ∫₀¹ (3x²y + 4xy³) dy = [3x²y² / 2 + xy⁴]₀¹ = 3x² / 2 + x

内层：∫₀¹ (3x²y + 4xy³) dy = [3x²y² / 2 + xy⁴]₀¹ = 3x² / 2 + x

Outer: ∫₀¹ (3x² / 2 + x) dx = [x³ / 2 + x² / 2]₀¹ = 1/2 + 1/2 = 1

外层：∫₀¹ (3x² / 2 + x) dx = [x³ / 2 + x² / 2]₀¹ = 1/2 + 1/2 = 1

7. Common Mistakes and Pitfalls | 常见错误与陷阱

Students frequently make the following errors when evaluating iterated integrals. Being aware of these pitfalls is essential for exam success.

学生在求值累次积分时经常会犯以下错误。了解这些陷阱对考试取得好成绩至关重要。

Treating the wrong variable as constant: When integrating with respect to y, never treat y as a constant — only x and all other variables are constants.
混淆积分变量：对 y 积分时，切勿将 y 视为常数——只有 x 及其他变量才被视为常数。

Swapping the order of differentials incorrectly: The differential dy dx means integrate y first, then x, not the reverse. Always read from the inside out.
错误交换微分顺序：dy dx 表示先对 y 积分，再对 x 积分，而非相反。务必由内向外阅读。

Omitting the constant of integration: In definite integrals, the constant cancels, but if you introduce it, ensure you subtract correctly. Many students prefer to simply omit it.
遗漏积分常数：在定积分中常数会被抵消，但若引入常数，务必正确相减。许多学生倾向于直接省略。

Writing the limits backwards: Remember that ∫ₐᵇ f(x) dx = F(b) − F(a). Applying the lower limit first and subtracting the upper limit gives a sign error.
积分限顺序颠倒：记住 ∫ₐᵇ f(x) dx = F(b) − F(a)。若先用下限代入后再减上限会导致正负号错误。

Arithmetic errors in evaluating antiderivatives: When integrating x² you must obtain x³/3, not x²/3. Double-check every antiderivative.
求原函数时的运算错误：对 x² 积分应得到 x³/3，而不是 x²/3。请逐一检查每个原函数。

8. Average Value and Applications | 平均值与应用

The average value of a function f(x, y) over a rectangular region R is given by the formula:

函数 f(x, y) 在矩形区域 R 上的平均值的计算公式为：

The average value of f over R = (1 / Area(R)) × ∬ᵣ f(x, y) dA

For a rectangle with dimensions (b − a) × (d − c), the area is simply (b − a)(d − c). This concept has important applications in physics (average temperature over a plate), economics (average profit over a production region), and statistics (expected value over a uniform distribution).

对于尺寸为 (b − a) × (d − c) 的矩形，其面积就是 (b − a)(d − c)。平均值的概念在物理学（平板上的平均温度）、经济学（生产区域内的平均利润）和统计学（均匀分布的期望值）中都有重要应用。

Example: Find the average value of f(x, y) = x + y over the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

例题：求函数 f(x, y) = x + y 在矩形 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 上的平均值。

We already computed ∫₀¹ ∫₀² (x + y) dy dx = 3. The area of the rectangle is 1 × 2 = 2. Therefore the average value is 3 / 2 = 1.5.

我们已经计算出 ∫₀¹ ∫₀² (x + y) dy dx = 3。矩形面积为 1 × 2 = 2。因此平均值为 3 / 2 = 1.5。

This matches the geometric intuition: the plane z = x + y has average height 1.5 over this region, exactly halfway between the minimum height 0 and maximum height 3.

这与几何直觉一致：平面 z = x + y 在此区域上的平均高度为 1.5，恰好处于最低高度 0 与最高高度 3 的中间。

9. Practice Problems | 练习题

Test your understanding with the following problems. Complete solutions are shown below.

请通过以下题目检验你的理解。答案和解答过程附后。

Problem 1: Evaluate ∫₀¹ ∫₀² (x²y + y) dy dx.	题目1：求 ∫₀¹ ∫₀² (x²y + y) dy dx。
Problem 2: Evaluate ∫₀^π ∫₀^π/2 cos(x) sin(y) dx dy.	题目2：求 ∫₀^π ∫₀^π/2 cos(x) sin(y) dx dy。
Problem 3: The temperature on a flat plate is T(x, y) = 3x + y, where 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1. Find the average temperature.	题目3：平板上的温度为 T(x, y) = 3x + y，其中 0 ≤ x ≤ 2，0 ≤ y ≤ 1。求平均温度。

Solution 1: Inner: ∫₀² (x²y + y) dy = [x²y²/2 + y²/2]₀² = 2x² + 2. Outer: ∫₀¹ (2x² + 2) dx = [2x³/3 + 2x]₀¹ = 2/3 + 2 = 8/3.

解答1：内层：∫₀² (x²y + y) dy = [x²y²/2 + y²/2]₀² = 2x² + 2。外层：∫₀¹ (2x² + 2) dx = [2x³/3 + 2x]₀¹ = 2/3 + 2 = 8/3。

Solution 2: ∫₀^π cos x dx × ∫₀^π/2 sin y dy = [sin x]₀^π × [−cos y]₀^π/2 = 0 × 1 = 0.

解答2：∫₀^π cos x dx × ∫₀^π/2 sin y dy = [sin x]₀^π × [−cos y]₀^π/2 = 0 × 1 = 0。

Solution 3: ∫₀² ∫₀¹ (3x + y) dy dx = ∫₀² [3xy + y²/2]₀¹ dx = ∫₀² (3x + 1/2) dx = [3x²/2 + x/2]₀² = 6 + 1 = 7. Area = 2 × 1 = 2, so average = 7/2 = 3.5.

解答3：∫₀² ∫₀¹ (3x + y) dy dx = ∫₀² [3xy + y²/2]₀¹ dx = ∫₀² (3x + 1/2) dx = [3x²/2 + x/2]₀² = 6 + 1 = 7。面积 = 2 × 1 = 2，所以平均值 = 7/2 = 3.5。

Mastering iterated integrals with constant limits is a foundational skill that opens the door to more advanced topics such as changing the order of integration, polar coordinates, and multiple integrals over general regions. Practice each step carefully, double-check your antiderivatives, and you will find these problems become routine.

掌握常数限下的累次积分是学习更高级主题（如交换积分顺序、极坐标变换、一般区域上的多重积分）的基础技能。仔细练习每一步，反复核验原函数，你会发现这类题目将变得得心应手。

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Iterated Integrals with Constant Limits | IB数学：常数限下的累次积分